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College TrigonometryVersion π = 3 by Carl Stitz, Ph.D. Jeff Zeager, Ph.D. Lakeland Community College Lorain County Community College Cover Design Dario Rampersad Photo Credits Jessica S. Novak, Lakeland Community College Ron Jantz, Lorain County Community College Lauren Potter, WebAssign July 15, 2011 ii Acknowledgements While the cover of this textbook lists only two names, the book as it stands today would simply not exist if not for the tireless work and dedication of several people. First and foremost, we wish to thank our families for their patience and support during the creative process. We would also like to thank our students - the sole inspiration for the work. Among our colleagues, we wish to thank Rich Basich, Bill Previts, and Irina Lomonosov, who not only were early adopters of the textbook, but also contributed materials to the project. Special thanks go to Katie Cimperman, Terry Dykstra, Frank LeMay, and Rich Hagen who provided valuable feedback from the classroom. Thanks also to David Stumpf, Ivana Gorgievska, Jorge Gerszonowicz, Kathryn Arocho, Heather Bubnick, and Florin Muscutariu for their unwaivering support (and sometimes defense!) of the project. From outside the classroom, we wish to thank Don Anthan and Ken White, who designed the electric circuit applications used in the text, as well as Drs. Wendy Marley and Marcia Ballinger for the Lorain CCC enrollment data used in the text. The authors are also indebted to the good folks at our schools’ bookstores, Gwen Sevtis (Lakeland CC) and Chris Callahan (Lorain CCC), for working with us to get printed copies to the students as inexpensively as possible. We would also like to thank Lakeland folks Jeri Dickinson, Mary Ann Blakeley, Jessica Novak, and Corrie Bergeron for their enthusiasm and promotion of the project. The administration at both schools have also been very supportive of the project, so from Lakeland, we wish to thank Dr. Morris W. Beverage, Jr., President, Dr. Fred Law, Provost, Deans Don Anthan and Dr. Steve Oluic, and the Board of Trustees. From Lorain County Community College, we which to thank Dr. Roy A. Church, Dr. Karen Wells, and the Board of Trustees. From the Ohio Board of Regents, we wish to thank former Chancellor Eric Fingerhut, Darlene McCoy, Associate Vice Chancellor of Affordability and Efficiency, and Kelly Bernard. From OhioLINK, we wish to thank Steve Acker, John Magill, and Stacy Brannan. We also wish to thank the good folks at WebAssign, most notably Chris Hall, COO, and Joel Hollenbeck (former VP of Sales.) Last, but certainly not least, we wish to thank all the folks who have contacted us over the interwebs, most notably Dimitri Moonen and Joel Wordsworth, who gave us great feedback, and Antonio Olivares who helped debug the source code. Table of Contents vii 10 Foundations of Trigonometry 693 10.1 Angles and their Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 693 10.1.1 Applications of Radian Measure: Circular Motion . . . . . . . . . . . . . . 706 10.1.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 709 10.1.3 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 712 10.2 The Unit Circle: Cosine and Sine . . . . . . . . . . . . . . . . . . . . . . . . . . . . 717 10.2.1 Beyond the Unit Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 730 10.2.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 736 10.2.3 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 740 10.3 The Six Circular Functions and Fundamental Identities . . . . . . . . . . . . . . . . 744 10.3.1 Beyond the Unit Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 752 10.3.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 759 10.3.3 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 766 10.4 Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 770 10.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 782 10.4.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 787 10.5 Graphs of the Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . 790 10.5.1 Graphs of the Cosine and Sine Functions . . . . . . . . . . . . . . . . . . . 790 10.5.2 Graphs of the Secant and Cosecant Functions . . . . . . . . . . . . . . . . 800 10.5.3 Graphs of the Tangent and Cotangent Functions . . . . . . . . . . . . . . . 804 10.5.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 809 10.5.5 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 811 10.6 The Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 819 10.6.1 Inverses of Secant and Cosecant: Trigonometry Friendly Approach . . . . . 827 10.6.2 Inverses of Secant and Cosecant: Calculus Friendly Approach . . . . . . . . 830 10.6.3 Calculators and the Inverse Circular Functions. . . . . . . . . . . . . . . . . 833 10.6.4 Solving Equations Using the Inverse Trigonometric Functions. . . . . . . . 838 10.6.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 841 10.6.6 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 849 10.7 Trigonometric Equations and Inequalities . . . . . . . . . . . . . . . . . . . . . . . 857 10.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 871 10.7.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 874 11 Applications of Trigonometry 879 11.1 Applications of Sinusoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 879 11.1.1 Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 883 11.1.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 889 11.1.3 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 892 11.2 The Law of Sines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 894 11.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 902 11.2.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 906 11.3 The Law of Cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 908 viii Table of Contents 11.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 914 11.3.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 916 11.4 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 917 11.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 928 11.4.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 930 11.5 Graphs of Polar Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 936 11.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 956 11.5.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 961 11.6 Hooked on Conics Again . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 971 11.6.1 Rotation of Axes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 971 11.6.2 The Polar Form of Conics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 979 11.6.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 984 11.6.4 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 985 11.7 Polar Form of Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 989 11.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1002 11.7.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1005 11.8 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1010 11.8.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1025 11.8.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1029 11.9 The Dot Product and Projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1032 11.9.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1041 11.9.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1043 11.10 Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1046 11.10.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1057 11.10.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1061 Index 1067 Preface Thank you for your interest in our book, but more importantly, thank you for taking the time to read the Preface. I always read the Prefaces of the textbooks which I use in my classes because I believe it is in the Preface where I begin to understand the authors - who they are, what their motivation for writing the book was, and what they hope the reader will get out of reading the text. Pedagogical issues such as content organization and how professors and students should best use a book can usually be gleaned out of its Table of Contents, but the reasons behind the choices authors make should be shared in the Preface. Also, I feel that the Preface of a textbook should demonstrate the authors’ love of their discipline and passion for teaching, so that I come away believing that they really want to help students and not just make money. Thus, I thank my fellow Preface-readers again for giving me the opportunity to share with you the need and vision which guided the creation of this book and passion which both Carl and I hold for Mathematics and the teaching of it. Carl and I are natives of Northeast Ohio. We met in graduate school at Kent State University in 1997. I finished my Ph.D in Pure Mathematics in August 1998 and started teaching at Lorain County Community College in Elyria, Ohio just two days after graduation. Carl earned his Ph.D in Pure Mathematics in August 2000 and started teaching at Lakeland Community College in Kirtland, Ohio that same month. Our schools are fairly similar in size and mission and each serves a similar population of students. The students range in age from about 16 (Ohio has a Post-Secondary Enrollment Option program which allows high school students to take college courses for free while still in high school.) to over 65. Many of the “non-traditional” students are returning to school in order to change careers. A majority of the students at both schools receive some sort of financial aid, be it scholarships from the schools’ foundations, state-funded grants or federal financial aid like student loans, and many of them have lives busied by family and job demands. Some will be taking their Associate degrees and entering (or re-entering) the workforce while others will be continuing on to a four-year college or university. Despite their many differences, our students share one common attribute: they do not want to spend $200 on a College Algebra book. The challenge of reducing the cost of textbooks is one that many states, including Ohio, are taking quite seriously. Indeed, state-level leaders have started to work with faculty from several of the colleges and universities in Ohio and with the major publishers as well. That process will take considerable time so Carl and I came up with a plan of our own. We decided that the best way to help our students right now was to write our own College Algebra book and give it away electronically for free. We were granted sabbaticals from our respective institutions for the Spring x Preface semester of 2009 and actually began writing the textbook on December 16, 2008. Using an open- source text editor called TexNicCenter and an open-source distribution of LaTeX called MikTex 2.7, Carl and I wrote and edited all of the text, exercises and answers and created all of the graphs (using Metapost within LaTeX) for Version 0.9 in about eight months. (We choose to create a text in only black and white to keep printing costs to a minimum for those students who prefer a printed edition. This somewhat Spartan page layout stands in sharp relief to the explosion of colors found in most other College Algebra texts, but neither Carl nor I believe the four-color print adds anything of value.) I used the book in three sections of College Algebra at Lorain County Community College in the Fall of 2009 and Carl’s colleague, Dr. Bill Previts, taught a section of College Algebra at Lakeland with the book that semester as well. Students had the option of downloading the book as a .pdf file from our website www.stitz-zeager.com or buying a low-cost printed version from our colleges’ respective bookstores. (By giving this book away for free electronically, we end the cycle of new editions appearing every 18 months to curtail the used book market.) During Thanksgiving break in November 2009, many additional exercises written by Dr. Previts were added and the typographical errors found by our students and others were corrected. On December 10, 2009, Version √ 2 was released. The book remains free for download at our website and by using Lulu.com as an on-demand printing service, our bookstores are now able to provide a printed edition for just under $19. Neither Carl nor I have, or will ever, receive any royalties from the printed editions. As a contribution back to the open-source community, all of the LaTeX files used to compile the book are available for free under a Creative Commons License on our website as well. That way, anyone who would like to rearrange or edit the content for their classes can do so as long as it remains free. The only disadvantage to not working for a publisher is that we don’t have a paid editorial staff. What we have instead, beyond ourselves, is friends, colleagues and unknown people in the open- source community who alert us to errors they find as they read the textbook. What we gain in not having to report to a publisher so dramatically outweighs the lack of the paid staff that we have turned down every offer to publish our book. (As of the writing of this Preface, we’ve had three offers.) By maintaining this book by ourselves, Carl and I retain all creative control and keep the book our own. We control the organization, depth and rigor of the content which means we can resist the pressure to diminish the rigor and homogenize the content so as to appeal to a mass market. A casual glance through the Table of Contents of most of the major publishers’ College Algebra books reveals nearly isomorphic content in both order and depth. Our Table of Contents shows a different approach, one that might be labeled “Functions First.” To truly use The Rule of Four, that is, in order to discuss each new concept algebraically, graphically, numerically and verbally, it seems completely obvious to us that one would need to introduce functions first. (Take a moment and compare our ordering to the classic “equations first, then the Cartesian Plane and THEN functions” approach seen in most of the major players.) We then introduce a class of functions and discuss the equations, inequalities (with a heavy emphasis on sign diagrams) and applications which involve functions in that class. The material is presented at a level that definitely prepares a student for Calculus while giving them relevant Mathematics which can be used in other classes as well. Graphing calculators are used sparingly and only as a tool to enhance the Mathematics, not to replace it. The answers to nearly all of the computational homework exercises are given in the xi text and we have gone to great lengths to write some very thought provoking discussion questions whose answers are not given. One will notice that our exercise sets are much shorter than the traditional sets of nearly 100 “drill and kill” questions which build skill devoid of understanding. Our experience has been that students can do about 15-20 homework exercises a night so we very carefully chose smaller sets of questions which cover all of the necessary skills and get the students thinking more deeply about the Mathematics involved. Critics of the Open Educational Resource movement might quip that “open-source is where bad content goes to die,” to which I say this: take a serious look at what we offer our students. Look through a few sections to see if what we’ve written is bad content in your opinion. I see this open- source book not as something which is “free and worth every penny”, but rather, as a high quality alternative to the business as usual of the textbook industry and I hope that you agree. If you have any comments, questions or concerns please feel free to contact me at jeff@stitz-zeager.com or Carl at [email protected]. Jeff Zeager Lorain County Community College January 25, 2010 xii Preface Chapter 10 Foundations of Trigonometry 10.1 Angles and their Measure This section begins our study of Trigonometry and to get started, we recall some basic definitions from Geometry. A ray is usually described as a ‘half-line’ and can be thought of as a line segment in which one of the two endpoints is pushed off infinitely distant from the other, as pictured below. The point from which the ray originates is called the initial point of the ray. P A ray with initial point P. When two rays share a common initial point they form an angle and the common initial point is called the vertex of the angle. Two examples of what are commonly thought of as angles are P An angle with vertex P. Q An angle with vertex Q. However, the two figures below also depict angles - albeit these are, in some sense, extreme cases. In the first case, the two rays are directly opposite each other forming what is known as a straight angle; in the second, the rays are identical so the ‘angle’ is indistinguishable from the ray itself. P A straight angle. Q The measure of an angle is a number which indicates the amount of rotation that separates the rays of the angle. There is one immediate problem with this, as pictured below. 694 Foundations of Trigonometry Which amount of rotation are we attempting to quantify? What we have just discovered is that we have at least two angles described by this diagram. 1 Clearly these two angles have different measures because one appears to represent a larger rotation than the other, so we must label them differently. In this book, we use lower case Greek letters such as α (alpha), β (beta), γ (gamma) and θ (theta) to label angles. So, for instance, we have α β One commonly used system to measure angles is degree measure. Quantities measured in degrees are denoted by the familiar ‘ ◦ ’ symbol. One complete revolution as shown below is 360 ◦ , and parts of a revolution are measured proportionately. 2 Thus half of a revolution (a straight angle) measures 1 2 (360 ◦ ) = 180 ◦ , a quarter of a revolution (a right angle) measures 1 4 (360 ◦ ) = 90 ◦ and so on. One revolution ↔360 ◦ 180 ◦ 90 ◦ Note that in the above figure, we have used the small square ‘ ’ to denote a right angle, as is commonplace in Geometry. Recall that if an angle measures strictly between 0 ◦ and 90 ◦ it is called an acute angle and if it measures strictly between 90 ◦ and 180 ◦ it is called an obtuse angle. It is important to note that, theoretically, we can know the measure of any angle as long as we 1 The phrase ‘at least’ will be justified in short order. 2 The choice of ‘360’ is most often attributed to the Babylonians. 10.1 Angles and their Measure 695 know the proportion it represents of entire revolution. 3 For instance, the measure of an angle which represents a rotation of 2 3 of a revolution would measure 2 3 (360 ◦ ) = 240 ◦ , the measure of an angle which constitutes only 1 12 of a revolution measures 1 12 (360 ◦ ) = 30 ◦ and an angle which indicates no rotation at all is measured as 0 ◦ . 240 ◦ 30 ◦ 0 ◦ Using our definition of degree measure, we have that 1 ◦ represents the measure of an angle which constitutes 1 360 of a revolution. Even though it may be hard to draw, it is nonetheless not difficult to imagine an angle with measure smaller than 1 ◦ . There are two ways to subdivide degrees. The first, and most familiar, is decimal degrees. For example, an angle with a measure of 30.5 ◦ would represent a rotation halfway between 30 ◦ and 31 ◦ , or equivalently, 30.5 360 = 61 720 of a full rotation. This can be taken to the limit using Calculus so that measures like √ 2 ◦ make sense. 4 The second way to divide degrees is the Degree - Minute - Second (DMS) system. In this system, one degree is divided equally into sixty minutes, and in turn, each minute is divided equally into sixty seconds. 5 In symbols, we write 1 ◦ = 60 and 1 = 60 , from which it follows that 1 ◦ = 3600 . To convert a measure of 42.125 ◦ to the DMS system, we start by noting that 42.125 ◦ = 42 ◦ +0.125 ◦ . Converting the partial amount of degrees to minutes, we find 0.125 ◦ _ 60 1 ◦ _ = 7.5 = 7 + 0.5 . Converting the partial amount of minutes to seconds gives 0.5 _ 60 1 _ = 30 . Putting it all together yields 42.125 ◦ = 42 ◦ + 0.125 ◦ = 42 ◦ + 7.5 = 42 ◦ + 7 + 0.5 = 42 ◦ + 7 + 30 = 42 ◦ 7 30 On the other hand, to convert 117 ◦ 15 45 to decimal degrees, we first compute 15 _ 1 ◦ 60 _ = 1 4 ◦ and 45 _ 1 ◦ 3600 _ = 1 80 ◦ . Then we find 3 This is how a protractor is graded. 4 Awesome math pun aside, this is the same idea behind defining irrational exponents in Section 6.1. 5 Does this kind of system seem familiar? 696 Foundations of Trigonometry 117 ◦ 15 45 = 117 ◦ + 15 + 45 = 117 ◦ + 1 4 ◦ + 1 80 ◦ = 9381 80 ◦ = 117.2625 ◦ Recall that two acute angles are called complementary angles if their measures add to 90 ◦ . Two angles, either a pair of right angles or one acute angle and one obtuse angle, are called supplementary angles if their measures add to 180 ◦ . In the diagram below, the angles α and β are supplementary angles while the pair γ and θ are complementary angles. α β Supplementary Angles γ θ Complementary Angles In practice, the distinction between the angle itself and its measure is blurred so that the sentence ‘α is an angle measuring 42 ◦ ’ is often abbreviated as ‘α = 42 ◦ .’ It is now time for an example. Example 10.1.1. Let α = 111.371 ◦ and β = 37 ◦ 28 17 . 1. Convert α to the DMS system. Round your answer to the nearest second. 2. Convert β to decimal degrees. Round your answer to the nearest thousandth of a degree. 3. Sketch α and β. 4. Find a supplementary angle for α. 5. Find a complementary angle for β. Solution. 1. To convert α to the DMS system, we start with 111.371 ◦ = 111 ◦ + 0.371 ◦ . Next we convert 0.371 ◦ _ 60 1 ◦ _ = 22.26 . Writing 22.26 = 22 + 0.26 , we convert 0.26 _ 60 1 _ = 15.6 . Hence, 111.371 ◦ = 111 ◦ + 0.371 ◦ = 111 ◦ + 22.26 = 111 ◦ + 22 + 0.26 = 111 ◦ + 22 + 15.6 = 111 ◦ 22 15.6 Rounding to seconds, we obtain α ≈ 111 ◦ 22 16 . 10.1 Angles and their Measure 697 2. To convert β to decimal degrees, we convert 28 _ 1 ◦ 60 _ = 7 15 ◦ and 17 _ 1 ◦ 3600 _ = 17 3600 ◦ . Putting it all together, we have 37 ◦ 28 17 = 37 ◦ + 28 + 17 = 37 ◦ + 7 15 ◦ + 17 3600 ◦ = 134897 3600 ◦ ≈ 37.471 ◦ 3. To sketch α, we first note that 90 ◦ < α < 180 ◦ . If we divide this range in half, we get 90 ◦ < α < 135 ◦ , and once more, we have 90 ◦ < α < 112.5 ◦ . This gives us a pretty good estimate for α, as shown below. 6 Proceeding similarly for β, we find 0 ◦ < β < 90 ◦ , then 0 ◦ < β < 45 ◦ , 22.5 ◦ < β < 45 ◦ , and lastly, 33.75 ◦ < β < 45 ◦ . Angle α Angle β 4. To find a supplementary angle for α, we seek an angle θ so that α + θ = 180 ◦ . We get θ = 180 ◦ −α = 180 ◦ −111.371 ◦ = 68.629 ◦ . 5. To find a complementary angle for β, we seek an angle γ so that β + γ = 90 ◦ . We get γ = 90 ◦ − β = 90 ◦ − 37 ◦ 28 17 . While we could reach for the calculator to obtain an approximate answer, we choose instead to do a bit of sexagesimal 7 arithmetic. We first rewrite 90 ◦ = 90 ◦ 0 0 = 89 ◦ 60 0 = 89 ◦ 59 60 . In essence, we are ‘borrowing’ 1 ◦ = 60 from the degree place, and then borrowing 1 = 60 from the minutes place. 8 This yields, γ = 90 ◦ −37 ◦ 28 17 = 89 ◦ 59 60 −37 ◦ 28 17 = 52 ◦ 31 43 . Up to this point, we have discussed only angles which measure between 0 ◦ and 360 ◦ , inclusive. Ultimately, we want to use the arsenal of Algebra which we have stockpiled in Chapters 1 through 9 to not only solve geometric problems involving angles, but also to extend their applicability to other real-world phenomena. A first step in this direction is to extend our notion of ‘angle’ from merely measuring an extent of rotation to quantities which can be associated with real numbers. To that end, we introduce the concept of an oriented angle. As its name suggests, in an oriented 6 If this process seems hauntingly familiar, it should. Compare this method to the Bisection Method introduced in Section 3.3. 7 Like ‘latus rectum,’ this is also a real math term. 8 This is the exact same kind of ‘borrowing’ you used to do in Elementary School when trying to find 300 − 125. Back then, you were working in a base ten system; here, it is base sixty. 698 Foundations of Trigonometry angle, the direction of the rotation is important. We imagine the angle being swept out starting from an initial side and ending at a terminal side, as shown below. When the rotation is counter-clockwise 9 from initial side to terminal side, we say that the angle is positive; when the rotation is clockwise, we say that the angle is negative. Initial Side T e r m i n a l S i d e Initial Side T e r m i n a l S i d e A positive angle, 45 ◦ A negative angle, −45 ◦ At this point, we also extend our allowable rotations to include angles which encompass more than one revolution. For example, to sketch an angle with measure 450 ◦ we start with an initial side, rotate counter-clockwise one complete revolution (to take care of the ‘first’ 360 ◦ ) then continue with an additional 90 ◦ counter-clockwise rotation, as seen below. 450 ◦ To further connect angles with the Algebra which has come before, we shall often overlay an angle diagram on the coordinate plane. An angle is said to be in standard position if its vertex is the origin and its initial side coincides with the positive x-axis. Angles in standard position are classified according to where their terminal side lies. For instance, an angle in standard position whose terminal side lies in Quadrant I is called a ‘Quadrant I angle’. If the terminal side of an angle lies on one of the coordinate axes, it is called a quadrantal angle. Two angles in standard position are called coterminal if they share the same terminal side. 10 In the figure below, α = 120 ◦ and β = −240 ◦ are two coterminal Quadrant II angles drawn in standard position. Note that α = β +360 ◦ , or equivalently, β = α −360 ◦ . We leave it as an exercise to the reader to verify that coterminal angles always differ by a multiple of 360 ◦ . 11 More precisely, if α and β are coterminal angles, then β = α + 360 ◦ k where k is an integer. 12 9 ‘widdershins’ 10 Note that by being in standard position they automatically share the same initial side which is the positive x-axis. 11 It is worth noting that all of the pathologies of Analytic Trigonometry result from this innocuous fact. 12 Recall that this means k = 0, ±1, ±2, . . .. 10.1 Angles and their Measure 699 x y α = 120 ◦ β = −240 ◦ −4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 Two coterminal angles, α = 120 ◦ and β = −240 ◦ , in standard position. Example 10.1.2. Graph each of the (oriented) angles below in standard position and classify them according to where their terminal side lies. Find three coterminal angles, at least one of which is positive and one of which is negative. 1. α = 60 ◦ 2. β = −225 ◦ 3. γ = 540 ◦ 4. φ = −750 ◦ Solution. 1. To graph α = 60 ◦ , we draw an angle with its initial side on the positive x-axis and rotate counter-clockwise 60 ◦ 360 ◦ = 1 6 of a revolution. We see that α is a Quadrant I angle. To find angles which are coterminal, we look for angles θ of the form θ = α + 360 ◦ k, for some integer k. When k = 1, we get θ = 60 ◦ +360 ◦ = 420 ◦ . Substituting k = −1 gives θ = 60 ◦ −360 ◦ = −300 ◦ . Finally, if we let k = 2, we get θ = 60 ◦ + 720 ◦ = 780 ◦ . 2. Since β = −225 ◦ is negative, we start at the positive x-axis and rotate clockwise 225 ◦ 360 ◦ = 5 8 of a revolution. We see that β is a Quadrant II angle. To find coterminal angles, we proceed as before and compute θ = −225 ◦ + 360 ◦ k for integer values of k. We find 135 ◦ , −585 ◦ and 495 ◦ are all coterminal with −225 ◦ . x y α = 60 ◦ −4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 x y β = −225 ◦ −4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 α = 60 ◦ in standard position. β = −225 ◦ in standard position. 700 Foundations of Trigonometry 3. Since γ = 540 ◦ is positive, we rotate counter-clockwise from the positive x-axis. One full revolution accounts for 360 ◦ , with 180 ◦ , or 1 2 of a revolution remaining. Since the terminal side of γ lies on the negative x-axis, γ is a quadrantal angle. All angles coterminal with γ are of the form θ = 540 ◦ + 360 ◦ k, where k is an integer. Working through the arithmetic, we find three such angles: 180 ◦ , −180 ◦ and 900 ◦ . 4. The Greek letter φ is pronounced ‘fee’ or ‘fie’ and since φ is negative, we begin our rotation clockwise from the positive x-axis. Two full revolutions account for 720 ◦ , with just 30 ◦ or 1 12 of a revolution to go. We find that φ is a Quadrant IV angle. To find coterminal angles, we compute θ = −750 ◦ + 360 ◦ k for a few integers k and obtain −390 ◦ , −30 ◦ and 330 ◦ . x y γ = 540 ◦ −4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 x y φ = −750 ◦ −4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 γ = 540 ◦ in standard position. φ = −750 ◦ in standard position. Note that since there are infinitely many integers, any given angle has infinitely many coterminal angles, and the reader is encouraged to plot the few sets of coterminal angles found in Example 10.1.2 to see this. We are now just one step away from completely marrying angles with the real numbers and the rest of Algebra. To that end, we recall this definition from Geometry. Definition 10.1. The real number π is defined to be the ratio of a circle’s circumference to its diameter. In symbols, given a circle of circumference C and diameter d, π = C d While Definition 10.1 is quite possibly the ‘standard’ definition of π, the authors would be remiss if we didn’t mention that buried in this definition is actually a theorem. As the reader is probably aware, the number π is a mathematical constant - that is, it doesn’t matter which circle is selected, the ratio of its circumference to its diameter will have the same value as any other circle. While this is indeed true, it is far from obvious and leads to a counterintuitive scenario which is explored in the Exercises. Since the diameter of a circle is twice its radius, we can quickly rearrange the equation in Definition 10.1 to get a formula more useful for our purposes, namely: 2π = C r 10.1 Angles and their Measure 701 This tells us that for any circle, the ratio of its circumference to its radius is also always constant; in this case the constant is 2π. Suppose now we take a portion of the circle, so instead of comparing the entire circumference C to the radius, we compare some arc measuring s units in length to the radius, as depicted below. Let θ be the central angle subtended by this arc, that is, an angle whose vertex is the center of the circle and whose determining rays pass through the endpoints of the arc. Using proportionality arguments, it stands to reason that the ratio s r should also be a constant among all circles, and it is this ratio which defines the radian measure of an angle. θ s r r The radian measure of θ is s r . To get a better feel for radian measure, we note that an angle with radian measure 1 means the corresponding arc length s equals the radius of the circle r, hence s = r. When the radian measure is 2, we have s = 2r; when the radian measure is 3, s = 3r, and so forth. Thus the radian measure of an angle θ tells us how many ‘radius lengths’ we need to sweep out along the circle to subtend the angle θ. α r r r β r r r r r r α has radian measure 1 β has radian measure 4 Since one revolution sweeps out the entire circumference 2πr, one revolution has radian measure 2πr r = 2π. From this we can find the radian measure of other central angles using proportions, 702 Foundations of Trigonometry just like we did with degrees. For instance, half of a revolution has radian measure 1 2 (2π) = π, a quarter revolution has radian measure 1 4 (2π) = π 2 , and so forth. Note that, by definition, the radian measure of an angle is a length divided by another length so that these measurements are actually dimensionless and are considered ‘pure’ numbers. For this reason, we do not use any symbols to denote radian measure, but we use the word ‘radians’ to denote these dimensionless units as needed. For instance, we say one revolution measures ‘2π radians,’ half of a revolution measures ‘π radians,’ and so forth. As with degree measure, the distinction between the angle itself and its measure is often blurred in practice, so when we write ‘θ = π 2 ’, we mean θ is an angle which measures π 2 radians. 13 We extend radian measure to oriented angles, just as we did with degrees beforehand, so that a positive measure indicates counter-clockwise rotation and a negative measure indicates clockwise rotation. 14 Much like before, two positive angles α and β are supplementary if α +β = π and complementary if α+β = π 2 . Finally, we leave it to the reader to show that when using radian measure, two angles α and β are coterminal if and only if β = α + 2πk for some integer k. Example 10.1.3. Graph each of the (oriented) angles below in standard position and classify them according to where their terminal side lies. Find three coterminal angles, at least one of which is positive and one of which is negative. 1. α = π 6 2. β = − 4π 3 3. γ = 9π 4 4. φ = − 5π 2 Solution. 1. The angle α = π 6 is positive, so we draw an angle with its initial side on the positive x-axis and rotate counter-clockwise (π/6) 2π = 1 12 of a revolution. Thus α is a Quadrant I angle. Coterminal angles θ are of the form θ = α + 2π k, for some integer k. To make the arithmetic a bit easier, we note that 2π = 12π 6 , thus when k = 1, we get θ = π 6 + 12π 6 = 13π 6 . Substituting k = −1 gives θ = π 6 − 12π 6 = − 11π 6 and when we let k = 2, we get θ = π 6 + 24π 6 = 25π 6 . 2. Since β = − 4π 3 is negative, we start at the positive x-axis and rotate clockwise (4π/3) 2π = 2 3 of a revolution. We find β to be a Quadrant II angle. To find coterminal angles, we proceed as before using 2π = 6π 3 , and compute θ = − 4π 3 + 6π 3 k for integer values of k. We obtain 2π 3 , − 10π 3 and 8π 3 as coterminal angles. 13 The authors are well aware that we are now identifying radians with real numbers. We will justify this shortly. 14 This, in turn, endows the subtended arcs with an orientation as well. We address this in short order. 10.1 Angles and their Measure 703 x y α = π 6 −4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 x y β = − 4π 3 −4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 α = π 6 in standard position. β = − 4π 3 in standard position. 3. Since γ = 9π 4 is positive, we rotate counter-clockwise from the positive x-axis. One full revolution accounts for 2π = 8π 4 of the radian measure with π 4 or 1 8 of a revolution remaining. We have γ as a Quadrant I angle. All angles coterminal with γ are of the form θ = 9π 4 + 8π 4 k, where k is an integer. Working through the arithmetic, we find: π 4 , − 7π 4 and 17π 4 . 4. To graph φ = − 5π 2 , we begin our rotation clockwise from the positive x-axis. As 2π = 4π 2 , after one full revolution clockwise, we have π 2 or 1 4 of a revolution remaining. Since the terminal side of φ lies on the negative y-axis, φ is a quadrantal angle. To find coterminal angles, we compute θ = − 5π 2 + 4π 2 k for a few integers k and obtain − π 2 , 3π 2 and 7π 2 . x y γ = 9π 4 −4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 x y φ = − 5π 2 −4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 γ = 9π 4 in standard position. φ = − 5π 2 in standard position. It is worth mentioning that we could have plotted the angles in Example 10.1.3 by first converting them to degree measure and following the procedure set forth in Example 10.1.2. While converting back and forth from degrees and radians is certainly a good skill to have, it is best that you learn to ‘think in radians’ as well as you can ‘think in degrees’. The authors would, however, be 704 Foundations of Trigonometry derelict in our duties if we ignored the basic conversion between these systems altogether. Since one revolution counter-clockwise measures 360 ◦ and the same angle measures 2π radians, we can use the proportion 2π radians 360 ◦ , or its reduced equivalent, π radians 180 ◦ , as the conversion factor between the two systems. For example, to convert 60 ◦ to radians we find 60 ◦ _ π radians 180 ◦ _ = π 3 radians, or simply π 3 . To convert from radian measure back to degrees, we multiply by the ratio 180 ◦ π radian . For example, − 5π 6 radians is equal to _ − 5π 6 radians _ _ 180 ◦ π radians _ = −150 ◦ . 15 Of particular interest is the fact that an angle which measures 1 in radian measure is equal to 180 ◦ π ≈ 57.2958 ◦ . We summarize these conversions below. Equation 10.1. Degree - Radian Conversion: • To convert degree measure to radian measure, multiply by π radians 180 ◦ • To convert radian measure to degree measure, multiply by 180 ◦ π radians In light of Example 10.1.3 and Equation 10.1, the reader may well wonder what the allure of radian measure is. The numbers involved are, admittedly, much more complicated than degree measure. The answer lies in how easily angles in radian measure can be identified with real numbers. Consider the Unit Circle, x 2 +y 2 = 1, as drawn below, the angle θ in standard position and the corresponding arc measuring s units in length. By definition, and the fact that the Unit Circle has radius 1, the radian measure of θ is s r = s 1 = s so that, once again blurring the distinction between an angle and its measure, we have θ = s. In order to identify real numbers with oriented angles, we make good use of this fact by essentially ‘wrapping’ the real number line around the Unit Circle and associating to each real number t an oriented arc on the Unit Circle with initial point (1, 0). Viewing the vertical line x = 1 as another real number line demarcated like the y-axis, given a real number t > 0, we ‘wrap’ the (vertical) interval [0, t] around the Unit Circle in a counter-clockwise fashion. The resulting arc has a length of t units and therefore the corresponding angle has radian measure equal to t. If t < 0, we wrap the interval [t, 0] clockwise around the Unit Circle. Since we have defined clockwise rotation as having negative radian measure, the angle determined by this arc has radian measure equal to t. If t = 0, we are at the point (1, 0) on the x-axis which corresponds to an angle with radian measure 0. In this way, we identify each real number t with the corresponding angle with radian measure t. 15 Note that the negative sign indicates clockwise rotation in both systems, and so it is carried along accordingly. 10.1 Angles and their Measure 705 x y 1 1 θ s x y 1 1 t t x y 1 1 t t On the Unit Circle, θ = s. Identifying t > 0 with an angle. Identifying t < 0 with an angle. Example 10.1.4. Sketch the oriented arc on the Unit Circle corresponding to each of the following real numbers. 1. t = 3π 4 2. t = −2π 3. t = −2 4. t = 117 Solution. 1. The arc associated with t = 3π 4 is the arc on the Unit Circle which subtends the angle 3π 4 in radian measure. Since 3π 4 is 3 8 of a revolution, we have an arc which begins at the point (1, 0) proceeds counter-clockwise up to midway through Quadrant II. 2. Since one revolution is 2π radians, and t = −2π is negative, we graph the arc which begins at (1, 0) and proceeds clockwise for one full revolution. x y 1 1 t = 3π 4 x y 1 1 t = −2π 3. Like t = −2π, t = −2 is negative, so we begin our arc at (1, 0) and proceed clockwise around the unit circle. Since π ≈ 3.14 and π 2 ≈ 1.57, we find that rotating 2 radians clockwise from the point (1, 0) lands us in Quadrant III. To more accurately place the endpoint, we proceed as we did in Example 10.1.1, successively halving the angle measure until we find 5π 8 ≈ 1.96 which tells us our arc extends just a bit beyond the quarter mark into Quadrant III. 706 Foundations of Trigonometry 4. Since 117 is positive, the arc corresponding to t = 117 begins at (1, 0) and proceeds counter- clockwise. As 117 is much greater than 2π, we wrap around the Unit Circle several times before finally reaching our endpoint. We approximate 117 2π as 18.62 which tells us we complete 18 revolutions counter-clockwise with 0.62, or just shy of 5 8 of a revolution to spare. In other words, the terminal side of the angle which measures 117 radians in standard position is just short of being midway through Quadrant III. x y 1 1 t = −2 x y 1 1 t = 117 10.1.1 Applications of Radian Measure: Circular Motion Now that we have paired angles with real numbers via radian measure, a whole world of applications awaits us. Our first excursion into this realm comes by way of circular motion. Suppose an object is moving as pictured below along a circular path of radius r from the point P to the point Q in an amount of time t. P Q r θ s Here s represents a displacement so that s > 0 means the object is traveling in a counter-clockwise direction and s < 0 indicates movement in a clockwise direction. Note that with this convention the formula we used to define radian measure, namely θ = s r , still holds since a negative value of s incurred from a clockwise displacement matches the negative we assign to θ for a clockwise rotation. In Physics, the average velocity of the object, denoted v and read as ‘v-bar’, is defined as the average rate of change of the position of the object with respect to time. 16 As a result, we 16 See Definition 2.3 in Section 2.1 for a review of this concept. 10.1 Angles and their Measure 707 have v = displacement time = s t . The quantity v has units of length time and conveys two ideas: the direction in which the object is moving and how fast the position of the object is changing. The contribution of direction in the quantity v is either to make it positive (in the case of counter-clockwise motion) or negative (in the case of clockwise motion), so that the quantity [v[ quantifies how fast the object is moving - it is the speed of the object. Measuring θ in radians we have θ = s r thus s = rθ and v = s t = rθ t = r θ t The quantity θ t is called the average angular velocity of the object. It is denoted by ω and is read ‘omega-bar’. The quantity ω is the average rate of change of the angle θ with respect to time and thus has units radians time . If ω is constant throughout the duration of the motion, then it can be shown 17 that the average velocities involved, namely v and ω, are the same as their instantaneous counterparts, v and ω, respectively. In this case, v is simply called the ‘velocity’ of the object and is the instantaneous rate of change of the position of the object with respect to time. 18 Similarly, ω is called the ‘angular velocity’ and is the instantaneous rate of change of the angle with respect to time. If the path of the object were ‘uncurled’ from a circle to form a line segment, then the velocity of the object on that line segment would be the same as the velocity on the circle. For this reason, the quantity v is often called the linear velocity of the object in order to distinguish it from the angular velocity, ω. Putting together the ideas of the previous paragraph, we get the following. Equation 10.2. Velocity for Circular Motion: For an object moving on a circular path of radius r with constant angular velocity ω, the (linear) velocity of the object is given by v = rω. We need to talk about units here. The units of v are length time , the units of r are length only, and the units of ω are radians time . Thus the left hand side of the equation v = rω has units length time , whereas the right hand side has units length radians time = length·radians time . The supposed contradiction in units is resolved by remembering that radians are a dimensionless quantity and angles in radian measure are identified with real numbers so that the units length·radians time reduce to the units length time . We are long overdue for an example. Example 10.1.5. Assuming that the surface of the Earth is a sphere, any point on the Earth can be thought of as an object traveling on a circle which completes one revolution in (approximately) 24 hours. The path traced out by the point during this 24 hour period is the Latitude of that point. Lakeland Community College is at 41.628 ◦ north latitude, and it can be shown 19 that the radius of the earth at this Latitude is approximately 2960 miles. Find the linear velocity, in miles per hour, of Lakeland Community College as the world turns. Solution. To use the formula v = rω, we first need to compute the angular velocity ω. The earth makes one revolution in 24 hours, and one revolution is 2π radians, so ω = 2π radians 24 hours = π 12 hours , 17 You guessed it, using Calculus . . . 18 See the discussion on Page 161 for more details on the idea of an ‘instantaneous’ rate of change. 19 We will discuss how we arrived at this approximation in Example 10.2.6. 708 Foundations of Trigonometry where, once again, we are using the fact that radians are real numbers and are dimensionless. (For simplicity’s sake, we are also assuming that we are viewing the rotation of the earth as counter- clockwise so ω > 0.) Hence, the linear velocity is v = 2960 miles π 12 hours ≈ 775 miles hour It is worth noting that the quantity 1 revolution 24 hours in Example 10.1.5 is called the ordinary frequency of the motion and is usually denoted by the variable f. The ordinary frequency is a measure of how often an object makes a complete cycle of the motion. The fact that ω = 2πf suggests that ω is also a frequency. Indeed, it is called the angular frequency of the motion. On a related note, the quantity T = 1 f is called the period of the motion and is the amount of time it takes for the object to complete one cycle of the motion. In the scenario of Example 10.1.5, the period of the motion is 24 hours, or one day. The concepts of frequency and period help frame the equation v = rω in a new light. That is, if ω is fixed, points which are farther from the center of rotation need to travel faster to maintain the same angular frequency since they have farther to travel to make one revolution in one period’s time. The distance of the object to the center of rotation is the radius of the circle, r, and is the ‘magnification factor’ which relates ω and v. We will have more to say about frequencies and periods in Section 11.1. While we have exhaustively discussed velocities associated with circular motion, we have yet to discuss a more natural question: if an object is moving on a circular path of radius r with a fixed angular velocity (frequency) ω, what is the position of the object at time t? The answer to this question is the very heart of Trigonometry and is answered in the next section. 10.1 Angles and their Measure 709 10.1.2 Exercises In Exercises 1 - 4, convert the angles into the DMS system. Round each of your answers to the nearest second. 1. 63.75 ◦ 2. 200.325 ◦ 3. −317.06 ◦ 4. 179.999 ◦ In Exercises 5 - 8, convert the angles into decimal degrees. Round each of your answers to three decimal places. 5. 125 ◦ 50 6. −32 ◦ 10 12 7. 502 ◦ 35 8. 237 ◦ 58 43 In Exercises 9 - 28, graph the oriented angle in standard position. Classify each angle according to where its terminal side lies and then give two coterminal angles, one of which is positive and the other negative. 9. 330 ◦ 10. −135 ◦ 11. 120 ◦ 12. 405 ◦ 13. −270 ◦ 14. 5π 6 15. − 11π 3 16. 5π 4 17. 3π 4 18. − π 3 19. 7π 2 20. π 4 21. − π 2 22. 7π 6 23. − 5π 3 24. 3π 25. −2π 26. − π 4 27. 15π 4 28. − 13π 6 In Exercises 29 - 36, convert the angle from degree measure into radian measure, giving the exact value in terms of π. 29. 0 ◦ 30. 240 ◦ 31. 135 ◦ 32. −270 ◦ 33. −315 ◦ 34. 150 ◦ 35. 45 ◦ 36. −225 ◦ In Exercises 37 - 44, convert the angle from radian measure into degree measure. 37. π 38. − 2π 3 39. 7π 6 40. 11π 6 41. π 3 42. 5π 3 43. − π 6 44. π 2 710 Foundations of Trigonometry In Exercises 45 - 49, sketch the oriented arc on the Unit Circle which corresponds to the given real number. 45. t = 5π 6 46. t = −π 47. t = 6 48. t = −2 49. t = 12 50. A yo-yo which is 2.25 inches in diameter spins at a rate of 4500 revolutions per minute. How fast is the edge of the yo-yo spinning in miles per hour? Round your answer to two decimal places. 51. How many revolutions per minute would the yo-yo in exercise 50 have to complete if the edge of the yo-yo is to be spinning at a rate of 42 miles per hour? Round your answer to two decimal places. 52. In the yo-yo trick ‘Around the World,’ the performer throws the yo-yo so it sweeps out a vertical circle whose radius is the yo-yo string. If the yo-yo string is 28 inches long and the yo-yo takes 3 seconds to complete one revolution of the circle, compute the speed of the yo-yo in miles per hour. Round your answer to two decimal places. 53. A computer hard drive contains a circular disk with diameter 2.5 inches and spins at a rate of 7200 RPM (revolutions per minute). Find the linear speed of a point on the edge of the disk in miles per hour. 54. A rock got stuck in the tread of my tire and when I was driving 70 miles per hour, the rock came loose and hit the inside of the wheel well of the car. How fast, in miles per hour, was the rock traveling when it came out of the tread? (The tire has a diameter of 23 inches.) 55. The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height is 136 feet. (Remember this from Exercise 17 in Section 7.2?) It completes two revolutions in 2 minutes and 7 seconds. 20 Assuming the riders are at the edge of the circle, how fast are they traveling in miles per hour? 56. Consider the circle of radius r pictured below with central angle θ, measured in radians, and subtended arc of length s. Prove that the area of the shaded sector is A = 1 2 r 2 θ. (Hint: Use the proportion A area of the circle = s circumference of the circle .) θ s r r 20 Source: Cedar Point’s webpage. 10.1 Angles and their Measure 711 In Exercises 57 - 62, use the result of Exercise 56 to compute the areas of the circular sectors with the given central angles and radii. 57. θ = π 6 , r = 12 58. θ = 5π 4 , r = 100 59. θ = 330 ◦ , r = 9.3 60. θ = π, r = 1 61. θ = 240 ◦ , r = 5 62. θ = 1 ◦ , r = 117 63. Imagine a rope tied around the Earth at the equator. Show that you need to add only 2π feet of length to the rope in order to lift it one foot above the ground around the entire equator. (You do NOT need to know the radius of the Earth to show this.) 64. With the help of your classmates, look for a proof that π is indeed a constant. 712 Foundations of Trigonometry 10.1.3 Answers 1. 63 ◦ 45 2. 200 ◦ 19 30 3. −317 ◦ 3 36 4. 179 ◦ 59 56 5. 125.833 ◦ 6. −32.17 ◦ 7. 502.583 ◦ 8. 237.979 ◦ 9. 330 ◦ is a Quadrant IV angle coterminal with 690 ◦ and −30 ◦ x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 10. −135 ◦ is a Quadrant III angle coterminal with 225 ◦ and −495 ◦ x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 11. 120 ◦ is a Quadrant II angle coterminal with 480 ◦ and −240 ◦ x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 12. 405 ◦ is a Quadrant I angle coterminal with 45 ◦ and −315 ◦ x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 13. −270 ◦ lies on the positive y-axis coterminal with 90 ◦ and −630 ◦ x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 14. 5π 6 is a Quadrant II angle coterminal with 17π 6 and − 7π 6 x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 10.1 Angles and their Measure 713 15. − 11π 3 is a Quadrant I angle coterminal with π 3 and − 5π 3 x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 16. 5π 4 is a Quadrant III angle coterminal with 13π 4 and − 3π 4 x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 17. 3π 4 is a Quadrant II angle coterminal with 11π 4 and − 5π 4 x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 18. − π 3 is a Quadrant IV angle coterminal with 5π 3 and − 7π 3 x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 19. 7π 2 lies on the negative y-axis coterminal with 3π 2 and − π 2 x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 20. π 4 is a Quadrant I angle coterminal with 9π 4 and − 7π 4 x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 714 Foundations of Trigonometry 21. − π 2 lies on the negative y-axis coterminal with 3π 2 and − 5π 2 x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 22. 7π 6 is a Quadrant III angle coterminal with 19π 6 and − 5π 6 x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 23. − 5π 3 is a Quadrant I angle coterminal with π 3 and − 11π 3 x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 24. 3π lies on the negative x-axis coterminal with π and −π x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 25. −2π lies on the positive x-axis coterminal with 2π and −4π x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 26. − π 4 is a Quadrant IV angle coterminal with 7π 4 and − 9π 4 x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 10.1 Angles and their Measure 715 27. 15π 4 is a Quadrant IV angle coterminal with 7π 4 and − π 4 x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 28. − 13π 6 is a Quadrant IV angle coterminal with 11π 6 and − π 6 x y −4−3−2−1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 29. 0 30. 4π 3 31. 3π 4 32. − 3π 2 33. − 7π 4 34. 5π 6 35. π 4 36. − 5π 4 37. 180 ◦ 38. −120 ◦ 39. 210 ◦ 40. 330 ◦ 41. 60 ◦ 42. 300 ◦ 43. −30 ◦ 44. 90 ◦ 45. t = 5π 6 x y 1 1 46. t = −π x y 1 1 47. t = 6 x y 1 1 48. t = −2 x y 1 1 716 Foundations of Trigonometry 49. t = 12 (between 1 and 2 revolutions) x y 1 1 50. About 30.12 miles per hour 51. About 6274.52 revolutions per minute 52. About 3.33 miles per hour 53. About 53.55 miles per hour 54. 70 miles per hour 55. About 4.32 miles per hour 57. 12π square units 58. 6250π square units 59. 79.2825π ≈ 249.07 square units 60. π 2 square units 61. 50π 3 square units 62. 38.025π ≈ 119.46 square units 10.2 The Unit Circle: Cosine and Sine 717 10.2 The Unit Circle: Cosine and Sine In Section 10.1.1, we introduced circular motion and derived a formula which describes the linear velocity of an object moving on a circular path at a constant angular velocity. One of the goals of this section is describe the position of such an object. To that end, consider an angle θ in standard position and let P denote the point where the terminal side of θ intersects the Unit Circle. By associating the point P with the angle θ, we are assigning a position on the Unit Circle to the angle θ. The x-coordinate of P is called the cosine of θ, written cos(θ), while the y-coordinate of P is called the sine of θ, written sin(θ). 1 The reader is encouraged to verify that these rules used to match an angle with its cosine and sine do, in fact, satisfy the definition of a function. That is, for each angle θ, there is only one associated value of cos(θ) and only one associated value of sin(θ). x y 1 1 θ x y 1 1 P(cos(θ), sin(θ)) θ Example 10.2.1. Find the cosine and sine of the following angles. 1. θ = 270 ◦ 2. θ = −π 3. θ = 45 ◦ 4. θ = π 6 5. θ = 60 ◦ Solution. 1. To find cos (270 ◦ ) and sin (270 ◦ ), we plot the angle θ = 270 ◦ in standard position and find the point on the terminal side of θ which lies on the Unit Circle. Since 270 ◦ represents 3 4 of a counter-clockwise revolution, the terminal side of θ lies along the negative y-axis. Hence, the point we seek is (0, −1) so that cos _ 3π 2 _ = 0 and sin _ 3π 2 _ = −1. 2. The angle θ = −π represents one half of a clockwise revolution so its terminal side lies on the negative x-axis. The point on the Unit Circle that lies on the negative x-axis is (−1, 0) which means cos(−π) = −1 and sin(−π) = 0. 1 The etymology of the name ‘sine’ is quite colorful, and the interested reader is invited to research it; the ‘co’ in ‘cosine’ is explained in Section 10.4. 718 Foundations of Trigonometry x y 1 1 P(0, −1) θ = 270 ◦ Finding cos (270 ◦ ) and sin (270 ◦ ) x y 1 1 P(−1, 0) θ = −π Finding cos (−π) and sin (−π) 3. When we sketch θ = 45 ◦ in standard position, we see that its terminal does not lie along any of the coordinate axes which makes our job of finding the cosine and sine values a bit more difficult. Let P(x, y) denote the point on the terminal side of θ which lies on the Unit Circle. By definition, x = cos (45 ◦ ) and y = sin (45 ◦ ). If we drop a perpendicular line segment from P to the x-axis, we obtain a 45 ◦ − 45 ◦ − 90 ◦ right triangle whose legs have lengths x and y units. From Geometry, 2 we get y = x. Since P(x, y) lies on the Unit Circle, we have x 2 + y 2 = 1. Substituting y = x into this equation yields 2x 2 = 1, or x = ± _ 1 2 = ± √ 2 2 . Since P(x, y) lies in the first quadrant, x > 0, so x = cos (45 ◦ ) = √ 2 2 and with y = x we have y = sin (45 ◦ ) = √ 2 2 . x y 1 1 P(x, y) θ = 45 ◦ θ = 45 ◦ 45 ◦ x y P(x, y) 2 Can you show this? 10.2 The Unit Circle: Cosine and Sine 719 4. As before, the terminal side of θ = π 6 does not lie on any of the coordinate axes, so we proceed using a triangle approach. Letting P(x, y) denote the point on the terminal side of θ which lies on the Unit Circle, we drop a perpendicular line segment from P to the x-axis to form a 30 ◦ − 60 ◦ − 90 ◦ right triangle. After a bit of Geometry 3 we find y = 1 2 so sin _ π 6 _ = 1 2 . Since P(x, y) lies on the Unit Circle, we substitute y = 1 2 into x 2 + y 2 = 1 to get x 2 = 3 4 , or x = ± √ 3 2 . Here, x > 0 so x = cos _ π 6 _ = √ 3 2 . x y 1 1 P(x, y) θ = π 6 θ = π 6 = 30 ◦ 60 ◦ x y P(x, y) 5. Plotting θ = 60 ◦ in standard position, we find it is not a quadrantal angle and set about using a triangle approach. Once again, we get a 30 ◦ −60 ◦ −90 ◦ right triangle and, after the usual computations, find x = cos (60 ◦ ) = 1 2 and y = sin (60 ◦ ) = √ 3 2 . x y 1 1 P(x, y) θ = 60 ◦ θ = 60 ◦ 30 ◦ x y P(x, y) 3 Again, can you show this? 720 Foundations of Trigonometry In Example 10.2.1, it was quite easy to find the cosine and sine of the quadrantal angles, but for non-quadrantal angles, the task was much more involved. In these latter cases, we made good use of the fact that the point P(x, y) = (cos(θ), sin(θ)) lies on the Unit Circle, x 2 + y 2 = 1. If we substitute x = cos(θ) and y = sin(θ) into x 2 + y 2 = 1, we get (cos(θ)) 2 + (sin(θ)) 2 = 1. An unfortunate 4 convention, which the authors are compelled to perpetuate, is to write (cos(θ)) 2 as cos 2 (θ) and (sin(θ)) 2 as sin 2 (θ). Rewriting the identity using this convention results in the following theorem, which is without a doubt one of the most important results in Trigonometry. Theorem 10.1. The Pythagorean Identity: For any angle θ, cos 2 (θ) + sin 2 (θ) = 1. The moniker ‘Pythagorean’ brings to mind the Pythagorean Theorem, from which both the Distance Formula and the equation for a circle are ultimately derived. 5 The word ‘Identity’ reminds us that, regardless of the angle θ, the equation in Theorem 10.1 is always true. If one of cos(θ) or sin(θ) is known, Theorem 10.1 can be used to determine the other, up to a (±) sign. If, in addition, we know where the terminal side of θ lies when in standard position, then we can remove the ambiguity of the (±) and completely determine the missing value as the next example illustrates. Example 10.2.2. Using the given information about θ, find the indicated value. 1. If θ is a Quadrant II angle with sin(θ) = 3 5 , find cos(θ). 2. If π < θ < 3π 2 with cos(θ) = − √ 5 5 , find sin(θ). 3. If sin(θ) = 1, find cos(θ). Solution. 1. When we substitute sin(θ) = 3 5 into The Pythagorean Identity, cos 2 (θ) + sin 2 (θ) = 1, we obtain cos 2 (θ) + 9 25 = 1. Solving, we find cos(θ) = ± 4 5 . Since θ is a Quadrant II angle, its terminal side, when plotted in standard position, lies in Quadrant II. Since the x-coordinates are negative in Quadrant II, cos(θ) is too. Hence, cos(θ) = − 4 5 . 2. Substituting cos(θ) = − √ 5 5 into cos 2 (θ) + sin 2 (θ) = 1 gives sin(θ) = ± 2 √ 5 = ± 2 √ 5 5 . Since we are given that π < θ < 3π 2 , we know θ is a Quadrant III angle. Hence both its sine and cosine are negative and we conclude sin(θ) = − 2 √ 5 5 . 3. When we substitute sin(θ) = 1 into cos 2 (θ) + sin 2 (θ) = 1, we find cos(θ) = 0. Another tool which helps immensely in determining cosines and sines of angles is the symmetry inherent in the Unit Circle. Suppose, for instance, we wish to know the cosine and sine of θ = 5π 6 . We plot θ in standard position below and, as usual, let P(x, y) denote the point on the terminal side of θ which lies on the Unit Circle. Note that the terminal side of θ lies π 6 radians short of one half revolution. In Example 10.2.1, we determined that cos _ π 6 _ = √ 3 2 and sin _ π 6 _ = 1 2 . This means 4 This is unfortunate from a ‘function notation’ perspective. See Section 10.6. 5 See Sections 1.1 and 7.2 for details. 10.2 The Unit Circle: Cosine and Sine 721 that the point on the terminal side of the angle π 6 , when plotted in standard position, is _ √ 3 2 , 1 2 _ . From the figure below, it is clear that the point P(x, y) we seek can be obtained by reflecting that point about the y-axis. Hence, cos _ 5π 6 _ = − √ 3 2 and sin _ 5π 6 _ = 1 2 . x y 1 1 P(x, y) θ = 5π 6 π 6 x y 1 1 _√ 3 2 , 1 2 _ P _ − √ 3 2 , 1 2 _ π 6 π 6 θ = 5π 6 In the above scenario, the angle π 6 is called the reference angle for the angle 5π 6 . In general, for a non-quadrantal angle θ, the reference angle for θ (usually denoted α) is the acute angle made between the terminal side of θ and the x-axis. If θ is a Quadrant I or IV angle, α is the angle between the terminal side of θ and the positive x-axis; if θ is a Quadrant II or III angle, α is the angle between the terminal side of θ and the negative x-axis. If we let P denote the point (cos(θ), sin(θ)), then P lies on the Unit Circle. Since the Unit Circle possesses symmetry with respect to the x-axis, y-axis and origin, regardless of where the terminal side of θ lies, there is a point Q symmetric with P which determines θ’s reference angle, α as seen below. x y 1 1 P = Q α x y 1 1 P Q α α Reference angle α for a Quadrant I angle Reference angle α for a Quadrant II angle 722 Foundations of Trigonometry x y 1 1 P Q α α x y 1 1 P Q α α Reference angle α for a Quadrant III angle Reference angle α for a Quadrant IV angle We have just outlined the proof of the following theorem. Theorem 10.2. Reference Angle Theorem. Suppose α is the reference angle for θ. Then cos(θ) = ±cos(α) and sin(θ) = ±sin(α), where the choice of the (±) depends on the quadrant in which the terminal side of θ lies. In light of Theorem 10.2, it pays to know the cosine and sine values for certain common angles. In the table below, we summarize the values which we consider essential and must be memorized. Cosine and Sine Values of Common Angles θ(degrees) θ(radians) cos(θ) sin(θ) 0 ◦ 0 1 0 30 ◦ π 6 √ 3 2 1 2 45 ◦ π 4 √ 2 2 √ 2 2 60 ◦ π 3 1 2 √ 3 2 90 ◦ π 2 0 1 Example 10.2.3. Find the cosine and sine of the following angles. 1. θ = 225 ◦ 2. θ = 11π 6 3. θ = − 5π 4 4. θ = 7π 3 Solution. 1. We begin by plotting θ = 225 ◦ in standard position and find its terminal side overshoots the negative x-axis to land in Quadrant III. Hence, we obtain θ’s reference angle α by subtracting: α = θ − 180 ◦ = 225 ◦ − 180 ◦ = 45 ◦ . Since θ is a Quadrant III angle, both cos(θ) < 0 and 10.2 The Unit Circle: Cosine and Sine 723 sin(θ) < 0. The Reference Angle Theorem yields: cos (225 ◦ ) = −cos (45 ◦ ) = − √ 2 2 and sin (225 ◦ ) = −sin (45 ◦ ) = − √ 2 2 . 2. The terminal side of θ = 11π 6 , when plotted in standard position, lies in Quadrant IV, just shy of the positive x-axis. To find θ’s reference angle α, we subtract: α = 2π −θ = 2π − 11π 6 = π 6 . Since θ is a Quadrant IV angle, cos(θ) > 0 and sin(θ) < 0, so the Reference Angle Theorem gives: cos _ 11π 6 _ = cos _ π 6 _ = √ 3 2 and sin _ 11π 6 _ = −sin _ π 6 _ = − 1 2 . x y 1 1 θ = 225 ◦ 45 ◦ Finding cos (225 ◦ ) and sin (225 ◦ ) x y 1 1 θ = 11π 6 π 6 Finding cos _ 11π 6 _ and sin _ 11π 6 _ 3. To plot θ = − 5π 4 , we rotate clockwise an angle of 5π 4 from the positive x-axis. The terminal side of θ, therefore, lies in Quadrant II making an angle of α = 5π 4 − π = π 4 radians with respect to the negative x-axis. Since θ is a Quadrant II angle, the Reference Angle Theorem gives: cos _ − 5π 4 _ = −cos _ π 4 _ = − √ 2 2 and sin _ − 5π 4 _ = sin _ π 4 _ = √ 2 2 . 4. Since the angle θ = 7π 3 measures more than 2π = 6π 3 , we find the terminal side of θ by rotating one full revolution followed by an additional α = 7π 3 − 2π = π 3 radians. Since θ and α are coterminal, cos _ 7π 3 _ = cos _ π 3 _ = 1 2 and sin _ 7π 3 _ = sin _ π 3 _ = √ 3 2 . x y 1 1 θ = − 5π 4 π 4 Finding cos _ − 5π 4 _ and sin _ − 5π 4 _ x y 1 1 θ = 7π 3 π 3 Finding cos _ 7π 3 _ and sin _ 7π 3 _ 724 Foundations of Trigonometry The reader may have noticed that when expressed in radian measure, the reference angle for a non-quadrantal angle is easy to spot. Reduced fraction multiples of π with a denominator of 6 have π 6 as a reference angle, those with a denominator of 4 have π 4 as their reference angle, and those with a denominator of 3 have π 3 as their reference angle. 6 The Reference Angle Theorem in conjunction with the table of cosine and sine values on Page 722 can be used to generate the following figure, which the authors feel should be committed to memory. x y (0, 1) (1, 0) (0, −1) (−1, 0) _ √ 2 2 , √ 2 2 _ _ √ 3 2 , 1 2 _ _ 1 2 , √ 3 2 _ _ − √ 2 2 , √ 2 2 _ _ − √ 3 2 , 1 2 _ _ − 1 2 , √ 3 2 _ _ √ 2 2 , − √ 2 2 _ _ √ 3 2 , − 1 2 _ _ 1 2 , − √ 3 2 _ _ − √ 2 2 , − √ 2 2 _ _ − √ 3 2 , − 1 2 _ _ − 1 2 , − √ 3 2 _ 0, 2π π 2 π 3π 2 π 4 π 6 π 3 3π 4 5π 6 2π 3 5π 4 7π 6 4π 3 7π 4 11π 6 5π 3 Important Points on the Unit Circle 6 For once, we have something convenient about using radian measure in contrast to the abstract theoretical nonsense about using them as a ‘natural’ way to match oriented angles with real numbers! 10.2 The Unit Circle: Cosine and Sine 725 The next example summarizes all of the important ideas discussed thus far in the section. Example 10.2.4. Suppose α is an acute angle with cos(α) = 5 13 . 1. Find sin(α) and use this to plot α in standard position. 2. Find the sine and cosine of the following angles: (a) θ = π +α (b) θ = 2π −α (c) θ = 3π −α (d) θ = π 2 +α Solution. 1. Proceeding as in Example 10.2.2, we substitute cos(α) = 5 13 into cos 2 (α) + sin 2 (α) = 1 and find sin(α) = ± 12 13 . Since α is an acute (and therefore Quadrant I) angle, sin(α) is positive. Hence, sin(α) = 12 13 . To plot α in standard position, we begin our rotation on the positive x-axis to the ray which contains the point (cos(α), sin(α)) = _ 5 13 , 12 13 _ . x y 1 1 _ 5 13 , 12 13 _ α Sketching α 2. (a) To find the cosine and sine of θ = π + α, we first plot θ in standard position. We can imagine the sum of the angles π+α as a sequence of two rotations: a rotation of π radians followed by a rotation of α radians. 7 We see that α is the reference angle for θ, so by The Reference Angle Theorem, cos(θ) = ±cos(α) = ± 5 13 and sin(θ) = ±sin(α) = ± 12 13 . Since the terminal side of θ falls in Quadrant III, both cos(θ) and sin(θ) are negative, hence, cos(θ) = − 5 13 and sin(θ) = − 12 13 . 7 Since π +α = α +π, θ may be plotted by reversing the order of rotations given here. You should do this. 726 Foundations of Trigonometry x y 1 1 θ π α Visualizing θ = π +α x y 1 1 θ α θ has reference angle α (b) Rewriting θ = 2π − α as θ = 2π + (−α), we can plot θ by visualizing one complete revolution counter-clockwise followed by a clockwise revolution, or ‘backing up,’ of α radians. We see that α is θ’s reference angle, and since θ is a Quadrant IV angle, the Reference Angle Theorem gives: cos(θ) = 5 13 and sin(θ) = − 12 13 . x y 1 1 θ 2π −α Visualizing θ = 2π −α x y 1 1 θ α θ has reference angle α (c) Taking a cue from the previous problem, we rewrite θ = 3π −α as θ = 3π + (−α). The angle 3π represents one and a half revolutions counter-clockwise, so that when we ‘back up’ α radians, we end up in Quadrant II. Using the Reference Angle Theorem, we get cos(θ) = − 5 13 and sin(θ) = 12 13 . 10.2 The Unit Circle: Cosine and Sine 727 x y 1 1 3π −α Visualizing 3π −α x y 1 1 θ α θ has reference angle α (d) To plot θ = π 2 +α, we first rotate π 2 radians and follow up with α radians. The reference angle here is not α, so The Reference Angle Theorem is not immediately applicable. (It’s important that you see why this is the case. Take a moment to think about this before reading on.) Let Q(x, y) be the point on the terminal side of θ which lies on the Unit Circle so that x = cos(θ) and y = sin(θ). Once we graph α in standard position, we use the fact that equal angles subtend equal chords to show that the dotted lines in the figure below are equal. Hence, x = cos(θ) = − 12 13 . Similarly, we find y = sin(θ) = 5 13 . x y 1 1 θ π 2 α Visualizing θ = π 2 +α x y 1 1 P _ 5 13 , 12 13 _ Q(x, y) α α Using symmetry to determine Q(x, y) 728 Foundations of Trigonometry Our next example asks us to solve some very basic trigonometric equations. 8 Example 10.2.5. Find all of the angles which satisfy the given equation. 1. cos(θ) = 1 2 2. sin(θ) = − 1 2 3. cos(θ) = 0. Solution. Since there is no context in the problem to indicate whether to use degrees or radians, we will default to using radian measure in our answers to each of these problems. This choice will be justified later in the text when we study what is known as Analytic Trigonometry. In those sections to come, radian measure will be the only appropriate angle measure so it is worth the time to become “fluent in radians” now. 1. If cos(θ) = 1 2 , then the terminal side of θ, when plotted in standard position, intersects the Unit Circle at x = 1 2 . This means θ is a Quadrant I or IV angle with reference angle π 3 . x y 1 1 2 1 π 3 x y 1 1 2 1 π 3 One solution in Quadrant I is θ = π 3 , and since all other Quadrant I solutions must be coterminal with π 3 , we find θ = π 3 +2πk for integers k. 9 Proceeding similarly for the Quadrant IV case, we find the solution to cos(θ) = 1 2 here is 5π 3 , so our answer in this Quadrant is θ = 5π 3 + 2πk for integers k. 2. If sin(θ) = − 1 2 , then when θ is plotted in standard position, its terminal side intersects the Unit Circle at y = − 1 2 . From this, we determine θ is a Quadrant III or Quadrant IV angle with reference angle π 6 . 8 We will more formally study of trigonometric equations in Section 10.7. Enjoy these relatively straightforward exercises while they last! 9 Recall in Section 10.1, two angles in radian measure are coterminal if and only if they differ by an integer multiple of 2π. Hence to describe all angles coterminal with a given angle, we add 2πk for integers k = 0, ±1, ±2, . . . . 10.2 The Unit Circle: Cosine and Sine 729 x y 1 − 1 2 1 π 6 x y 1 − 1 2 1 π 6 In Quadrant III, one solution is 7π 6 , so we capture all Quadrant III solutions by adding integer multiples of 2π: θ = 7π 6 + 2πk. In Quadrant IV, one solution is 11π 6 so all the solutions here are of the form θ = 11π 6 + 2πk for integers k. 3. The angles with cos(θ) = 0 are quadrantal angles whose terminal sides, when plotted in standard position, lie along the y-axis. x y 1 1 π 2 x y 1 1 π 2 π 2 π While, technically speaking, π 2 isn’t a reference angle we can nonetheless use it to find our answers. If we follow the procedure set forth in the previous examples, we find θ = π 2 + 2πk and θ = 3π 2 + 2πk for integers, k. While this solution is correct, it can be shortened to θ = π 2 +πk for integers k. (Can you see why this works from the diagram?) One of the key items to take from Example 10.2.5 is that, in general, solutions to trigonometric equations consist of infinitely many answers. To get a feel for these answers, the reader is encouraged to follow our mantra from Chapter 9 - that is, ‘When in doubt, write it out!’ This is especially important when checking answers to the exercises. For example, another Quadrant IV solution to sin(θ) = − 1 2 is θ = − π 6 . Hence, the family of Quadrant IV answers to number 2 above could just have easily been written θ = − π 6 + 2πk for integers k. While on the surface, this family may look 730 Foundations of Trigonometry different than the stated solution of θ = 11π 6 +2πk for integers k, we leave it to the reader to show they represent the same list of angles. 10.2.1 Beyond the Unit Circle We began the section with a quest to describe the position of a particle experiencing circular motion. In defining the cosine and sine functions, we assigned to each angle a position on the Unit Circle. In this subsection, we broaden our scope to include circles of radius r centered at the origin. Consider for the moment the acute angle θ drawn below in standard position. Let Q(x, y) be the point on the terminal side of θ which lies on the circle x 2 + y 2 = r 2 , and let P(x , y ) be the point on the terminal side of θ which lies on the Unit Circle. Now consider dropping perpendiculars from P and Q to create two right triangles, ∆OPA and ∆OQB. These triangles are similar, 10 thus it follows that x x = r 1 = r, so x = rx and, similarly, we find y = ry . Since, by definition, x = cos(θ) and y = sin(θ), we get the coordinates of Q to be x = r cos(θ) and y = r sin(θ). By reflecting these points through the x-axis, y-axis and origin, we obtain the result for all non-quadrantal angles θ, and we leave it to the reader to verify these formulas hold for the quadrantal angles. x y 1 1 r r Q(x, y) P (x , y ) θ θ x y 1 O B(x, 0) A(x , 0) P(x , y ) Q(x, y) = (r cos(θ), r sin(θ)) Not only can we describe the coordinates of Q in terms of cos(θ) and sin(θ) but since the radius of the circle is r = _ x 2 +y 2 , we can also express cos(θ) and sin(θ) in terms of the coordinates of Q. These results are summarized in the following theorem. Theorem 10.3. If Q(x, y) is the point on the terminal side of an angle θ, plotted in standard position, which lies on the circle x 2 +y 2 = r 2 then x = r cos(θ) and y = r sin(θ). Moreover, cos(θ) = x r = x _ x 2 +y 2 and sin(θ) = y r = y _ x 2 +y 2 10 Do you remember why? 10.2 The Unit Circle: Cosine and Sine 731 Note that in the case of the Unit Circle we have r = _ x 2 +y 2 = 1, so Theorem 10.3 reduces to our definitions of cos(θ) and sin(θ). Example 10.2.6. 1. Suppose that the terminal side of an angle θ, when plotted in standard position, contains the point Q(4, −2). Find sin(θ) and cos(θ). 2. In Example 10.1.5 in Section 10.1, we approximated the radius of the earth at 41.628 ◦ north latitude to be 2960 miles. Justify this approximation if the radius of the Earth at the Equator is approximately 3960 miles. Solution. 1. Using Theorem 10.3 with x = 4 and y = −2, we find r = _ (4) 2 + (−2) 2 = √ 20 = 2 √ 5 so that cos(θ) = x r = 4 2 √ 5 = 2 √ 5 5 and y = y r = −2 2 √ 5 = − √ 5 5 . 2. Assuming the Earth is a sphere, a cross-section through the poles produces a circle of radius 3960 miles. Viewing the Equator as the x-axis, the value we seek is the x-coordinate of the point Q(x, y) indicated in the figure below. x y Q(4, −2) −4 −2 2 4 −2 −4 2 4 The terminal side of θ contains Q(4, −2) x y 3960 3960 Q(x, y) 41.628 ◦ A point on the Earth at 41.628 ◦ N Using Theorem 10.3, we get x = 3960 cos (41.628 ◦ ). Using a calculator in ‘degree’ mode, we find 3960 cos (41.628 ◦ ) ≈ 2960. Hence, the radius of the Earth at North Latitude 41.628 ◦ is approximately 2960 miles. 732 Foundations of Trigonometry Theorem 10.3 gives us what we need to describe the position of an object traveling in a circular path of radius r with constant angular velocity ω. Suppose that at time t, the object has swept out an angle measuring θ radians. If we assume that the object is at the point (r, 0) when t = 0, the angle θ is in standard position. By definition, ω = θ t which we rewrite as θ = ωt. According to Theorem 10.3, the location of the object Q(x, y) on the circle is found using the equations x = r cos(θ) = r cos(ωt) and y = r sin(θ) = r sin(ωt). Hence, at time t, the object is at the point (r cos(ωt), r sin(ωt)). We have just argued the following. Equation 10.3. Suppose an object is traveling in a circular path of radius r centered at the origin with constant angular velocity ω. If t = 0 corresponds to the point (r, 0), then the x and y coordinates of the object are functions of t and are given by x = r cos(ωt) and y = r sin(ωt). Here, ω > 0 indicates a counter-clockwise direction and ω < 0 indicates a clockwise direction. x y 1 1 r r Q(x, y) = (r cos(ωt), r sin(ωt)) θ = ωt Equations for Circular Motion Example 10.2.7. Suppose we are in the situation of Example 10.1.5. Find the equations of motion of Lakeland Community College as the earth rotates. Solution. From Example 10.1.5, we take r = 2960 miles and and ω = π 12 hours . Hence, the equations of motion are x = r cos(ωt) = 2960 cos _ π 12 t _ and y = r sin(ωt) = 2960 sin _ π 12 t _ , where x and y are measured in miles and t is measured in hours. In addition to circular motion, Theorem 10.3 is also the key to developing what is usually called ‘right triangle’ trigonometry. 11 As we shall see in the sections to come, many applications in trigonometry involve finding the measures of the angles in, and lengths of the sides of, right triangles. Indeed, we made good use of some properties of right triangles to find the exact values of the cosine and sine of many of the angles in Example 10.2.1, so the following development shouldn’t be that much of a surprise. Consider the generic right triangle below with corresponding acute angle θ. The side with length a is called the side of the triangle adjacent to θ; the side with length b is called the side of the triangle opposite θ; and the remaining side of length c (the side opposite the 11 You may have been exposed to this in High School. 10.2 The Unit Circle: Cosine and Sine 733 right angle) is called the hypotenuse. We now imagine drawing this triangle in Quadrant I so that the angle θ is in standard position with the adjacent side to θ lying along the positive x-axis. θ a b c x y c c P(a, b) θ According to the Pythagorean Theorem, a 2 + b 2 = c 2 , so that the point P(a, b) lies on a circle of radius c. Theorem 10.3 tells us that cos(θ) = a c and sin(θ) = b c , so we have determined the cosine and sine of θ in terms of the lengths of the sides of the right triangle. Thus we have the following theorem. Theorem 10.4. Suppose θ is an acute angle residing in a right triangle. If the length of the side adjacent to θ is a, the length of the side opposite θ is b, and the length of the hypotenuse is c, then cos(θ) = a c and sin(θ) = b c . Example 10.2.8. Find the measure of the missing angle and the lengths of the missing sides of: 30 ◦ 7 Solution. The first and easiest task is to find the measure of the missing angle. Since the sum of angles of a triangle is 180 ◦ , we know that the missing angle has measure 180 ◦ − 30 ◦ − 90 ◦ = 60 ◦ . We now proceed to find the lengths of the remaining two sides of the triangle. Let c denote the length of the hypotenuse of the triangle. By Theorem 10.4, we have cos (30 ◦ ) = 7 c , or c = 7 cos(30 ◦ ) . Since cos (30 ◦ ) = √ 3 2 , we have, after the usual fraction gymnastics, c = 14 √ 3 3 . At this point, we have two ways to proceed to find the length of the side opposite the 30 ◦ angle, which we’ll denote b. We know the length of the adjacent side is 7 and the length of the hypotenuse is 14 √ 3 3 , so we 734 Foundations of Trigonometry could use the Pythagorean Theorem to find the missing side and solve (7) 2 + b 2 = _ 14 √ 3 3 _ 2 for b. Alternatively, we could use Theorem 10.4, namely that sin (30 ◦ ) = b c . Choosing the latter, we find b = c sin (30 ◦ ) = 14 √ 3 3 1 2 = 7 √ 3 3 . The triangle with all of its data is recorded below. 30 ◦ 7 b = 7 √ 3 3 c = 14 √ 3 3 60 ◦ We close this section by noting that we can easily extend the functions cosine and sine to real numbers by identifying a real number t with the angle θ = t radians. Using this identification, we define cos(t) = cos(θ) and sin(t) = sin(θ). In practice this means expressions like cos(π) and sin(2) can be found by regarding the inputs as angles in radian measure or real numbers; the choice is the reader’s. If we trace the identification of real numbers t with angles θ in radian measure to its roots on page 704, we can spell out this correspondence more precisely. For each real number t, we associate an oriented arc t units in length with initial point (1, 0) and endpoint P(cos(t), sin(t)). x y 1 1 θ = t t x y 1 1 P(cos(t), sin(t)) θ = t In the same way we studied polynomial, rational, exponential, and logarithmic functions, we will study the trigonometric functions f(t) = cos(t) and g(t) = sin(t). The first order of business is to find the domains and ranges of these functions. Whether we think of identifying the real number t with the angle θ = t radians, or think of wrapping an oriented arc around the Unit Circle to find coordinates on the Unit Circle, it should be clear that both the cosine and sine functions are defined for all real numbers t. In other words, the domain of f(t) = cos(t) and of g(t) = sin(t) is (−∞, ∞). Since cos(t) and sin(t) represent x- and y-coordinates, respectively, of points on the Unit Circle, they both take on all of the values between −1 an 1, inclusive. In other words, the range of f(t) = cos(t) and of g(t) = sin(t) is the interval [−1, 1]. To summarize: 10.2 The Unit Circle: Cosine and Sine 735 Theorem 10.5. Domain and Range of the Cosine and Sine Functions: • The function f(t) = cos(t) • The function g(t) = sin(t) – has domain (−∞, ∞) – has domain (−∞, ∞) – has range [−1, 1] – has range [−1, 1] Suppose, as in the Exercises, we are asked to solve an equation such as sin(t) = − 1 2 . As we have already mentioned, the distinction between t as a real number and as an angle θ = t radians is often blurred. Indeed, we solve sin(t) = − 1 2 in the exact same manner 12 as we did in Example 10.2.5 number 2. Our solution is only cosmetically different in that the variable used is t rather than θ: t = 7π 6 +2πk or t = 11π 6 +2πk for integers, k. We will study the cosine and sine functions in greater detail in Section 10.5. Until then, keep in mind that any properties of cosine and sine developed in the following sections which regard them as functions of angles in radian measure apply equally well if the inputs are regarded as real numbers. 12 Well, to be pedantic, we would be technically using ‘reference numbers’ or ‘reference arcs’ instead of ‘reference angles’ – but the idea is the same. 736 Foundations of Trigonometry 10.2.2 Exercises In Exercises 1 - 20, find the exact value of the cosine and sine of the given angle. 1. θ = 0 2. θ = π 4 3. θ = π 3 4. θ = π 2 5. θ = 2π 3 6. θ = 3π 4 7. θ = π 8. θ = 7π 6 9. θ = 5π 4 10. θ = 4π 3 11. θ = 3π 2 12. θ = 5π 3 13. θ = 7π 4 14. θ = 23π 6 15. θ = − 13π 2 16. θ = − 43π 6 17. θ = − 3π 4 18. θ = − π 6 19. θ = 10π 3 20. θ = 117π In Exercises 21 - 30, use the results developed throughout the section to find the requested value. 21. If sin(θ) = − 7 25 with θ in Quadrant IV, what is cos(θ)? 22. If cos(θ) = 4 9 with θ in Quadrant I, what is sin(θ)? 23. If sin(θ) = 5 13 with θ in Quadrant II, what is cos(θ)? 24. If cos(θ) = − 2 11 with θ in Quadrant III, what is sin(θ)? 25. If sin(θ) = − 2 3 with θ in Quadrant III, what is cos(θ)? 26. If cos(θ) = 28 53 with θ in Quadrant IV, what is sin(θ)? 27. If sin(θ) = 2 √ 5 5 and π 2 < θ < π, what is cos(θ)? 28. If cos(θ) = √ 10 10 and 2π < θ < 5π 2 , what is sin(θ)? 29. If sin(θ) = −0.42 and π < θ < 3π 2 , what is cos(θ)? 30. If cos(θ) = −0.98 and π 2 < θ < π, what is sin(θ)? 10.2 The Unit Circle: Cosine and Sine 737 In Exercises 31 - 39, find all of the angles which satisfy the given equation. 31. sin(θ) = 1 2 32. cos(θ) = − √ 3 2 33. sin(θ) = 0 34. cos(θ) = √ 2 2 35. sin(θ) = √ 3 2 36. cos(θ) = −1 37. sin(θ) = −1 38. cos(θ) = √ 3 2 39. cos(θ) = −1.001 In Exercises 40 - 48, solve the equation for t. (See the comments following Theorem 10.5.) 40. cos(t) = 0 41. sin(t) = − √ 2 2 42. cos(t) = 3 43. sin(t) = − 1 2 44. cos(t) = 1 2 45. sin(t) = −2 46. cos(t) = 1 47. sin(t) = 1 48. cos(t) = − √ 2 2 In Exercises 49 - 54, use your calculator to approximate the given value to three decimal places. Make sure your calculator is in the proper angle measurement mode! 49. sin(78.95 ◦ ) 50. cos(−2.01) 51. sin(392.994) 52. cos(207 ◦ ) 53. sin (π ◦ ) 54. cos(e) In Exercises 55 - 58, find the measurement of the missing angle and the lengths of the missing sides. (See Example 10.2.8) 55. Find θ, b, and c. 30 ◦ 1 b c θ 56. Find θ, a, and c. 45 ◦ 3 a c θ 738 Foundations of Trigonometry 57. Find α, a, and b. 33 ◦ 8 b a α 58. Find β, a, and c. 48 ◦ 6 a c β In Exercises 59 - 64, assume that θ is an acute angle in a right triangle and use Theorem 10.4 to find the requested side. 59. If θ = 12 ◦ and the side adjacent to θ has length 4, how long is the hypotenuse? 60. If θ = 78.123 ◦ and the hypotenuse has length 5280, how long is the side adjacent to θ? 61. If θ = 59 ◦ and the side opposite θ has length 117.42, how long is the hypotenuse? 62. If θ = 5 ◦ and the hypotenuse has length 10, how long is the side opposite θ? 63. If θ = 5 ◦ and the hypotenuse has length 10, how long is the side adjacent to θ? 64. If θ = 37.5 ◦ and the side opposite θ has length 306, how long is the side adjacent to θ? In Exercises 65 - 68, let θ be the angle in standard position whose terminal side contains the given point then compute cos(θ) and sin(θ). 65. P(−7, 24) 66. Q(3, 4) 67. R(5, −9) 68. T(−2, −11) In Exercises 69 - 72, find the equations of motion for the given scenario. Assume that the center of the motion is the origin, the motion is counter-clockwise and that t = 0 corresponds to a position along the positive x-axis. (See Equation 10.3 and Example 10.1.5.) 69. A point on the edge of the spinning yo-yo in Exercise 50 from Section 10.1. Recall: The diameter of the yo-yo is 2.25 inches and it spins at 4500 revolutions per minute. 70. The yo-yo in exercise 52 from Section 10.1. Recall: The radius of the circle is 28 inches and it completes one revolution in 3 seconds. 71. A point on the edge of the hard drive in Exercise 53 from Section 10.1. Recall: The diameter of the hard disk is 2.5 inches and it spins at 7200 revolutions per minute. 10.2 The Unit Circle: Cosine and Sine 739 72. A passenger on the Big Wheel in Exercise 55 from Section 10.1. Recall: The diameter is 128 feet and completes 2 revolutions in 2 minutes, 7 seconds. 73. Consider the numbers: 0, 1, 2, 3, 4. Take the square root of each of these numbers, then divide each by 2. The resulting numbers should look hauntingly familiar. (See the values in the table on 722.) 74. Let α and β be the two acute angles of a right triangle. (Thus α and β are complementary angles.) Show that sin(α) = cos(β) and sin(β) = cos(α). The fact that co-functions of complementary angles are equal in this case is not an accident and a more general result will be given in Section 10.4. 75. In the scenario of Equation 10.3, we assumed that at t = 0, the object was at the point (r, 0). If this is not the case, we can adjust the equations of motion by introducing a ‘time delay.’ If t 0 > 0 is the first time the object passes through the point (r, 0), show, with the help of your classmates, the equations of motion are x = r cos(ω(t −t 0 )) and y = r sin(ω(t −t 0 )). 740 Foundations of Trigonometry 10.2.3 Answers 1. cos(0) = 1, sin(0) = 0 2. cos _ π 4 _ = √ 2 2 , sin _ π 4 _ = √ 2 2 3. cos _ π 3 _ = 1 2 , sin _ π 3 _ = √ 3 2 4. cos _ π 2 _ = 0, sin _ π 2 _ = 1 5. cos _ 2π 3 _ = − 1 2 , sin _ 2π 3 _ = √ 3 2 6. cos _ 3π 4 _ = − √ 2 2 , sin _ 3π 4 _ = √ 2 2 7. cos(π) = −1, sin(π) = 0 8. cos _ 7π 6 _ = − √ 3 2 , sin _ 7π 6 _ = − 1 2 9. cos _ 5π 4 _ = − √ 2 2 , sin _ 5π 4 _ = − √ 2 2 10. cos _ 4π 3 _ = − 1 2 , sin _ 4π 3 _ = − √ 3 2 11. cos _ 3π 2 _ = 0, sin _ 3π 2 _ = −1 12. cos _ 5π 3 _ = 1 2 , sin _ 5π 3 _ = − √ 3 2 13. cos _ 7π 4 _ = √ 2 2 , sin _ 7π 4 _ = − √ 2 2 14. cos _ 23π 6 _ = √ 3 2 , sin _ 23π 6 _ = − 1 2 15. cos _ − 13π 2 _ = 0, sin _ − 13π 2 _ = −1 16. cos _ − 43π 6 _ = − √ 3 2 , sin _ − 43π 6 _ = 1 2 17. cos _ − 3π 4 _ = − √ 2 2 , sin _ − 3π 4 _ = − √ 2 2 18. cos _ − π 6 _ = √ 3 2 , sin _ − π 6 _ = − 1 2 19. cos _ 10π 3 _ = − 1 2 , sin _ 10π 3 _ = − √ 3 2 20. cos(117π) = −1, sin(117π) = 0 21. If sin(θ) = − 7 25 with θ in Quadrant IV, then cos(θ) = 24 25 . 22. If cos(θ) = 4 9 with θ in Quadrant I, then sin(θ) = √ 65 9 . 23. If sin(θ) = 5 13 with θ in Quadrant II, then cos(θ) = − 12 13 . 24. If cos(θ) = − 2 11 with θ in Quadrant III, then sin(θ) = − √ 117 11 . 25. If sin(θ) = − 2 3 with θ in Quadrant III, then cos(θ) = − √ 5 3 . 26. If cos(θ) = 28 53 with θ in Quadrant IV, then sin(θ) = − 45 53 . 10.2 The Unit Circle: Cosine and Sine 741 27. If sin(θ) = 2 √ 5 5 and π 2 < θ < π, then cos(θ) = − √ 5 5 . 28. If cos(θ) = √ 10 10 and 2π < θ < 5π 2 , then sin(θ) = 3 √ 10 10 . 29. If sin(θ) = −0.42 and π < θ < 3π 2 , then cos(θ) = − √ 0.8236 ≈ −0.9075. 30. If cos(θ) = −0.98 and π 2 < θ < π, then sin(θ) = √ 0.0396 ≈ 0.1990. 31. sin(θ) = 1 2 when θ = π 6 + 2πk or θ = 5π 6 + 2πk for any integer k. 32. cos(θ) = − √ 3 2 when θ = 5π 6 + 2πk or θ = 7π 6 + 2πk for any integer k. 33. sin(θ) = 0 when θ = πk for any integer k. 34. cos(θ) = √ 2 2 when θ = π 4 + 2πk or θ = 7π 4 + 2πk for any integer k. 35. sin(θ) = √ 3 2 when θ = π 3 + 2πk or θ = 2π 3 + 2πk for any integer k. 36. cos(θ) = −1 when θ = (2k + 1)π for any integer k. 37. sin(θ) = −1 when θ = 3π 2 + 2πk for any integer k. 38. cos(θ) = √ 3 2 when θ = π 6 + 2πk or θ = 11π 6 + 2πk for any integer k. 39. cos(θ) = −1.001 never happens 40. cos(t) = 0 when t = π 2 +πk for any integer k. 41. sin(t) = − √ 2 2 when t = 5π 4 + 2πk or t = 7π 4 + 2πk for any integer k. 42. cos(t) = 3 never happens. 43. sin(t) = − 1 2 when t = 7π 6 + 2πk or t = 11π 6 + 2πk for any integer k. 44. cos(t) = 1 2 when t = π 3 + 2πk or t = 5π 3 + 2πk for any integer k. 45. sin(t) = −2 never happens 46. cos(t) = 1 when t = 2πk for any integer k. 742 Foundations of Trigonometry 47. sin(t) = 1 when t = π 2 + 2πk for any integer k. 48. cos(t) = − √ 2 2 when t = 3π 4 + 2πk or t = 5π 4 + 2πk for any integer k. 49. sin(78.95 ◦ ) ≈ 0.981 50. cos(−2.01) ≈ −0.425 51. sin(392.994) ≈ −0.291 52. cos(207 ◦ ) ≈ −0.891 53. sin (π ◦ ) ≈ 0.055 54. cos(e) ≈ −0.912 55. θ = 60 ◦ , b = √ 3 3 , c = 2 √ 3 3 56. θ = 45 ◦ , a = 3, c = 3 √ 2 57. α = 57 ◦ , a = 8 cos(33 ◦ ) ≈ 6.709, b = 8 sin(33 ◦ ) ≈ 4.357 58. β = 42 ◦ , c = 6 sin(48 ◦ ) ≈ 8.074, a = √ c 2 −6 2 ≈ 5.402 59. The hypotenuse has length 4 cos(12 ◦ ) ≈ 4.089. 60. The side adjacent to θ has length 5280 cos(78.123 ◦ ) ≈ 1086.68. 61. The hypotenuse has length 117.42 sin(59 ◦ ) ≈ 136.99. 62. The side opposite θ has length 10 sin(5 ◦ ) ≈ 0.872. 63. The side adjacent to θ has length 10 cos(5 ◦ ) ≈ 9.962. 64. The hypotenuse has length c = 306 sin(37.5 ◦ ) ≈ 502.660, so the side adjacent to θ has length √ c 2 −306 2 ≈ 398.797. 65. cos(θ) = − 7 25 , sin(θ) = 24 25 66. cos(θ) = 3 5 , sin(θ) = 4 5 67. cos(θ) = 5 √ 106 106 , sin(θ) = − 9 √ 106 106 68. cos(θ) = − 2 √ 5 25 , sin(θ) = − 11 √ 5 25 69. r = 1.125 inches, ω = 9000π radians minute , x = 1.125 cos(9000π t), y = 1.125 sin(9000π t). Here x and y are measured in inches and t is measured in minutes. 10.2 The Unit Circle: Cosine and Sine 743 70. r = 28 inches, ω = 2π 3 radians second , x = 28 cos _ 2π 3 t _ , y = 28 sin _ 2π 3 t _ . Here x and y are measured in inches and t is measured in seconds. 71. r = 1.25 inches, ω = 14400π radians minute , x = 1.25 cos(14400π t), y = 1.25 sin(14400π t). Here x and y are measured in inches and t is measured in minutes. 72. r = 64 feet, ω = 4π 127 radians second , x = 64 cos _ 4π 127 t _ , y = 64 sin _ 4π 127 t _ . Here x and y are measured in feet and t is measured in seconds 744 Foundations of Trigonometry 10.3 The Six Circular Functions and Fundamental Identities In section 10.2, we defined cos(θ) and sin(θ) for angles θ using the coordinate values of points on the Unit Circle. As such, these functions earn the moniker circular functions. 1 It turns out that cosine and sine are just two of the six commonly used circular functions which we define below. Definition 10.2. The Circular Functions: Suppose θ is an angle plotted in standard position and P(x, y) is the point on the terminal side of θ which lies on the Unit Circle. • The cosine of θ, denoted cos(θ), is defined by cos(θ) = x. • The sine of θ, denoted sin(θ), is defined by sin(θ) = y. • The secant of θ, denoted sec(θ), is defined by sec(θ) = 1 x , provided x ,= 0. • The cosecant of θ, denoted csc(θ), is defined by csc(θ) = 1 y , provided y ,= 0. • The tangent of θ, denoted tan(θ), is defined by tan(θ) = y x , provided x ,= 0. • The cotangent of θ, denoted cot(θ), is defined by cot(θ) = x y , provided y ,= 0. While we left the history of the name ‘sine’ as an interesting research project in Section 10.2, the names ‘tangent’ and ‘secant’ can be explained using the diagram below. Consider the acute angle θ below in standard position. Let P(x, y) denote, as usual, the point on the terminal side of θ which lies on the Unit Circle and let Q(1, y ) denote the point on the terminal side of θ which lies on the vertical line x = 1. θ x y 1 O B(1, 0) A(x, 0) P(x, y) Q(1, y ) = (1, tan(θ)) 1 In Theorem 10.4 we also showed cosine and sine to be functions of an angle residing in a right triangle so we could just as easily call them trigonometric functions. In later sections, you will find that we do indeed use the phrase ‘trigonometric function’ interchangeably with the term ‘circular function’. 10.3 The Six Circular Functions and Fundamental Identities 745 The word ‘tangent’ comes from the Latin meaning ‘to touch,’ and for this reason, the line x = 1 is called a tangent line to the Unit Circle since it intersects, or ‘touches’, the circle at only one point, namely (1, 0). Dropping perpendiculars from P and Q creates a pair of similar triangles ∆OPA and ∆OQB. Thus y y = 1 x which gives y = y x = tan(θ), where this last equality comes from applying Definition 10.2. We have just shown that for acute angles θ, tan(θ) is the y-coordinate of the point on the terminal side of θ which lies on the line x = 1 which is tangent to the Unit Circle. Now the word ‘secant’ means ‘to cut’, so a secant line is any line that ‘cuts through’ a circle at two points. 2 The line containing the terminal side of θ is a secant line since it intersects the Unit Circle in Quadrants I and III. With the point P lying on the Unit Circle, the length of the hypotenuse of ∆OPA is 1. If we let h denote the length of the hypotenuse of ∆OQB, we have from similar triangles that h 1 = 1 x , or h = 1 x = sec(θ). Hence for an acute angle θ, sec(θ) is the length of the line segment which lies on the secant line determined by the terminal side of θ and ‘cuts off’ the tangent line x = 1. Not only do these observations help explain the names of these functions, they serve as the basis for a fundamental inequality needed for Calculus which we’ll explore in the Exercises. Of the six circular functions, only cosine and sine are defined for all angles. Since cos(θ) = x and sin(θ) = y in Definition 10.2, it is customary to rephrase the remaining four circular functions in terms of cosine and sine. The following theorem is a result of simply replacing x with cos(θ) and y with sin(θ) in Definition 10.2. Theorem 10.6. Reciprocal and Quotient Identities: • sec(θ) = 1 cos(θ) , provided cos(θ) ,= 0; if cos(θ) = 0, sec(θ) is undefined. • csc(θ) = 1 sin(θ) , provided sin(θ) ,= 0; if sin(θ) = 0, csc(θ) is undefined. • tan(θ) = sin(θ) cos(θ) , provided cos(θ) ,= 0; if cos(θ) = 0, tan(θ) is undefined. • cot(θ) = cos(θ) sin(θ) , provided sin(θ) ,= 0; if sin(θ) = 0, cot(θ) is undefined. It is high time for an example. Example 10.3.1. Find the indicated value, if it exists. 1. sec (60 ◦ ) 2. csc _ 7π 4 _ 3. cot(3) 4. tan (θ), where θ is any angle coterminal with 3π 2 . 5. cos (θ), where csc(θ) = − √ 5 and θ is a Quadrant IV angle. 6. sin (θ), where tan(θ) = 3 and π < θ < 3π 2 . 2 Compare this with the definition given in Section 2.1. 746 Foundations of Trigonometry Solution. 1. According to Theorem 10.6, sec (60 ◦ ) = 1 cos(60 ◦ ) . Hence, sec (60 ◦ ) = 1 (1/2) = 2. 2. Since sin _ 7π 4 _ = − √ 2 2 , csc _ 7π 4 _ = 1 sin( 7π 4 ) = 1 − √ 2/2 = − 2 √ 2 = − √ 2. 3. Since θ = 3 radians is not one of the ‘common angles’ from Section 10.2, we resort to the calculator for a decimal approximation. Ensuring that the calculator is in radian mode, we find cot(3) = cos(3) sin(3) ≈ −7.015. 4. If θ is coterminal with 3π 2 , then cos(θ) = cos _ 3π 2 _ = 0 and sin(θ) = sin _ 3π 2 _ = −1. Attempting to compute tan(θ) = sin(θ) cos(θ) results in −1 0 , so tan(θ) is undefined. 5. We are given that csc(θ) = 1 sin(θ) = − √ 5 so sin(θ) = − 1 √ 5 = − √ 5 5 . As we saw in Section 10.2, we can use the Pythagorean Identity, cos 2 (θ) +sin 2 (θ) = 1, to find cos(θ) by knowing sin(θ). Substituting, we get cos 2 (θ) + _ − √ 5 5 _ 2 = 1, which gives cos 2 (θ) = 4 5 , or cos(θ) = ± 2 √ 5 5 . Since θ is a Quadrant IV angle, cos(θ) > 0, so cos(θ) = 2 √ 5 5 . 6. If tan(θ) = 3, then sin(θ) cos(θ) = 3. Be careful - this does NOT mean we can take sin(θ) = 3 and cos(θ) = 1. Instead, from sin(θ) cos(θ) = 3 we get: sin(θ) = 3 cos(θ). To relate cos(θ) and sin(θ), we once again employ the Pythagorean Identity, cos 2 (θ) + sin 2 (θ) = 1. Solving sin(θ) = 3 cos(θ) for cos(θ), we find cos(θ) = 1 3 sin(θ). Substituting this into the Pythagorean Identity, we find sin 2 (θ) + _ 1 3 sin(θ) _ 2 = 1. Solving, we get sin 2 (θ) = 9 10 so sin(θ) = ± 3 √ 10 10 . Since π < θ < 3π 2 , θ is a Quadrant III angle. This means sin(θ) < 0, so our final answer is sin(θ) = − 3 √ 10 10 . While the Reciprocal and Quotient Identities presented in Theorem 10.6 allow us to always reduce problems involving secant, cosecant, tangent and cotangent to problems involving cosine and sine, it is not always convenient to do so. 3 It is worth taking the time to memorize the tangent and cotangent values of the common angles summarized below. 3 As we shall see shortly, when solving equations involving secant and cosecant, we usually convert back to cosines and sines. However, when solving for tangent or cotangent, we usually stick with what we’re dealt. 10.3 The Six Circular Functions and Fundamental Identities 747 Tangent and Cotangent Values of Common Angles θ(degrees) θ(radians) tan(θ) cot(θ) 0 ◦ 0 0 undefined 30 ◦ π 6 √ 3 3 √ 3 45 ◦ π 4 1 1 60 ◦ π 3 √ 3 √ 3 3 90 ◦ π 2 undefined 0 Coupling Theorem 10.6 with the Reference Angle Theorem, Theorem 10.2, we get the following. Theorem 10.7. Generalized Reference Angle Theorem. The values of the circular functions of an angle, if they exist, are the same, up to a sign, of the corresponding circu- lar functions of its reference angle. More specifically, if α is the reference angle for θ, then: cos(θ) = ±cos(α), sin(θ) = ±sin(α), sec(θ) = ±sec(α), csc(θ) = ±csc(α), tan(θ) = ±tan(α) and cot(θ) = ±cot(α). The choice of the (±) depends on the quadrant in which the terminal side of θ lies. We put Theorem 10.7 to good use in the following example. Example 10.3.2. Find all angles which satisfy the given equation. 1. sec(θ) = 2 2. tan(θ) = √ 3 3. cot(θ) = −1. Solution. 1. To solve sec(θ) = 2, we convert to cosines and get 1 cos(θ) = 2 or cos(θ) = 1 2 . This is the exact same equation we solved in Example 10.2.5, number 1, so we know the answer is: θ = π 3 +2πk or θ = 5π 3 + 2πk for integers k. 2. From the table of common values, we see tan _ π 3 _ = √ 3. According to Theorem 10.7, we know the solutions to tan(θ) = √ 3 must, therefore, have a reference angle of π 3 . Our next task is to determine in which quadrants the solutions to this equation lie. Since tangent is defined as the ratio y x of points (x, y) on the Unit Circle with x ,= 0, tangent is positive when x and y have the same sign (i.e., when they are both positive or both negative.) This happens in Quadrants I and III. In Quadrant I, we get the solutions: θ = π 3 +2πk for integers k, and for Quadrant III, we get θ = 4π 3 + 2πk for integers k. While these descriptions of the solutions are correct, they can be combined into one list as θ = π 3 +πk for integers k. The latter form of the solution is best understood looking at the geometry of the situation in the diagram below. 4 4 See Example 10.2.5 number 3 in Section 10.2 for another example of this kind of simplification of the solution. 748 Foundations of Trigonometry x y 1 1 π 3 x y 1 1 π 3 π 3 π 3. From the table of common values, we see that π 4 has a cotangent of 1, which means the solutions to cot(θ) = −1 have a reference angle of π 4 . To find the quadrants in which our solutions lie, we note that cot(θ) = x y for a point (x, y) on the Unit Circle where y ,= 0. If cot(θ) is negative, then x and y must have different signs (i.e., one positive and one negative.) Hence, our solutions lie in Quadrants II and IV. Our Quadrant II solution is θ = 3π 4 + 2πk, and for Quadrant IV, we get θ = 7π 4 +2πk for integers k. Can these lists be combined? Indeed they can - one such way to capture all the solutions is: θ = 3π 4 +πk for integers k. x y 1 1 π 4 x y 1 1 π 4 π π 4 We have already seen the importance of identities in trigonometry. Our next task is to use use the Reciprocal and Quotient Identities found in Theorem 10.6 coupled with the Pythagorean Identity found in Theorem 10.1 to derive new Pythagorean-like identities for the remaining four circular functions. Assuming cos(θ) ,= 0, we may start with cos 2 (θ) + sin 2 (θ) = 1 and divide both sides by cos 2 (θ) to obtain 1 + sin 2 (θ) cos 2 (θ) = 1 cos 2 (θ) . Using properties of exponents along with the Reciprocal and Quotient Identities, this reduces to 1 + tan 2 (θ) = sec 2 (θ). If sin(θ) ,= 0, we can divide both sides of the identity cos 2 (θ) + sin 2 (θ) = 1 by sin 2 (θ), apply Theorem 10.6 once again, and obtain cot 2 (θ) + 1 = csc 2 (θ). These three Pythagorean Identities are worth memorizing and they, along with some of their other common forms, are summarized in the following theorem. 10.3 The Six Circular Functions and Fundamental Identities 749 Theorem 10.8. The Pythagorean Identities: 1. cos 2 (θ) + sin 2 (θ) = 1. Common Alternate Forms: • 1 −sin 2 (θ) = cos 2 (θ) • 1 −cos 2 (θ) = sin 2 (θ) 2. 1 + tan 2 (θ) = sec 2 (θ), provided cos(θ) ,= 0. Common Alternate Forms: • sec 2 (θ) −tan 2 (θ) = 1 • sec 2 (θ) −1 = tan 2 (θ) 3. 1 + cot 2 (θ) = csc 2 (θ), provided sin(θ) ,= 0. Common Alternate Forms: • csc 2 (θ) −cot 2 (θ) = 1 • csc 2 (θ) −1 = cot 2 (θ) Trigonometric identities play an important role in not just Trigonometry, but in Calculus as well. We’ll use them in this book to find the values of the circular functions of an angle and solve equations and inequalities. In Calculus, they are needed to simplify otherwise complicated expressions. In the next example, we make good use of the Theorems 10.6 and 10.8. Example 10.3.3. Verify the following identities. Assume that all quantities are defined. 1. 1 csc(θ) = sin(θ) 2. tan(θ) = sin(θ) sec(θ) 3. (sec(θ) −tan(θ))(sec(θ) + tan(θ)) = 1 4. sec(θ) 1 −tan(θ) = 1 cos(θ) −sin(θ) 5. 6 sec(θ) tan(θ) = 3 1 −sin(θ) − 3 1 + sin(θ) 6. sin(θ) 1 −cos(θ) = 1 + cos(θ) sin(θ) Solution. In verifying identities, we typically start with the more complicated side of the equation and use known identities to transform it into the other side of the equation. 1. To verify 1 csc(θ) = sin(θ), we start with the left side. Using csc(θ) = 1 sin(θ) , we get: 1 csc(θ) = 1 1 sin(θ) = sin(θ), which is what we were trying to prove. 750 Foundations of Trigonometry 2. Starting with the right hand side of tan(θ) = sin(θ) sec(θ), we use sec(θ) = 1 cos(θ) and find: sin(θ) sec(θ) = sin(θ) 1 cos(θ) = sin(θ) cos(θ) = tan(θ), where the last equality is courtesy of Theorem 10.6. 3. Expanding the left hand side of the equation gives: (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = sec 2 (θ) −tan 2 (θ). According to Theorem 10.8, sec 2 (θ) −tan 2 (θ) = 1. Putting it all together, (sec(θ) −tan(θ))(sec(θ) + tan(θ)) = sec 2 (θ) −tan 2 (θ) = 1. 4. While both sides of our last identity contain fractions, the left side affords us more opportu- nities to use our identities. 5 Substituting sec(θ) = 1 cos(θ) and tan(θ) = sin(θ) cos(θ) , we get: sec(θ) 1 −tan(θ) = 1 cos(θ) 1 − sin(θ) cos(θ) = 1 cos(θ) 1 − sin(θ) cos(θ) cos(θ) cos(θ) = _ 1 cos(θ) _ (cos(θ)) _ 1 − sin(θ) cos(θ) _ (cos(θ)) = 1 (1)(cos(θ)) − _ sin(θ) cos(θ) _ (cos(θ)) = 1 cos(θ) −sin(θ) , which is exactly what we had set out to show. 5. The right hand side of the equation seems to hold more promise. We get common denomina- tors and add: 3 1 −sin(θ) − 3 1 + sin(θ) = 3(1 + sin(θ)) (1 −sin(θ))(1 + sin(θ)) − 3(1 −sin(θ)) (1 + sin(θ))(1 −sin(θ)) = 3 + 3 sin(θ) 1 −sin 2 (θ) − 3 −3 sin(θ) 1 −sin 2 (θ) = (3 + 3 sin(θ)) −(3 −3 sin(θ)) 1 −sin 2 (θ) = 6 sin(θ) 1 −sin 2 (θ) 5 Or, to put to another way, earn more partial credit if this were an exam question! 10.3 The Six Circular Functions and Fundamental Identities 751 At this point, it is worth pausing to remind ourselves of our goal. We wish to trans- form this expression into 6 sec(θ) tan(θ). Using a reciprocal and quotient identity, we find 6 sec(θ) tan(θ) = 6 _ 1 cos(θ) __ sin(θ) cos(θ) _ . In other words, we need to get cosines in our denomina- tor. Theorem 10.8 tells us 1 −sin 2 (θ) = cos 2 (θ) so we get: 3 1 −sin(θ) − 3 1 + sin(θ) = 6 sin(θ) 1 −sin 2 (θ) = 6 sin(θ) cos 2 (θ) = 6 _ 1 cos(θ) __ sin(θ) cos(θ) _ = 6 sec(θ) tan(θ) 6. It is debatable which side of the identity is more complicated. One thing which stands out is that the denominator on the left hand side is 1 −cos(θ), while the numerator of the right hand side is 1 + cos(θ). This suggests the strategy of starting with the left hand side and multiplying the numerator and denominator by the quantity 1 + cos(θ): sin(θ) 1 −cos(θ) = sin(θ) (1 −cos(θ)) (1 + cos(θ)) (1 + cos(θ)) = sin(θ)(1 + cos(θ)) (1 −cos(θ))(1 + cos(θ)) = sin(θ)(1 + cos(θ)) 1 −cos 2 (θ) = sin(θ)(1 + cos(θ)) sin 2 (θ) = $ $ $ sin(θ)(1 + cos(θ)) $ $ $ sin(θ) sin(θ) = 1 + cos(θ) sin(θ) In Example 10.3.3 number 6 above, we see that multiplying 1 − cos(θ) by 1 + cos(θ) produces a difference of squares that can be simplified to one term using Theorem 10.8. This is exactly the same kind of phenomenon that occurs when we multiply expressions such as 1 − √ 2 by 1 + √ 2 or 3 − 4i by 3 + 4i. (Can you recall instances from Algebra where we did such things?) For this reason, the quantities (1 −cos(θ)) and (1 + cos(θ)) are called ‘Pythagorean Conjugates.’ Below is a list of other common Pythagorean Conjugates. Pythagorean Conjugates • 1 −cos(θ) and 1 + cos(θ): (1 −cos(θ))(1 + cos(θ)) = 1 −cos 2 (θ) = sin 2 (θ) • 1 −sin(θ) and 1 + sin(θ): (1 −sin(θ))(1 + sin(θ)) = 1 −sin 2 (θ) = cos 2 (θ) • sec(θ) −1 and sec(θ) + 1: (sec(θ) −1)(sec(θ) + 1) = sec 2 (θ) −1 = tan 2 (θ) • sec(θ)−tan(θ) and sec(θ)+tan(θ): (sec(θ)−tan(θ))(sec(θ)+tan(θ)) = sec 2 (θ)−tan 2 (θ) = 1 • csc(θ) −1 and csc(θ) + 1: (csc(θ) −1)(csc(θ) + 1) = csc 2 (θ) −1 = cot 2 (θ) • csc(θ)−cot(θ) and csc(θ)+cot(θ): (csc(θ)−cot(θ))(csc(θ)+cot(θ)) = csc 2 (θ)−cot 2 (θ) = 1 752 Foundations of Trigonometry Verifying trigonometric identities requires a healthy mix of tenacity and inspiration. You will need to spend many hours struggling with them just to become proficient in the basics. Like many things in life, there is no short-cut here – there is no complete algorithm for verifying identities. Nevertheless, a summary of some strategies which may be helpful (depending on the situation) is provided below and ample practice is provided for you in the Exercises. Strategies for Verifying Identities • Try working on the more complicated side of the identity. • Use the Reciprocal and Quotient Identities in Theorem 10.6 to write functions on one side of the identity in terms of the functions on the other side of the identity. Simplify the resulting complex fractions. • Add rational expressions with unlike denominators by obtaining common denominators. • Use the Pythagorean Identities in Theorem 10.8 to ‘exchange’ sines and cosines, secants and tangents, cosecants and cotangents, and simplify sums or differences of squares to one term. • Multiply numerator and denominator by Pythagorean Conjugates in order to take advan- tage of the Pythagorean Identities in Theorem 10.8. • If you find yourself stuck working with one side of the identity, try starting with the other side of the identity and see if you can find a way to bridge the two parts of your work. 10.3.1 Beyond the Unit Circle In Section 10.2, we generalized the cosine and sine functions from coordinates on the Unit Circle to coordinates on circles of radius r. Using Theorem 10.3 in conjunction with Theorem 10.8, we generalize the remaining circular functions in kind. Theorem 10.9. Suppose Q(x, y) is the point on the terminal side of an angle θ (plotted in standard position) which lies on the circle of radius r, x 2 +y 2 = r 2 . Then: • sec(θ) = r x = _ x 2 +y 2 x , provided x ,= 0. • csc(θ) = r y = _ x 2 +y 2 y , provided y ,= 0. • tan(θ) = y x , provided x ,= 0. • cot(θ) = x y , provided y ,= 0. 10.3 The Six Circular Functions and Fundamental Identities 753 Example 10.3.4. 1. Suppose the terminal side of θ, when plotted in standard position, contains the point Q(3, −4). Find the values of the six circular functions of θ. 2. Suppose θ is a Quadrant IV angle with cot(θ) = −4. Find the values of the five remaining circular functions of θ. Solution. 1. Since x = 3 and y = −4, r = _ x 2 +y 2 = _ (3) 2 + (−4) 2 = √ 25 = 5. Theorem 10.9 tells us cos(θ) = 3 5 , sin(θ) = − 4 5 , sec(θ) = 5 3 , csc(θ) = − 5 4 , tan(θ) = − 4 3 and cot(θ) = − 3 4 . 2. In order to use Theorem 10.9, we need to find a point Q(x, y) which lies on the terminal side of θ, when θ is plotted in standard position. We have that cot(θ) = −4 = x y , and since θ is a Quadrant IV angle, we also know x > 0 and y < 0. Viewing −4 = 4 −1 , we may choose 6 x = 4 and y = −1 so that r = _ x 2 +y 2 = _ (4) 2 + (−1) 2 = √ 17. Applying Theorem 10.9 once more, we find cos(θ) = 4 √ 17 = 4 √ 17 17 , sin(θ) = − 1 √ 17 = − √ 17 17 , sec(θ) = √ 17 4 , csc(θ) = − √ 17 and tan(θ) = − 1 4 . We may also specialize Theorem 10.9 to the case of acute angles θ which reside in a right triangle, as visualized below. θ a b c Theorem 10.10. Suppose θ is an acute angle residing in a right triangle. If the length of the side adjacent to θ is a, the length of the side opposite θ is b, and the length of the hypotenuse is c, then tan(θ) = b a sec(θ) = c a csc(θ) = c b cot(θ) = a b The following example uses Theorem 10.10 as well as the concept of an ‘angle of inclination.’ The angle of inclination (or angle of elevation) of an object refers to the angle whose initial side is some kind of base-line (say, the ground), and whose terminal side is the line-of-sight to an object above the base-line. This is represented schematically below. 6 We may choose any values x and y so long as x > 0, y < 0 and x y = −4. For example, we could choose x = 8 and y = −2. The fact that all such points lie on the terminal side of θ is a consequence of the fact that the terminal side of θ is the portion of the line with slope − 1 4 which extends from the origin into Quadrant IV. 754 Foundations of Trigonometry θ ‘base line’ object The angle of inclination from the base line to the object is θ Example 10.3.5. 1. The angle of inclination from a point on the ground 30 feet away to the top of Lakeland’s Armington Clocktower 7 is 60 ◦ . Find the height of the Clocktower to the nearest foot. 2. In order to determine the height of a California Redwood tree, two sightings from the ground, one 200 feet directly behind the other, are made. If the angles of inclination were 45 ◦ and 30 ◦ , respectively, how tall is the tree to the nearest foot? Solution. 1. We can represent the problem situation using a right triangle as shown below. If we let h denote the height of the tower, then Theorem 10.10 gives tan (60 ◦ ) = h 30 . From this we get h = 30 tan (60 ◦ ) = 30 √ 3 ≈ 51.96. Hence, the Clocktower is approximately 52 feet tall. 60 ◦ 30 ft. h ft. Finding the height of the Clocktower 2. Sketching the problem situation below, we find ourselves with two unknowns: the height h of the tree and the distance x from the base of the tree to the first observation point. 7 Named in honor of Raymond Q. Armington, Lakeland’s Clocktower has been a part of campus since 1972. 10.3 The Six Circular Functions and Fundamental Identities 755 45 ◦ 30 ◦ 200 ft. x ft. h ft. Finding the height of a California Redwood Using Theorem 10.10, we get a pair of equations: tan (45 ◦ ) = h x and tan (30 ◦ ) = h x+200 . Since tan (45 ◦ ) = 1, the first equation gives h x = 1, or x = h. Substituting this into the second equation gives h h+200 = tan (30 ◦ ) = √ 3 3 . Clearing fractions, we get 3h = (h + 200) √ 3. The result is a linear equation for h, so we proceed to expand the right hand side and gather all the terms involving h to one side. 3h = (h + 200) √ 3 3h = h √ 3 + 200 √ 3 3h −h √ 3 = 200 √ 3 (3 − √ 3)h = 200 √ 3 h = 200 √ 3 3 − √ 3 ≈ 273.20 Hence, the tree is approximately 273 feet tall. As we did in Section 10.2.1, we may consider all six circular functions as functions of real numbers. At this stage, there are three equivalent ways to define the functions sec(t), csc(t), tan(t) and cot(t) for real numbers t. First, we could go through the formality of the wrapping function on page 704 and define these functions as the appropriate ratios of x and y coordinates of points on the Unit Circle; second, we could define them by associating the real number t with the angle θ = t radians so that the value of the trigonometric function of t coincides with that of θ; lastly, we could simply define them using the Reciprocal and Quotient Identities as combinations of the functions f(t) = cos(t) and g(t) = sin(t). Presently, we adopt the last approach. We now set about determining the domains and ranges of the remaining four circular functions. Consider the function F(t) = sec(t) defined as F(t) = sec(t) = 1 cos(t) . We know F is undefined whenever cos(t) = 0. From Example 10.2.5 number 3, we know cos(t) = 0 whenever t = π 2 + πk for integers k. Hence, our domain for F(t) = sec(t), in set builder notation is ¦t : t ,= π 2 +πk, for integers k¦. To get a better understanding what set of real numbers we’re dealing with, it pays to write out and graph this set. Running through a few values of k, we find the domain to be ¦t : t ,= ± π 2 , ± 3π 2 , ± 5π 2 , . . .¦. Graphing this set on the number line we get 756 Foundations of Trigonometry − 5π 2 − 3π 2 − π 2 0 π 2 3π 2 5π 2 Using interval notation to describe this set, we get . . . ∪ _ − 5π 2 , − 3π 2 _ ∪ _ − 3π 2 , − π 2 _ ∪ _ − π 2 , π 2 _ ∪ _ π 2 , 3π 2 _ ∪ _ 3π 2 , 5π 2 _ ∪ . . . This is cumbersome, to say the least! In order to write this in a more compact way, we note that from the set-builder description of the domain, the kth point excluded from the domain, which we’ll call x k , can be found by the formula x k = π 2 +πk. (We are using sequence notation from Chapter 9.) Getting a common denominator and factoring out the π in the numerator, we get x k = (2k+1)π 2 . The domain consists of the intervals determined by successive points x k : (x k , x k + 1 ) = _ (2k+1)π 2 , (2k+3)π 2 _ . In order to capture all of the intervals in the domain, k must run through all of the integers, that is, k = 0, ±1, ±2, . . . . The way we denote taking the union of infinitely many intervals like this is to use what we call in this text extended interval notation. The domain of F(t) = sec(t) can now be written as ∞ _ k=−∞ _ (2k + 1)π 2 , (2k + 3)π 2 _ The reader should compare this notation with summation notation introduced in Section 9.2, in particular the notation used to describe geometric series in Theorem 9.2. In the same way the index k in the series ∞ k=1 ar k−1 can never equal the upper limit ∞, but rather, ranges through all of the natural numbers, the index k in the union ∞ _ k=−∞ _ (2k + 1)π 2 , (2k + 3)π 2 _ can never actually be ∞ or −∞, but rather, this conveys the idea that k ranges through all of the integers. Now that we have painstakingly determined the domain of F(t) = sec(t), it is time to discuss the range. Once again, we appeal to the definition F(t) = sec(t) = 1 cos(t) . The range of f(t) = cos(t) is [−1, 1], and since F(t) = sec(t) is undefined when cos(t) = 0, we split our discussion into two cases: when 0 < cos(t) ≤ 1 and when −1 ≤ cos(t) < 0. If 0 < cos(t) ≤ 1, then we can divide the inequality cos(t) ≤ 1 by cos(t) to obtain sec(t) = 1 cos(t) ≥ 1. Moreover, using the notation introduced in Section 4.2, we have that as cos(t) →0 + , sec(t) = 1 cos(t) ≈ 1 very small (+) ≈ very big (+). In other words, as cos(t) →0 + , sec(t) →∞. If, on the other hand, if −1 ≤ cos(t) < 0, then dividing by cos(t) causes a reversal of the inequality so that sec(t) = 1 sec(t) ≤ −1. In this case, as cos(t) →0 − , sec(t) = 1 cos(t) ≈ 1 very small (−) ≈ very big (−), so that as cos(t) → 0 − , we get sec(t) → −∞. Since 10.3 The Six Circular Functions and Fundamental Identities 757 f(t) = cos(t) admits all of the values in [−1, 1], the function F(t) = sec(t) admits all of the values in (−∞, −1] ∪ [1, ∞). Using set-builder notation, the range of F(t) = sec(t) can be written as ¦u : u ≤ −1 or u ≥ 1¦, or, more succinctly, 8 as ¦u : [u[ ≥ 1¦. 9 Similar arguments can be used to determine the domains and ranges of the remaining three circular functions: csc(t), tan(t) and cot(t). The reader is encouraged to do so. (See the Exercises.) For now, we gather these facts into the theorem below. Theorem 10.11. Domains and Ranges of the Circular Functions • The function f(t) = cos(t) • The function g(t) = sin(t) – has domain (−∞, ∞) – has domain (−∞, ∞) – has range [−1, 1] – has range [−1, 1] • The function F(t) = sec(t) = 1 cos(t) – has domain ¦t : t ,= π 2 +πk, for integers k¦ = ∞ _ k=−∞ _ (2k + 1)π 2 , (2k + 3)π 2 _ – has range ¦u : [u[ ≥ 1¦ = (−∞, −1] ∪ [1, ∞) • The function G(t) = csc(t) = 1 sin(t) – has domain ¦t : t ,= πk, for integers k¦ = ∞ _ k=−∞ (kπ, (k + 1)π) – has range ¦u : [u[ ≥ 1¦ = (−∞, −1] ∪ [1, ∞) • The function J(t) = tan(t) = sin(t) cos(t) – has domain ¦t : t ,= π 2 +πk, for integers k¦ = ∞ _ k=−∞ _ (2k + 1)π 2 , (2k + 3)π 2 _ – has range (−∞, ∞) • The function K(t) = cot(t) = cos(t) sin(t) – has domain ¦t : t ,= πk, for integers k¦ = ∞ _ k=−∞ (kπ, (k + 1)π) – has range (−∞, ∞) 8 Using Theorem 2.4 from Section 2.4. 9 Notice we have used the variable ‘u’ as the ‘dummy variable’ to describe the range elements. While there is no mathematical reason to do this (we are describing a set of real numbers, and, as such, could use t again) we choose u to help solidify the idea that these real numbers are the outputs from the inputs, which we have been calling t. 758 Foundations of Trigonometry We close this section with a few notes about solving equations which involve the circular functions. First, the discussion on page 735 in Section 10.2.1 concerning solving equations applies to all six circular functions, not just f(t) = cos(t) and g(t) = sin(t). In particular, to solve the equation cot(t) = −1 for real numbers t, we can use the same thought process we used in Example 10.3.2, number 3 to solve cot(θ) = −1 for angles θ in radian measure – we just need to remember to write our answers using the variable t as opposed to θ. Next, it is critical that you know the domains and ranges of the six circular functions so that you know which equations have no solutions. For example, sec(t) = 1 2 has no solution because 1 2 is not in the range of secant. Finally, you will need to review the notions of reference angles and coterminal angles so that you can see why csc(t) = −42 has an infinite set of solutions in Quadrant III and another infinite set of solutions in Quadrant IV. 10.3 The Six Circular Functions and Fundamental Identities 759 10.3.2 Exercises In Exercises 1 - 20, find the exact value or state that it is undefined. 1. tan _ π 4 _ 2. sec _ π 6 _ 3. csc _ 5π 6 _ 4. cot _ 4π 3 _ 5. tan _ − 11π 6 _ 6. sec _ − 3π 2 _ 7. csc _ − π 3 _ 8. cot _ 13π 2 _ 9. tan (117π) 10. sec _ − 5π 3 _ 11. csc (3π) 12. cot (−5π) 13. tan _ 31π 2 _ 14. sec _ π 4 _ 15. csc _ − 7π 4 _ 16. cot _ 7π 6 _ 17. tan _ 2π 3 _ 18. sec (−7π) 19. csc _ π 2 _ 20. cot _ 3π 4 _ In Exercises 21 - 34, use the given the information to find the exact values of the remaining circular functions of θ. 21. sin(θ) = 3 5 with θ in Quadrant II 22. tan(θ) = 12 5 with θ in Quadrant III 23. csc(θ) = 25 24 with θ in Quadrant I 24. sec(θ) = 7 with θ in Quadrant IV 25. csc(θ) = − 10 √ 91 91 with θ in Quadrant III 26. cot(θ) = −23 with θ in Quadrant II 27. tan(θ) = −2 with θ in Quadrant IV. 28. sec(θ) = −4 with θ in Quadrant II. 29. cot(θ) = √ 5 with θ in Quadrant III. 30. cos(θ) = 1 3 with θ in Quadrant I. 31. cot(θ) = 2 with 0 < θ < π 2 . 32. csc(θ) = 5 with π 2 < θ < π. 33. tan(θ) = √ 10 with π < θ < 3π 2 . 34. sec(θ) = 2 √ 5 with 3π 2 < θ < 2π. In Exercises 35 - 42, use your calculator to approximate the given value to three decimal places. Make sure your calculator is in the proper angle measurement mode! 35. csc(78.95 ◦ ) 36. tan(−2.01) 37. cot(392.994) 38. sec(207 ◦ ) 39. csc(5.902) 40. tan(39.672 ◦ ) 41. cot(3 ◦ ) 42. sec(0.45) 760 Foundations of Trigonometry In Exercises 43 - 57, find all of the angles which satisfy the equation. 43. tan(θ) = √ 3 44. sec(θ) = 2 45. csc(θ) = −1 46. cot(θ) = √ 3 3 47. tan(θ) = 0 48. sec(θ) = 1 49. csc(θ) = 2 50. cot(θ) = 0 51. tan(θ) = −1 52. sec(θ) = 0 53. csc(θ) = − 1 2 54. sec(θ) = −1 55. tan(θ) = − √ 3 56. csc(θ) = −2 57. cot(θ) = −1 In Exercises 58 - 65, solve the equation for t. Give exact values. 58. cot(t) = 1 59. tan(t) = √ 3 3 60. sec(t) = − 2 √ 3 3 61. csc(t) = 0 62. cot(t) = − √ 3 63. tan(t) = − √ 3 3 64. sec(t) = 2 √ 3 3 65. csc(t) = 2 √ 3 3 In Exercises 66 - 69, use Theorem 10.10 to find the requested quantities. 66. Find θ, a, and c. θ 9 a c 60 ◦ 67. Find α, b, and c. 34 ◦ c b 12 α 68. Find θ, a, and c. 47 ◦ 6 a c θ 69. Find β, b, and c. β 2.5 b c 50 ◦ 10.3 The Six Circular Functions and Fundamental Identities 761 In Exercises 70 - 75, use Theorem 10.10 to answer the question. Assume that θ is an angle in a right triangle. 70. If θ = 30 ◦ and the side opposite θ has length 4, how long is the side adjacent to θ? 71. If θ = 15 ◦ and the hypotenuse has length 10, how long is the side opposite θ? 72. If θ = 87 ◦ and the side adjacent to θ has length 2, how long is the side opposite θ? 73. If θ = 38.2 ◦ and the side opposite θ has lengh 14, how long is the hypoteneuse? 74. If θ = 2.05 ◦ and the hypotenuse has length 3.98, how long is the side adjacent to θ? 75. If θ = 42 ◦ and the side adjacent to θ has length 31, how long is the side opposite θ? 76. A tree standing vertically on level ground casts a 120 foot long shadow. The angle of elevation from the end of the shadow to the top of the tree is 21.4 ◦ . Find the height of the tree to the nearest foot. With the help of your classmates, research the term umbra versa and see what it has to do with the shadow in this problem. 77. The broadcast tower for radio station WSAZ (Home of “Algebra in the Morning with Carl and Jeff”) has two enormous flashing red lights on it: one at the very top and one a few feet below the top. From a point 5000 feet away from the base of the tower on level ground the angle of elevation to the top light is 7.970 ◦ and to the second light is 7.125 ◦ . Find the distance between the lights to the nearest foot. 78. On page 753 we defined the angle of inclination (also known as the angle of elevation) and in this exercise we introduce a related angle - the angle of depression (also known as the angle of declination). The angle of depression of an object refers to the angle whose initial side is a horizontal line above the object and whose terminal side is the line-of-sight to the object below the horizontal. This is represented schematically below. θ horizontal observer object The angle of depression from the horizontal to the object is θ (a) Show that if the horizontal is above and parallel to level ground then the angle of depression (from observer to object) and the angle of inclination (from object to observer) will be congruent because they are alternate interior angles. 762 Foundations of Trigonometry (b) From a firetower 200 feet above level ground in the Sasquatch National Forest, a ranger spots a fire off in the distance. The angle of depression to the fire is 2.5 ◦ . How far away from the base of the tower is the fire? (c) The ranger in part 78b sees a Sasquatch running directly from the fire towards the firetower. The ranger takes two sightings. At the first sighthing, the angle of depression from the tower to the Sasquatch is 6 ◦ . The second sighting, taken just 10 seconds later, gives the the angle of depression as 6.5 ◦ . How far did the Saquatch travel in those 10 seconds? Round your answer to the nearest foot. How fast is it running in miles per hour? Round your answer to the nearest mile per hour. If the Sasquatch keeps up this pace, how long will it take for the Sasquatch to reach the firetower from his location at the second sighting? Round your answer to the nearest minute. 79. When I stand 30 feet away from a tree at home, the angle of elevation to the top of the tree is 50 ◦ and the angle of depression to the base of the tree is 10 ◦ . What is the height of the tree? Round your answer to the nearest foot. 80. From the observation deck of the lighthouse at Sasquatch Point 50 feet above the surface of Lake Ippizuti, a lifeguard spots a boat out on the lake sailing directly toward the lighthouse. The first sighting had an angle of depression of 8.2 ◦ and the second sighting had an angle of depression of 25.9 ◦ . How far had the boat traveled between the sightings? 81. A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it makes a 43 ◦ angle with the ground. How tall is the tower? How far away from the base of the tower does the wire hit the ground? In Exercises 82 - 128, verify the identity. Assume that all quantities are defined. 82. cos(θ) sec(θ) = 1 83. tan(θ) cos(θ) = sin(θ) 84. sin(θ) csc(θ) = 1 85. tan(θ) cot(θ) = 1 86. csc(θ) cos(θ) = cot(θ) 87. sin(θ) cos 2 (θ) = sec(θ) tan(θ) 88. cos(θ) sin 2 (θ) = csc(θ) cot(θ) 89. 1 + sin(θ) cos(θ) = sec(θ) + tan(θ) 90. 1 −cos(θ) sin(θ) = csc(θ) −cot(θ) 91. cos(θ) 1 −sin 2 (θ) = sec(θ) 92. sin(θ) 1 −cos 2 (θ) = csc(θ) 93. sec(θ) 1 + tan 2 (θ) = cos(θ) 10.3 The Six Circular Functions and Fundamental Identities 763 94. csc(θ) 1 + cot 2 (θ) = sin(θ) 95. tan(θ) sec 2 (θ) −1 = cot(θ) 96. cot(θ) csc 2 (θ) −1 = tan(θ) 97. 4 cos 2 (θ) + 4 sin 2 (θ) = 4 98. 9 −cos 2 (θ) −sin 2 (θ) = 8 99. tan 3 (θ) = tan(θ) sec 2 (θ) −tan(θ) 100. sin 5 (θ) = _ 1 −cos 2 (θ) _ 2 sin(θ) 101. sec 10 (θ) = _ 1 + tan 2 (θ) _ 4 sec 2 (θ) 102. cos 2 (θ) tan 3 (θ) = tan(θ) −sin(θ) cos(θ) 103. sec 4 (θ) −sec 2 (θ) = tan 2 (θ) + tan 4 (θ) 104. cos(θ) + 1 cos(θ) −1 = 1 + sec(θ) 1 −sec(θ) 105. sin(θ) + 1 sin(θ) −1 = 1 + csc(θ) 1 −csc(θ) 106. 1 −cot(θ) 1 + cot(θ) = tan(θ) −1 tan(θ) + 1 107. 1 −tan(θ) 1 + tan(θ) = cos(θ) −sin(θ) cos(θ) + sin(θ) 108. tan(θ) + cot(θ) = sec(θ) csc(θ) 109. csc(θ) −sin(θ) = cot(θ) cos(θ) 110. cos(θ) −sec(θ) = −tan(θ) sin(θ) 111. cos(θ)(tan(θ) + cot(θ)) = csc(θ) 112. sin(θ)(tan(θ) + cot(θ)) = sec(θ) 113. 1 1 −cos(θ) + 1 1 + cos(θ) = 2 csc 2 (θ) 114. 1 sec(θ) + 1 + 1 sec(θ) −1 = 2 csc(θ) cot(θ) 115. 1 csc(θ) + 1 + 1 csc(θ) −1 = 2 sec(θ) tan(θ) 116. 1 csc(θ) −cot(θ) − 1 csc(θ) + cot(θ) = 2 cot(θ) 117. cos(θ) 1 −tan(θ) + sin(θ) 1 −cot(θ) = sin(θ) + cos(θ) 118. 1 sec(θ) + tan(θ) = sec(θ) −tan(θ) 119. 1 sec(θ) −tan(θ) = sec(θ) + tan(θ) 120. 1 csc(θ) −cot(θ) = csc(θ) + cot(θ) 121. 1 csc(θ) + cot(θ) = csc(θ) −cot(θ) 122. 1 1 −sin(θ) = sec 2 (θ) + sec(θ) tan(θ) 123. 1 1 + sin(θ) = sec 2 (θ) −sec(θ) tan(θ) 124. 1 1 −cos(θ) = csc 2 (θ) + csc(θ) cot(θ) 125. 1 1 + cos(θ) = csc 2 (θ) −csc(θ) cot(θ) 126. cos(θ) 1 + sin(θ) = 1 −sin(θ) cos(θ) 127. csc(θ) −cot(θ) = sin(θ) 1 + cos(θ) 128. 1 −sin(θ) 1 + sin(θ) = (sec(θ) −tan(θ)) 2 764 Foundations of Trigonometry In Exercises 129 - 132, verify the identity. You may need to consult Sections 2.2 and 6.2 for a review of the properties of absolute value and logarithms before proceeding. 129. ln [ sec(θ)[ = −ln [ cos(θ)[ 130. −ln [ csc(θ)[ = ln [ sin(θ)[ 131. −ln [ sec(θ) −tan(θ)[ = ln [ sec(θ) +tan(θ)[ 132. −ln [ csc(θ) +cot(θ)[ = ln [ csc(θ) −cot(θ)[ 133. Verify the domains and ranges of the tangent, cosecant and cotangent functions as presented in Theorem 10.11. 134. As we did in Exercise 74 in Section 10.2, let α and β be the two acute angles of a right triangle. (Thus α and β are complementary angles.) Show that sec(α) = csc(β) and tan(α) = cot(β). The fact that co-functions of complementary angles are equal in this case is not an accident and a more general result will be given in Section 10.4. 135. We wish to establish the inequality cos(θ) < sin(θ) θ < 1 for 0 < θ < π 2 . Use the diagram from the beginning of the section, partially reproduced below, to answer the following. θ x y 1 O B(1, 0) P Q (a) Show that triangle OPB has area 1 2 sin(θ). (b) Show that the circular sector OPB with central angle θ has area 1 2 θ. (c) Show that triangle OQB has area 1 2 tan(θ). (d) Comparing areas, show that sin(θ) < θ < tan(θ) for 0 < θ < π 2 . (e) Use the inequality sin(θ) < θ to show that sin(θ) θ < 1 for 0 < θ < π 2 . (f) Use the inequality θ < tan(θ) to show that cos(θ) < sin(θ) θ for 0 < θ < π 2 . Combine this with the previous part to complete the proof. 10.3 The Six Circular Functions and Fundamental Identities 765 136. Show that cos(θ) < sin(θ) θ < 1 also holds for − π 2 < θ < 0. 137. Explain why the fact that tan(θ) = 3 = 3 1 does not mean sin(θ) = 3 and cos(θ) = 1? (See the solution to number 6 in Example 10.3.1.) 766 Foundations of Trigonometry 10.3.3 Answers 1. tan _ π 4 _ = 1 2. sec _ π 6 _ = 2 √ 3 3 3. csc _ 5π 6 _ = 2 4. cot _ 4π 3 _ = √ 3 3 5. tan _ − 11π 6 _ = √ 3 3 6. sec _ − 3π 2 _ is undefined 7. csc _ − π 3 _ = − 2 √ 3 3 8. cot _ 13π 2 _ = 0 9. tan (117π) = 0 10. sec _ − 5π 3 _ = 2 11. csc (3π) is undefined 12. cot (−5π) is undefined 13. tan _ 31π 2 _ is undefined 14. sec _ π 4 _ = √ 2 15. csc _ − 7π 4 _ = √ 2 16. cot _ 7π 6 _ = √ 3 17. tan _ 2π 3 _ = − √ 3 18. sec (−7π) = −1 19. csc _ π 2 _ = 1 20. cot _ 3π 4 _ = −1 21. sin(θ) = 3 5 , cos(θ) = − 4 5 , tan(θ) = − 3 4 , csc(θ) = 5 3 , sec(θ) = − 5 4 , cot(θ) = − 4 3 22. sin(θ) = − 12 13 , cos(θ) = − 5 13 , tan(θ) = 12 5 , csc(θ) = − 13 12 , sec(θ) = − 13 5 , cot(θ) = 5 12 23. sin(θ) = 24 25 , cos(θ) = 7 25 , tan(θ) = 24 7 , csc(θ) = 25 24 , sec(θ) = 25 7 , cot(θ) = 7 24 24. sin(θ) = −4 √ 3 7 , cos(θ) = 1 7 , tan(θ) = −4 √ 3, csc(θ) = − 7 √ 3 12 , sec(θ) = 7, cot(θ) = − √ 3 12 25. sin(θ) = − √ 91 10 , cos(θ) = − 3 10 , tan(θ) = √ 91 3 , csc(θ) = − 10 √ 91 91 , sec(θ) = − 10 3 , cot(θ) = 3 √ 91 91 26. sin(θ) = √ 530 530 , cos(θ) = − 23 √ 530 530 , tan(θ) = − 1 23 , csc(θ) = √ 530, sec(θ) = − √ 530 23 , cot(θ) = −23 27. sin(θ) = − 2 √ 5 5 , cos(θ) = √ 5 5 , tan(θ) = −2, csc(θ) = − √ 5 2 , sec(θ) = √ 5, cot(θ) = − 1 2 28. sin(θ) = √ 15 4 , cos(θ) = − 1 4 , tan(θ) = − √ 15, csc(θ) = 4 √ 15 15 , sec(θ) = −4, cot(θ) = − √ 15 15 29. sin(θ) = − √ 6 6 , cos(θ) = − √ 30 6 , tan(θ) = √ 5 5 , csc(θ) = − √ 6, sec(θ) = − √ 30 5 , cot(θ) = √ 5 30. sin(θ) = 2 √ 2 3 , cos(θ) = 1 3 , tan(θ) = 2 √ 2, csc(θ) = 3 √ 2 4 , sec(θ) = 3, cot(θ) = √ 2 4 31. sin(θ) = √ 5 5 , cos(θ) = 2 √ 5 5 , tan(θ) = 1 2 , csc(θ) = √ 5, sec(θ) = √ 5 2 , cot(θ) = 2 32. sin(θ) = 1 5 , cos(θ) = − 2 √ 6 5 , tan(θ) = − √ 6 12 , csc(θ) = 5, sec(θ) = − 5 √ 6 12 , cot(θ) = −2 √ 6 33. sin(θ) = − √ 110 11 , cos(θ) = − √ 11 11 , tan(θ) = √ 10, csc(θ) = − √ 110 10 , sec(θ) = − √ 11, cot(θ) = √ 10 10 34. sin(θ) = − √ 95 10 , cos(θ) = √ 5 10 , tan(θ) = − √ 19, csc(θ) = − 2 √ 95 19 , sec(θ) = 2 √ 5, cot(θ) = − √ 19 19 10.3 The Six Circular Functions and Fundamental Identities 767 35. csc(78.95 ◦ ) ≈ 1.019 36. tan(−2.01) ≈ 2.129 37. cot(392.994) ≈ 3.292 38. sec(207 ◦ ) ≈ −1.122 39. csc(5.902) ≈ −2.688 40. tan(39.672 ◦ ) ≈ 0.829 41. cot(3 ◦ ) ≈ 19.081 42. sec(0.45) ≈ 1.111 43. tan(θ) = √ 3 when θ = π 3 +πk for any integer k 44. sec(θ) = 2 when θ = π 3 + 2πk or θ = 5π 3 + 2πk for any integer k 45. csc(θ) = −1 when θ = 3π 2 + 2πk for any integer k. 46. cot(θ) = √ 3 3 when θ = π 3 +πk for any integer k 47. tan(θ) = 0 when θ = πk for any integer k 48. sec(θ) = 1 when θ = 2πk for any integer k 49. csc(θ) = 2 when θ = π 6 + 2πk or θ = 5π 6 + 2πk for any integer k. 50. cot(θ) = 0 when θ = π 2 +πk for any integer k 51. tan(θ) = −1 when θ = 3π 4 +πk for any integer k 52. sec(θ) = 0 never happens 53. csc(θ) = − 1 2 never happens 54. sec(θ) = −1 when θ = π + 2πk = (2k + 1)π for any integer k 55. tan(θ) = − √ 3 when θ = 2π 3 +πk for any integer k 56. csc(θ) = −2 when θ = 7π 6 + 2πk or θ = 11π 6 + 2πk for any integer k 57. cot(θ) = −1 when θ = 3π 4 +πk for any integer k 58. cot(t) = 1 when t = π 4 +πk for any integer k 59. tan(t) = √ 3 3 when t = π 6 +πk for any integer k 768 Foundations of Trigonometry 60. sec(t) = − 2 √ 3 3 when t = 5π 6 + 2πk or t = 7π 6 + 2πk for any integer k 61. csc(t) = 0 never happens 62. cot(t) = − √ 3 when t = 5π 6 +πk for any integer k 63. tan(t) = − √ 3 3 when t = 5π 6 +πk for any integer k 64. sec(t) = 2 √ 3 3 when t = π 6 + 2πk or t = 11π 6 + 2πk for any integer k 65. csc(t) = 2 √ 3 3 when t = π 3 + 2πk or t = 2π 3 + 2πk for any integer k 66. θ = 30 ◦ , a = 3 √ 3, c = √ 108 = 6 √ 3 67. α = 56 ◦ , b = 12 tan(34 ◦ ) = 8.094, c = 12 sec(34 ◦ ) = 12 cos(34 ◦ ) ≈ 14.475 68. θ = 43 ◦ , a = 6 cot(47 ◦ ) = 6 tan(47 ◦ ) ≈ 5.595, c = 6 csc(47 ◦ ) = 6 sin(47 ◦ ) ≈ 8.204 69. β = 40 ◦ , b = 2.5 tan(50 ◦ ) ≈ 2.979, c = 2.5 sec(50 ◦ ) = 2.5 cos(50 ◦ ) ≈ 3.889 70. The side adjacent to θ has length 4 √ 3 ≈ 6.928 71. The side opposite θ has length 10 sin(15 ◦ ) ≈ 2.588 72. The side opposite θ is 2 tan(87 ◦ ) ≈ 38.162 73. The hypoteneuse has length 14 csc(38.2 ◦ ) = 14 sin(38.2 ◦ ) ≈ 22.639 74. The side adjacent to θ has length 3.98 cos(2.05 ◦ ) ≈ 3.977 75. The side opposite θ has length 31 tan(42 ◦ ) ≈ 27.912 76. The tree is about 47 feet tall. 77. The lights are about 75 feet apart. 78. (b) The fire is about 4581 feet from the base of the tower. (c) The Sasquatch ran 200 cot(6 ◦ ) − 200 cot(6.5 ◦ ) ≈ 147 feet in those 10 seconds. This translates to ≈ 10 miles per hour. At the scene of the second sighting, the Sasquatch was ≈ 1755 feet from the tower, which means, if it keeps up this pace, it will reach the tower in about 2 minutes. 10.3 The Six Circular Functions and Fundamental Identities 769 79. The tree is about 41 feet tall. 80. The boat has traveled about 244 feet. 81. The tower is about 682 feet tall. The guy wire hits the ground about 731 feet away from the base of the tower. 770 Foundations of Trigonometry 10.4 Trigonometric Identities In Section 10.3, we saw the utility of the Pythagorean Identities in Theorem 10.8 along with the Quotient and Reciprocal Identities in Theorem 10.6. Not only did these identities help us compute the values of the circular functions for angles, they were also useful in simplifying expressions involving the circular functions. In this section, we introduce several collections of identities which have uses in this course and beyond. Our first set of identities is the ‘Even / Odd’ identities. 1 Theorem 10.12. Even / Odd Identities: For all applicable angles θ, • cos(−θ) = cos(θ) • sec(−θ) = sec(θ) • sin(−θ) = −sin(θ) • csc(−θ) = −csc(θ) • tan(−θ) = −tan(θ) • cot(−θ) = −cot(θ) In light of the Quotient and Reciprocal Identities, Theorem 10.6, it suffices to show cos(−θ) = cos(θ) and sin(−θ) = −sin(θ). The remaining four circular functions can be expressed in terms of cos(θ) and sin(θ) so the proofs of their Even / Odd Identities are left as exercises. Consider an angle θ plotted in standard position. Let θ 0 be the angle coterminal with θ with 0 ≤ θ 0 < 2π. (We can construct the angle θ 0 by rotating counter-clockwise from the positive x-axis to the terminal side of θ as pictured below.) Since θ and θ 0 are coterminal, cos(θ) = cos(θ 0 ) and sin(θ) = sin(θ 0 ). x y 1 1 θ θ 0 x y 1 1 θ 0 −θ 0 P(cos(θ 0 ), sin(θ 0 )) Q(cos(−θ 0 ), sin(−θ 0 )) We now consider the angles −θ and −θ 0 . Since θ is coterminal with θ 0 , there is some integer k so that θ = θ 0 + 2π k. Therefore, −θ = −θ 0 − 2π k = −θ 0 + 2π (−k). Since k is an integer, so is (−k), which means −θ is coterminal with −θ 0 . Hence, cos(−θ) = cos(−θ 0 ) and sin(−θ) = sin(−θ 0 ). Let P and Q denote the points on the terminal sides of θ 0 and −θ 0 , respectively, which lie on the Unit Circle. By definition, the coordinates of P are (cos(θ 0 ), sin(θ 0 )) and the coordinates of Q are (cos(−θ 0 ), sin(−θ 0 )). Since θ 0 and −θ 0 sweep out congruent central sectors of the Unit Circle, it 1 As mentioned at the end of Section 10.2, properties of the circular functions when thought of as functions of angles in radian measure hold equally well if we view these functions as functions of real numbers. Not surprisingly, the Even / Odd properties of the circular functions are so named because they identify cosine and secant as even functions, while the remaining four circular functions are odd. (See Section 1.6.) 10.4 Trigonometric Identities 771 follows that the points P and Q are symmetric about the x-axis. Thus, cos(−θ 0 ) = cos(θ 0 ) and sin(−θ 0 ) = −sin(θ 0 ). Since the cosines and sines of θ 0 and −θ 0 are the same as those for θ and −θ, respectively, we get cos(−θ) = cos(θ) and sin(−θ) = −sin(θ), as required. The Even / Odd Identities are readily demonstrated using any of the ‘common angles’ noted in Section 10.2. Their true utility, however, lies not in computation, but in simplifying expressions involving the circular functions. In fact, our next batch of identities makes heavy use of the Even / Odd Identities. Theorem 10.13. Sum and Difference Identities for Cosine: For all angles α and β, • cos(α +β) = cos(α) cos(β) −sin(α) sin(β) • cos(α −β) = cos(α) cos(β) + sin(α) sin(β) We first prove the result for differences. As in the proof of the Even / Odd Identities, we can reduce the proof for general angles α and β to angles α 0 and β 0 , coterminal with α and β, respectively, each of which measure between 0 and 2π radians. Since α and α 0 are coterminal, as are β and β 0 , it follows that α −β is coterminal with α 0 −β 0 . Consider the case below where α 0 ≥ β 0 . α 0 β 0 x y 1 O P(cos(α 0 ), sin(α 0 )) Q(cos(β 0 ), sin(β 0 )) α 0 −β 0 x y 1 O A(cos(α 0 −β 0 ), sin(α 0 −β 0 )) B(1, 0) α 0 −β 0 Since the angles POQ and AOB are congruent, the distance between P and Q is equal to the distance between A and B. 2 The distance formula, Equation 1.1, yields _ (cos(α 0 ) −cos(β 0 )) 2 + (sin(α 0 ) −sin(β 0 )) 2 = _ (cos(α 0 −β 0 ) −1) 2 + (sin(α 0 −β 0 ) −0) 2 Squaring both sides, we expand the left hand side of this equation as (cos(α 0 ) −cos(β 0 )) 2 + (sin(α 0 ) −sin(β 0 )) 2 = cos 2 (α 0 ) −2 cos(α 0 ) cos(β 0 ) + cos 2 (β 0 ) +sin 2 (α 0 ) −2 sin(α 0 ) sin(β 0 ) + sin 2 (β 0 ) = cos 2 (α 0 ) + sin 2 (α 0 ) + cos 2 (β 0 ) + sin 2 (β 0 ) −2 cos(α 0 ) cos(β 0 ) −2 sin(α 0 ) sin(β 0 ) 2 In the picture we’ve drawn, the triangles POQ and AOB are congruent, which is even better. However, α0 −β0 could be 0 or it could be π, neither of which makes a triangle. It could also be larger than π, which makes a triangle, just not the one we’ve drawn. You should think about those three cases. 772 Foundations of Trigonometry From the Pythagorean Identities, cos 2 (α 0 ) + sin 2 (α 0 ) = 1 and cos 2 (β 0 ) + sin 2 (β 0 ) = 1, so (cos(α 0 ) −cos(β 0 )) 2 + (sin(α 0 ) −sin(β 0 )) 2 = 2 −2 cos(α 0 ) cos(β 0 ) −2 sin(α 0 ) sin(β 0 ) Turning our attention to the right hand side of our equation, we find (cos(α 0 −β 0 ) −1) 2 + (sin(α 0 −β 0 ) −0) 2 = cos 2 (α 0 −β 0 ) −2 cos(α 0 −β 0 ) + 1 + sin 2 (α 0 −β 0 ) = 1 + cos 2 (α 0 −β 0 ) + sin 2 (α 0 −β 0 ) −2 cos(α 0 −β 0 ) Once again, we simplify cos 2 (α 0 −β 0 ) + sin 2 (α 0 −β 0 ) = 1, so that (cos(α 0 −β 0 ) −1) 2 + (sin(α 0 −β 0 ) −0) 2 = 2 −2 cos(α 0 −β 0 ) Putting it all together, we get 2 − 2 cos(α 0 ) cos(β 0 ) − 2 sin(α 0 ) sin(β 0 ) = 2 − 2 cos(α 0 − β 0 ), which simplifies to: cos(α 0 −β 0 ) = cos(α 0 ) cos(β 0 ) + sin(α 0 ) sin(β 0 ). Since α and α 0 , β and β 0 and α −β and α 0 −β 0 are all coterminal pairs of angles, we have cos(α −β) = cos(α) cos(β) + sin(α) sin(β). For the case where α 0 ≤ β 0 , we can apply the above argument to the angle β 0 − α 0 to obtain the identity cos(β 0 − α 0 ) = cos(β 0 ) cos(α 0 ) + sin(β 0 ) sin(α 0 ). Applying the Even Identity of cosine, we get cos(β 0 −α 0 ) = cos(−(α 0 −β 0 )) = cos(α 0 −β 0 ), and we get the identity in this case, too. To get the sum identity for cosine, we use the difference formula along with the Even/Odd Identities cos(α +β) = cos(α −(−β)) = cos(α) cos(−β) + sin(α) sin(−β) = cos(α) cos(β) −sin(α) sin(β) We put these newfound identities to good use in the following example. Example 10.4.1. 1. Find the exact value of cos (15 ◦ ). 2. Verify the identity: cos _ π 2 −θ _ = sin(θ). Solution. 1. In order to use Theorem 10.13 to find cos (15 ◦ ), we need to write 15 ◦ as a sum or difference of angles whose cosines and sines we know. One way to do so is to write 15 ◦ = 45 ◦ −30 ◦ . cos (15 ◦ ) = cos (45 ◦ −30 ◦ ) = cos (45 ◦ ) cos (30 ◦ ) + sin (45 ◦ ) sin (30 ◦ ) = _ √ 2 2 __ √ 3 2 _ + _ √ 2 2 _ _ 1 2 _ = √ 6 + √ 2 4 10.4 Trigonometric Identities 773 2. In a straightforward application of Theorem 10.13, we find cos _ π 2 −θ _ = cos _ π 2 _ cos (θ) + sin _ π 2 _ sin (θ) = (0) (cos(θ)) + (1) (sin(θ)) = sin(θ) The identity verified in Example 10.4.1, namely, cos _ π 2 −θ _ = sin(θ), is the first of the celebrated ‘cofunction’ identities. These identities were first hinted at in Exercise 74 in Section 10.2. From sin(θ) = cos _ π 2 −θ _ , we get: sin _ π 2 −θ _ = cos _ π 2 − _ π 2 −θ __ = cos(θ), which says, in words, that the ‘co’sine of an angle is the sine of its ‘co’mplement. Now that these identities have been established for cosine and sine, the remaining circular functions follow suit. The remaining proofs are left as exercises. Theorem 10.14. Cofunction Identities: For all applicable angles θ, • cos _ π 2 −θ _ = sin(θ) • sin _ π 2 −θ _ = cos(θ) • sec _ π 2 −θ _ = csc(θ) • csc _ π 2 −θ _ = sec(θ) • tan _ π 2 −θ _ = cot(θ) • cot _ π 2 −θ _ = tan(θ) With the Cofunction Identities in place, we are now in the position to derive the sum and difference formulas for sine. To derive the sum formula for sine, we convert to cosines using a cofunction identity, then expand using the difference formula for cosine sin(α +β) = cos _ π 2 −(α +β) _ = cos __ π 2 −α _ −β _ = cos _ π 2 −α _ cos(β) + sin _ π 2 −α _ sin(β) = sin(α) cos(β) + cos(α) sin(β) We can derive the difference formula for sine by rewriting sin(α − β) as sin(α + (−β)) and using the sum formula and the Even / Odd Identities. Again, we leave the details to the reader. Theorem 10.15. Sum and Difference Identities for Sine: For all angles α and β, • sin(α +β) = sin(α) cos(β) + cos(α) sin(β) • sin(α −β) = sin(α) cos(β) −cos(α) sin(β) 774 Foundations of Trigonometry Example 10.4.2. 1. Find the exact value of sin _ 19π 12 _ 2. If α is a Quadrant II angle with sin(α) = 5 13 , and β is a Quadrant III angle with tan(β) = 2, find sin(α −β). 3. Derive a formula for tan(α +β) in terms of tan(α) and tan(β). Solution. 1. As in Example 10.4.1, we need to write the angle 19π 12 as a sum or difference of common angles. The denominator of 12 suggests a combination of angles with denominators 3 and 4. One such combination is 19π 12 = 4π 3 + π 4 . Applying Theorem 10.15, we get sin _ 19π 12 _ = sin _ 4π 3 + π 4 _ = sin _ 4π 3 _ cos _ π 4 _ + cos _ 4π 3 _ sin _ π 4 _ = _ − √ 3 2 __ √ 2 2 _ + _ − 1 2 _ _ √ 2 2 _ = − √ 6 − √ 2 4 2. In order to find sin(α − β) using Theorem 10.15, we need to find cos(α) and both cos(β) and sin(β). To find cos(α), we use the Pythagorean Identity cos 2 (α) + sin 2 (α) = 1. Since sin(α) = 5 13 , we have cos 2 (α) + _ 5 13 _ 2 = 1, or cos(α) = ± 12 13 . Since α is a Quadrant II angle, cos(α) = − 12 13 . We now set about finding cos(β) and sin(β). We have several ways to proceed, but the Pythagorean Identity 1 + tan 2 (β) = sec 2 (β) is a quick way to get sec(β), and hence, cos(β). With tan(β) = 2, we get 1 + 2 2 = sec 2 (β) so that sec(β) = ± √ 5. Since β is a Quadrant III angle, we choose sec(β) = − √ 5 so cos(β) = 1 sec(β) = 1 − √ 5 = − √ 5 5 . We now need to determine sin(β). We could use The Pythagorean Identity cos 2 (β) + sin 2 (β) = 1, but we opt instead to use a quotient identity. From tan(β) = sin(β) cos(β) , we have sin(β) = tan(β) cos(β) so we get sin(β) = (2) _ − √ 5 5 _ = − 2 √ 5 5 . We now have all the pieces needed to find sin(α−β): sin(α −β) = sin(α) cos(β) −cos(α) sin(β) = _ 5 13 _ _ − √ 5 5 _ − _ − 12 13 _ _ − 2 √ 5 5 _ = − 29 √ 5 65 10.4 Trigonometric Identities 775 3. We can start expanding tan(α +β) using a quotient identity and our sum formulas tan(α +β) = sin(α +β) cos(α +β) = sin(α) cos(β) + cos(α) sin(β) cos(α) cos(β) −sin(α) sin(β) Since tan(α) = sin(α) cos(α) and tan(β) = sin(β) cos(β) , it looks as though if we divide both numerator and denominator by cos(α) cos(β) we will have what we want tan(α +β) = sin(α) cos(β) + cos(α) sin(β) cos(α) cos(β) −sin(α) sin(β) 1 cos(α) cos(β) 1 cos(α) cos(β) = sin(α) cos(β) cos(α) cos(β) + cos(α) sin(β) cos(α) cos(β) cos(α) cos(β) cos(α) cos(β) − sin(α) sin(β) cos(α) cos(β) = sin(α) $ $ $$ cos(β) cos(α) $ $ $$ cos(β) + $ $ $$ cos(α) sin(β) $ $ $$ cos(α) cos(β) $ $ $$ cos(α) $ $ $$ cos(β) $ $ $$ cos(α) $ $ $$ cos(β) − sin(α) sin(β) cos(α) cos(β) = tan(α) + tan(β) 1 −tan(α) tan(β) Naturally, this formula is limited to those cases where all of the tangents are defined. The formula developed in Exercise 10.4.2 for tan(α+β) can be used to find a formula for tan(α−β) by rewriting the difference as a sum, tan(α+(−β)), and the reader is encouraged to fill in the details. Below we summarize all of the sum and difference formulas for cosine, sine and tangent. Theorem 10.16. Sum and Difference Identities: For all applicable angles α and β, • cos(α ±β) = cos(α) cos(β) ∓sin(α) sin(β) • sin(α ±β) = sin(α) cos(β) ±cos(α) sin(β) • tan(α ±β) = tan(α) ±tan(β) 1 ∓tan(α) tan(β) In the statement of Theorem 10.16, we have combined the cases for the sum ‘+’ and difference ‘−’ of angles into one formula. The convention here is that if you want the formula for the sum ‘+’ of 776 Foundations of Trigonometry two angles, you use the top sign in the formula; for the difference, ‘−’, use the bottom sign. For example, tan(α −β) = tan(α) −tan(β) 1 + tan(α) tan(β) If we specialize the sum formulas in Theorem 10.16 to the case when α = β, we obtain the following ‘Double Angle’ Identities. Theorem 10.17. Double Angle Identities: For all applicable angles θ, • cos(2θ) = _ ¸ _ ¸ _ cos 2 (θ) −sin 2 (θ) 2 cos 2 (θ) −1 1 −2 sin 2 (θ) • sin(2θ) = 2 sin(θ) cos(θ) • tan(2θ) = 2 tan(θ) 1 −tan 2 (θ) The three different forms for cos(2θ) can be explained by our ability to ‘exchange’ squares of cosine and sine via the Pythagorean Identity cos 2 (θ) +sin 2 (θ) = 1 and we leave the details to the reader. It is interesting to note that to determine the value of cos(2θ), only one piece of information is required: either cos(θ) or sin(θ). To determine sin(2θ), however, it appears that we must know both sin(θ) and cos(θ). In the next example, we show how we can find sin(2θ) knowing just one piece of information, namely tan(θ). Example 10.4.3. 1. Suppose P(−3, 4) lies on the terminal side of θ when θ is plotted in standard position. Find cos(2θ) and sin(2θ) and determine the quadrant in which the terminal side of the angle 2θ lies when it is plotted in standard position. 2. If sin(θ) = x for − π 2 ≤ θ ≤ π 2 , find an expression for sin(2θ) in terms of x. 3. Verify the identity: sin(2θ) = 2 tan(θ) 1 + tan 2 (θ) . 4. Express cos(3θ) as a polynomial in terms of cos(θ). Solution. 1. Using Theorem 10.3 from Section 10.2 with x = −3 and y = 4, we find r = _ x 2 +y 2 = 5. Hence, cos(θ) = − 3 5 and sin(θ) = 4 5 . Applying Theorem 10.17, we get cos(2θ) = cos 2 (θ) − sin 2 (θ) = _ − 3 5 _ 2 − _ 4 5 _ 2 = − 7 25 , and sin(2θ) = 2 sin(θ) cos(θ) = 2 _ 4 5 _ _ − 3 5 _ = − 24 25 . Since both cosine and sine of 2θ are negative, the terminal side of 2θ, when plotted in standard position, lies in Quadrant III. 10.4 Trigonometric Identities 777 2. If your first reaction to ‘sin(θ) = x’ is ‘No it’s not, cos(θ) = x!’ then you have indeed learned something, and we take comfort in that. However, context is everything. Here, ‘x’ is just a variable - it does not necessarily represent the x-coordinate of the point on The Unit Circle which lies on the terminal side of θ, assuming θ is drawn in standard position. Here, x represents the quantity sin(θ), and what we wish to know is how to express sin(2θ) in terms of x. We will see more of this kind of thing in Section 10.6, and, as usual, this is something we need for Calculus. Since sin(2θ) = 2 sin(θ) cos(θ), we need to write cos(θ) in terms of x to finish the problem. We substitute x = sin(θ) into the Pythagorean Identity, cos 2 (θ) + sin 2 (θ) = 1, to get cos 2 (θ) + x 2 = 1, or cos(θ) = ± √ 1 −x 2 . Since − π 2 ≤ θ ≤ π 2 , cos(θ) ≥ 0, and thus cos(θ) = √ 1 −x 2 . Our final answer is sin(2θ) = 2 sin(θ) cos(θ) = 2x √ 1 −x 2 . 3. We start with the right hand side of the identity and note that 1 + tan 2 (θ) = sec 2 (θ). From this point, we use the Reciprocal and Quotient Identities to rewrite tan(θ) and sec(θ) in terms of cos(θ) and sin(θ): 2 tan(θ) 1 + tan 2 (θ) = 2 tan(θ) sec 2 (θ) = 2 _ sin(θ) cos(θ) _ 1 cos 2 (θ) = 2 _ sin(θ) cos(θ) _ cos 2 (θ) = 2 _ sin(θ) $ $ $$ cos(θ) _ $ $ $$ cos(θ) cos(θ) = 2 sin(θ) cos(θ) = sin(2θ) 4. In Theorem 10.17, one of the formulas for cos(2θ), namely cos(2θ) = 2 cos 2 (θ) −1, expresses cos(2θ) as a polynomial in terms of cos(θ). We are now asked to find such an identity for cos(3θ). Using the sum formula for cosine, we begin with cos(3θ) = cos(2θ +θ) = cos(2θ) cos(θ) −sin(2θ) sin(θ) Our ultimate goal is to express the right hand side in terms of cos(θ) only. We substitute cos(2θ) = 2 cos 2 (θ) −1 and sin(2θ) = 2 sin(θ) cos(θ) which yields cos(3θ) = cos(2θ) cos(θ) −sin(2θ) sin(θ) = _ 2 cos 2 (θ) −1 _ cos(θ) −(2 sin(θ) cos(θ)) sin(θ) = 2 cos 3 (θ) −cos(θ) −2 sin 2 (θ) cos(θ) Finally, we exchange sin 2 (θ) for 1 −cos 2 (θ) courtesy of the Pythagorean Identity, and get cos(3θ) = 2 cos 3 (θ) −cos(θ) −2 sin 2 (θ) cos(θ) = 2 cos 3 (θ) −cos(θ) −2 _ 1 −cos 2 (θ) _ cos(θ) = 2 cos 3 (θ) −cos(θ) −2 cos(θ) + 2 cos 3 (θ) = 4 cos 3 (θ) −3 cos(θ) and we are done. 778 Foundations of Trigonometry In the last problem in Example 10.4.3, we saw how we could rewrite cos(3θ) as sums of powers of cos(θ). In Calculus, we have occasion to do the reverse; that is, reduce the power of cosine and sine. Solving the identity cos(2θ) = 2 cos 2 (θ) −1 for cos 2 (θ) and the identity cos(2θ) = 1−2 sin 2 (θ) for sin 2 (θ) results in the aptly-named ‘Power Reduction’ formulas below. Theorem 10.18. Power Reduction Formulas: For all angles θ, • cos 2 (θ) = 1 + cos(2θ) 2 • sin 2 (θ) = 1 −cos(2θ) 2 Example 10.4.4. Rewrite sin 2 (θ) cos 2 (θ) as a sum and difference of cosines to the first power. Solution. We begin with a straightforward application of Theorem 10.18 sin 2 (θ) cos 2 (θ) = _ 1 −cos(2θ) 2 __ 1 + cos(2θ) 2 _ = 1 4 _ 1 −cos 2 (2θ) _ = 1 4 − 1 4 cos 2 (2θ) Next, we apply the power reduction formula to cos 2 (2θ) to finish the reduction sin 2 (θ) cos 2 (θ) = 1 4 − 1 4 cos 2 (2θ) = 1 4 − 1 4 _ 1 + cos(2(2θ)) 2 _ = 1 4 − 1 8 − 1 8 cos(4θ) = 1 8 − 1 8 cos(4θ) Another application of the Power Reduction Formulas is the Half Angle Formulas. To start, we apply the Power Reduction Formula to cos 2 _ θ 2 _ cos 2 _ θ 2 _ = 1 + cos _ 2 _ θ 2 __ 2 = 1 + cos(θ) 2 . We can obtain a formula for cos _ θ 2 _ by extracting square roots. In a similar fashion, we may obtain a half angle formula for sine, and by using a quotient formula, obtain a half angle formula for tangent. We summarize these formulas below. 10.4 Trigonometric Identities 779 Theorem 10.19. Half Angle Formulas: For all applicable angles θ, • cos _ θ 2 _ = ± _ 1 + cos(θ) 2 • sin _ θ 2 _ = ± _ 1 −cos(θ) 2 • tan _ θ 2 _ = ± ¸ 1 −cos(θ) 1 + cos(θ) where the choice of ± depends on the quadrant in which the terminal side of θ 2 lies. Example 10.4.5. 1. Use a half angle formula to find the exact value of cos (15 ◦ ). 2. Suppose −π ≤ θ ≤ 0 with cos(θ) = − 3 5 . Find sin _ θ 2 _ . 3. Use the identity given in number 3 of Example 10.4.3 to derive the identity tan _ θ 2 _ = sin(θ) 1 + cos(θ) Solution. 1. To use the half angle formula, we note that 15 ◦ = 30 ◦ 2 and since 15 ◦ is a Quadrant I angle, its cosine is positive. Thus we have cos (15 ◦ ) = + _ 1 + cos (30 ◦ ) 2 = ¸ 1 + √ 3 2 2 = ¸ 1 + √ 3 2 2 2 2 = _ 2 + √ 3 4 = _ 2 + √ 3 2 Back in Example 10.4.1, we found cos (15 ◦ ) by using the difference formula for cosine. In that case, we determined cos (15 ◦ ) = √ 6+ √ 2 4 . The reader is encouraged to prove that these two expressions are equal. 2. If −π ≤ θ ≤ 0, then − π 2 ≤ θ 2 ≤ 0, which means sin _ θ 2 _ < 0. Theorem 10.19 gives sin _ θ 2 _ = − _ 1 −cos (θ) 2 = − _ 1 − _ − 3 5 _ 2 = − _ 1 + 3 5 2 5 5 = − _ 8 10 = − 2 √ 5 5 780 Foundations of Trigonometry 3. Instead of our usual approach to verifying identities, namely starting with one side of the equation and trying to transform it into the other, we will start with the identity we proved in number 3 of Example 10.4.3 and manipulate it into the identity we are asked to prove. The identity we are asked to start with is sin(2θ) = 2 tan(θ) 1+tan 2 (θ) . If we are to use this to derive an identity for tan _ θ 2 _ , it seems reasonable to proceed by replacing each occurrence of θ with θ 2 sin _ 2 _ θ 2 __ = 2 tan _ θ 2 _ 1 + tan 2 _ θ 2 _ sin(θ) = 2 tan _ θ 2 _ 1 + tan 2 _ θ 2 _ We now have the sin(θ) we need, but we somehow need to get a factor of 1 +cos(θ) involved. To get cosines involved, recall that 1 + tan 2 _ θ 2 _ = sec 2 _ θ 2 _ . We continue to manipulate our given identity by converting secants to cosines and using a power reduction formula sin(θ) = 2 tan _ θ 2 _ 1 + tan 2 _ θ 2 _ sin(θ) = 2 tan _ θ 2 _ sec 2 _ θ 2 _ sin(θ) = 2 tan _ θ 2 _ cos 2 _ θ 2 _ sin(θ) = 2 tan _ θ 2 _ _ 1 + cos _ 2 _ θ 2 __ 2 _ sin(θ) = tan _ θ 2 _ (1 + cos(θ)) tan _ θ 2 _ = sin(θ) 1 + cos(θ) Our next batch of identities, the Product to Sum Formulas, 3 are easily verified by expanding each of the right hand sides in accordance with Theorem 10.16 and as you should expect by now we leave the details as exercises. They are of particular use in Calculus, and we list them here for reference. Theorem 10.20. Product to Sum Formulas: For all angles α and β, • cos(α) cos(β) = 1 2 [cos(α −β) + cos(α +β)] • sin(α) sin(β) = 1 2 [cos(α −β) −cos(α +β)] • sin(α) cos(β) = 1 2 [sin(α −β) + sin(α +β)] 3 These are also known as the Prosthaphaeresis Formulas and have a rich history. The authors recommend that you conduct some research on them as your schedule allows. 10.4 Trigonometric Identities 781 Related to the Product to Sum Formulas are the Sum to Product Formulas, which we will have need of in Section 10.7. These are easily verified using the Product to Sum Formulas, and as such, their proofs are left as exercises. Theorem 10.21. Sum to Product Formulas: For all angles α and β, • cos(α) + cos(β) = 2 cos _ α +β 2 _ cos _ α −β 2 _ • cos(α) −cos(β) = −2 sin _ α +β 2 _ sin _ α −β 2 _ • sin(α) ±sin(β) = 2 sin _ α ±β 2 _ cos _ α ∓β 2 _ Example 10.4.6. 1. Write cos(2θ) cos(6θ) as a sum. 2. Write sin(θ) −sin(3θ) as a product. Solution. 1. Identifying α = 2θ and β = 6θ, we find cos(2θ) cos(6θ) = 1 2 [cos(2θ −6θ) + cos(2θ + 6θ)] = 1 2 cos(−4θ) + 1 2 cos(8θ) = 1 2 cos(4θ) + 1 2 cos(8θ), where the last equality is courtesy of the even identity for cosine, cos(−4θ) = cos(4θ). 2. Identifying α = θ and β = 3θ yields sin(θ) −sin(3θ) = 2 sin _ θ −3θ 2 _ cos _ θ + 3θ 2 _ = 2 sin (−θ) cos (2θ) = −2 sin (θ) cos (2θ) , where the last equality is courtesy of the odd identity for sine, sin(−θ) = −sin(θ). The reader is reminded that all of the identities presented in this section which regard the circular functions as functions of angles (in radian measure) apply equally well to the circular (trigonometric) functions regarded as functions of real numbers. In Exercises 38 - 43 in Section 10.5, we see how some of these identities manifest themselves geometrically as we study the graphs of the these functions. In the upcoming Exercises, however, you need to do all of your work analytically without graphs. 782 Foundations of Trigonometry 10.4.1 Exercises In Exercises 1 - 6, use the Even / Odd Identities to verify the identity. Assume all quantities are defined. 1. sin(3π −2θ) = −sin(2θ −3π) 2. cos _ − π 4 −5t _ = cos _ 5t + π 4 _ 3. tan(−t 2 + 1) = −tan(t 2 −1) 4. csc(−θ −5) = −csc(θ + 5) 5. sec(−6t) = sec(6t) 6. cot(9 −7θ) = −cot(7θ −9) In Exercises 7 - 21, use the Sum and Difference Identities to find the exact value. You may have need of the Quotient, Reciprocal or Even / Odd Identities as well. 7. cos(75 ◦ ) 8. sec(165 ◦ ) 9. sin(105 ◦ ) 10. csc(195 ◦ ) 11. cot(255 ◦ ) 12. tan(375 ◦ ) 13. cos _ 13π 12 _ 14. sin _ 11π 12 _ 15. tan _ 13π 12 _ 16. cos _ 7π 12 _ 17. tan _ 17π 12 _ 18. sin _ π 12 _ 19. cot _ 11π 12 _ 20. csc _ 5π 12 _ 21. sec _ − π 12 _ 22. If α is a Quadrant IV angle with cos(α) = √ 5 5 , and sin(β) = √ 10 10 , where π 2 < β < π, find (a) cos(α +β) (b) sin(α +β) (c) tan(α +β) (d) cos(α −β) (e) sin(α −β) (f) tan(α −β) 23. If csc(α) = 3, where 0 < α < π 2 , and β is a Quadrant II angle with tan(β) = −7, find (a) cos(α +β) (b) sin(α +β) (c) tan(α +β) (d) cos(α −β) (e) sin(α −β) (f) tan(α −β) 24. If sin(α) = 3 5 , where 0 < α < π 2 , and cos(β) = 12 13 where 3π 2 < β < 2π, find (a) sin(α +β) (b) cos(α −β) (c) tan(α −β) 10.4 Trigonometric Identities 783 25. If sec(α) = − 5 3 , where π 2 < α < π, and tan(β) = 24 7 , where π < β < 3π 2 , find (a) csc(α −β) (b) sec(α +β) (c) cot(α +β) In Exercises 26 - 38, verify the identity. 26. cos(θ −π) = −cos(θ) 27. sin(π −θ) = sin(θ) 28. tan _ θ + π 2 _ = −cot(θ) 29. sin(α +β) + sin(α −β) = 2 sin(α) cos(β) 30. sin(α +β) −sin(α −β) = 2 cos(α) sin(β) 31. cos(α +β) + cos(α −β) = 2 cos(α) cos(β) 32. cos(α+β) −cos(α−β) = −2 sin(α) sin(β) 33. sin(α +β) sin(α −β) = 1 + cot(α) tan(β) 1 −cot(α) tan(β) 34. cos(α +β) cos(α −β) = 1 −tan(α) tan(β) 1 + tan(α) tan(β) 35. tan(α +β) tan(α −β) = sin(α) cos(α) + sin(β) cos(β) sin(α) cos(α) −sin(β) cos(β) 36. sin(t +h) −sin(t) h = cos(t) _ sin(h) h _ + sin(t) _ cos(h) −1 h _ 37. cos(t +h) −cos(t) h = cos(t) _ cos(h) −1 h _ −sin(t) _ sin(h) h _ 38. tan(t +h) −tan(t) h = _ tan(h) h __ sec 2 (t) 1 −tan(t) tan(h) _ In Exercises 39 - 48, use the Half Angle Formulas to find the exact value. You may have need of the Quotient, Reciprocal or Even / Odd Identities as well. 39. cos(75 ◦ ) (compare with Exercise 7) 40. sin(105 ◦ ) (compare with Exercise 9) 41. cos(67.5 ◦ ) 42. sin(157.5 ◦ ) 43. tan(112.5 ◦ ) 44. cos _ 7π 12 _ (compare with Exercise 16) 45. sin _ π 12 _ (compare with Exercise 18) 46. cos _ π 8 _ 47. sin _ 5π 8 _ 48. tan _ 7π 8 _ 784 Foundations of Trigonometry In Exercises 49 - 58, use the given information about θ to find the exact values of • sin(2θ) • sin _ θ 2 _ • cos(2θ) • cos _ θ 2 _ • tan(2θ) • tan _ θ 2 _ 49. sin(θ) = − 7 25 where 3π 2 < θ < 2π 50. cos(θ) = 28 53 where 0 < θ < π 2 51. tan(θ) = 12 5 where π < θ < 3π 2 52. csc(θ) = 4 where π 2 < θ < π 53. cos(θ) = 3 5 where 0 < θ < π 2 54. sin(θ) = − 4 5 where π < θ < 3π 2 55. cos(θ) = 12 13 where 3π 2 < θ < 2π 56. sin(θ) = 5 13 where π 2 < θ < π 57. sec(θ) = √ 5 where 3π 2 < θ < 2π 58. tan(θ) = −2 where π 2 < θ < π In Exercises 59 - 73, verify the identity. Assume all quantities are defined. 59. (cos(θ) + sin(θ)) 2 = 1 + sin(2θ) 60. (cos(θ) −sin(θ)) 2 = 1 −sin(2θ) 61. tan(2θ) = 1 1 −tan(θ) − 1 1 + tan(θ) 62. csc(2θ) = cot(θ) + tan(θ) 2 63. 8 sin 4 (θ) = cos(4θ) −4 cos(2θ) + 3 64. 8 cos 4 (θ) = cos(4θ) + 4 cos(2θ) + 3 65. sin(3θ) = 3 sin(θ) −4 sin 3 (θ) 66. sin(4θ) = 4 sin(θ) cos 3 (θ) −4 sin 3 (θ) cos(θ) 67. 32 sin 2 (θ) cos 4 (θ) = 2 + cos(2θ) −2 cos(4θ) −cos(6θ) 68. 32 sin 4 (θ) cos 2 (θ) = 2 −cos(2θ) −2 cos(4θ) + cos(6θ) 69. cos(4θ) = 8 cos 4 (θ) −8 cos 2 (θ) + 1 70. cos(8θ) = 128 cos 8 (θ) −256 cos 6 (θ) +160 cos 4 (θ) −32 cos 2 (θ) +1 (HINT: Use the result to 69.) 71. sec(2θ) = cos(θ) cos(θ) + sin(θ) + sin(θ) cos(θ) −sin(θ) 72. 1 cos(θ) −sin(θ) + 1 cos(θ) + sin(θ) = 2 cos(θ) cos(2θ) 73. 1 cos(θ) −sin(θ) − 1 cos(θ) + sin(θ) = 2 sin(θ) cos(2θ) 10.4 Trigonometric Identities 785 In Exercises 74 - 79, write the given product as a sum. You may need to use an Even/Odd Identity. 74. cos(3θ) cos(5θ) 75. sin(2θ) sin(7θ) 76. sin(9θ) cos(θ) 77. cos(2θ) cos(6θ) 78. sin(3θ) sin(2θ) 79. cos(θ) sin(3θ) In Exercises 80 - 85, write the given sum as a product. You may need to use an Even/Odd or Cofunction Identity. 80. cos(3θ) + cos(5θ) 81. sin(2θ) −sin(7θ) 82. cos(5θ) −cos(6θ) 83. sin(9θ) −sin(−θ) 84. sin(θ) + cos(θ) 85. cos(θ) −sin(θ) 86. Suppose θ is a Quadrant I angle with sin(θ) = x. Verify the following formulas (a) cos(θ) = √ 1 −x 2 (b) sin(2θ) = 2x √ 1 −x 2 (c) cos(2θ) = 1 −2x 2 87. Discuss with your classmates how each of the formulas, if any, in Exercise 86 change if we change assume θ is a Quadrant II, III, or IV angle. 88. Suppose θ is a Quadrant I angle with tan(θ) = x. Verify the following formulas (a) cos(θ) = 1 √ x 2 + 1 (b) sin(θ) = x √ x 2 + 1 (c) sin(2θ) = 2x x 2 + 1 (d) cos(2θ) = 1 −x 2 x 2 + 1 89. Discuss with your classmates how each of the formulas, if any, in Exercise 88 change if we change assume θ is a Quadrant II, III, or IV angle. 90. If sin(θ) = x 2 for − π 2 < θ < π 2 , find an expression for cos(2θ) in terms of x. 91. If tan(θ) = x 7 for − π 2 < θ < π 2 , find an expression for sin(2θ) in terms of x. 92. If sec(θ) = x 4 for 0 < θ < π 2 , find an expression for ln [ sec(θ) + tan(θ)[ in terms of x. 93. Show that cos 2 (θ) −sin 2 (θ) = 2 cos 2 (θ) −1 = 1 −2 sin 2 (θ) for all θ. 94. Let θ be a Quadrant III angle with cos(θ) = − 1 5 . Show that this is not enough information to determine the sign of sin _ θ 2 _ by first assuming 3π < θ < 7π 2 and then assuming π < θ < 3π 2 and computing sin _ θ 2 _ in both cases. 786 Foundations of Trigonometry 95. Without using your calculator, show that _ 2 + √ 3 2 = √ 6 + √ 2 4 96. In part 4 of Example 10.4.3, we wrote cos(3θ) as a polynomial in terms of cos(θ). In Exercise 69, we had you verify an identity which expresses cos(4θ) as a polynomial in terms of cos(θ). Can you find a polynomial in terms of cos(θ) for cos(5θ)? cos(6θ)? Can you find a pattern so that cos(nθ) could be written as a polynomial in cosine for any natural number n? 97. In Exercise 65, we has you verify an identity which expresses sin(3θ) as a polynomial in terms of sin(θ). Can you do the same for sin(5θ)? What about for sin(4θ)? If not, what goes wrong? 98. Verify the Even / Odd Identities for tangent, secant, cosecant and cotangent. 99. Verify the Cofunction Identities for tangent, secant, cosecant and cotangent. 100. Verify the Difference Identities for sine and tangent. 101. Verify the Product to Sum Identities. 102. Verify the Sum to Product Identities. 10.4 Trigonometric Identities 787 10.4.2 Answers 7. cos(75 ◦ ) = √ 6 − √ 2 4 8. sec(165 ◦ ) = − 4 √ 2 + √ 6 = √ 2 − √ 6 9. sin(105 ◦ ) = √ 6 + √ 2 4 10. csc(195 ◦ ) = 4 √ 2 − √ 6 = −( √ 2 + √ 6) 11. cot(255 ◦ ) = √ 3 −1 √ 3 + 1 = 2 − √ 3 12. tan(375 ◦ ) = 3 − √ 3 3 + √ 3 = 2 − √ 3 13. cos _ 13π 12 _ = − √ 6 + √ 2 4 14. sin _ 11π 12 _ = √ 6 − √ 2 4 15. tan _ 13π 12 _ = 3 − √ 3 3 + √ 3 = 2 − √ 3 16. cos _ 7π 12 _ = √ 2 − √ 6 4 17. tan _ 17π 12 _ = 2 + √ 3 18. sin _ π 12 _ = √ 6 − √ 2 4 19. cot _ 11π 12 _ = −(2 + √ 3) 20. csc _ 5π 12 _ = √ 6 − √ 2 21. sec _ − π 12 _ = √ 6 − √ 2 22. (a) cos(α +β) = − √ 2 10 (b) sin(α +β) = 7 √ 2 10 (c) tan(α +β) = −7 (d) cos(α −β) = − √ 2 2 (e) sin(α −β) = √ 2 2 (f) tan(α −β) = −1 23. (a) cos(α +β) = − 4 + 7 √ 2 30 (b) sin(α +β) = 28 − √ 2 30 (c) tan(α +β) = −28 + √ 2 4 + 7 √ 2 = 63 −100 √ 2 41 (d) cos(α −β) = −4 + 7 √ 2 30 (e) sin(α −β) = − 28 + √ 2 30 (f) tan(α −β) = 28 + √ 2 4 −7 √ 2 = − 63 + 100 √ 2 41 24. (a) sin(α +β) = 16 65 (b) cos(α −β) = 33 65 (c) tan(α −β) = 56 33 788 Foundations of Trigonometry 25. (a) csc(α −β) = − 5 4 (b) sec(α +β) = 125 117 (c) cot(α +β) = 117 44 39. cos(75 ◦ ) = _ 2 − √ 3 2 40. sin(105 ◦ ) = _ 2 + √ 3 2 41. cos(67.5 ◦ ) = _ 2 − √ 2 2 42. sin(157.5 ◦ ) = _ 2 − √ 2 2 43. tan(112.5 ◦ ) = − ¸ 2 + √ 2 2 − √ 2 = −1 − √ 2 44. cos _ 7π 12 _ = − _ 2 − √ 3 2 45. sin _ π 12 _ = _ 2 − √ 3 2 46. cos _ π 8 _ = _ 2 + √ 2 2 47. sin _ 5π 8 _ = _ 2 + √ 2 2 48. tan _ 7π 8 _ = − ¸ 2 − √ 2 2 + √ 2 = 1 − √ 2 49. • sin(2θ) = − 336 625 • sin _ θ 2 _ = √ 2 10 • cos(2θ) = 527 625 • cos _ θ 2 _ = − 7 √ 2 10 • tan(2θ) = − 336 527 • tan _ θ 2 _ = − 1 7 50. • sin(2θ) = 2520 2809 • sin _ θ 2 _ = 5 √ 106 106 • cos(2θ) = − 1241 2809 • cos _ θ 2 _ = 9 √ 106 106 • tan(2θ) = − 2520 1241 • tan _ θ 2 _ = 5 9 51. • sin(2θ) = 120 169 • sin _ θ 2 _ = 3 √ 13 13 • cos(2θ) = − 119 169 • cos _ θ 2 _ = − 2 √ 13 13 • tan(2θ) = − 120 119 • tan _ θ 2 _ = − 3 2 52. • sin(2θ) = − √ 15 8 • sin _ θ 2 _ = _ 8 + 2 √ 15 4 • cos(2θ) = 7 8 • cos _ θ 2 _ = _ 8 −2 √ 15 4 • tan(2θ) = − √ 15 7 • tan _ θ 2 _ = ¸ 8 + 2 √ 15 8 −2 √ 15 tan _ θ 2 _ = 4 + √ 15 53. • sin(2θ) = 24 25 • sin _ θ 2 _ = √ 5 5 • cos(2θ) = − 7 25 • cos _ θ 2 _ = 2 √ 5 5 • tan(2θ) = − 24 7 • tan _ θ 2 _ = 1 2 10.4 Trigonometric Identities 789 54. • sin(2θ) = 24 25 • sin _ θ 2 _ = 2 √ 5 5 • cos(2θ) = − 7 25 • cos _ θ 2 _ = − √ 5 5 • tan(2θ) = − 24 7 • tan _ θ 2 _ = −2 55. • sin(2θ) = − 120 169 • sin _ θ 2 _ = √ 26 26 • cos(2θ) = 119 169 • cos _ θ 2 _ = − 5 √ 26 26 • tan(2θ) = − 120 119 • tan _ θ 2 _ = − 1 5 56. • sin(2θ) = − 120 169 • sin _ θ 2 _ = 5 √ 26 26 • cos(2θ) = 119 169 • cos _ θ 2 _ = √ 26 26 • tan(2θ) = − 120 119 • tan _ θ 2 _ = 5 57. • sin(2θ) = − 4 5 • sin _ θ 2 _ = _ 50 −10 √ 5 10 • cos(2θ) = − 3 5 • cos _ θ 2 _ = − _ 50 + 10 √ 5 10 • tan(2θ) = 4 3 • tan _ θ 2 _ = − ¸ 5 − √ 5 5 + √ 5 tan _ θ 2 _ = 5 −5 √ 5 10 58. • sin(2θ) = − 4 5 • sin _ θ 2 _ = _ 50 + 10 √ 5 10 • cos(2θ) = − 3 5 • cos _ θ 2 _ = _ 50 −10 √ 5 10 • tan(2θ) = 4 3 • tan _ θ 2 _ = ¸ 5 + √ 5 5 − √ 5 tan _ θ 2 _ = 5 + 5 √ 5 10 74. cos(2θ) + cos(8θ) 2 75. cos(5θ) −cos(9θ) 2 76. sin(8θ) + sin(10θ) 2 77. cos(4θ) + cos(8θ) 2 78. cos(θ) −cos(5θ) 2 79. sin(2θ) + sin(4θ) 2 80. 2 cos(4θ) cos(θ) 81. −2 cos _ 9 2 θ _ sin _ 5 2 θ _ 82. 2 sin _ 11 2 θ _ sin _ 1 2 θ _ 83. 2 cos(4θ) sin(5θ) 84. √ 2 cos _ θ − π 4 _ 85. − √ 2 sin _ θ − π 4 _ 90. 1 − x 2 2 91. 14x x 2 + 49 92. ln [x + √ x 2 + 16[ −ln(4) 790 Foundations of Trigonometry 10.5 Graphs of the Trigonometric Functions In this section, we return to our discussion of the circular (trigonometric) functions as functions of real numbers and pick up where we left off in Sections 10.2.1 and 10.3.1. As usual, we begin our study with the functions f(t) = cos(t) and g(t) = sin(t). 10.5.1 Graphs of the Cosine and Sine Functions From Theorem 10.5 in Section 10.2.1, we know that the domain of f(t) = cos(t) and of g(t) = sin(t) is all real numbers, (−∞, ∞), and the range of both functions is [−1, 1]. The Even / Odd Identities in Theorem 10.12 tell us cos(−t) = cos(t) for all real numbers t and sin(−t) = −sin(t) for all real numbers t. This means f(t) = cos(t) is an even function, while g(t) = sin(t) is an odd function. 1 Another important property of these functions is that for coterminal angles α and β, cos(α) = cos(β) and sin(α) = sin(β). Said differently, cos(t+2πk) = cos(t) and sin(t+2πk) = sin(t) for all real numbers t and any integer k. This last property is given a special name. Definition 10.3. Periodic Functions: A function f is said to be periodic if there is a real number c so that f(t +c) = f(t) for all real numbers t in the domain of f. The smallest positive number p for which f(t +p) = f(t) for all real numbers t in the domain of f, if it exists, is called the period of f. We have already seen a family of periodic functions in Section 2.1: the constant functions. However, despite being periodic a constant function has no period. (We’ll leave that odd gem as an exercise for you.) Returning to the circular functions, we see that by Definition 10.3, f(t) = cos(t) is periodic, since cos(t + 2πk) = cos(t) for any integer k. To determine the period of f, we need to find the smallest real number p so that f(t + p) = f(t) for all real numbers t or, said differently, the smallest positive real number p such that cos(t + p) = cos(t) for all real numbers t. We know that cos(t +2π) = cos(t) for all real numbers t but the question remains if any smaller real number will do the trick. Suppose p > 0 and cos(t +p) = cos(t) for all real numbers t. Then, in particular, cos(0 + p) = cos(0) so that cos(p) = 1. From this we know p is a multiple of 2π and, since the smallest positive multiple of 2π is 2π itself, we have the result. Similarly, we can show g(t) = sin(t) is also periodic with 2π as its period. 2 Having period 2π essentially means that we can completely understand everything about the functions f(t) = cos(t) and g(t) = sin(t) by studying one interval of length 2π, say [0, 2π]. 3 One last property of the functions f(t) = cos(t) and g(t) = sin(t) is worth pointing out: both of these functions are continuous and smooth. Recall from Section 3.1 that geometrically this means the graphs of the cosine and sine functions have no jumps, gaps, holes in the graph, asymptotes, 1 See section 1.6 for a review of these concepts. 2 Alternatively, we can use the Cofunction Identities in Theorem 10.14 to show that g(t) = sin(t) is periodic with period 2π since g(t) = sin(t) = cos π 2 −t = f π 2 −t . 3 Technically, we should study the interval [0, 2π), 4 since whatever happens at t = 2π is the same as what happens at t = 0. As we will see shortly, t = 2π gives us an extra ‘check’ when we go to graph these functions. 4 In some advanced texts, the interval of choice is [−π, π). 10.5 Graphs of the Trigonometric Functions 791 corners or cusps. As we shall see, the graphs of both f(t) = cos(t) and g(t) = sin(t) meander nicely and don’t cause any trouble. We summarize these facts in the following theorem. Theorem 10.22. Properties of the Cosine and Sine Functions • The function f(x) = cos(x) • The function g(x) = sin(x) – has domain (−∞, ∞) – has domain (−∞, ∞) – has range [−1, 1] – has range [−1, 1] – is continuous and smooth – is continuous and smooth – is even – is odd – has period 2π – has period 2π In the chart above, we followed the convention established in Section 1.6 and used x as the indepen- dent variable and y as the dependent variable. 5 This allows us to turn our attention to graphing the cosine and sine functions in the Cartesian Plane. To graph y = cos(x), we make a table as we did in Section 1.6 using some of the ‘common values’ of x in the interval [0, 2π]. This generates a portion of the cosine graph, which we call the ‘fundamental cycle’ of y = cos(x). x cos(x) (x, cos(x)) 0 1 (0, 1) π 4 √ 2 2 _ π 4 , √ 2 2 _ π 2 0 _ π 2 , 0 _ 3π 4 − √ 2 2 _ 3π 4 , − √ 2 2 _ π −1 (π, −1) 5π 4 − √ 2 2 _ 5π 4 , − √ 2 2 _ 3π 2 0 _ 3π 2 , 0 _ 7π 4 √ 2 2 _ 7π 4 , √ 2 2 _ 2π 1 (2π, 1) x y π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π −1 1 The ‘fundamental cycle’ of y = cos(x). A few things about the graph above are worth mentioning. First, this graph represents only part of the graph of y = cos(x). To get the entire graph, we imagine ‘copying and pasting’ this graph end to end infinitely in both directions (left and right) on the x-axis. Secondly, the vertical scale here has been greatly exaggerated for clarity and aesthetics. Below is an accurate-to-scale graph of y = cos(x) showing several cycles with the ‘fundamental cycle’ plotted thicker than the others. The 5 The use of x and y in this context is not to be confused with the x- and y-coordinates of points on the Unit Circle which define cosine and sine. Using the term ‘trigonometric function’ as opposed to ‘circular function’ can help with that, but one could then ask, “Hey, where’s the triangle?” 792 Foundations of Trigonometry graph of y = cos(x) is usually described as ‘wavelike’ – indeed, many of the applications involving the cosine and sine functions feature modeling wavelike phenomena. x y An accurately scaled graph of y = cos(x). We can plot the fundamental cycle of the graph of y = sin(x) similarly, with similar results. x sin(x) (x, sin(x)) 0 0 (0, 0) π 4 √ 2 2 _ π 4 , √ 2 2 _ π 2 1 _ π 2 , 1 _ 3π 4 √ 2 2 _ 3π 4 , √ 2 2 _ π 0 (π, 0) 5π 4 − √ 2 2 _ 5π 4 , − √ 2 2 _ 3π 2 −1 _ 3π 2 , −1 _ 7π 4 − √ 2 2 _ 7π 4 , − √ 2 2 _ 2π 0 (2π, 0) x y π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π −1 1 The ‘fundamental cycle’ of y = sin(x). As with the graph of y = cos(x), we provide an accurately scaled graph of y = sin(x) below with the fundamental cycle highlighted. x y An accurately scaled graph of y = sin(x). It is no accident that the graphs of y = cos(x) and y = sin(x) are so similar. Using a cofunction identity along with the even property of cosine, we have sin(x) = cos _ π 2 −x _ = cos _ − _ x − π 2 __ = cos _ x − π 2 _ Recalling Section 1.7, we see from this formula that the graph of y = sin(x) is the result of shifting the graph of y = cos(x) to the right π 2 units. A visual inspection confirms this. Now that we know the basic shapes of the graphs of y = cos(x) and y = sin(x), we can use Theorem 1.7 in Section 1.7 to graph more complicated curves. To do so, we need to keep track of 10.5 Graphs of the Trigonometric Functions 793 the movement of some key points on the original graphs. We choose to track the values x = 0, π 2 , π, 3π 2 and 2π. These ‘quarter marks’ correspond to quadrantal angles, and as such, mark the location of the zeros and the local extrema of these functions over exactly one period. Before we begin our next example, we need to review the concept of the ‘argument’ of a function as first introduced in Section 1.4. For the function f(x) = 1 − 5 cos(2x − π), the argument of f is x. We shall have occasion, however, to refer to the argument of the cosine, which in this case is 2x − π. Loosely stated, the argument of a trigonometric function is the expression ‘inside’ the function. Example 10.5.1. Graph one cycle of the following functions. State the period of each. 1. f(x) = 3 cos _ πx−π 2 _ + 1 2. g(x) = 1 2 sin(π −2x) + 3 2 Solution. 1. We set the argument of the cosine, πx−π 2 , equal to each of the values: 0, π 2 , π, 3π 2 , 2π and solve for x. We summarize the results below. a πx−π 2 = a x 0 πx−π 2 = 0 1 π 2 πx−π 2 = π 2 2 π πx−π 2 = π 3 3π 2 πx−π 2 = 3π 2 4 2π πx−π 2 = 2π 5 Next, we substitute each of these x values into f(x) = 3 cos _ πx−π 2 _ + 1 to determine the corresponding y-values and connect the dots in a pleasing wavelike fashion. x f(x) (x, f(x)) 1 4 (1, 4) 2 1 (2, 1) 3 −2 (3, −2) 4 1 (4, 1) 5 4 (5, 4) x y 1 2 3 4 5 −2 −1 1 2 3 4 One cycle of y = f(x). One cycle is graphed on [1, 5] so the period is the length of that interval which is 4. 2. Proceeding as above, we set the argument of the sine, π − 2x, equal to each of our quarter marks and solve for x. 794 Foundations of Trigonometry a π −2x = a x 0 π −2x = 0 π 2 π 2 π −2x = π 2 π 4 π π −2x = π 0 3π 2 π −2x = 3π 2 − π 4 2π π −2x = 2π − π 2 We now find the corresponding y-values on the graph by substituting each of these x-values into g(x) = 1 2 sin(π −2x) + 3 2 . Once again, we connect the dots in a wavelike fashion. x g(x) (x, g(x)) π 2 3 2 _ π 2 , 3 2 _ π 4 2 _ π 4 , 2 _ 0 3 2 _ 0, 3 2 _ − π 4 1 _ − π 4 , 1 _ − π 2 3 2 _ − π 2 , 3 2 _ x y − π 2 − π 4 π 4 π 2 1 2 One cycle of y = g(x). One cycle was graphed on the interval _ − π 2 , π 2 ¸ so the period is π 2 − _ − π 2 _ = π. The functions in Example 10.5.1 are examples of sinusoids. Roughly speaking, a sinusoid is the result of taking the basic graph of f(x) = cos(x) or g(x) = sin(x) and performing any of the transformations 6 mentioned in Section 1.7. Sinusoids can be characterized by four properties: period, amplitude, phase shift and vertical shift. We have already discussed period, that is, how long it takes for the sinusoid to complete one cycle. The standard period of both f(x) = cos(x) and g(x) = sin(x) is 2π, but horizontal scalings will change the period of the resulting sinusoid. The amplitude of the sinusoid is a measure of how ‘tall’ the wave is, as indicated in the figure below. The amplitude of the standard cosine and sine functions is 1, but vertical scalings can alter this. 6 We have already seen how the Even/Odd and Cofunction Identities can be used to rewrite g(x) = sin(x) as a transformed version of f(x) = cos(x), so of course, the reverse is true: f(x) = cos(x) can be written as a transformed version of g(x) = sin(x). The authors have seen some instances where sinusoids are always converted to cosine functions while in other disciplines, the sinusoids are always written in terms of sine functions. We will discuss the applications of sinusoids in greater detail in Chapter 11. Until then, we will keep our options open. 10.5 Graphs of the Trigonometric Functions 795 amplitude baseline period The phase shift of the sinusoid is the horizontal shift experienced by the fundamental cycle. We have seen that a phase (horizontal) shift of π 2 to the right takes f(x) = cos(x) to g(x) = sin(x) since cos _ x − π 2 _ = sin(x). As the reader can verify, a phase shift of π 2 to the left takes g(x) = sin(x) to f(x) = cos(x). The vertical shift of a sinusoid is exactly the same as the vertical shifts in Section 1.7. In most contexts, the vertical shift of a sinusoid is assumed to be 0, but we state the more general case below. The following theorem, which is reminiscent of Theorem 1.7 in Section 1.7, shows how to find these four fundamental quantities from the formula of the given sinusoid. Theorem 10.23. For ω > 0, the functions C(x) = Acos(ωx +φ) +B and S(x) = Asin(ωx +φ) +B • have period 2π ω • have amplitude [A[ • have phase shift − φ ω • have vertical shift B We note that in some scientific and engineering circles, the quantity φ mentioned in Theorem 10.23 is called the phase of the sinusoid. Since our interest in this book is primarily with graphing sinusoids, we focus our attention on the horizontal shift − φ ω induced by φ. The proof of Theorem 10.23 is a direct application of Theorem 1.7 in Section 1.7 and is left to the reader. The parameter ω, which is stipulated to be positive, is called the (angular) frequency of the sinusoid and is the number of cycles the sinusoid completes over a 2π interval. We can always ensure ω > 0 using the Even/Odd Identities. 7 We now test out Theorem 10.23 using the functions f and g featured in Example 10.5.1. First, we write f(x) in the form prescribed in Theorem 10.23, f(x) = 3 cos _ πx −π 2 _ + 1 = 3 cos _ π 2 x + _ − π 2 __ + 1, 7 Try using the formulas in Theorem 10.23 applied to C(x) = cos(−x +π) to see why we need ω > 0. 796 Foundations of Trigonometry so that A = 3, ω = π 2 , φ = − π 2 and B = 1. According to Theorem 10.23, the period of f is 2π ω = 2π π/2 = 4, the amplitude is [A[ = [3[ = 3, the phase shift is − φ ω = − −π/2 π/2 = 1 (indicating a shift to the right 1 unit) and the vertical shift is B = 1 (indicating a shift up 1 unit.) All of these match with our graph of y = f(x). Moreover, if we start with the basic shape of the cosine graph, shift it 1 unit to the right, 1 unit up, stretch the amplitude to 3 and shrink the period to 4, we will have reconstructed one period of the graph of y = f(x). In other words, instead of tracking the five ‘quarter marks’ through the transformations to plot y = f(x), we can use five other pieces of information: the phase shift, vertical shift, amplitude, period and basic shape of the cosine curve. Turning our attention now to the function g in Example 10.5.1, we first need to use the odd property of the sine function to write it in the form required by Theorem 10.23 g(x) = 1 2 sin(π −2x) + 3 2 = 1 2 sin(−(2x −π)) + 3 2 = − 1 2 sin(2x −π) + 3 2 = − 1 2 sin(2x + (−π)) + 3 2 We find A = − 1 2 , ω = 2, φ = −π and B = 3 2 . The period is then 2π 2 = π, the amplitude is ¸ ¸ − 1 2 ¸ ¸ = 1 2 , the phase shift is − −π 2 = π 2 (indicating a shift right π 2 units) and the vertical shift is up 3 2 . Note that, in this case, all of the data match our graph of y = g(x) with the exception of the phase shift. Instead of the graph starting at x = π 2 , it ends there. Remember, however, that the graph presented in Example 10.5.1 is only one portion of the graph of y = g(x). Indeed, another complete cycle begins at x = π 2 , and this is the cycle Theorem 10.23 is detecting. The reason for the discrepancy is that, in order to apply Theorem 10.23, we had to rewrite the formula for g(x) using the odd property of the sine function. Note that whether we graph y = g(x) using the ‘quarter marks’ approach or using the Theorem 10.23, we get one complete cycle of the graph, which means we have completely determined the sinusoid. Example 10.5.2. Below is the graph of one complete cycle of a sinusoid y = f(x). −1, 5 2 1 2 , 1 2 2, − 3 2 7 2 , 1 2 5, 5 2 x y −1 1 2 3 4 5 −2 −1 1 2 3 One cycle of y = f(x). 1. Find a cosine function whose graph matches the graph of y = f(x). 10.5 Graphs of the Trigonometric Functions 797 2. Find a sine function whose graph matches the graph of y = f(x). Solution. 1. We fit the data to a function of the form C(x) = Acos(ωx + φ) + B. Since one cycle is graphed over the interval [−1, 5], its period is 5 − (−1) = 6. According to Theorem 10.23, 6 = 2π ω , so that ω = π 3 . Next, we see that the phase shift is −1, so we have − φ ω = −1, or φ = ω = π 3 . To find the amplitude, note that the range of the sinusoid is _ − 3 2 , 5 2 ¸ . As a result, the amplitude A = 1 2 _ 5 2 − _ − 3 2 _¸ = 1 2 (4) = 2. Finally, to determine the vertical shift, we average the endpoints of the range to find B = 1 2 _ 5 2 + _ − 3 2 _¸ = 1 2 (1) = 1 2 . Our final answer is C(x) = 2 cos _ π 3 x + π 3 _ + 1 2 . 2. Most of the work to fit the data to a function of the form S(x) = Asin(ωx +φ) +B is done. The period, amplitude and vertical shift are the same as before with ω = π 3 , A = 2 and B = 1 2 . The trickier part is finding the phase shift. To that end, we imagine extending the graph of the given sinusoid as in the figure below so that we can identify a cycle beginning at _ 7 2 , 1 2 _ . Taking the phase shift to be 7 2 , we get − φ ω = 7 2 , or φ = − 7 2 ω = − 7 2 _ π 3 _ = − 7π 6 . Hence, our answer is S(x) = 2 sin _ π 3 x − 7π 6 _ + 1 2 . 7 2 , 1 2 5, 5 2 13 2 , 1 2 8, − 3 2 19 2 , 5 2 x y −1 1 2 3 4 5 6 7 8 9 10 −2 −1 1 2 3 Extending the graph of y = f(x). Note that each of the answers given in Example 10.5.2 is one choice out of many possible answers. For example, when fitting a sine function to the data, we could have chosen to start at _ 1 2 , 1 2 _ taking A = −2. In this case, the phase shift is 1 2 so φ = − π 6 for an answer of S(x) = −2 sin _ π 3 x − π 6 _ + 1 2 . Alternatively, we could have extended the graph of y = f(x) to the left and considered a sine function starting at _ − 5 2 , 1 2 _ , and so on. Each of these formulas determine the same sinusoid curve and their formulas are all equivalent using identities. Speaking of identities, if we use the sum identity for cosine, we can expand the formula to yield C(x) = Acos(ωx +φ) +B = Acos(ωx) cos(φ) −Asin(ωx) sin(φ) +B. 798 Foundations of Trigonometry Similarly, using the sum identity for sine, we get S(x) = Asin(ωx +φ) +B = Asin(ωx) cos(φ) +Acos(ωx) sin(φ) +B. Making these observations allows us to recognize (and graph) functions as sinusoids which, at first glance, don’t appear to fit the forms of either C(x) or S(x). Example 10.5.3. Consider the function f(x) = cos(2x) − √ 3 sin(2x). Find a formula for f(x): 1. in the form C(x) = Acos(ωx +φ) +B for ω > 0 2. in the form S(x) = Asin(ωx +φ) +B for ω > 0 Check your answers analytically using identities and graphically using a calculator. Solution. 1. The key to this problem is to use the expanded forms of the sinusoid formulas and match up corresponding coefficients. Equating f(x) = cos(2x) − √ 3 sin(2x) with the expanded form of C(x) = Acos(ωx +φ) +B, we get cos(2x) − √ 3 sin(2x) = Acos(ωx) cos(φ) −Asin(ωx) sin(φ) +B It should be clear that we can take ω = 2 and B = 0 to get cos(2x) − √ 3 sin(2x) = Acos(2x) cos(φ) −Asin(2x) sin(φ) To determine A and φ, a bit more work is involved. We get started by equating the coefficients of the trigonometric functions on either side of the equation. On the left hand side, the coefficient of cos(2x) is 1, while on the right hand side, it is Acos(φ). Since this equation is to hold for all real numbers, we must have 8 that Acos(φ) = 1. Similarly, we find by equating the coefficients of sin(2x) that Asin(φ) = √ 3. What we have here is a system of nonlinear equations! We can temporarily eliminate the dependence on φ by using the Pythagorean Identity. We know cos 2 (φ) + sin 2 (φ) = 1, so multiplying this by A 2 gives A 2 cos 2 (φ)+A 2 sin 2 (φ) = A 2 . Since Acos(φ) = 1 and Asin(φ) = √ 3, we get A 2 = 1 2 +( √ 3) 2 = 4 or A = ±2. Choosing A = 2, we have 2 cos(φ) = 1 and 2 sin(φ) = √ 3 or, after some rearrangement, cos(φ) = 1 2 and sin(φ) = √ 3 2 . One such angle φ which satisfies this criteria is φ = π 3 . Hence, one way to write f(x) as a sinusoid is f(x) = 2 cos _ 2x + π 3 _ . We can easily check our answer using the sum formula for cosine f(x) = 2 cos _ 2x + π 3 _ = 2 _ cos(2x) cos _ π 3 _ −sin(2x) sin _ π 3 _¸ = 2 _ cos(2x) _ 1 2 _ −sin(2x) _ √ 3 2 __ = cos(2x) − √ 3 sin(2x) 8 This should remind you of equation coefficients of like powers of x in Section 8.6. 10.5 Graphs of the Trigonometric Functions 799 2. Proceeding as before, we equate f(x) = cos(2x) − √ 3 sin(2x) with the expanded form of S(x) = Asin(ωx +φ) +B to get cos(2x) − √ 3 sin(2x) = Asin(ωx) cos(φ) +Acos(ωx) sin(φ) +B Once again, we may take ω = 2 and B = 0 so that cos(2x) − √ 3 sin(2x) = Asin(2x) cos(φ) +Acos(2x) sin(φ) We equate 9 the coefficients of cos(2x) on either side and get Asin(φ) = 1 and Acos(φ) = − √ 3. Using A 2 cos 2 (φ) + A 2 sin 2 (φ) = A 2 as before, we get A = ±2, and again we choose A = 2. This means 2 sin(φ) = 1, or sin(φ) = 1 2 , and 2 cos(φ) = − √ 3, which means cos(φ) = − √ 3 2 . One such angle which meets these criteria is φ = 5π 6 . Hence, we have f(x) = 2 sin _ 2x + 5π 6 _ . Checking our work analytically, we have f(x) = 2 sin _ 2x + 5π 6 _ = 2 _ sin(2x) cos _ 5π 6 _ + cos(2x) sin _ 5π 6 _¸ = 2 _ sin(2x) _ − √ 3 2 _ + cos(2x) _ 1 2 _ _ = cos(2x) − √ 3 sin(2x) Graphing the three formulas for f(x) result in the identical curve, verifying our analytic work. It is important to note that in order for the technique presented in Example 10.5.3 to fit a function into one of the forms in Theorem 10.23, the arguments of the cosine and sine function much match. That is, while f(x) = cos(2x) − √ 3 sin(2x) is a sinusoid, g(x) = cos(2x) − √ 3 sin(3x) is not. 10 It is also worth mentioning that, had we chosen A = −2 instead of A = 2 as we worked through Example 10.5.3, our final answers would have looked different. The reader is encouraged to rework Example 10.5.3 using A = −2 to see what these differences are, and then for a challenging exercise, use identities to show that the formulas are all equivalent. The general equations to fit a function of the form f(x) = a cos(ωx) +b sin(ωx) +B into one of the forms in Theorem 10.23 are explored in Exercise 35. 9 Be careful here! 10 This graph does, however, exhibit sinusoid-like characteristics! Check it out! 800 Foundations of Trigonometry 10.5.2 Graphs of the Secant and Cosecant Functions We now turn our attention to graphing y = sec(x). Since sec(x) = 1 cos(x) , we can use our table of values for the graph of y = cos(x) and take reciprocals. We know from Section 10.3.1 that the domain of F(x) = sec(x) excludes all odd multiples of π 2 , and sure enough, we run into trouble at x = π 2 and x = 3π 2 since cos(x) = 0 at these values. Using the notation introduced in Section 4.2, we have that as x → π 2 − , cos(x) → 0 + , so sec(x) → ∞. (See Section 10.3.1 for a more detailed analysis.) Similarly, we find that as x → π 2 + , sec(x) → −∞; as x → 3π 2 − , sec(x) → −∞; and as x → 3π 2 + , sec(x) →∞. This means we have a pair of vertical asymptotes to the graph of y = sec(x), x = π 2 and x = 3π 2 . Since cos(x) is periodic with period 2π, it follows that sec(x) is also. 11 Below we graph a fundamental cycle of y = sec(x) along with a more complete graph obtained by the usual ‘copying and pasting.’ 12 x cos(x) sec(x) (x, sec(x)) 0 1 1 (0, 1) π 4 √ 2 2 √ 2 _ π 4 , √ 2 _ π 2 0 undefined 3π 4 − √ 2 2 − √ 2 _ 3π 4 , − √ 2 _ π −1 −1 (π, −1) 5π 4 − √ 2 2 − √ 2 _ 5π 4 , − √ 2 _ 3π 2 0 undefined 7π 4 √ 2 2 √ 2 _ 7π 4 , √ 2 _ 2π 1 1 (2π, 1) x y π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π −3 −2 −1 1 2 3 The ‘fundamental cycle’ of y = sec(x). x y The graph of y = sec(x). 11 Provided sec(α) and sec(β) are defined, sec(α) = sec(β) if and only if cos(α) = cos(β). Hence, sec(x) inherits its period from cos(x). 12 In Section 10.3.1, we argued the range of F(x) = sec(x) is (−∞, −1] ∪ [1, ∞). We can now see this graphically. 10.5 Graphs of the Trigonometric Functions 801 As one would expect, to graph y = csc(x) we begin with y = sin(x) and take reciprocals of the corresponding y-values. Here, we encounter issues at x = 0, x = π and x = 2π. Proceeding with the usual analysis, we graph the fundamental cycle of y = csc(x) below along with the dotted graph of y = sin(x) for reference. Since y = sin(x) and y = cos(x) are merely phase shifts of each other, so too are y = csc(x) and y = sec(x). x sin(x) csc(x) (x, csc(x)) 0 0 undefined π 4 √ 2 2 √ 2 _ π 4 , √ 2 _ π 2 1 1 _ π 2 , 1 _ 3π 4 √ 2 2 √ 2 _ 3π 4 , √ 2 _ π 0 undefined 5π 4 − √ 2 2 − √ 2 _ 5π 4 , − √ 2 _ 3π 2 −1 −1 _ 3π 2 , −1 _ 7π 4 − √ 2 2 − √ 2 _ 7π 4 , − √ 2 _ 2π 0 undefined x y π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π −3 −2 −1 1 2 3 The ‘fundamental cycle’ of y = csc(x). Once again, our domain and range work in Section 10.3.1 is verified geometrically in the graph of y = G(x) = csc(x). x y The graph of y = csc(x). Note that, on the intervals between the vertical asymptotes, both F(x) = sec(x) and G(x) = csc(x) are continuous and smooth. In other words, they are continuous and smooth on their domains. 13 The following theorem summarizes the properties of the secant and cosecant functions. Note that 13 Just like the rational functions in Chapter 4 are continuous and smooth on their domains because polynomials are continuous and smooth everywhere, the secant and cosecant functions are continuous and smooth on their domains since the cosine and sine functions are continuous and smooth everywhere. 802 Foundations of Trigonometry all of these properties are direct results of them being reciprocals of the cosine and sine functions, respectively. Theorem 10.24. Properties of the Secant and Cosecant Functions • The function F(x) = sec(x) – has domain _ x : x ,= π 2 +πk, k is an integer _ = ∞ _ k=−∞ _ (2k + 1)π 2 , (2k + 3)π 2 _ – has range ¦y : [y[ ≥ 1¦ = (−∞, −1] ∪ [1, ∞) – is continuous and smooth on its domain – is even – has period 2π • The function G(x) = csc(x) – has domain ¦x : x ,= πk, k is an integer¦ = ∞ _ k=−∞ (kπ, (k + 1)π) – has range ¦y : [y[ ≥ 1¦ = (−∞, −1] ∪ [1, ∞) – is continuous and smooth on its domain – is odd – has period 2π In the next example, we discuss graphing more general secant and cosecant curves. Example 10.5.4. Graph one cycle of the following functions. State the period of each. 1. f(x) = 1 −2 sec(2x) 2. g(x) = csc(π −πx) −5 3 Solution. 1. To graph y = 1 −2 sec(2x), we follow the same procedure as in Example 10.5.1. First, we set the argument of secant, 2x, equal to the ‘quarter marks’ 0, π 2 , π, 3π 2 and 2π and solve for x. a 2x = a x 0 2x = 0 0 π 2 2x = π 2 π 4 π 2x = π π 2 3π 2 2x = 3π 2 3π 4 2π 2x = 2π π 10.5 Graphs of the Trigonometric Functions 803 Next, we substitute these x values into f(x). If f(x) exists, we have a point on the graph; otherwise, we have found a vertical asymptote. In addition to these points and asymptotes, we have graphed the associated cosine curve – in this case y = 1 −2 cos(2x) – dotted in the picture below. Since one cycle is graphed over the interval [0, π], the period is π −0 = π. x f(x) (x, f(x)) 0 −1 (0, −1) π 4 undefined π 2 3 _ π 2 , 3 _ 3π 4 undefined π −1 (π, −1) x y π 4 π 2 3π 4 π −1 1 2 3 One cycle of y = 1 −2 sec(2x). 2. Proceeding as before, we set the argument of cosecant in g(x) = csc(π−πx)−5 3 equal to the quarter marks and solve for x. a π −πx = a x 0 π −πx = 0 1 π 2 π −πx = π 2 1 2 π π −πx = π 0 3π 2 π −πx = 3π 2 − 1 2 2π π −πx = 2π −1 Substituting these x-values into g(x), we generate the graph below and find the period to be 1 −(−1) = 2. The associated sine curve, y = sin(π−πx)−5 3 , is dotted in as a reference. x g(x) (x, g(x)) 1 undefined 1 2 − 4 3 _ 1 2 , − 4 3 _ 0 undefined − 1 2 −2 _ − 1 2 , −2 _ −1 undefined x y −1 − 1 2 1 2 1 −2 −1 One cycle of y = csc(π−πx)−5 3 . 804 Foundations of Trigonometry Before moving on, we note that it is possible to speak of the period, phase shift and vertical shift of secant and cosecant graphs and use even/odd identities to put them in a form similar to the sinusoid forms mentioned in Theorem 10.23. Since these quantities match those of the corresponding cosine and sine curves, we do not spell this out explicitly. Finally, since the ranges of secant and cosecant are unbounded, there is no amplitude associated with these curves. 10.5.3 Graphs of the Tangent and Cotangent Functions Finally, we turn our attention to the graphs of the tangent and cotangent functions. When con- structing a table of values for the tangent function, we see that J(x) = tan(x) is undefined at x = π 2 and x = 3π 2 , in accordance with our findings in Section 10.3.1. As x → π 2 − , sin(x) → 1 − and cos(x) → 0 + , so that tan(x) = sin(x) cos(x) → ∞ producing a vertical asymptote at x = π 2 . Using a similar analysis, we get that as x → π 2 + , tan(x) →−∞; as x → 3π 2 − , tan(x) →∞; and as x → 3π 2 + , tan(x) →−∞. Plotting this information and performing the usual ‘copy and paste’ produces: x tan(x) (x, tan(x)) 0 0 (0, 0) π 4 1 _ π 4 , 1 _ π 2 undefined 3π 4 −1 _ 3π 4 , −1 _ π 0 (π, 0) 5π 4 1 _ 5π 4 , 1 _ 3π 2 undefined 7π 4 −1 _ 7π 4 , −1 _ 2π 0 (2π, 0) x y π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π −1 1 The graph of y = tan(x) over [0, 2π]. x y The graph of y = tan(x). 10.5 Graphs of the Trigonometric Functions 805 From the graph, it appears as if the tangent function is periodic with period π. To prove that this is the case, we appeal to the sum formula for tangents. We have: tan(x +π) = tan(x) + tan(π) 1 −tan(x) tan(π) = tan(x) + 0 1 −(tan(x))(0) = tan(x), which tells us the period of tan(x) is at most π. To show that it is exactly π, suppose p is a positive real number so that tan(x + p) = tan(x) for all real numbers x. For x = 0, we have tan(p) = tan(0 +p) = tan(0) = 0, which means p is a multiple of π. The smallest positive multiple of π is π itself, so we have established the result. We take as our fundamental cycle for y = tan(x) the interval _ − π 2 , π 2 _ , and use as our ‘quarter marks’ x = − π 2 , − π 4 , 0, π 4 and π 2 . From the graph, we see confirmation of our domain and range work in Section 10.3.1. It should be no surprise that K(x) = cot(x) behaves similarly to J(x) = tan(x). Plotting cot(x) over the interval [0, 2π] results in the graph below. x cot(x) (x, cot(x)) 0 undefined π 4 1 _ π 4 , 1 _ π 2 0 _ π 2 , 0 _ 3π 4 −1 _ 3π 4 , −1 _ π undefined 5π 4 1 _ 5π 4 , 1 _ 3π 2 0 _ 3π 2 , 0 _ 7π 4 −1 _ 7π 4 , −1 _ 2π undefined x y π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π −1 1 The graph of y = cot(x) over [0, 2π]. From these data, it clearly appears as if the period of cot(x) is π, and we leave it to the reader to prove this. 14 We take as one fundamental cycle the interval (0, π) with quarter marks: x = 0, π 4 , π 2 , 3π 4 and π. A more complete graph of y = cot(x) is below, along with the fundamental cycle highlighted as usual. Once again, we see the domain and range of K(x) = cot(x) as read from the graph matches with what we found analytically in Section 10.3.1. 14 Certainly, mimicking the proof that the period of tan(x) is an option; for another approach, consider transforming tan(x) to cot(x) using identities. 806 Foundations of Trigonometry x y The graph of y = cot(x). The properties of the tangent and cotangent functions are summarized below. As with Theorem 10.24, each of the results below can be traced back to properties of the cosine and sine functions and the definition of the tangent and cotangent functions as quotients thereof. Theorem 10.25. Properties of the Tangent and Cotangent Functions • The function J(x) = tan(x) – has domain _ x : x ,= π 2 +πk, k is an integer _ = ∞ _ k=−∞ _ (2k + 1)π 2 , (2k + 3)π 2 _ – has range (−∞, ∞) – is continuous and smooth on its domain – is odd – has period π • The function K(x) = cot(x) – has domain ¦x : x ,= πk, k is an integer¦ = ∞ _ k=−∞ (kπ, (k + 1)π) – has range (−∞, ∞) – is continuous and smooth on its domain – is odd – has period π 10.5 Graphs of the Trigonometric Functions 807 Example 10.5.5. Graph one cycle of the following functions. Find the period. 1. f(x) = 1 −tan _ x 2 _ . 2. g(x) = 2 cot _ π 2 x +π _ + 1. Solution. 1. We proceed as we have in all of the previous graphing examples by setting the argument of tangent in f(x) = 1 −tan _ x 2 _ , namely x 2 , equal to each of the ‘quarter marks’ − π 2 , − π 4 , 0, π 4 and π 2 , and solving for x. a x 2 = a x − π 2 x 2 = − π 2 −π − π 4 x 2 = − π 4 − π 2 0 x 2 = 0 0 π 4 x 2 = π 4 π 2 π 2 x 2 = π 2 π Substituting these x-values into f(x), we find points on the graph and the vertical asymptotes. x f(x) (x, f(x)) −π undefined − π 2 2 _ − π 2 , 2 _ 0 1 (0, 1) π 2 0 _ π 2 , 0 _ π undefined x y −π − π 2 π 2 π −2 −1 1 2 One cycle of y = 1 −tan _ x 2 _ . We see that the period is π −(−π) = 2π. 2. The ‘quarter marks’ for the fundamental cycle of the cotangent curve are 0, π 4 , π 2 , 3π 4 and π. To graph g(x) = 2 cot _ π 2 x +π _ + 1, we begin by setting π 2 x + π equal to each quarter mark and solving for x. 808 Foundations of Trigonometry a π 2 x +π = a x 0 π 2 x +π = 0 −2 π 4 π 2 x +π = π 4 − 3 2 π 2 π 2 x +π = π 2 −1 3π 4 π 2 x +π = 3π 4 − 1 2 π π 2 x +π = π 0 We now use these x-values to generate our graph. x g(x) (x, g(x)) −2 undefined − 3 2 3 _ − 3 2 , 3 _ −1 1 (−1, 1) − 1 2 −1 _ − 1 2 , −1 _ 0 undefined x y −2 −1 −1 1 2 3 One cycle of y = 2 cot _ π 2 x + π _ + 1. We find the period to be 0 −(−2) = 2. As with the secant and cosecant functions, it is possible to extend the notion of period, phase shift and vertical shift to the tangent and cotangent functions as we did for the cosine and sine functions in Theorem 10.23. Since the number of classical applications involving sinusoids far outnumber those involving tangent and cotangent functions, we omit this. The ambitious reader is invited to formulate such a theorem, however. 10.5 Graphs of the Trigonometric Functions 809 10.5.4 Exercises In Exercises 1 - 12, graph one cycle of the given function. State the period, amplitude, phase shift and vertical shift of the function. 1. y = 3 sin(x) 2. y = sin(3x) 3. y = −2 cos(x) 4. y = cos _ x − π 2 _ 5. y = −sin _ x + π 3 _ 6. y = sin(2x −π) 7. y = − 1 3 cos _ 1 2 x + π 3 _ 8. y = cos(3x −2π) + 4 9. y = sin _ −x − π 4 _ −2 10. y = 2 3 cos _ π 2 −4x _ + 1 11. y = − 3 2 cos _ 2x + π 3 _ − 1 2 12. y = 4 sin(−2πx +π) In Exercises 13 - 24, graph one cycle of the given function. State the period of the function. 13. y = tan _ x − π 3 _ 14. y = 2 tan _ 1 4 x _ −3 15. y = 1 3 tan(−2x −π) + 1 16. y = sec _ x − π 2 _ 17. y = −csc _ x + π 3 _ 18. y = − 1 3 sec _ 1 2 x + π 3 _ 19. y = csc(2x −π) 20. y = sec(3x −2π) + 4 21. y = csc _ −x − π 4 _ −2 22. y = cot _ x + π 6 _ 23. y = −11 cot _ 1 5 x _ 24. y = 1 3 cot _ 2x + 3π 2 _ + 1 In Exercises 25 - 34, use Example 10.5.3 as a guide to show that the function is a sinusoid by rewriting it in the forms C(x) = Acos(ωx + φ) + B and S(x) = Asin(ωx + φ) + B for ω > 0 and 0 ≤ φ < 2π. 25. f(x) = √ 2 sin(x) + √ 2 cos(x) + 1 26. f(x) = 3 √ 3 sin(3x) −3 cos(3x) 27. f(x) = −sin(x) + cos(x) −2 28. f(x) = − 1 2 sin(2x) − √ 3 2 cos(2x) 29. f(x) = 2 √ 3 cos(x) −2 sin(x) 30. f(x) = 3 2 cos(2x) − 3 √ 3 2 sin(2x) + 6 31. f(x) = − 1 2 cos(5x) − √ 3 2 sin(5x) 32. f(x) = −6 √ 3 cos(3x) −6 sin(3x) −3 810 Foundations of Trigonometry 33. f(x) = 5 √ 2 2 sin(x) − 5 √ 2 2 cos(x) 34. f(x) = 3 sin _ x 6 _ −3 √ 3 cos _ x 6 _ 35. In Exercises 25 - 34, you should have noticed a relationship between the phases φ for the S(x) and C(x). Show that if f(x) = Asin(ωx + α) + B, then f(x) = Acos(ωx + β) + B where β = α − π 2 . 36. Let φ be an angle measured in radians and let P(a, b) be a point on the terminal side of φ when it is drawn in standard position. Use Theorem 10.3 and the sum identity for sine in Theorem 10.15 to show that f(x) = a sin(ωx) +b cos(ωx) +B (with ω > 0) can be rewritten as f(x) = √ a 2 +b 2 sin(ωx +φ) +B. 37. With the help of your classmates, express the domains of the functions in Examples 10.5.4 and 10.5.5 using extended interval notation. (We will revisit this in Section 10.7.) In Exercises 38 - 43, verify the identity by graphing the right and left hand sides on a calculator. 38. sin 2 (x) + cos 2 (x) = 1 39. sec 2 (x) −tan 2 (x) = 1 40. cos(x) = sin _ π 2 −x _ 41. tan(x +π) = tan(x) 42. sin(2x) = 2 sin(x) cos(x) 43. tan _ x 2 _ = sin(x) 1 + cos(x) In Exercises 44 - 50, graph the function with the help of your calculator and discuss the given questions with your classmates. 44. f(x) = cos(3x) + sin(x). Is this function periodic? If so, what is the period? 45. f(x) = sin(x) x . What appears to be the horizontal asymptote of the graph? 46. f(x) = xsin(x). Graph y = ±x on the same set of axes and describe the behavior of f. 47. f(x) = sin _ 1 x _ . What’s happening as x →0? 48. f(x) = x −tan(x). Graph y = x on the same set of axes and describe the behavior of f. 49. f(x) = e −0.1x (cos(2x) + sin(2x)). Graph y = ±e −0.1x on the same set of axes and describe the behavior of f. 50. f(x) = e −0.1x (cos(2x) + 2 sin(x)). Graph y = ±e −0.1x on the same set of axes and describe the behavior of f. 51. Show that a constant function f is periodic by showing that f(x + 117) = f(x) for all real numbers x. Then show that f has no period by showing that you cannot find a smallest number p such that f(x + p) = f(x) for all real numbers x. Said another way, show that f(x +p) = f(x) for all real numbers x for ALL values of p > 0, so no smallest value exists to satisfy the definition of ‘period’. 10.5 Graphs of the Trigonometric Functions 811 10.5.5 Answers 1. y = 3 sin(x) Period: 2π Amplitude: 3 Phase Shift: 0 Vertical Shift: 0 x y π 2 π 3π 2 2π −3 3 2. y = sin(3x) Period: 2π 3 Amplitude: 1 Phase Shift: 0 Vertical Shift: 0 x y π 6 π 3 π 2 2π 3 −1 1 3. y = −2 cos(x) Period: 2π Amplitude: 2 Phase Shift: 0 Vertical Shift: 0 x y π 2 π 3π 2 2π −2 2 4. y = cos _ x − π 2 _ Period: 2π Amplitude: 1 Phase Shift: π 2 Vertical Shift: 0 x y π 2 π 3π 2 2π 5π 2 −1 1 812 Foundations of Trigonometry 5. y = −sin _ x + π 3 _ Period: 2π Amplitude: 1 Phase Shift: − π 3 Vertical Shift: 0 x y − π 3 π 6 2π 3 7π 6 5π 3 −1 1 6. y = sin(2x −π) Period: π Amplitude: 1 Phase Shift: π 2 Vertical Shift: 0 x y π 2 3π 4 π 5π 4 3π 2 −1 1 7. y = − 1 3 cos _ 1 2 x + π 3 _ Period: 4π Amplitude: 1 3 Phase Shift: − 2π 3 Vertical Shift: 0 x y − 2π 3 π 3 4π 3 7π 3 10π 3 − 1 3 1 3 8. y = cos(3x −2π) + 4 Period: 2π 3 Amplitude: 1 Phase Shift: 2π 3 Vertical Shift: 4 x y 2π 3 5π 6 π 7π 6 4π 3 3 4 5 10.5 Graphs of the Trigonometric Functions 813 9. y = sin _ −x − π 4 _ −2 Period: 2π Amplitude: 1 Phase Shift: − π 4 (You need to use y = −sin _ x + π 4 _ −2 to find this.) 15 Vertical Shift: −2 x y − 9π 4 − 7π 4 − 5π 4 − 3π 4 − π 4 π 4 3π 4 5π 4 7π 4 −3 −2 −1 10. y = 2 3 cos _ π 2 −4x _ + 1 Period: π 2 Amplitude: 2 3 Phase Shift: π 8 (You need to use y = 2 3 cos _ 4x − π 2 _ + 1 to find this.) 16 Vertical Shift: 1 x y − 3π 8 − π 4 − π 8 π 8 π 4 3π 8 π 2 5π 8 1 3 1 5 3 11. y = − 3 2 cos _ 2x + π 3 _ − 1 2 Period: π Amplitude: 3 2 Phase Shift: − π 6 Vertical Shift: − 1 2 x y − π 6 π 12 π 3 7π 12 5π 6 −2 − 1 2 1 12. y = 4 sin(−2πx +π) Period: 1 Amplitude: 4 Phase Shift: 1 2 (You need to use y = −4 sin(2πx −π) to find this.) 17 Vertical Shift: 0 x y − 1 2 − 1 4 1 4 1 2 3 4 1 5 4 3 2 −4 4 15 Two cycles of the graph are shown to illustrate the discrepancy discussed on page 796. 16 Again, we graph two cycles to illustrate the discrepancy discussed on page 796. 17 This will be the last time we graph two cycles to illustrate the discrepancy discussed on page 796. 814 Foundations of Trigonometry 13. y = tan _ x − π 3 _ Period: π x y − π 6 π 12 π 3 7π 12 5π 6 −1 1 14. y = 2 tan _ 1 4 x _ −3 Period: 4π x y −2π −π π 2π −5 −3 −1 15. y = 1 3 tan(−2x −π) + 1 is equivalent to y = − 1 3 tan(2x +π) + 1 via the Even / Odd identity for tangent. Period: π 2 x y − 3π 4 − 5π 8 − π 2 − 3π 8 − π 4 4 3 1 2 3 10.5 Graphs of the Trigonometric Functions 815 16. y = sec _ x − π 2 _ Start with y = cos _ x − π 2 _ Period: 2π x y π 2 π 3π 2 2π 5π 2 −1 1 17. y = −csc _ x + π 3 _ Start with y = −sin _ x + π 3 _ Period: 2π x y − π 3 π 6 2π 3 7π 6 5π 3 −1 1 18. y = − 1 3 sec _ 1 2 x + π 3 _ Start with y = − 1 3 cos _ 1 2 x + π 3 _ Period: 4π x y − 2π 3 π 3 4π 3 7π 3 10π 3 − 1 3 1 3 816 Foundations of Trigonometry 19. y = csc(2x −π) Start with y = sin(2x −π) Period: π x y π 2 3π 4 π 5π 4 3π 2 −1 1 20. y = sec(3x −2π) + 4 Start with y = cos(3x −2π) + 4 Period: 2π 3 x y 2π 3 5π 6 π 7π 6 4π 3 3 4 5 21. y = csc _ −x − π 4 _ −2 Start with y = sin _ −x − π 4 _ −2 Period: 2π x y − π 4 π 4 3π 4 5π 4 7π 4 −3 −2 −1 10.5 Graphs of the Trigonometric Functions 817 22. y = cot _ x + π 6 _ Period: π x y − π 6 π 12 π 3 7π 12 5π 6 −1 1 23. y = −11 cot _ 1 5 x _ Period: 5π x y 5π 4 5π 2 15π 4 5π −11 11 24. y = 1 3 cot _ 2x + 3π 2 _ + 1 Period: π 2 x y − 3π 4 − 5π 8 − π 2 − 3π 8 − π 4 4 3 1 2 3 818 Foundations of Trigonometry 25. f(x) = √ 2 sin(x) + √ 2 cos(x) + 1 = 2 sin _ x + π 4 _ + 1 = 2 cos _ x + 7π 4 _ + 1 26. f(x) = 3 √ 3 sin(3x) −3 cos(3x) = 6 sin _ 3x + 11π 6 _ = 6 cos _ 3x + 4π 3 _ 27. f(x) = −sin(x) + cos(x) −2 = √ 2 sin _ x + 3π 4 _ −2 = √ 2 cos _ x + π 4 _ −2 28. f(x) = − 1 2 sin(2x) − √ 3 2 cos(2x) = sin _ 2x + 4π 3 _ = cos _ 2x + 5π 6 _ 29. f(x) = 2 √ 3 cos(x) −2 sin(x) = 4 sin _ x + 2π 3 _ = 4 cos _ x + π 6 _ 30. f(x) = 3 2 cos(2x) − 3 √ 3 2 sin(2x) + 6 = 3 sin _ 2x + 5π 6 _ + 6 = 3 cos _ 2x + π 3 _ + 6 31. f(x) = − 1 2 cos(5x) − √ 3 2 sin(5x) = sin _ 5x + 7π 6 _ = cos _ 5x + 2π 3 _ 32. f(x) = −6 √ 3 cos(3x) −6 sin(3x) −3 = 12 sin _ 3x + 4π 3 _ −3 = 12 cos _ 3x + 5π 6 _ −3 33. f(x) = 5 √ 2 2 sin(x) − 5 √ 2 2 cos(x) = 5 sin _ x + 7π 4 _ = 5 cos _ x + 5π 4 _ 34. f(x) = 3 sin _ x 6 _ −3 √ 3 cos _ x 6 _ = 6 sin _ x 6 + 5π 3 _ = 6 cos _ x 6 + 7π 6 _ 10.6 The Inverse Trigonometric Functions 819 10.6 The Inverse Trigonometric Functions As the title indicates, in this section we concern ourselves with finding inverses of the (circular) trigonometric functions. Our immediate problem is that, owing to their periodic nature, none of the six circular functions is one-to-one. To remedy this, we restrict the domains of the circular functions in the same way we restricted the domain of the quadratic function in Example 5.2.3 in Section 5.2 to obtain a one-to-one function. We first consider f(x) = cos(x). Choosing the interval [0, π] allows us to keep the range as [−1, 1] as well as the properties of being smooth and continuous. x y Restricting the domain of f(x) = cos(x) to [0, π]. Recall from Section 5.2 that the inverse of a function f is typically denoted f −1 . For this reason, some textbooks use the notation f −1 (x) = cos −1 (x) for the inverse of f(x) = cos(x). The obvious pitfall here is our convention of writing (cos(x)) 2 as cos 2 (x), (cos(x)) 3 as cos 3 (x) and so on. It is far too easy to confuse cos −1 (x) with 1 cos(x) = sec(x) so we will not use this notation in our text. 1 Instead, we use the notation f −1 (x) = arccos(x), read ‘arc-cosine of x’. To understand the ‘arc’ in ‘arccosine’, recall that an inverse function, by definition, reverses the process of the original function. The function f(t) = cos(t) takes a real number input t, associates it with the angle θ = t radians, and returns the value cos(θ). Digging deeper, 2 we have that cos(θ) = cos(t) is the x-coordinate of the terminal point on the Unit Circle of an oriented arc of length [t[ whose initial point is (1, 0). Hence, we may view the inputs to f(t) = cos(t) as oriented arcs and the outputs as x-coordinates on the Unit Circle. The function f −1 , then, would take x-coordinates on the Unit Circle and return oriented arcs, hence the ‘arc’ in arccosine. Below are the graphs of f(x) = cos(x) and f −1 (x) = arccos(x), where we obtain the latter from the former by reflecting it across the line y = x, in accordance with Theorem 5.3. x y π 2 π −1 1 f(x) = cos(x), 0 ≤ x ≤ π reflect across y = x −−−−−−−−−−−−→ switch x and y coordinates x y π 2 π −1 1 f −1 (x) = arccos(x). 1 But be aware that many books do! As always, be sure to check the context! 2 See page 704 if you need a review of how we associate real numbers with angles in radian measure. 820 Foundations of Trigonometry We restrict g(x) = sin(x) in a similar manner, although the interval of choice is _ − π 2 , π 2 ¸ . x y Restricting the domain of f(x) = sin(x) to _ − π 2 , π 2 ¸ . It should be no surprise that we call g −1 (x) = arcsin(x), which is read ‘arc-sine of x’. x y − π 2 π 2 −1 1 g(x) = sin(x), − π 2 ≤ x ≤ π 2 . reflect across y = x −−−−−−−−−−−−→ switch x and y coordinates x y − π 2 π 2 −1 1 g −1 (x) = arcsin(x). We list some important facts about the arccosine and arcsine functions in the following theorem. Theorem 10.26. Properties of the Arccosine and Arcsine Functions • Properties of F(x) = arccos(x) – Domain: [−1, 1] – Range: [0, π] – arccos(x) = t if and only if 0 ≤ t ≤ π and cos(t) = x – cos(arccos(x)) = x provided −1 ≤ x ≤ 1 – arccos(cos(x)) = x provided 0 ≤ x ≤ π • Properties of G(x) = arcsin(x) – Domain: [−1, 1] – Range: _ − π 2 , π 2 ¸ – arcsin(x) = t if and only if − π 2 ≤ t ≤ π 2 and sin(t) = x – sin(arcsin(x)) = x provided −1 ≤ x ≤ 1 – arcsin(sin(x)) = x provided − π 2 ≤ x ≤ π 2 – additionally, arcsine is odd 10.6 The Inverse Trigonometric Functions 821 Everything in Theorem 10.26 is a direct consequence of the facts that f(x) = cos(x) for 0 ≤ x ≤ π and F(x) = arccos(x) are inverses of each other as are g(x) = sin(x) for − π 2 ≤ x ≤ π 2 and G(x) = arcsin(x). It’s about time for an example. Example 10.6.1. 1. Find the exact values of the following. (a) arccos _ 1 2 _ (b) arcsin _ √ 2 2 _ (c) arccos _ − √ 2 2 _ (d) arcsin _ − 1 2 _ (e) arccos _ cos _ π 6 __ (f) arccos _ cos _ 11π 6 __ (g) cos _ arccos _ − 3 5 __ (h) sin _ arccos _ − 3 5 __ 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan (arccos (x)) (b) cos (2 arcsin(x)) Solution. 1. (a) To find arccos _ 1 2 _ , we need to find the real number t (or, equivalently, an angle measuring t radians) which lies between 0 and π with cos(t) = 1 2 . We know t = π 3 meets these criteria, so arccos _ 1 2 _ = π 3 . (b) The value of arcsin _ √ 2 2 _ is a real number t between − π 2 and π 2 with sin(t) = √ 2 2 . The number we seek is t = π 4 . Hence, arcsin _ √ 2 2 _ = π 4 . (c) The number t = arccos _ − √ 2 2 _ lies in the interval [0, π] with cos(t) = − √ 2 2 . Our answer is arccos _ − √ 2 2 _ = 3π 4 . (d) To find arcsin _ − 1 2 _ , we seek the number t in the interval _ − π 2 , π 2 ¸ with sin(t) = − 1 2 . The answer is t = − π 6 so that arcsin _ − 1 2 _ = − π 6 . (e) Since 0 ≤ π 6 ≤ π, we could simply invoke Theorem 10.26 to get arccos _ cos _ π 6 __ = π 6 . However, in order to make sure we understand why this is the case, we choose to work the example through using the definition of arccosine. Working from the inside out, arccos _ cos _ π 6 __ = arccos _ √ 3 2 _ . Now, arccos _ √ 3 2 _ is the real number t with 0 ≤ t ≤ π and cos(t) = √ 3 2 . We find t = π 6 , so that arccos _ cos _ π 6 __ = π 6 . 822 Foundations of Trigonometry (f) Since 11π 6 does not fall between 0 and π, Theorem 10.26 does not apply. We are forced to work through from the inside out starting with arccos _ cos _ 11π 6 __ = arccos _ √ 3 2 _ . From the previous problem, we know arccos _ √ 3 2 _ = π 6 . Hence, arccos _ cos _ 11π 6 __ = π 6 . (g) One way to simplify cos _ arccos _ − 3 5 __ is to use Theorem 10.26 directly. Since − 3 5 is between −1 and 1, we have that cos _ arccos _ − 3 5 __ = − 3 5 and we are done. However, as before, to really understand why this cancellation occurs, we let t = arccos _ − 3 5 _ . Then, by definition, cos(t) = − 3 5 . Hence, cos _ arccos _ − 3 5 __ = cos(t) = − 3 5 , and we are finished in (nearly) the same amount of time. (h) As in the previous example, we let t = arccos _ − 3 5 _ so that cos(t) = − 3 5 for some t where 0 ≤ t ≤ π. Since cos(t) < 0, we can narrow this down a bit and conclude that π 2 < t < π, so that t corresponds to an angle in Quadrant II. In terms of t, then, we need to find sin _ arccos _ − 3 5 __ = sin(t). Using the Pythagorean Identity cos 2 (t) + sin 2 (t) = 1, we get _ − 3 5 _ 2 + sin 2 (t) = 1 or sin(t) = ± 4 5 . Since t corresponds to a Quadrants II angle, we choose sin(t) = 4 5 . Hence, sin _ arccos _ − 3 5 __ = 4 5 . 2. (a) We begin this problem in the same manner we began the previous two problems. To help us see the forest for the trees, we let t = arccos(x), so our goal is to find a way to express tan (arccos (x)) = tan(t) in terms of x. Since t = arccos(x), we know cos(t) = x where 0 ≤ t ≤ π, but since we are after an expression for tan(t), we know we need to throw out t = π 2 from consideration. Hence, either 0 ≤ t < π 2 or π 2 < t ≤ π so that, geometrically, t corresponds to an angle in Quadrant I or Quadrant II. One approach 3 to finding tan(t) is to use the quotient identity tan(t) = sin(t) cos(t) . Substituting cos(t) = x into the Pythagorean Identity cos 2 (t) +sin 2 (t) = 1 gives x 2 +sin 2 (t) = 1, from which we get sin(t) = ± √ 1 −x 2 . Since t corresponds to angles in Quadrants I and II, sin(t) ≥ 0, so we choose sin(t) = √ 1 −x 2 . Thus, tan(t) = sin(t) cos(t) = √ 1 −x 2 x To determine the values of x for which this equivalence is valid, we consider our sub- stitution t = arccos(x). Since the domain of arccos(x) is [−1, 1], we know we must restrict −1 ≤ x ≤ 1. Additionally, since we had to discard t = π 2 , we need to discard x = cos _ π 2 _ = 0. Hence, tan (arccos (x)) = √ 1−x 2 x is valid for x in [−1, 0) ∪ (0, 1]. (b) We proceed as in the previous problem by writing t = arcsin(x) so that t lies in the interval _ − π 2 , π 2 ¸ with sin(t) = x. We aim to express cos (2 arcsin(x)) = cos(2t) in terms of x. Since cos(2t) is defined everywhere, we get no additional restrictions on t as we did in the previous problem. We have three choices for rewriting cos(2t): cos 2 (t) − sin 2 (t), 2 cos 2 (t) −1 and 1−2 sin 2 (t). Since we know x = sin(t), it is easiest to use the last form: cos (2 arcsin(x)) = cos(2t) = 1 −2 sin 2 (t) = 1 −2x 2 3 Alternatively, we could use the identity: 1 + tan 2 (t) = sec 2 (t). Since x = cos(t), sec(t) = 1 cos(t) = 1 x . The reader is invited to work through this approach to see what, if any, difficulties arise. 10.6 The Inverse Trigonometric Functions 823 To find the restrictions on x, we once again appeal to our substitution t = arcsin(x). Since arcsin(x) is defined only for −1 ≤ x ≤ 1, the equivalence cos (2 arcsin(x)) = 1−2x 2 is valid only on [−1, 1]. A few remarks about Example 10.6.1 are in order. Most of the common errors encountered in dealing with the inverse circular functions come from the need to restrict the domains of the original functions so that they are one-to-one. One instance of this phenomenon is the fact that arccos _ cos _ 11π 6 __ = π 6 as opposed to 11π 6 . This is the exact same phenomenon discussed in Section 5.2 when we saw _ (−2) 2 = 2 as opposed to −2. Additionally, even though the expression we arrived at in part 2b above, namely 1 − 2x 2 , is defined for all real numbers, the equivalence cos (2 arcsin(x)) = 1 − 2x 2 is valid for only −1 ≤ x ≤ 1. This is akin to the fact that while the expression x is defined for all real numbers, the equivalence ( √ x) 2 = x is valid only for x ≥ 0. For this reason, it pays to be careful when we determine the intervals where such equivalences are valid. The next pair of functions we wish to discuss are the inverses of tangent and cotangent, which are named arctangent and arccotangent, respectively. First, we restrict f(x) = tan(x) to its fundamental cycle on _ − π 2 , π 2 _ to obtain f −1 (x) = arctan(x). Among other things, note that the vertical asymptotes x = − π 2 and x = π 2 of the graph of f(x) = tan(x) become the horizontal asymptotes y = − π 2 and y = π 2 of the graph of f −1 (x) = arctan(x). x y − π 4 − π 2 π 4 π 2 −1 1 f(x) = tan(x), − π 2 < x < π 2 . reflect across y = x −−−−−−−−−−−−→ switch x and y coordinates x y − π 4 − π 2 π 4 π 2 −1 1 f −1 (x) = arctan(x). Next, we restrict g(x) = cot(x) to its fundamental cycle on (0, π) to obtain g −1 (x) = arccot(x). Once again, the vertical asymptotes x = 0 and x = π of the graph of g(x) = cot(x) become the horizontal asymptotes y = 0 and y = π of the graph of g −1 (x) = arccot(x). We show these graphs on the next page and list some of the basic properties of the arctangent and arccotangent functions. 824 Foundations of Trigonometry x y π 4 π 2 3π 4 π −1 1 g(x) = cot(x), 0 < x < π. reflect across y = x −−−−−−−−−−−−→ switch x and y coordinates x y π 4 π 2 3π 4 π −1 1 g −1 (x) = arccot(x). Theorem 10.27. Properties of the Arctangent and Arccotangent Functions • Properties of F(x) = arctan(x) – Domain: (−∞, ∞) – Range: _ − π 2 , π 2 _ – as x →−∞, arctan(x) →− π 2 + ; as x →∞, arctan(x) → π 2 − – arctan(x) = t if and only if − π 2 < t < π 2 and tan(t) = x – arctan(x) = arccot _ 1 x _ for x > 0 – tan (arctan(x)) = x for all real numbers x – arctan(tan(x)) = x provided − π 2 < x < π 2 – additionally, arctangent is odd • Properties of G(x) = arccot(x) – Domain: (−∞, ∞) – Range: (0, π) – as x →−∞, arccot(x) →π − ; as x →∞, arccot(x) →0 + – arccot(x) = t if and only if 0 < t < π and cot(t) = x – arccot(x) = arctan _ 1 x _ for x > 0 – cot (arccot(x)) = x for all real numbers x – arccot(cot(x)) = x provided 0 < x < π 10.6 The Inverse Trigonometric Functions 825 Example 10.6.2. 1. Find the exact values of the following. (a) arctan( √ 3) (b) arccot(− √ 3) (c) cot(arccot(−5)) (d) sin _ arctan _ − 3 4 __ 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan(2 arctan(x)) (b) cos(arccot(2x)) Solution. 1. (a) We know arctan( √ 3) is the real number t between − π 2 and π 2 with tan(t) = √ 3. We find t = π 3 , so arctan( √ 3) = π 3 . (b) The real number t = arccot(− √ 3) lies in the interval (0, π) with cot(t) = − √ 3. We get arccot(− √ 3) = 5π 6 . (c) We can apply Theorem 10.27 directly and obtain cot(arccot(−5)) = −5. However, working it through provides us with yet another opportunity to understand why this is the case. Letting t = arccot(−5), we have that t belongs to the interval (0, π) and cot(t) = −5. Hence, cot(arccot(−5)) = cot(t) = −5. (d) We start simplifying sin _ arctan _ − 3 4 __ by letting t = arctan _ − 3 4 _ . Then tan(t) = − 3 4 for some − π 2 < t < π 2 . Since tan(t) < 0, we know, in fact, − π 2 < t < 0. One way to proceed is to use The Pythagorean Identity, 1+cot 2 (t) = csc 2 (t), since this relates the reciprocals of tan(t) and sin(t) and is valid for all t under consideration. 4 From tan(t) = − 3 4 , we get cot(t) = − 4 3 . Substituting, we get 1 + _ − 4 3 _ 2 = csc 2 (t) so that csc(t) = ± 5 3 . Since − π 2 < t < 0, we choose csc(t) = − 5 3 , so sin(t) = − 3 5 . Hence, sin _ arctan _ − 3 4 __ = − 3 5 . 2. (a) If we let t = arctan(x), then − π 2 < t < π 2 and tan(t) = x. We look for a way to express tan(2 arctan(x)) = tan(2t) in terms of x. Before we get started using identities, we note that tan(2t) is undefined when 2t = π 2 + πk for integers k. Dividing both sides of this equation by 2 tells us we need to exclude values of t where t = π 4 + π 2 k, where k is an integer. The only members of this family which lie in _ − π 2 , π 2 _ are t = ± π 4 , which means the values of t under consideration are _ − π 2 , − π 4 _ ∪ _ − π 4 , π 4 _ ∪ _ π 4 , π 2 _ . Returning to arctan(2t), we note the double angle identity tan(2t) = 2 tan(t) 1−tan 2 (t) , is valid for all the values of t under consideration, hence we get tan(2 arctan(x)) = tan(2t) = 2 tan(t) 1 −tan 2 (t) = 2x 1 −x 2 4 It’s always a good idea to make sure the identities used in these situations are valid for all values t under consideration. Check our work back in Example 10.6.1. Were the identities we used there valid for all t under consideration? A pedantic point, to be sure, but what else do you expect from this book? 826 Foundations of Trigonometry To find where this equivalence is valid we check back with our substitution t = arctan(x). Since the domain of arctan(x) is all real numbers, the only exclusions come from the values of t we discarded earlier, t = ± π 4 . Since x = tan(t), this means we exclude x = tan _ ± π 4 _ = ±1. Hence, the equivalence tan(2 arctan(x)) = 2x 1−x 2 holds for all x in (−∞, −1) ∪ (−1, 1) ∪ (1, ∞). (b) To get started, we let t = arccot(2x) so that cot(t) = 2x where 0 < t < π. In terms of t, cos(arccot(2x)) = cos(t), and our goal is to express the latter in terms of x. Since cos(t) is always defined, there are no additional restrictions on t, so we can begin using identities to relate cot(t) to cos(t). The identity cot(t) = cos(t) sin(t) is valid for t in (0, π), so our strategy is to obtain sin(t) in terms of x, then write cos(t) = cot(t) sin(t). The identity 1+cot 2 (t) = csc 2 (t) holds for all t in (0, π) and relates cot(t) and csc(t) = 1 sin(t) . Substituting cot(t) = 2x, we get 1 + (2x) 2 = csc 2 (t), or csc(t) = ± √ 4x 2 + 1. Since t is between 0 and π, csc(t) > 0, so csc(t) = √ 4x 2 + 1 which gives sin(t) = 1 √ 4x 2 +1 . Hence, cos(arccot(2x)) = cos(t) = cot(t) sin(t) = 2x √ 4x 2 + 1 Since arccot(2x) is defined for all real numbers x and we encountered no additional restrictions on t, we have cos (arccot(2x)) = 2x √ 4x 2 +1 for all real numbers x. The last two functions to invert are secant and cosecant. A portion of each of their graphs, which were first discussed in Subsection 10.5.2, are given below with the fundamental cycles highlighted. x y The graph of y = sec(x). x y The graph of y = csc(x). It is clear from the graph of secant that we cannot find one single continuous piece of its graph which covers its entire range of (−∞, −1] ∪[1, ∞) and restricts the domain of the function so that it is one-to-one. The same is true for cosecant. Thus in order to define the arcsecant and arccosecant functions, we must settle for a piecewise approach wherein we choose one piece to cover the top of the range, namely [1, ∞), and another piece to cover the bottom, namely (−∞, −1]. There are two generally accepted ways make these choices which restrict the domains of these functions so that they are one-to-one. One approach simplifies the Trigonometry associated with the inverse functions, but complicates the Calculus; the other makes the Calculus easier, but the Trigonometry less so. We present both points of view. 10.6 The Inverse Trigonometric Functions 827 10.6.1 Inverses of Secant and Cosecant: Trigonometry Friendly Approach In this subsection, we restrict the secant and cosecant functions to coincide with the restrictions on cosine and sine, respectively. For f(x) = sec(x), we restrict the domain to _ 0, π 2 _ ∪ _ π 2 , π ¸ x y π 2 π −1 1 f(x) = sec(x) on _ 0, π 2 _ ∪ _ π 2 , π _ reflect across y = x −−−−−−−−−−−−→ switch x and y coordinates x y π 2 π −1 1 f −1 (x) = arcsec(x) and we restrict g(x) = csc(x) to _ − π 2 , 0 _ ∪ _ 0, π 2 ¸ . x y − π 2 π 2 −1 1 g(x) = csc(x) on _ − π 2 , 0 _ ∪ _ 0, π 2 _ reflect across y = x −−−−−−−−−−−−→ switch x and y coordinates x y − π 2 π 2 −1 1 g −1 (x) = arccsc(x) Note that for both arcsecant and arccosecant, the domain is (−∞, −1] ∪ [1, ∞). Taking a page from Section 2.2, we can rewrite this as ¦x : [x[ ≥ 1¦. This is often done in Calculus textbooks, so we include it here for completeness. Using these definitions, we get the following properties of the arcsecant and arccosecant functions. 828 Foundations of Trigonometry Theorem 10.28. Properties of the Arcsecant and Arccosecant Functions a • Properties of F(x) = arcsec(x) – Domain: ¦x : [x[ ≥ 1¦ = (−∞, −1] ∪ [1, ∞) – Range: _ 0, π 2 _ ∪ _ π 2 , π ¸ – as x →−∞, arcsec(x) → π 2 + ; as x →∞, arcsec(x) → π 2 − – arcsec(x) = t if and only if 0 ≤ t < π 2 or π 2 < t ≤ π and sec(t) = x – arcsec(x) = arccos _ 1 x _ provided [x[ ≥ 1 – sec (arcsec(x)) = x provided [x[ ≥ 1 – arcsec(sec(x)) = x provided 0 ≤ x < π 2 or π 2 < x ≤ π • Properties of G(x) = arccsc(x) – Domain: ¦x : [x[ ≥ 1¦ = (−∞, −1] ∪ [1, ∞) – Range: _ − π 2 , 0 _ ∪ _ 0, π 2 ¸ – as x →−∞, arccsc(x) →0 − ; as x →∞, arccsc(x) →0 + – arccsc(x) = t if and only if − π 2 ≤ t < 0 or 0 < t ≤ π 2 and csc(t) = x – arccsc(x) = arcsin _ 1 x _ provided [x[ ≥ 1 – csc (arccsc(x)) = x provided [x[ ≥ 1 – arccsc(csc(x)) = x provided − π 2 ≤ x < 0 or 0 < x ≤ π 2 – additionally, arccosecant is odd a . . . assuming the “Trigonometry Friendly” ranges are used. Example 10.6.3. 1. Find the exact values of the following. (a) arcsec(2) (b) arccsc(−2) (c) arcsec _ sec _ 5π 4 __ (d) cot (arccsc (−3)) 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan(arcsec(x)) (b) cos(arccsc(4x)) 10.6 The Inverse Trigonometric Functions 829 Solution. 1. (a) Using Theorem 10.28, we have arcsec(2) = arccos _ 1 2 _ = π 3 . (b) Once again, Theorem 10.28 comes to our aid giving arccsc(−2) = arcsin _ − 1 2 _ = − π 6 . (c) Since 5π 4 doesn’t fall between 0 and π 2 or π 2 and π, we cannot use the inverse property stated in Theorem 10.28. We can, nevertheless, begin by working ‘inside out’ which yields arcsec _ sec _ 5π 4 __ = arcsec(− √ 2) = arccos _ − √ 2 2 _ = 3π 4 . (d) One way to begin to simplify cot (arccsc (−3)) is to let t = arccsc(−3). Then, csc(t) = −3 and, since this is negative, we have that t lies in the interval _ − π 2 , 0 _ . We are after cot (arccsc (−3)) = cot(t), so we use the Pythagorean Identity 1 + cot 2 (t) = csc 2 (t). Substituting, we have 1 +cot 2 (t) = (−3) 2 , or cot(t) = ± √ 8 = ±2 √ 2. Since − π 2 ≤ t < 0, cot(t) < 0, so we get cot (arccsc (−3)) = −2 √ 2. 2. (a) We begin simplifying tan(arcsec(x)) by letting t = arcsec(x). Then, sec(t) = x for t in _ 0, π 2 _ ∪ _ π 2 , π ¸ , and we seek a formula for tan(t). Since tan(t) is defined for all t values under consideration, we have no additional restrictions on t. To relate sec(t) to tan(t), we use the identity 1+tan 2 (t) = sec 2 (t). This is valid for all values of t under consideration, and when we substitute sec(t) = x, we get 1 +tan 2 (t) = x 2 . Hence, tan(t) = ± √ x 2 −1. If t belongs to _ 0, π 2 _ then tan(t) ≥ 0; if, on the the other hand, t belongs to _ π 2 , π ¸ then tan(t) ≤ 0. As a result, we get a piecewise defined function for tan(t) tan(t) = _ √ x 2 −1, if 0 ≤ t < π 2 − √ x 2 −1, if π 2 < t ≤ π Now we need to determine what these conditions on t mean for x. Since x = sec(t), when 0 ≤ t < π 2 , x ≥ 1, and when π 2 < t ≤ π, x ≤ −1. Since we encountered no further restrictions on t, the equivalence below holds for all x in (−∞, −1] ∪ [1, ∞). tan(arcsec(x)) = _ √ x 2 −1, if x ≥ 1 − √ x 2 −1, if x ≤ −1 (b) To simplify cos(arccsc(4x)), we start by letting t = arccsc(4x). Then csc(t) = 4x for t in _ − π 2 , 0 _ ∪ _ 0, π 2 ¸ , and we now set about finding an expression for cos(arccsc(4x)) = cos(t). Since cos(t) is defined for all t, we do not encounter any additional restrictions on t. From csc(t) = 4x, we get sin(t) = 1 4x , so to find cos(t), we can make use if the identity cos 2 (t) +sin 2 (t) = 1. Substituting sin(t) = 1 4x gives cos 2 (t) + _ 1 4x _ 2 = 1. Solving, we get cos(t) = ± _ 16x 2 −1 16x 2 = ± √ 16x 2 −1 4[x[ Since t belongs to _ − π 2 , 0 _ ∪ _ 0, π 2 ¸ , we know cos(t) ≥ 0, so we choose cos(t) = √ 16−x 2 4|x| . (The absolute values here are necessary, since x could be negative.) To find the values for 830 Foundations of Trigonometry which this equivalence is valid, we look back at our original substution, t = arccsc(4x). Since the domain of arccsc(x) requires its argument x to satisfy [x[ ≥ 1, the domain of arccsc(4x) requires [4x[ ≥ 1. Using Theorem 2.4, we rewrite this inequality and solve to get x ≤ − 1 4 or x ≥ 1 4 . Since we had no additional restrictions on t, the equivalence cos(arccsc(4x)) = √ 16x 2 −1 4|x| holds for all x in _ −∞, − 1 4 ¸ ∪ _ 1 4 , ∞ _ . 10.6.2 Inverses of Secant and Cosecant: Calculus Friendly Approach In this subsection, we restrict f(x) = sec(x) to _ 0, π 2 _ ∪ _ π, 3π 2 _ x y π 2 π 3π 2 −1 1 f(x) = sec(x) on _ 0, π 2 _ ∪ _ π, 3π 2 _ reflect across y = x −−−−−−−−−−−−→ switch x and y coordinates x y π 2 π 3π 2 −1 1 f −1 (x) = arcsec(x) and we restrict g(x) = csc(x) to _ 0, π 2 ¸ ∪ _ π, 3π 2 ¸ . x y π 2 π 3π 2 −1 1 g(x) = csc(x) on _ 0, π 2 _ ∪ _ π, 3π 2 _ reflect across y = x −−−−−−−−−−−−→ switch x and y coordinates x y π 2 π 3π 2 −1 1 g −1 (x) = arccsc(x) Using these definitions, we get the following result. 10.6 The Inverse Trigonometric Functions 831 Theorem 10.29. Properties of the Arcsecant and Arccosecant Functions a • Properties of F(x) = arcsec(x) – Domain: ¦x : [x[ ≥ 1¦ = (−∞, −1] ∪ [1, ∞) – Range: _ 0, π 2 _ ∪ _ π, 3π 2 _ – as x →−∞, arcsec(x) → 3π 2 − ; as x →∞, arcsec(x) → π 2 − – arcsec(x) = t if and only if 0 ≤ t < π 2 or π ≤ t < 3π 2 and sec(t) = x – arcsec(x) = arccos _ 1 x _ for x ≥ 1 only b – sec (arcsec(x)) = x provided [x[ ≥ 1 – arcsec(sec(x)) = x provided 0 ≤ x < π 2 or π ≤ x < 3π 2 • Properties of G(x) = arccsc(x) – Domain: ¦x : [x[ ≥ 1¦ = (−∞, −1] ∪ [1, ∞) – Range: _ 0, π 2 ¸ ∪ _ π, 3π 2 ¸ – as x →−∞, arccsc(x) →π + ; as x →∞, arccsc(x) →0 + – arccsc(x) = t if and only if 0 < t ≤ π 2 or π < t ≤ 3π 2 and csc(t) = x – arccsc(x) = arcsin _ 1 x _ for x ≥ 1 only c – csc (arccsc(x)) = x provided [x[ ≥ 1 – arccsc(csc(x)) = x provided 0 < x ≤ π 2 or π < x ≤ 3π 2 a . . . assuming the “Calculus Friendly” ranges are used. b Compare this with the similar result in Theorem 10.28. c Compare this with the similar result in Theorem 10.28. Our next example is a duplicate of Example 10.6.3. The interested reader is invited to compare and contrast the solution to each. Example 10.6.4. 1. Find the exact values of the following. (a) arcsec(2) (b) arccsc(−2) (c) arcsec _ sec _ 5π 4 __ (d) cot (arccsc (−3)) 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan(arcsec(x)) (b) cos(arccsc(4x)) 832 Foundations of Trigonometry Solution. 1. (a) Since 2 ≥ 1, we may invoke Theorem 10.29 to get arcsec(2) = arccos _ 1 2 _ = π 3 . (b) Unfortunately, −2 is not greater to or equal to 1, so we cannot apply Theorem 10.29 to arccsc(−2) and convert this into an arcsine problem. Instead, we appeal to the definition. The real number t = arccsc(−2) lies in _ 0, π 2 ¸ ∪ _ π, 3π 2 ¸ and satisfies csc(t) = −2. The t we’re after is t = 7π 6 , so arccsc(−2) = 7π 6 . (c) Since 5π 4 lies between π and 3π 2 , we may apply Theorem 10.29 directly to simplify arcsec _ sec _ 5π 4 __ = 5π 4 . We encourage the reader to work this through using the defini- tion as we have done in the previous examples to see how it goes. (d) To simplify cot (arccsc (−3)) we let t = arccsc (−3) so that cot (arccsc (−3)) = cot(t). We know csc(t) = −3, and since this is negative, t lies in _ π, 3π 2 ¸ . Using the identity 1 + cot 2 (t) = csc 2 (t), we find 1 + cot 2 (t) = (−3) 2 so that cot(t) = ± √ 8 = ±2 √ 2. Since t is in the interval _ π, 3π 2 ¸ , we know cot(t) > 0. Our answer is cot (arccsc (−3)) = 2 √ 2. 2. (a) We begin simplifying tan(arcsec(x)) by letting t = arcsec(x). Then, sec(t) = x for t in _ 0, π 2 _ ∪ _ π, 3π 2 _ , and we seek a formula for tan(t). Since tan(t) is defined for all t values under consideration, we have no additional restrictions on t. To relate sec(t) to tan(t), we use the identity 1+tan 2 (t) = sec 2 (t). This is valid for all values of t under consideration, and when we substitute sec(t) = x, we get 1 +tan 2 (t) = x 2 . Hence, tan(t) = ± √ x 2 −1. Since t lies in _ 0, π 2 _ ∪ _ π, 3π 2 _ , tan(t) ≥ 0, so we choose tan(t) = √ x 2 −1. Since we found no additional restrictions on t, the equivalence tan(arcsec(x)) = √ x 2 −1 holds for all x in the domain of t = arcsec(x), namely (−∞, −1] ∪ [1, ∞). (b) To simplify cos(arccsc(4x)), we start by letting t = arccsc(4x). Then csc(t) = 4x for t in _ 0, π 2 ¸ ∪ _ π, 3π 2 ¸ , and we now set about finding an expression for cos(arccsc(4x)) = cos(t). Since cos(t) is defined for all t, we do not encounter any additional restrictions on t. From csc(t) = 4x, we get sin(t) = 1 4x , so to find cos(t), we can make use if the identity cos 2 (t) +sin 2 (t) = 1. Substituting sin(t) = 1 4x gives cos 2 (t) + _ 1 4x _ 2 = 1. Solving, we get cos(t) = ± _ 16x 2 −1 16x 2 = ± √ 16x 2 −1 4[x[ If t lies in _ 0, π 2 ¸ , then cos(t) ≥ 0, and we choose cos(t) = √ 16x 2 −1 4|x| . Otherwise, t belongs to _ π, 3π 2 ¸ in which case cos(t) ≤ 0, so, we choose cos(t) = − √ 16x 2 −1 4|x| This leads us to a (momentarily) piecewise defined function for cos(t) cos(t) = _ ¸ ¸ _ ¸ ¸ _ √ 16x 2 −1 4[x[ , if 0 ≤ t ≤ π 2 − √ 16x 2 −1 4[x[ , if π < t ≤ 3π 2 10.6 The Inverse Trigonometric Functions 833 We now see what these restrictions mean in terms of x. Since 4x = csc(t), we get that for 0 ≤ t ≤ π 2 , 4x ≥ 1, or x ≥ 1 4 . In this case, we can simplify [x[ = x so cos(t) = √ 16x 2 −1 4[x[ = √ 16x 2 −1 4x Similarly, for π < t ≤ 3π 2 , we get 4x ≤ −1, or x ≤ − 1 4 . In this case, [x[ = −x, so we also get cos(t) = − √ 16x 2 −1 4[x[ = − √ 16x 2 −1 4(−x) = √ 16x 2 −1 4x Hence, in all cases, cos(arccsc(4x)) = √ 16x 2 −1 4x , and this equivalence is valid for all x in the domain of t = arccsc(4x), namely _ −∞, − 1 4 ¸ ∪ _ 1 4 , ∞ _ 10.6.3 Calculators and the Inverse Circular Functions. In the sections to come, we will have need to approximate the values of the inverse circular functions. On most calculators, only the arcsine, arccosine and arctangent functions are available and they are usually labeled as sin −1 , cos −1 and tan −1 , respectively. If we are asked to approximate these values, it is a simple matter to punch up the appropriate decimal on the calculator. If we are asked for an arccotangent, arcsecant or arccosecant, however, we often need to employ some ingenuity, as our next example illustrates. Example 10.6.5. 1. Use a calculator to approximate the following values to four decimal places. (a) arccot(2) (b) arcsec(5) (c) arccot(−2) (d) arccsc _ − 3 2 _ 2. Find the domain and range of the following functions. Check your answers using a calculator. (a) f(x) = π 2 −arccos _ x 5 _ (b) f(x) = 3 arctan (4x). (c) f(x) = arccot _ x 2 _ +π Solution. 1. (a) Since 2 > 0, we can use the property listed in Theorem 10.27 to rewrite arccot(2) as arccot(2) = arctan _ 1 2 _ . In ‘radian’ mode, we find arccot(2) = arctan _ 1 2 _ ≈ 0.4636. (b) Since 5 ≥ 1, we can use the property from either Theorem 10.28 or Theorem 10.29 to write arcsec(5) = arccos _ 1 5 _ ≈ 1.3694. 834 Foundations of Trigonometry (c) Since the argument −2 is negative, we cannot directly apply Theorem 10.27 to help us find arccot(−2). Let t = arccot(−2). Then t is a real number such that 0 < t < π and cot(t) = −2. Moreover, since cot(t) < 0, we know π 2 < t < π. Geometrically, this means t corresponds to a Quadrant II angle θ = t radians. This allows us to proceed using a ‘reference angle’ approach. Consider α, the reference angle for θ, as pictured below. By definition, α is an acute angle so 0 < α < π 2 , and the Reference Angle Theorem, Theorem 10.2, tells us that cot(α) = 2. This means α = arccot(2) radians. Since the argument of arccotangent is now a positive 2, we can use Theorem 10.27 to get α = arccot(2) = arctan _ 1 2 _ radians. Since θ = π −α = π −arctan _ 1 2 _ ≈ 2.6779 radians, we get arccot(−2) ≈ 2.6779. x y 1 1 α θ = arccot(−2) radians Another way to attack the problem is to use arctan _ − 1 2 _ . By definition, the real number t = arctan _ − 1 2 _ satisfies tan(t) = − 1 2 with − π 2 < t < π 2 . Since tan(t) < 0, we know more specifically that − π 2 < t < 0, so t corresponds to an angle β in Quadrant IV. To find the value of arccot(−2), we once again visualize the angle θ = arccot(−2) radians and note that it is a Quadrant II angle with tan(θ) = − 1 2 . This means it is exactly π units away from β, and we get θ = π + β = π + arctan _ − 1 2 _ ≈ 2.6779 radians. Hence, as before, arccot(−2) ≈ 2.6779. 10.6 The Inverse Trigonometric Functions 835 x y 1 1 π β θ = arccot(−2) radians (d) If the range of arccosecant is taken to be _ − π 2 , 0 _ ∪ _ 0, π 2 ¸ , we can use Theorem 10.28 to get arccsc _ − 3 2 _ = arcsin _ − 2 3 _ ≈ −0.7297. If, on the other hand, the range of arccosecant is taken to be _ 0, π 2 ¸ ∪ _ π, 3π 2 ¸ , then we proceed as in the previous problem by letting t = arccsc _ − 3 2 _ . Then t is a real number with csc(t) = − 3 2 . Since csc(t) < 0, we have that π < θ ≤ 3π 2 , so t corresponds to a Quadrant III angle, θ. As above, we let α be the reference angle for θ. Then 0 < α < π 2 and csc(α) = 3 2 , which means α = arccsc _ 3 2 _ radians. Since the argument of arccosecant is now positive, we may use Theorem 10.29 to get α = arccsc _ 3 2 _ = arcsin _ 2 3 _ radians. Since θ = π + α = π + arcsin _ 2 3 _ ≈ 3.8713 radians, arccsc _ − 3 2 _ ≈ 3.8713. x y 1 1 α θ = arccsc _ − 3 2 _ radians 836 Foundations of Trigonometry 2. (a) Since the domain of F(x) = arccos(x) is −1 ≤ x ≤ 1, we can find the domain of f(x) = π 2 −arccos _ x 5 _ by setting the argument of the arccosine, in this case x 5 , between −1 and 1. Solving −1 ≤ x 5 ≤ 1 gives −5 ≤ x ≤ 5, so the domain is [−5, 5]. To determine the range of f, we take a cue from Section 1.7. Three ‘key’ points on the graph of F(x) = arccos(x) are (−1, π), _ 0, π 2 _ and (1, 0) . Following the procedure outlined in Theorem 1.7, we track these points to _ −5, − π 2 _ , (0, 0) and _ 5, π 2 _ . Plotting these values tells us that the range 5 of f is _ − π 2 , π 2 ¸ . Our graph confirms our results. (b) To find the domain and range of f(x) = 3 arctan (4x), we note that since the domain of F(x) = arctan(x) is all real numbers, the only restrictions, if any, on the domain of f(x) = 3 arctan (4x) come from the argument of the arctangent, in this case, 4x. Since 4x is defined for all real numbers, we have established that the domain of f is all real numbers. To determine the range of f, we can, once again, appeal to Theorem 1.7. Choosing our ‘key’ point to be (0, 0) and tracking the horizontal asymptotes y = − π 2 and y = π 2 , we find that the graph of y = f(x) = 3 arctan (4x) differs from the graph of y = F(x) = arctan(x) by a horizontal compression by a factor of 4 and a vertical stretch by a factor of 3. It is the latter which affects the range, producing a range of _ − 3π 2 , 3π 2 _ . We confirm our findings on the calculator below. y = f(x) = π 2 −arccos _ x 5 _ y = f(x) = 3 arctan (4x) (c) To find the domain of g(x) = arccot _ x 2 _ +π, we proceed as above. Since the domain of G(x) = arccot(x) is (−∞, ∞), and x 2 is defined for all x, we get that the domain of g is (−∞, ∞) as well. As for the range, we note that the range of G(x) = arccot(x), like that of F(x) = arctan(x), is limited by a pair of horizontal asymptotes, in this case y = 0 and y = π. Following Theorem 1.7, we graph y = g(x) = arccot _ x 2 _ + π starting with y = G(x) = arccot(x) and first performing a horizontal expansion by a factor of 2 and following that with a vertical shift upwards by π. This latter transformation is the one which affects the range, making it now (π, 2π). To check this graphically, we encounter a bit of a problem, since on many calculators, there is no shortcut button corresponding to the arccotangent function. Taking a cue from number 1c, we attempt to rewrite g(x) = arccot _ x 2 _ +π in terms of the arctangent function. Using Theorem 10.27, we have that arccot _ x 2 _ = arctan _ 2 x _ when x 2 > 0, or, in this case, when x > 0. Hence, for x > 0, we have g(x) = arctan _ 2 x _ +π. When x 2 < 0, we can use the same argument in number 1c that gave us arccot(−2) = π + arctan _ − 1 2 _ to give us arccot _ x 2 _ = π + arctan _ 2 x _ . 5 It also confirms our domain! 10.6 The Inverse Trigonometric Functions 837 Hence, for x < 0, g(x) = π +arctan _ 2 x _ +π = arctan _ 2 x _ +2π. What about x = 0? We know g(0) = arccot(0) + π = π, and neither of the formulas for g involving arctangent will produce this result. 6 Hence, in order to graph y = g(x) on our calculators, we need to write it as a piecewise defined function: g(x) = arccot _ x 2 _ +π = _ ¸ _ ¸ _ arctan _ 2 x _ + 2π, if x < 0 π, if x = 0 arctan _ 2 x _ +π, if x > 0 We show the input and the result below. y = g(x) in terms of arctangent y = g(x) = arccot _ x 2 _ +π The inverse trigonometric functions are typically found in applications whenever the measure of an angle is required. One such scenario is presented in the following example. Example 10.6.6. 7 The roof on the house below has a ‘6/12 pitch’. This means that when viewed from the side, the roof line has a rise of 6 feet over a run of 12 feet. Find the angle of inclination from the bottom of the roof to the top of the roof. Express your answer in decimal degrees, rounded to the nearest hundredth of a degree. Front View Side View Solution. If we divide the side view of the house down the middle, we find that the roof line forms the hypotenuse of a right triangle with legs of length 6 feet and 12 feet. Using Theorem 10.10, we 6 Without Calculus, of course . . . 7 The authors would like to thank Dan Stitz for this problem and associated graphics. 838 Foundations of Trigonometry find the angle of inclination, labeled θ below, satisfies tan(θ) = 6 12 = 1 2 . Since θ is an acute angle, we can use the arctangent function and we find θ = arctan _ 1 2 _ radians ≈ 26.56 ◦ . 12 feet 6 feet θ 10.6.4 Solving Equations Using the Inverse Trigonometric Functions. In Sections 10.2 and 10.3, we learned how to solve equations like sin(θ) = 1 2 for angles θ and tan(t) = −1 for real numbers t. In each case, we ultimately appealed to the Unit Circle and relied on the fact that the answers corresponded to a set of ‘common angles’ listed on page 724. If, on the other hand, we had been asked to find all angles with sin(θ) = 1 3 or solve tan(t) = −2 for real numbers t, we would have been hard-pressed to do so. With the introduction of the inverse trigonometric functions, however, we are now in a position to solve these equations. A good parallel to keep in mind is how the square root function can be used to solve certain quadratic equations. The equation x 2 = 4 is a lot like sin(θ) = 1 2 in that it has friendly, ‘common value’ answers x = ±2. The equation x 2 = 7, on the other hand, is a lot like sin(θ) = 1 3 . We know 8 there are answers, but we can’t express them using ‘friendly’ numbers. 9 To solve x 2 = 7, we make use of the square root function and write x = ± √ 7. We can certainly approximate these answers using a calculator, but as far as exact answers go, we leave them as x = ± √ 7. In the same way, we will use the arcsine function to solve sin(θ) = 1 3 , as seen in the following example. Example 10.6.7. Solve the following equations. 1. Find all angles θ for which sin(θ) = 1 3 . 2. Find all real numbers t for which tan(t) = −2 3. Solve sec(x) = − 5 3 for x. Solution. 1. If sin(θ) = 1 3 , then the terminal side of θ, when plotted in standard position, intersects the Unit Circle at y = 1 3 . Geometrically, we see that this happens at two places: in Quadrant I and Quadrant II. If we let α denote the acute solution to the equation, then all the solutions 8 How do we know this again? 9 This is all, of course, a matter of opinion. For the record, the authors find ± √ 7 just as ‘nice’ as ±2. 10.6 The Inverse Trigonometric Functions 839 to this equation in Quadrant I are coterminal with α, and α serves as the reference angle for all of the solutions to this equation in Quadrant II. x y 1 1 1 3 α = arcsin _ 1 3 _ radians x y 1 1 1 3 α Since 1 3 isn’t the sine of any of the ‘common angles’ discussed earlier, we use the arcsine functions to express our answers. The real number t = arcsin _ 1 3 _ is defined so it satisfies 0 < t < π 2 with sin(t) = 1 3 . Hence, α = arcsin _ 1 3 _ radians. Since the solutions in Quadrant I are all coterminal with α, we get part of our solution to be θ = α + 2πk = arcsin _ 1 3 _ + 2πk for integers k. Turning our attention to Quadrant II, we get one solution to be π −α. Hence, the Quadrant II solutions are θ = π −α + 2πk = π −arcsin _ 1 3 _ + 2πk, for integers k. 2. We may visualize the solutions to tan(t) = −2 as angles θ with tan(θ) = −2. Since tangent is negative only in Quadrants II and IV, we focus our efforts there. x y 1 1 β = arctan(−2) radians x y 1 1 π β Since −2 isn’t the tangent of any of the ‘common angles’, we need to use the arctangent function to express our answers. The real number t = arctan(−2) satisfies tan(t) = −2 and − π 2 < t < 0. If we let β = arctan(−2) radians, we see that all of the Quadrant IV solutions 840 Foundations of Trigonometry to tan(θ) = −2 are coterminal with β. Moreover, the solutions from Quadrant II differ by exactly π units from the solutions in Quadrant IV, so all the solutions to tan(θ) = −2 are of the form θ = β +πk = arctan(−2) +πk for some integer k. Switching back to the variable t, we record our final answer to tan(t) = −2 as t = arctan(−2) +πk for integers k. 3. The last equation we are asked to solve, sec(x) = − 5 3 , poses two immediate problems. First, we are not told whether or not x represents an angle or a real number. We assume the latter, but note that we will use angles and the Unit Circle to solve the equation regardless. Second, as we have mentioned, there is no universally accepted range of the arcsecant function. For that reason, we adopt the advice given in Section 10.3 and convert this to the cosine problem cos(x) = − 3 5 . Adopting an angle approach, we consider the equation cos(θ) = − 3 5 and note that our solutions lie in Quadrants II and III. Since − 3 5 isn’t the cosine of any of the ‘common angles’, we’ll need to express our solutions in terms of the arccosine function. The real number t = arccos _ − 3 5 _ is defined so that π 2 < t < π with cos(t) = − 3 5 . If we let β = arccos _ − 3 5 _ radians, we see that β is a Quadrant II angle. To obtain a Quadrant III angle solution, we may simply use −β = −arccos _ − 3 5 _ . Since all angle solutions are coterminal with β or −β, we get our solutions to cos(θ) = − 3 5 to be θ = β + 2πk = arccos _ − 3 5 _ + 2πk or θ = −β + 2πk = −arccos _ − 3 5 _ + 2πk for integers k. Switching back to the variable x, we record our final answer to sec(x) = − 5 3 as x = arccos _ − 3 5 _ +2πk or x = −arccos _ − 3 5 _ +2πk for integers k. x y 1 1 β = arccos _ − 3 5 _ radians x y 1 1 β = arccos _ − 3 5 _ radians −β = −arccos _ − 3 5 _ radians The reader is encouraged to check the answers found in Example 10.6.7 - both analytically and with the calculator (see Section 10.6.3). With practice, the inverse trigonometric functions will become as familiar to you as the square root function. Speaking of practice . . . 10.6 The Inverse Trigonometric Functions 841 10.6.5 Exercises In Exercises 1 - 40, find the exact value. 1. arcsin (−1) 2. arcsin _ − √ 3 2 _ 3. arcsin _ − √ 2 2 _ 4. arcsin _ − 1 2 _ 5. arcsin (0) 6. arcsin _ 1 2 _ 7. arcsin _ √ 2 2 _ 8. arcsin _ √ 3 2 _ 9. arcsin (1) 10. arccos (−1) 11. arccos _ − √ 3 2 _ 12. arccos _ − √ 2 2 _ 13. arccos _ − 1 2 _ 14. arccos (0) 15. arccos _ 1 2 _ 16. arccos _ √ 2 2 _ 17. arccos _ √ 3 2 _ 18. arccos (1) 19. arctan _ − √ 3 _ 20. arctan (−1) 21. arctan _ − √ 3 3 _ 22. arctan (0) 23. arctan _ √ 3 3 _ 24. arctan (1) 25. arctan _√ 3 _ 26. arccot _ − √ 3 _ 27. arccot (−1) 28. arccot _ − √ 3 3 _ 29. arccot (0) 30. arccot _ √ 3 3 _ 31. arccot (1) 32. arccot _√ 3 _ 33. arcsec (2) 34. arccsc (2) 35. arcsec _√ 2 _ 36. arccsc _√ 2 _ 37. arcsec _ 2 √ 3 3 _ 38. arccsc _ 2 √ 3 3 _ 39. arcsec (1) 40. arccsc (1) In Exercises 41 - 48, assume that the range of arcsecant is _ 0, π 2 _ ∪ _ π, 3π 2 _ and that the range of arccosecant is _ 0, π 2 ¸ ∪ _ π, 3π 2 ¸ when finding the exact value. 41. arcsec (−2) 42. arcsec _ − √ 2 _ 43. arcsec _ − 2 √ 3 3 _ 44. arcsec (−1) 45. arccsc (−2) 46. arccsc _ − √ 2 _ 47. arccsc _ − 2 √ 3 3 _ 48. arccsc (−1) 842 Foundations of Trigonometry In Exercises 49 - 56, assume that the range of arcsecant is _ 0, π 2 _ ∪ _ π 2 , π ¸ and that the range of arccosecant is _ − π 2 , 0 _ ∪ _ 0, π 2 ¸ when finding the exact value. 49. arcsec (−2) 50. arcsec _ − √ 2 _ 51. arcsec _ − 2 √ 3 3 _ 52. arcsec (−1) 53. arccsc (−2) 54. arccsc _ − √ 2 _ 55. arccsc _ − 2 √ 3 3 _ 56. arccsc (−1) In Exercises 57 - 86, find the exact value or state that it is undefined. 57. sin _ arcsin _ 1 2 __ 58. sin _ arcsin _ − √ 2 2 __ 59. sin _ arcsin _ 3 5 __ 60. sin (arcsin (−0.42)) 61. sin _ arcsin _ 5 4 __ 62. cos _ arccos _ √ 2 2 __ 63. cos _ arccos _ − 1 2 __ 64. cos _ arccos _ 5 13 __ 65. cos (arccos (−0.998)) 66. cos (arccos (π)) 67. tan (arctan (−1)) 68. tan _ arctan _√ 3 __ 69. tan _ arctan _ 5 12 __ 70. tan (arctan (0.965)) 71. tan (arctan (3π)) 72. cot (arccot (1)) 73. cot _ arccot _ − √ 3 __ 74. cot _ arccot _ − 7 24 __ 75. cot (arccot (−0.001)) 76. cot _ arccot _ 17π 4 __ 77. sec (arcsec (2)) 78. sec (arcsec (−1)) 79. sec _ arcsec _ 1 2 __ 80. sec (arcsec (0.75)) 81. sec (arcsec (117π)) 82. csc _ arccsc _√ 2 __ 83. csc _ arccsc _ − 2 √ 3 3 __ 84. csc _ arccsc _ √ 2 2 __ 85. csc (arccsc (1.0001)) 86. csc _ arccsc _ π 4 __ In Exercises 87 - 106, find the exact value or state that it is undefined. 87. arcsin _ sin _ π 6 __ 88. arcsin _ sin _ − π 3 __ 89. arcsin _ sin _ 3π 4 __ 10.6 The Inverse Trigonometric Functions 843 90. arcsin _ sin _ 11π 6 __ 91. arcsin _ sin _ 4π 3 __ 92. arccos _ cos _ π 4 __ 93. arccos _ cos _ 2π 3 __ 94. arccos _ cos _ 3π 2 __ 95. arccos _ cos _ − π 6 __ 96. arccos _ cos _ 5π 4 __ 97. arctan _ tan _ π 3 __ 98. arctan _ tan _ − π 4 __ 99. arctan (tan (π)) 100. arctan _ tan _ π 2 __ 101. arctan _ tan _ 2π 3 __ 102. arccot _ cot _ π 3 __ 103. arccot _ cot _ − π 4 __ 104. arccot (cot (π)) 105. arccot _ cot _ π 2 __ 106. arccot _ cot _ 2π 3 __ In Exercises 107 - 118, assume that the range of arcsecant is _ 0, π 2 _ ∪ _ π, 3π 2 _ and that the range of arccosecant is _ 0, π 2 ¸ ∪ _ π, 3π 2 ¸ when finding the exact value. 107. arcsec _ sec _ π 4 __ 108. arcsec _ sec _ 4π 3 __ 109. arcsec _ sec _ 5π 6 __ 110. arcsec _ sec _ − π 2 __ 111. arcsec _ sec _ 5π 3 __ 112. arccsc _ csc _ π 6 __ 113. arccsc _ csc _ 5π 4 __ 114. arccsc _ csc _ 2π 3 __ 115. arccsc _ csc _ − π 2 __ 116. arccsc _ csc _ 11π 6 __ 117. arcsec _ sec _ 11π 12 __ 118. arccsc _ csc _ 9π 8 __ In Exercises 119 - 130, assume that the range of arcsecant is _ 0, π 2 _ ∪ _ π 2 , π ¸ and that the range of arccosecant is _ − π 2 , 0 _ ∪ _ 0, π 2 ¸ when finding the exact value. 119. arcsec _ sec _ π 4 __ 120. arcsec _ sec _ 4π 3 __ 121. arcsec _ sec _ 5π 6 __ 122. arcsec _ sec _ − π 2 __ 123. arcsec _ sec _ 5π 3 __ 124. arccsc _ csc _ π 6 __ 125. arccsc _ csc _ 5π 4 __ 126. arccsc _ csc _ 2π 3 __ 127. arccsc _ csc _ − π 2 __ 128. arccsc _ csc _ 11π 6 __ 129. arcsec _ sec _ 11π 12 __ 130. arccsc _ csc _ 9π 8 __ 844 Foundations of Trigonometry In Exercises 131 - 154, find the exact value or state that it is undefined. 131. sin _ arccos _ − 1 2 __ 132. sin _ arccos _ 3 5 __ 133. sin (arctan (−2)) 134. sin _ arccot _√ 5 __ 135. sin (arccsc (−3)) 136. cos _ arcsin _ − 5 13 __ 137. cos _ arctan _√ 7 __ 138. cos (arccot (3)) 139. cos (arcsec (5)) 140. tan _ arcsin _ − 2 √ 5 5 __ 141. tan _ arccos _ − 1 2 __ 142. tan _ arcsec _ 5 3 __ 143. tan (arccot (12)) 144. cot _ arcsin _ 12 13 __ 145. cot _ arccos _ √ 3 2 __ 146. cot _ arccsc _√ 5 __ 147. cot (arctan (0.25)) 148. sec _ arccos _ √ 3 2 __ 149. sec _ arcsin _ − 12 13 __ 150. sec (arctan (10)) 151. sec _ arccot _ − √ 10 10 __ 152. csc (arccot (9)) 153. csc _ arcsin _ 3 5 __ 154. csc _ arctan _ − 2 3 __ In Exercises 155 - 164, find the exact value or state that it is undefined. 155. sin _ arcsin _ 5 13 _ + π 4 _ 156. cos (arcsec(3) + arctan(2)) 157. tan _ arctan(3) + arccos _ − 3 5 __ 158. sin _ 2 arcsin _ − 4 5 __ 159. sin _ 2arccsc _ 13 5 __ 160. sin (2 arctan (2)) 161. cos _ 2 arcsin _ 3 5 __ 162. cos _ 2arcsec _ 25 7 __ 163. cos _ 2arccot _ − √ 5 __ 164. sin _ arctan(2) 2 _ 10.6 The Inverse Trigonometric Functions 845 In Exercises 165 - 184, rewrite the quantity as algebraic expressions of x and state the domain on which the equivalence is valid. 165. sin (arccos (x)) 166. cos (arctan (x)) 167. tan (arcsin (x)) 168. sec (arctan (x)) 169. csc (arccos (x)) 170. sin (2 arctan (x)) 171. sin (2 arccos (x)) 172. cos (2 arctan (x)) 173. sin(arccos(2x)) 174. sin _ arccos _ x 5 __ 175. cos _ arcsin _ x 2 __ 176. cos (arctan (3x)) 177. sin(2 arcsin(7x)) 178. sin _ 2 arcsin _ x √ 3 3 __ 179. cos(2 arcsin(4x)) 180. sec(arctan(2x)) tan(arctan(2x)) 181. sin (arcsin(x) + arccos(x)) 182. cos (arcsin(x) + arctan(x)) 183. tan (2 arcsin(x)) 184. sin _ 1 2 arctan(x) _ 185. If sin(θ) = x 2 for − π 2 < θ < π 2 , find an expression for θ + sin(2θ) in terms of x. 186. If tan(θ) = x 7 for − π 2 < θ < π 2 , find an expression for 1 2 θ − 1 2 sin(2θ) in terms of x. 187. If sec(θ) = x 4 for 0 < θ < π 2 , find an expression for 4 tan(θ) −4θ in terms of x. In Exercises 188 - 207, solve the equation using the techniques discussed in Example 10.6.7 then approximate the solutions which lie in the interval [0, 2π) to four decimal places. 188. sin(x) = 7 11 189. cos(x) = − 2 9 190. sin(x) = −0.569 191. cos(x) = 0.117 192. sin(x) = 0.008 193. cos(x) = 359 360 194. tan(x) = 117 195. cot(x) = −12 196. sec(x) = 3 2 197. csc(x) = − 90 17 198. tan(x) = − √ 10 199. sin(x) = 3 8 200. cos(x) = − 7 16 201. tan(x) = 0.03 202. sin(x) = 0.3502 846 Foundations of Trigonometry 203. sin(x) = −0.721 204. cos(x) = 0.9824 205. cos(x) = −0.5637 206. cot(x) = 1 117 207. tan(x) = −0.6109 In Exercises 208 - 210, find the two acute angles in the right triangle whose sides have the given lengths. Express your answers using degree measure rounded to two decimal places. 208. 3, 4 and 5 209. 5, 12 and 13 210. 336, 527 and 625 211. A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it touches level ground 360 feet from the base of the tower. What angle does the wire make with the ground? Express your answer using degree measure rounded to one decimal place. 212. At Cliffs of Insanity Point, The Great Sasquatch Canyon is 7117 feet deep. From that point, a fire is seen at a location known to be 10 miles away from the base of the sheer canyon wall. What angle of depression is made by the line of sight from the canyon edge to the fire? Express your answer using degree measure rounded to one decimal place. 213. Shelving is being built at the Utility Muffin Research Library which is to be 14 inches deep. An 18-inch rod will be attached to the wall and the underside of the shelf at its edge away from the wall, forming a right triangle under the shelf to support it. What angle, to the nearest degree, will the rod make with the wall? 214. A parasailor is being pulled by a boat on Lake Ippizuti. The cable is 300 feet long and the parasailor is 100 feet above the surface of the water. What is the angle of elevation from the boat to the parasailor? Express your answer using degree measure rounded to one decimal place. 215. A tag-and-release program to study the Sasquatch population of the eponymous Sasquatch National Park is begun. From a 200 foot tall tower, a ranger spots a Sasquatch lumbering through the wilderness directly towards the tower. Let θ denote the angle of depression from the top of the tower to a point on the ground. If the range of the rifle with a tranquilizer dart is 300 feet, find the smallest value of θ for which the corresponding point on the ground is in range of the rifle. Round your answer to the nearest hundreth of a degree. In Exercises 216 - 221, rewrite the given function as a sinusoid of the form S(x) = Asin(ωx + φ) using Exercises 35 and 36 in Section 10.5 for reference. Approximate the value of φ (which is in radians, of course) to four decimal places. 216. f(x) = 5 sin(3x) + 12 cos(3x) 217. f(x) = 3 cos(2x) + 4 sin(2x) 218. f(x) = cos(x) −3 sin(x) 219. f(x) = 7 sin(10x) −24 cos(10x) 10.6 The Inverse Trigonometric Functions 847 220. f(x) = −cos(x) −2 √ 2 sin(x) 221. f(x) = 2 sin(x) −cos(x) In Exercises 222 - 233, find the domain of the given function. Write your answers in interval notation. 222. f(x) = arcsin(5x) 223. f(x) = arccos _ 3x −1 2 _ 224. f(x) = arcsin _ 2x 2 _ 225. f(x) = arccos _ 1 x 2 −4 _ 226. f(x) = arctan(4x) 227. f(x) = arccot _ 2x x 2 −9 _ 228. f(x) = arctan(ln(2x −1)) 229. f(x) = arccot( √ 2x −1) 230. f(x) = arcsec(12x) 231. f(x) = arccsc(x + 5) 232. f(x) = arcsec _ x 3 8 _ 233. f(x) = arccsc _ e 2x _ 234. Show that arcsec(x) = arccos _ 1 x _ for [x[ ≥ 1 as long as we use _ 0, π 2 _ ∪ _ π 2 , π _ as the range of f(x) = arcsec(x). 235. Show that arccsc(x) = arcsin _ 1 x _ for [x[ ≥ 1 as long as we use _ − π 2 , 0 _ ∪ _ 0, π 2 _ as the range of f(x) = arccsc(x). 236. Show that arcsin(x) + arccos(x) = π 2 for −1 ≤ x ≤ 1. 237. Discuss with your classmates why arcsin _ 1 2 _ ,= 30 ◦ . 238. Use the following picture and the series of exercises on the next page to show that arctan(1) + arctan(2) + arctan(3) = π x y A(0, 1) O(0, 0) B(1, 0) C(2, 0) D(2, 3) α β γ 848 Foundations of Trigonometry (a) Clearly ´AOB and ´BCD are right triangles because the line through O and A and the line through C and D are perpendicular to the x-axis. Use the distance formula to show that ´BAD is also a right triangle (with ∠BAD being the right angle) by showing that the sides of the triangle satisfy the Pythagorean Theorem. (b) Use ´AOB to show that α = arctan(1) (c) Use ´BAD to show that β = arctan(2) (d) Use ´BCD to show that γ = arctan(3) (e) Use the fact that O, B and C all lie on the x-axis to conclude that α+β +γ = π. Thus arctan(1) + arctan(2) + arctan(3) = π. 10.6 The Inverse Trigonometric Functions 849 10.6.6 Answers 1. arcsin (−1) = − π 2 2. arcsin _ − √ 3 2 _ = − π 3 3. arcsin _ − √ 2 2 _ = − π 4 4. arcsin _ − 1 2 _ = − π 6 5. arcsin (0) = 0 6. arcsin _ 1 2 _ = π 6 7. arcsin _ √ 2 2 _ = π 4 8. arcsin _ √ 3 2 _ = π 3 9. arcsin (1) = π 2 10. arccos (−1) = π 11. arccos _ − √ 3 2 _ = 5π 6 12. arccos _ − √ 2 2 _ = 3π 4 13. arccos _ − 1 2 _ = 2π 3 14. arccos (0) = π 2 15. arccos _ 1 2 _ = π 3 16. arccos _ √ 2 2 _ = π 4 17. arccos _ √ 3 2 _ = π 6 18. arccos (1) = 0 19. arctan _ − √ 3 _ = − π 3 20. arctan (−1) = − π 4 21. arctan _ − √ 3 3 _ = − π 6 22. arctan (0) = 0 23. arctan _ √ 3 3 _ = π 6 24. arctan (1) = π 4 25. arctan _√ 3 _ = π 3 26. arccot _ − √ 3 _ = 5π 6 27. arccot (−1) = 3π 4 28. arccot _ − √ 3 3 _ = 2π 3 29. arccot (0) = π 2 30. arccot _ √ 3 3 _ = π 3 31. arccot (1) = π 4 32. arccot _√ 3 _ = π 6 33. arcsec (2) = π 3 34. arccsc (2) = π 6 35. arcsec _√ 2 _ = π 4 36. arccsc _√ 2 _ = π 4 37. arcsec _ 2 √ 3 3 _ = π 6 38. arccsc _ 2 √ 3 3 _ = π 3 39. arcsec (1) = 0 40. arccsc (1) = π 2 41. arcsec (−2) = 4π 3 42. arcsec _ − √ 2 _ = 5π 4 850 Foundations of Trigonometry 43. arcsec _ − 2 √ 3 3 _ = 7π 6 44. arcsec (−1) = π 45. arccsc (−2) = 7π 6 46. arccsc _ − √ 2 _ = 5π 4 47. arccsc _ − 2 √ 3 3 _ = 4π 3 48. arccsc (−1) = 3π 2 49. arcsec (−2) = 2π 3 50. arcsec _ − √ 2 _ = 3π 4 51. arcsec _ − 2 √ 3 3 _ = 5π 6 52. arcsec (−1) = π 53. arccsc (−2) = − π 6 54. arccsc _ − √ 2 _ = − π 4 55. arccsc _ − 2 √ 3 3 _ = − π 3 56. arccsc (−1) = − π 2 57. sin _ arcsin _ 1 2 __ = 1 2 58. sin _ arcsin _ − √ 2 2 __ = − √ 2 2 59. sin _ arcsin _ 3 5 __ = 3 5 60. sin (arcsin (−0.42)) = −0.42 61. sin _ arcsin _ 5 4 __ is undefined. 62. cos _ arccos _ √ 2 2 __ = √ 2 2 63. cos _ arccos _ − 1 2 __ = − 1 2 64. cos _ arccos _ 5 13 __ = 5 13 65. cos (arccos (−0.998)) = −0.998 66. cos (arccos (π)) is undefined. 67. tan (arctan (−1)) = −1 68. tan _ arctan _√ 3 __ = √ 3 69. tan _ arctan _ 5 12 __ = 5 12 70. tan (arctan (0.965)) = 0.965 71. tan (arctan (3π)) = 3π 72. cot (arccot (1)) = 1 73. cot _ arccot _ − √ 3 __ = − √ 3 74. cot _ arccot _ − 7 24 __ = − 7 24 75. cot (arccot (−0.001)) = −0.001 76. cot _ arccot _ 17π 4 __ = 17π 4 77. sec (arcsec (2)) = 2 78. sec (arcsec (−1)) = −1 10.6 The Inverse Trigonometric Functions 851 79. sec _ arcsec _ 1 2 __ is undefined. 80. sec (arcsec (0.75)) is undefined. 81. sec _ arcsec _ π 2 __ = π 2 82. csc _ arccsc _√ 2 __ = √ 2 83. csc _ arccsc _ − 2 √ 3 3 __ = − 2 √ 3 3 84. csc _ arccsc _ √ 2 2 __ is undefined. 85. csc (arccsc (1.0001)) = 1.0001 86. csc _ arccsc _ π 4 __ is undefined. 87. arcsin _ sin _ π 6 __ = π 6 88. arcsin _ sin _ − π 3 __ = − π 3 89. arcsin _ sin _ 3π 4 __ = π 4 90. arcsin _ sin _ 11π 6 __ = − π 6 91. arcsin _ sin _ 4π 3 __ = − π 3 92. arccos _ cos _ π 4 __ = π 4 93. arccos _ cos _ 2π 3 __ = 2π 3 94. arccos _ cos _ 3π 2 __ = π 2 95. arccos _ cos _ − π 6 __ = π 6 96. arccos _ cos _ 5π 4 __ = 3π 4 97. arctan _ tan _ π 3 __ = π 3 98. arctan _ tan _ − π 4 __ = − π 4 99. arctan (tan (π)) = 0 100. arctan _ tan _ π 2 __ is undefined 101. arctan _ tan _ 2π 3 __ = − π 3 102. arccot _ cot _ π 3 __ = π 3 103. arccot _ cot _ − π 4 __ = 3π 4 104. arccot (cot (π)) is undefined 105. arccot _ cot _ 3π 2 __ = π 2 106. arccot _ cot _ 2π 3 __ = 2π 3 107. arcsec _ sec _ π 4 __ = π 4 108. arcsec _ sec _ 4π 3 __ = 4π 3 109. arcsec _ sec _ 5π 6 __ = 7π 6 110. arcsec _ sec _ − π 2 __ is undefined. 111. arcsec _ sec _ 5π 3 __ = π 3 112. arccsc _ csc _ π 6 __ = π 6 852 Foundations of Trigonometry 113. arccsc _ csc _ 5π 4 __ = 5π 4 114. arccsc _ csc _ 2π 3 __ = π 3 115. arccsc _ csc _ − π 2 __ = 3π 2 116. arccsc _ csc _ 11π 6 __ = 7π 6 117. arcsec _ sec _ 11π 12 __ = 13π 12 118. arccsc _ csc _ 9π 8 __ = 9π 8 119. arcsec _ sec _ π 4 __ = π 4 120. arcsec _ sec _ 4π 3 __ = 2π 3 121. arcsec _ sec _ 5π 6 __ = 5π 6 122. arcsec _ sec _ − π 2 __ is undefined. 123. arcsec _ sec _ 5π 3 __ = π 3 124. arccsc _ csc _ π 6 __ = π 6 125. arccsc _ csc _ 5π 4 __ = − π 4 126. arccsc _ csc _ 2π 3 __ = π 3 127. arccsc _ csc _ − π 2 __ = − π 2 128. arccsc _ csc _ 11π 6 __ = − π 6 129. arcsec _ sec _ 11π 12 __ = 11π 12 130. arccsc _ csc _ 9π 8 __ = − π 8 131. sin _ arccos _ − 1 2 __ = √ 3 2 132. sin _ arccos _ 3 5 __ = 4 5 133. sin (arctan (−2)) = − 2 √ 5 5 134. sin _ arccot _√ 5 __ = √ 6 6 135. sin (arccsc (−3)) = − 1 3 136. cos _ arcsin _ − 5 13 __ = 12 13 137. cos _ arctan _√ 7 __ = √ 2 4 138. cos (arccot (3)) = 3 √ 10 10 139. cos (arcsec (5)) = 1 5 140. tan _ arcsin _ − 2 √ 5 5 __ = −2 141. tan _ arccos _ − 1 2 __ = − √ 3 142. tan _ arcsec _ 5 3 __ = 4 3 143. tan (arccot (12)) = 1 12 144. cot _ arcsin _ 12 13 __ = 5 12 10.6 The Inverse Trigonometric Functions 853 145. cot _ arccos _ √ 3 2 __ = √ 3 146. cot _ arccsc _√ 5 __ = 2 147. cot (arctan (0.25)) = 4 148. sec _ arccos _ √ 3 2 __ = 2 √ 3 3 149. sec _ arcsin _ − 12 13 __ = 13 5 150. sec (arctan (10)) = √ 101 151. sec _ arccot _ − √ 10 10 __ = − √ 11 152. csc (arccot (9)) = √ 82 153. csc _ arcsin _ 3 5 __ = 5 3 154. csc _ arctan _ − 2 3 __ = − √ 13 2 155. sin _ arcsin _ 5 13 _ + π 4 _ = 17 √ 2 26 156. cos (arcsec(3) + arctan(2)) = √ 5 −4 √ 10 15 157. tan _ arctan(3) + arccos _ − 3 5 __ = 1 3 158. sin _ 2 arcsin _ − 4 5 __ = − 24 25 159. sin _ 2arccsc _ 13 5 __ = 120 169 160. sin (2 arctan (2)) = 4 5 161. cos _ 2 arcsin _ 3 5 __ = 7 25 162. cos _ 2arcsec _ 25 7 __ = − 527 625 163. cos _ 2arccot _ − √ 5 __ = 2 3 164. sin _ arctan(2) 2 _ = _ 5 − √ 5 10 165. sin (arccos (x)) = √ 1 −x 2 for −1 ≤ x ≤ 1 166. cos (arctan (x)) = 1 √ 1 +x 2 for all x 167. tan (arcsin (x)) = x √ 1 −x 2 for −1 < x < 1 168. sec (arctan (x)) = √ 1 +x 2 for all x 169. csc (arccos (x)) = 1 √ 1 −x 2 for −1 < x < 1 170. sin (2 arctan (x)) = 2x x 2 + 1 for all x 171. sin (2 arccos (x)) = 2x √ 1 −x 2 for −1 ≤ x ≤ 1 854 Foundations of Trigonometry 172. cos (2 arctan (x)) = 1 −x 2 1 +x 2 for all x 173. sin(arccos(2x)) = √ 1 −4x 2 for − 1 2 ≤ x ≤ 1 2 174. sin _ arccos _ x 5 __ = √ 25 −x 2 5 for −5 ≤ x ≤ 5 175. cos _ arcsin _ x 2 __ = √ 4 −x 2 2 for −2 ≤ x ≤ 2 176. cos (arctan (3x)) = 1 √ 1 + 9x 2 for all x 177. sin(2 arcsin(7x)) = 14x √ 1 −49x 2 for − 1 7 ≤ x ≤ 1 7 178. sin _ 2 arcsin _ x √ 3 3 __ = 2x √ 3 −x 2 3 for − √ 3 ≤ x ≤ √ 3 179. cos(2 arcsin(4x)) = 1 −32x 2 for − 1 4 ≤ x ≤ 1 4 180. sec(arctan(2x)) tan(arctan(2x)) = 2x √ 1 + 4x 2 for all x 181. sin (arcsin(x) + arccos(x)) = 1 for −1 ≤ x ≤ 1 182. cos (arcsin(x) + arctan(x)) = √ 1 −x 2 −x 2 √ 1 +x 2 for −1 ≤ x ≤ 1 183. 10 tan (2 arcsin(x)) = 2x √ 1 −x 2 1 −2x 2 for x in _ −1, − √ 2 2 _ ∪ _ − √ 2 2 , √ 2 2 _ ∪ _ √ 2 2 , 1 _ 184. sin _ 1 2 arctan(x) _ = _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ ¸ √ x 2 + 1 −1 2 √ x 2 + 1 for x ≥ 0 − ¸ √ x 2 + 1 −1 2 √ x 2 + 1 for x < 0 185. If sin(θ) = x 2 for − π 2 < θ < π 2 , then θ + sin(2θ) = arcsin _ x 2 _ + x √ 4 −x 2 2 186. If tan(θ) = x 7 for − π 2 < θ < π 2 , then 1 2 θ − 1 2 sin(2θ) = 1 2 arctan _ x 7 _ − 7x x 2 + 49 10 The equivalence for x = ±1 can be verified independently of the derivation of the formula, but Calculus is required to fully understand what is happening at those x values. You’ll see what we mean when you work through the details of the identity for tan(2t). For now, we exclude x = ±1 from our answer. 10.6 The Inverse Trigonometric Functions 855 187. If sec(θ) = x 4 for 0 < θ < π 2 , then 4 tan(θ) −4θ = √ x 2 −16 −4arcsec _ x 4 _ 188. x = arcsin _ 7 11 _ + 2πk or x = π −arcsin _ 7 11 _ + 2πk, in [0, 2π), x ≈ 0.6898, 2.4518 189. x = arccos _ − 2 9 _ + 2πk or x = −arccos _ − 2 9 _ + 2πk, in [0, 2π), x ≈ 1.7949, 4.4883 190. x = π + arcsin(0.569) + 2πk or x = 2π −arcsin(0.569) + 2πk, in [0, 2π), x ≈ 3.7469, 5.6779 191. x = arccos(0.117) + 2πk or x = 2π −arccos(0.117) + 2πk, in [0, 2π), x ≈ 1.4535, 4.8297 192. x = arcsin(0.008) + 2πk or x = π −arcsin(0.008) + 2πk, in [0, 2π), x ≈ 0.0080, 3.1336 193. x = arccos _ 359 360 _ + 2πk or x = 2π −arccos _ 359 360 _ + 2πk, in [0, 2π), x ≈ 0.0746, 6.2086 194. x = arctan(117) +πk, in [0, 2π), x ≈ 1.56225, 4.70384 195. x = arctan _ − 1 12 _ +πk, in [0, 2π), x ≈ 3.0585, 6.2000 196. x = arccos _ 2 3 _ + 2πk or x = 2π −arccos _ 2 3 _ + 2πk, in [0, 2π), x ≈ 0.8411, 5.4422 197. x = π + arcsin _ 17 90 _ + 2πk or x = 2π −arcsin _ 17 90 _ + 2πk, in [0, 2π), x ≈ 3.3316, 6.0932 198. x = arctan _ − √ 10 _ +πk, in [0, 2π), x ≈ 1.8771, 5.0187 199. x = arcsin _ 3 8 _ + 2πk or x = π −arcsin _ 3 8 _ + 2πk, in [0, 2π), x ≈ 0.3844, 2.7572 200. x = arccos _ − 7 16 _ + 2πk or x = −arccos _ − 7 16 _ + 2πk, in [0, 2π), x ≈ 2.0236, 4.2596 201. x = arctan(0.03) +πk, in [0, 2π), x ≈ 0.0300, 3.1716 202. x = arcsin(0.3502) + 2πk or x = π −arcsin(0.3502) + 2πk, in [0, 2π), x ≈ 0.3578, 2.784 203. x = π + arcsin(0.721) + 2πk or x = 2π −arcsin(0.721) + 2πk, in [0, 2π), x ≈ 3.9468, 5.4780 204. x = arccos(0.9824) + 2πk or x = 2π −arccos(0.9824) + 2πk, in [0, 2π), x ≈ 0.1879, 6.0953 205. x = arccos(−0.5637) + 2πk or x = −arccos(−0.5637) + 2πk, in [0, 2π), x ≈ 2.1697, 4.1135 206. x = arctan(117) +πk, in [0, 2π), x ≈ 1.5622, 4.7038 207. x = arctan(−0.6109) +πk, in [0, 2π), x ≈ 2.5932, 5.7348 856 Foundations of Trigonometry 208. 36.87 ◦ and 53.13 ◦ 209. 22.62 ◦ and 67.38 ◦ 210. 32.52 ◦ and 57.48 ◦ 211. 68.9 ◦ 212. 7.7 ◦ 213. 51 ◦ 214. 19.5 ◦ 215. 41.81 ◦ 216. f(x) = 5 sin(3x) + 12 cos(3x) = 13 sin _ 3x + arcsin _ 12 13 __ ≈ 13 sin(3x + 1.1760) 217. f(x) = 3 cos(2x) + 4 sin(2x) = 5 sin _ 2x + arcsin _ 3 5 __ ≈ 5 sin(2x + 0.6435) 218. f(x) = cos(x) −3 sin(x) = √ 10 sin _ x + arccos _ − 3 √ 10 10 __ ≈ √ 10 sin(x + 2.8198) 219. f(x) = 7 sin(10x) −24 cos(10x) = 25 sin _ 10x + arcsin _ − 24 25 __ ≈ 25 sin(10x −1.2870) 220. f(x) = −cos(x) −2 √ 2 sin(x) = 3 sin _ x +π + arcsin _ 1 3 __ ≈ 3 sin(x + 3.4814) 221. f(x) = 2 sin(x) −cos(x) = √ 5 sin _ x + arcsin _ − √ 5 5 __ ≈ √ 5 sin(x −0.4636) 222. _ − 1 5 , 1 5 _ 223. _ − 1 3 , 1 _ 224. _ − √ 2 2 , √ 2 2 _ 225. (−∞, − √ 5] ∪ [− √ 3, √ 3] ∪ [ √ 5, ∞) 226. (−∞, ∞) 227. (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) 228. _ 1 2 , ∞ _ 229. _ 1 2 , ∞ _ 230. _ −∞, − 1 12 _ ∪ _ 1 12 , ∞ _ 231. (−∞, −6] ∪ [−4, ∞) 232. (−∞, −2] ∪ [2, ∞) 233. [0, ∞) 10.7 Trigonometric Equations and Inequalities 857 10.7 Trigonometric Equations and Inequalities In Sections 10.2, 10.3 and most recently 10.6, we solved some basic equations involving the trigono- metric functions. Below we summarize the techniques we’ve employed thus far. Note that we use the neutral letter ‘u’ as the argument 1 of each circular function for generality. Strategies for Solving Basic Equations Involving Trigonometric Functions • To solve cos(u) = c or sin(u) = c for −1 ≤ c ≤ 1, first solve for u in the interval [0, 2π) and add integer multiples of the period 2π. If c < −1 or of c > 1, there are no real solutions. • To solve sec(u) = c or csc(u) = c for c ≤ −1 or c ≥ 1, convert to cosine or sine, respectively, and solve as above. If −1 < c < 1, there are no real solutions. • To solve tan(u) = c for any real number c, first solve for u in the interval _ − π 2 , π 2 _ and add integer multiples of the period π. • To solve cot(u) = c for c ,= 0, convert to tangent and solve as above. If c = 0, the solution to cot(u) = 0 is u = π 2 +πk for integers k. Using the above guidelines, we can comfortably solve sin(x) = 1 2 and find the solution x = π 6 +2πk or x = 5π 6 + 2πk for integers k. How do we solve something like sin(3x) = 1 2 ? Since this equation has the form sin(u) = 1 2 , we know the solutions take the form u = π 6 + 2πk or u = 5π 6 + 2πk for integers k. Since the argument of sine here is 3x, we have 3x = π 6 + 2πk or 3x = 5π 6 + 2πk for integers k. To solve for x, we divide both sides 2 of these equations by 3, and obtain x = π 18 + 2π 3 k or x = 5π 18 + 2π 3 k for integers k. This is the technique employed in the example below. Example 10.7.1. Solve the following equations and check your answers analytically. List the solutions which lie in the interval [0, 2π) and verify them using a graphing utility. 1. cos(2x) = − √ 3 2 2. csc _ 1 3 x −π _ = √ 2 3. cot (3x) = 0 4. sec 2 (x) = 4 5. tan _ x 2 _ = −3 6. sin(2x) = 0.87 Solution. 1. The solutions to cos(u) = − √ 3 2 are u = 5π 6 + 2πk or u = 7π 6 + 2πk for integers k. Since the argument of cosine here is 2x, this means 2x = 5π 6 + 2πk or 2x = 7π 6 + 2πk for integers k. Solving for x gives x = 5π 12 + πk or x = 7π 12 + πk for integers k. To check these answers analytically, we substitute them into the original equation. For any integer k we have cos _ 2 _ 5π 12 +πk ¸_ = cos _ 5π 6 + 2πk _ = cos _ 5π 6 _ (the period of cosine is 2π) = − √ 3 2 1 See the comments at the beginning of Section 10.5 for a review of this concept. 2 Don’t forget to divide the 2πk by 3 as well! 858 Foundations of Trigonometry Similarly, we find cos _ 2 _ 7π 12 +πk ¸_ = cos _ 7π 6 + 2πk _ = cos _ 7π 6 _ = − √ 3 2 . To determine which of our solutions lie in [0, 2π), we substitute integer values for k. The solutions we keep come from the values of k = 0 and k = 1 and are x = 5π 12 , 7π 12 , 17π 12 and 19π 12 . Using a calculator, we graph y = cos(2x) and y = − √ 3 2 over [0, 2π) and examine where these two graphs intersect. We see that the x-coordinates of the intersection points correspond to the decimal representations of our exact answers. 2. Since this equation has the form csc(u) = √ 2, we rewrite this as sin(u) = √ 2 2 and find u = π 4 +2πk or u = 3π 4 +2πk for integers k. Since the argument of cosecant here is _ 1 3 x −π _ , 1 3 x −π = π 4 + 2πk or 1 3 x −π = 3π 4 + 2πk To solve 1 3 x −π = π 4 + 2πk, we first add π to both sides 1 3 x = π 4 + 2πk +π A common error is to treat the ‘2πk’ and ‘π’ terms as ‘like’ terms and try to combine them when they are not. 3 We can, however, combine the ‘π’ and ‘ π 4 ’ terms to get 1 3 x = 5π 4 + 2πk We now finish by multiplying both sides by 3 to get x = 3 _ 5π 4 + 2πk _ = 15π 4 + 6πk Solving the other equation, 1 3 x − π = 3π 4 + 2πk produces x = 21π 4 + 6πk for integers k. To check the first family of answers, we substitute, combine line terms, and simplify. csc _ 1 3 _ 15π 4 + 6πk ¸ −π _ = csc _ 5π 4 + 2πk −π _ = csc _ π 4 + 2πk _ = csc _ π 4 _ (the period of cosecant is 2π) = √ 2 The family x = 21π 4 +6πk checks similarly. Despite having infinitely many solutions, we find that none of them lie in [0, 2π). To verify this graphically, we use a reciprocal identity to rewrite the cosecant as a sine and we find that y = 1 sin( 1 3 x−π) and y = √ 2 do not intersect at all over the interval [0, 2π). 3 Do you see why? 10.7 Trigonometric Equations and Inequalities 859 y = cos(2x) and y = − √ 3 2 y = 1 sin( 1 3 x−π) and y = √ 2 3. Since cot(3x) = 0 has the form cot(u) = 0, we know u = π 2 +πk, so, in this case, 3x = π 2 +πk for integers k. Solving for x yields x = π 6 + π 3 k. Checking our answers, we get cot _ 3 _ π 6 + π 3 k ¸_ = cot _ π 2 +πk _ = cot _ π 2 _ (the period of cotangent is π) = 0 As k runs through the integers, we obtain six answers, corresponding to k = 0 through k = 5, which lie in [0, 2π): x = π 6 , π 2 , 5π 6 , 7π 6 , 3π 2 and 11π 6 . To confirm these graphically, we must be careful. On many calculators, there is no function button for cotangent. We choose 4 to use the quotient identity cot(3x) = cos(3x) sin(3x) . Graphing y = cos(3x) sin(3x) and y = 0 (the x-axis), we see that the x-coordinates of the intersection points approximately match our solutions. 4. The complication in solving an equation like sec 2 (x) = 4 comes not from the argument of secant, which is just x, but rather, the fact the secant is being squared. To get this equation to look like one of the forms listed on page 857, we extract square roots to get sec(x) = ±2. Converting to cosines, we have cos(x) = ± 1 2 . For cos(x) = 1 2 , we get x = π 3 + 2πk or x = 5π 3 + 2πk for integers k. For cos(x) = − 1 2 , we get x = 2π 3 + 2πk or x = 4π 3 + 2πk for integers k. If we take a step back and think of these families of solutions geometrically, we see we are finding the measures of all angles with a reference angle of π 3 . As a result, these solutions can be combined and we may write our solutions as x = π 3 + πk and x = 2π 3 + πk for integers k. To check the first family of solutions, we note that, depending on the integer k, sec _ π 3 +πk _ doesn’t always equal sec _ π 3 _ . However, it is true that for all integers k, sec _ π 3 +πk _ = ±sec _ π 3 _ = ±2. (Can you show this?) As a result, sec 2 _ π 3 +πk _ = _ ±sec _ π 3 __ 2 = (±2) 2 = 4 The same holds for the family x = 2π 3 + πk. The solutions which lie in [0, 2π) come from the values k = 0 and k = 1, namely x = π 3 , 2π 3 , 4π 3 and 5π 3 . To confirm graphically, we use 4 The reader is encouraged to see what happens if we had chosen the reciprocal identity cot(3x) = 1 tan(3x) instead. The graph on the calculator appears identical, but what happens when you try to find the intersection points? 860 Foundations of Trigonometry a reciprocal identity to rewrite the secant as cosine. The x-coordinates of the intersection points of y = 1 (cos(x)) 2 and y = 4 verify our answers. y = cos(3x) sin(3x) and y = 0 y = 1 cos 2 (x) and y = 4 5. The equation tan _ x 2 _ = −3 has the form tan(u) = −3, whose solution is u = arctan(−3)+πk. Hence, x 2 = arctan(−3) +πk, so x = 2 arctan(−3) + 2πk for integers k. To check, we note tan _ 2 arctan(−3)+2πk 2 _ = tan (arctan(−3) +πk) = tan (arctan(−3)) (the period of tangent is π) = −3 (See Theorem 10.27) To determine which of our answers lie in the interval [0, 2π), we first need to get an idea of the value of 2 arctan(−3). While we could easily find an approximation using a calculator, 5 we proceed analytically. Since −3 < 0, it follows that − π 2 < arctan(−3) < 0. Multiplying through by 2 gives −π < 2 arctan(−3) < 0. We are now in a position to argue which of the solutions x = 2 arctan(−3) + 2πk lie in [0, 2π). For k = 0, we get x = 2 arctan(−3) < 0, so we discard this answer and all answers x = 2 arctan(−3) + 2πk where k < 0. Next, we turn our attention to k = 1 and get x = 2 arctan(−3) + 2π. Starting with the inequality −π < 2 arctan(−3) < 0, we add 2π and get π < 2 arctan(−3) + 2π < 2π. This means x = 2 arctan(−3) +2π lies in [0, 2π). Advancing k to 2 produces x = 2 arctan(−3) +4π. Once again, we get from −π < 2 arctan(−3) < 0 that 3π < 2 arctan(−3) + 4π < 4π. Since this is outside the interval [0, 2π), we discard x = 2 arctan(−3) + 4π and all solutions of the form x = 2 arctan(−3) +2πk for k > 2. Graphically, we see y = tan _ x 2 _ and y = −3 intersect only once on [0, 2π) at x = 2 arctan(−3) + 2π ≈ 3.7851. 6. To solve sin(2x) = 0.87, we first note that it has the form sin(u) = 0.87, which has the family of solutions u = arcsin(0.87) + 2πk or u = π − arcsin(0.87) + 2πk for integers k. Since the argument of sine here is 2x, we get 2x = arcsin(0.87) + 2πk or 2x = π − arcsin(0.87) + 2πk which gives x = 1 2 arcsin(0.87) +πk or x = π 2 − 1 2 arcsin(0.87) +πk for integers k. To check, 5 Your instructor will let you know if you should abandon the analytic route at this point and use your calculator. But seriously, what fun would that be? 10.7 Trigonometric Equations and Inequalities 861 sin _ 2 _ 1 2 arcsin(0.87) +πk ¸_ = sin (arcsin(0.87) + 2πk) = sin (arcsin(0.87)) (the period of sine is 2π) = 0.87 (See Theorem 10.26) For the family x = π 2 − 1 2 arcsin(0.87) +πk , we get sin _ 2 _ π 2 − 1 2 arcsin(0.87) +πk ¸_ = sin (π −arcsin(0.87) + 2πk) = sin (π −arcsin(0.87)) (the period of sine is 2π) = sin (arcsin(0.87)) (sin(π −t) = sin(t)) = 0.87 (See Theorem 10.26) To determine which of these solutions lie in [0, 2π), we first need to get an idea of the value of x = 1 2 arcsin(0.87). Once again, we could use the calculator, but we adopt an analytic route here. By definition, 0 < arcsin(0.87) < π 2 so that multiplying through by 1 2 gives us 0 < 1 2 arcsin(0.87) < π 4 . Starting with the family of solutions x = 1 2 arcsin(0.87) +πk, we use the same kind of arguments as in our solution to number 5 above and find only the solutions corresponding to k = 0 and k = 1 lie in [0, 2π): x = 1 2 arcsin(0.87) and x = 1 2 arcsin(0.87) +π. Next, we move to the family x = π 2 − 1 2 arcsin(0.87) + πk for integers k. Here, we need to get a better estimate of π 2 − 1 2 arcsin(0.87). From the inequality 0 < 1 2 arcsin(0.87) < π 4 , we first multiply through by −1 and then add π 2 to get π 2 > π 2 − 1 2 arcsin(0.87) > π 4 , or π 4 < π 2 − 1 2 arcsin(0.87) < π 2 . Proceeding with the usual arguments, we find the only solutions which lie in [0, 2π) correspond to k = 0 and k = 1, namely x = π 2 − 1 2 arcsin(0.87) and x = 3π 2 − 1 2 arcsin(0.87). All told, we have found four solutions to sin(2x) = 0.87 in [0, 2π): x = 1 2 arcsin(0.87), x = 1 2 arcsin(0.87) +π, x = π 2 − 1 2 arcsin(0.87) and x = 3π 2 − 1 2 arcsin(0.87). By graphing y = sin(2x) and y = 0.87, we confirm our results. y = tan _ x 2 _ and y = −3 y = sin(2x) and y = 0.87 862 Foundations of Trigonometry Each of the problems in Example 10.7.1 featured one trigonometric function. If an equation involves two different trigonometric functions or if the equation contains the same trigonometric function but with different arguments, we will need to use identities and Algebra to reduce the equation to the same form as those given on page 857. Example 10.7.2. Solve the following equations and list the solutions which lie in the interval [0, 2π). Verify your solutions on [0, 2π) graphically. 1. 3 sin 3 (x) = sin 2 (x) 2. sec 2 (x) = tan(x) + 3 3. cos(2x) = 3 cos(x) −2 4. cos(3x) = 2 −cos(x) 5. cos(3x) = cos(5x) 6. sin(2x) = √ 3 cos(x) 7. sin(x) cos _ x 2 _ + cos(x) sin _ x 2 _ = 1 8. cos(x) − √ 3 sin(x) = 2 Solution. 1. We resist the temptation to divide both sides of 3 sin 3 (x) = sin 2 (x) by sin 2 (x) (What goes wrong if you do?) and instead gather all of the terms to one side of the equation and factor. 3 sin 3 (x) = sin 2 (x) 3 sin 3 (x) −sin 2 (x) = 0 sin 2 (x)(3 sin(x) −1) = 0 Factor out sin 2 (x) from both terms. We get sin 2 (x) = 0 or 3 sin(x) − 1 = 0. Solving for sin(x), we find sin(x) = 0 or sin(x) = 1 3 . The solution to the first equation is x = πk, with x = 0 and x = π being the two solutions which lie in [0, 2π). To solve sin(x) = 1 3 , we use the arcsine function to get x = arcsin _ 1 3 _ +2πk or x = π −arcsin _ 1 3 _ +2πk for integers k. We find the two solutions here which lie in [0, 2π) to be x = arcsin _ 1 3 _ and x = π −arcsin _ 1 3 _ . To check graphically, we plot y = 3(sin(x)) 3 and y = (sin(x)) 2 and find the x-coordinates of the intersection points of these two curves. Some extra zooming is required near x = 0 and x = π to verify that these two curves do in fact intersect four times. 6 2. Analysis of sec 2 (x) = tan(x) + 3 reveals two different trigonometric functions, so an identity is in order. Since sec 2 (x) = 1 + tan 2 (x), we get sec 2 (x) = tan(x) + 3 1 + tan 2 (x) = tan(x) + 3 (Since sec 2 (x) = 1 + tan 2 (x).) tan 2 (x) −tan(x) −2 = 0 u 2 −u −2 = 0 Let u = tan(x). (u + 1)(u −2) = 0 6 Note that we are not counting the point (2π, 0) in our solution set since x = 2π is not in the interval [0, 2π). In the forthcoming solutions, remember that while x = 2π may be a solution to the equation, it isn’t counted among the solutions in [0, 2π). 10.7 Trigonometric Equations and Inequalities 863 This gives u = −1 or u = 2. Since u = tan(x), we have tan(x) = −1 or tan(x) = 2. From tan(x) = −1, we get x = − π 4 + πk for integers k. To solve tan(x) = 2, we employ the arctangent function and get x = arctan(2) +πk for integers k. From the first set of solutions, we get x = 3π 4 and x = 5π 4 as our answers which lie in [0, 2π). Using the same sort of argument we saw in Example 10.7.1, we get x = arctan(2) and x = π + arctan(2) as answers from our second set of solutions which lie in [0, 2π). Using a reciprocal identity, we rewrite the secant as a cosine and graph y = 1 (cos(x)) 2 and y = tan(x) +3 to find the x-values of the points where they intersect. y = 3(sin(x)) 3 and y = (sin(x)) 2 y = 1 (cos(x)) 2 and y = tan(x) + 3 3. In the equation cos(2x) = 3 cos(x) −2, we have the same circular function, namely cosine, on both sides but the arguments differ. Using the identity cos(2x) = 2 cos 2 (x) −1, we obtain a ‘quadratic in disguise’ and proceed as we have done in the past. cos(2x) = 3 cos(x) −2 2 cos 2 (x) −1 = 3 cos(x) −2 (Since cos(2x) = 2 cos 2 (x) −1.) 2 cos 2 (x) −3 cos(x) + 1 = 0 2u 2 −3u + 1 = 0 Let u = cos(x). (2u −1)(u −1) = 0 This gives u = 1 2 or u = 1. Since u = cos(x), we get cos(x) = 1 2 or cos(x) = 1. Solving cos(x) = 1 2 , we get x = π 3 + 2πk or x = 5π 3 + 2πk for integers k. From cos(x) = 1, we get x = 2πk for integers k. The answers which lie in [0, 2π) are x = 0, π 3 , and 5π 3 . Graphing y = cos(2x) and y = 3 cos(x) −2, we find, after a little extra effort, that the curves intersect in three places on [0, 2π), and the x-coordinates of these points confirm our results. 4. To solve cos(3x) = 2 − cos(x), we use the same technique as in the previous problem. From Example 10.4.3, number 4, we know that cos(3x) = 4 cos 3 (x) −3 cos(x). This transforms the equation into a polynomial in terms of cos(x). cos(3x) = 2 −cos(x) 4 cos 3 (x) −3 cos(x) = 2 −cos(x) 2 cos 3 (x) −2 cos(x) −2 = 0 4u 3 −2u −2 = 0 Let u = cos(x). 864 Foundations of Trigonometry To solve 4u 3 − 2u − 2 = 0, we need the techniques in Chapter 3 to factor 4u 3 − 2u − 2 into (u−1) _ 4u 2 + 4u + 2 _ . We get either u−1 = 0 or 4u 2 +2u+2 = 0, and since the discriminant of the latter is negative, the only real solution to 4u 3 −2u−2 = 0 is u = 1. Since u = cos(x), we get cos(x) = 1, so x = 2πk for integers k. The only solution which lies in [0, 2π) is x = 0. Graphing y = cos(3x) and y = 2 −cos(x) on the same set of axes over [0, 2π) shows that the graphs intersect at what appears to be (0, 1), as required. y = cos(2x) and y = 3 cos(x) −2 y = cos(3x) and y = 2 −cos(x) 5. While we could approach cos(3x) = cos(5x) in the same manner as we did the previous two problems, we choose instead to showcase the utility of the Sum to Product Identities. From cos(3x) = cos(5x), we get cos(5x) − cos(3x) = 0, and it is the presence of 0 on the right hand side that indicates a switch to a product would be a good move. 7 Using Theorem 10.21, we have that cos(5x) − cos(3x) = −2 sin _ 5x+3x 2 _ sin _ 5x−3x 2 _ = −2 sin(4x) sin(x). Hence, the equation cos(5x) = cos(3x) is equivalent to −2 sin(4x) sin(x) = 0. From this, we get sin(4x) = 0 or sin(x) = 0. Solving sin(4x) = 0 gives x = π 4 k for integers k, and the solution to sin(x) = 0 is x = πk for integers k. The second set of solutions is contained in the first set of solutions, 8 so our final solution to cos(5x) = cos(3x) is x = π 4 k for integers k. There are eight of these answers which lie in [0, 2π): x = 0, π 4 , π 2 , 3π 4 , π, 5π 4 , 3π 2 and 7π 4 . Our plot of the graphs of y = cos(3x) and y = cos(5x) below (after some careful zooming) bears this out. 6. In examining the equation sin(2x) = √ 3 cos(x), not only do we have different circular func- tions involved, namely sine and cosine, we also have different arguments to contend with, namely 2x and x. Using the identity sin(2x) = 2 sin(x) cos(x) makes all of the arguments the same and we proceed as we would solving any nonlinear equation – gather all of the nonzero terms on one side of the equation and factor. sin(2x) = √ 3 cos(x) 2 sin(x) cos(x) = √ 3 cos(x) (Since sin(2x) = 2 sin(x) cos(x).) 2 sin(x) cos(x) − √ 3 cos(x) = 0 cos(x)(2 sin(x) − √ 3) = 0 from which we get cos(x) = 0 or sin(x) = √ 3 2 . From cos(x) = 0, we obtain x = π 2 + πk for integers k. From sin(x) = √ 3 2 , we get x = π 3 +2πk or x = 2π 3 +2πk for integers k. The answers 7 As always, experience is the greatest teacher here! 8 As always, when in doubt, write it out! 10.7 Trigonometric Equations and Inequalities 865 which lie in [0, 2π) are x = π 2 , 3π 2 , π 3 and 2π 3 . We graph y = sin(2x) and y = √ 3 cos(x) and, after some careful zooming, verify our answers. y = cos(3x) and y = cos(5x) y = sin(2x) and y = √ 3 cos(x) 7. Unlike the previous problem, there seems to be no quick way to get the circular functions or their arguments to match in the equation sin(x) cos _ x 2 _ + cos(x) sin _ x 2 _ = 1. If we stare at it long enough, however, we realize that the left hand side is the expanded form of the sum formula for sin _ x + x 2 _ . Hence, our original equation is equivalent to sin _ 3 2 x _ = 1. Solving, we find x = π 3 + 4π 3 k for integers k. Two of these solutions lie in [0, 2π): x = π 3 and x = 5π 3 . Graphing y = sin(x) cos _ x 2 _ + cos(x) sin _ x 2 _ and y = 1 validates our solutions. 8. With the absence of double angles or squares, there doesn’t seem to be much we can do. However, since the arguments of the cosine and sine are the same, we can rewrite the left hand side of this equation as a sinusoid. 9 To fit f(x) = cos(x) − √ 3 sin(x) to the form Asin(ωt + φ) + B, we use what we learned in Example 10.5.3 and find A = 2, B = 0, ω = 1 and φ = 5π 6 . Hence, we can rewrite the equation cos(x) − √ 3 sin(x) = 2 as 2 sin _ x + 5π 6 _ = 2, or sin _ x + 5π 6 _ = 1. Solving the latter, we get x = − π 3 + 2πk for integers k. Only one of these solutions, x = 5π 3 , which corresponds to k = 1, lies in [0, 2π). Geometrically, we see that y = cos(x) − √ 3 sin(x) and y = 2 intersect just once, supporting our answer. y = sin(x) cos _ x 2 _ + cos(x) sin _ x 2 _ and y = 1 y = cos(x) − √ 3 sin(x) and y = 2 We repeat here the advice given when solving systems of nonlinear equations in section 8.7 – when it comes to solving equations involving the trigonometric functions, it helps to just try something. 9 We are essentially ‘undoing’ the sum / difference formula for cosine or sine, depending on which form we use, so this problem is actually closely related to the previous one! 866 Foundations of Trigonometry Next, we focus on solving inequalities involving the trigonometric functions. Since these functions are continuous on their domains, we may use the sign diagram technique we’ve used in the past to solve the inequalities. 10 Example 10.7.3. Solve the following inequalities on [0, 2π). Express your answers using interval notation and verify your answers graphically. 1. 2 sin(x) ≤ 1 2. sin(2x) > cos(x) 3. tan(x) ≥ 3 Solution. 1. We begin solving 2 sin(x) ≤ 1 by collecting all of the terms on one side of the equation and zero on the other to get 2 sin(x) −1 ≤ 0. Next, we let f(x) = 2 sin(x) −1 and note that our original inequality is equivalent to solving f(x) ≤ 0. We now look to see where, if ever, f is undefined and where f(x) = 0. Since the domain of f is all real numbers, we can immediately set about finding the zeros of f. Solving f(x) = 0, we have 2 sin(x) − 1 = 0 or sin(x) = 1 2 . The solutions here are x = π 6 +2πk and x = 5π 6 +2πk for integers k. Since we are restricting our attention to [0, 2π), only x = π 6 and x = 5π 6 are of concern to us. Next, we choose test values in [0, 2π) other than the zeros and determine if f is positive or negative there. For x = 0 we have f(0) = −1, for x = π 2 we get f _ π 2 _ = 1 and for x = π we get f(π) = −1. Since our original inequality is equivalent to f(x) ≤ 0, we are looking for where the function is negative (−) or 0, and we get the intervals _ 0, π 6 ¸ ∪ _ 5π 6 , 2π _ . We can confirm our answer graphically by seeing where the graph of y = 2 sin(x) crosses or is below the graph of y = 1. 0 (−) π 6 0 (+) 5π 6 0 (−) 2π y = 2 sin(x) and y = 1 2. We first rewrite sin(2x) > cos(x) as sin(2x) − cos(x) > 0 and let f(x) = sin(2x) − cos(x). Our original inequality is thus equivalent to f(x) > 0. The domain of f is all real numbers, so we can advance to finding the zeros of f. Setting f(x) = 0 yields sin(2x) − cos(x) = 0, which, by way of the double angle identity for sine, becomes 2 sin(x) cos(x) − cos(x) = 0 or cos(x)(2 sin(x)−1) = 0. From cos(x) = 0, we get x = π 2 +πk for integers k of which only x = π 2 and x = 3π 2 lie in [0, 2π). For 2 sin(x) −1 = 0, we get sin(x) = 1 2 which gives x = π 6 + 2πk or x = 5π 6 + 2πk for integers k. Of those, only x = π 6 and x = 5π 6 lie in [0, 2π). Next, we choose 10 See page 214, Example 3.1.5, page 321, page 399, Example 6.3.2 and Example 6.4.2 for discussion of this technique. 10.7 Trigonometric Equations and Inequalities 867 our test values. For x = 0 we find f(0) = −1; when x = π 4 we get f _ π 4 _ = 1 − √ 2 2 = 2− √ 2 2 ; for x = 3π 4 we get f _ 3π 4 _ = −1 + √ 2 2 = √ 2−2 2 ; when x = π we have f(π) = 1, and lastly, for x = 7π 4 we get f _ 7π 4 _ = −1 − √ 2 2 = −2− √ 2 2 . We see f(x) > 0 on _ π 6 , π 2 _ ∪ _ 5π 6 , 3π 2 _ , so this is our answer. We can use the calculator to check that the graph of y = sin(2x) is indeed above the graph of y = cos(x) on those intervals. 0 (−) π 6 0 (+) π 2 0 (−) 5π 6 0 (+) 3π 2 0 (−) 2π y = sin(2x) and y = cos(x) 3. Proceeding as in the last two problems, we rewrite tan(x) ≥ 3 as tan(x) − 3 ≥ 0 and let f(x) = tan(x) − 3. We note that on [0, 2π), f is undefined at x = π 2 and 3π 2 , so those values will need the usual disclaimer on the sign diagram. 11 Moving along to zeros, solving f(x) = tan(x) − 3 = 0 requires the arctangent function. We find x = arctan(3) + πk for integers k and of these, only x = arctan(3) and x = arctan(3) + π lie in [0, 2π). Since 3 > 0, we know 0 < arctan(3) < π 2 which allows us to position these zeros correctly on the sign diagram. To choose test values, we begin with x = 0 and find f(0) = −3. Finding a convenient test value in the interval _ arctan(3), π 2 _ is a bit more challenging. Keep in mind that the arctangent function is increasing and is bounded above by π 2 . This means that the number x = arctan(117) is guaranteed 12 to lie between arctan(3) and π 2 . We see that f(arctan(117)) = tan(arctan(117)) − 3 = 114. For our next test value, we take x = π and find f(π) = −3. To find our next test value, we note that since arctan(3) < arctan(117) < π 2 , it follows 13 that arctan(3) +π < arctan(117) +π < 3π 2 . Evaluating f at x = arctan(117) +π yields f(arctan(117) + π) = tan(arctan(117) + π) − 3 = tan(arctan(117)) − 3 = 114. We choose our last test value to be x = 7π 4 and find f _ 7π 4 _ = −4. Since we want f(x) ≥ 0, we see that our answer is _ arctan(3), π 2 _ ∪ _ arctan(3) +π, 3π 2 _ . Using the graphs of y = tan(x) and y = 3, we see when the graph of the former is above (or meets) the graph of the latter. 11 See page 321 for a discussion of the non-standard character known as the interrobang. 12 We could have chosen any value arctan(t) where t > 3. 13 . . . by adding π through the inequality . . . 868 Foundations of Trigonometry 0 (−) arctan(3) 0 (+) π 2 ‽ (−) (arctan(3) + π) 0 (+) 3π 2 ‽ (−) 2π y = tan(x) and y = 3 We close this section with an example that puts solving equations and inequalities to good use – finding domains of functions. Example 10.7.4. Express the domain of the following functions using extended interval notation. 14 1. f(x) = csc _ 2x + π 3 _ 2. f(x) = sin(x) 2 cos(x) −1 3. f(x) = _ 1 −cot(x) Solution. 1. To find the domain of f(x) = csc _ 2x + π 3 _ , we rewrite f in terms of sine as f(x) = 1 sin(2x+ π 3 ) . Since the sine function is defined everywhere, our only concern comes from zeros in the denom- inator. Solving sin _ 2x + π 3 _ = 0, we get x = − π 6 + π 2 k for integers k. In set-builder notation, our domain is _ x : x ,= − π 6 + π 2 k for integers k _ . To help visualize the domain, we follow the old mantra ‘When in doubt, write it out!’ We get _ x : x ,= − π 6 , 2π 6 , − 4π 6 , 5π 6 , − 7π 6 , 8π 6 , . . . _ , where we have kept the denominators 6 throughout to help see the pattern. Graphing the situation on a numberline, we have − 7π 6 − 4π 6 − π 6 2π 6 5π 6 8π 6 Proceeding as we did on page 756 in Section 10.3.1, we let x k denote the kth number excluded from the domain and we have x k = − π 6 + π 2 k = (3k−1)π 6 for integers k. The intervals which comprise the domain are of the form (x k , x k + 1 ) = _ (3k−1)π 6 , (3k+2)π 6 _ as k runs through the integers. Using extended interval notation, we have that the domain is ∞ _ k=−∞ _ (3k −1)π 6 , (3k + 2)π 6 _ We can check our answer by substituting in values of k to see that it matches our diagram. 14 See page 756 for details about this notation. 10.7 Trigonometric Equations and Inequalities 869 2. Since the domains of sin(x) and cos(x) are all real numbers, the only concern when finding the domain of f(x) = sin(x) 2 cos(x)−1 is division by zero so we set the denominator equal to zero and solve. From 2 cos(x)−1 = 0 we get cos(x) = 1 2 so that x = π 3 +2πk or x = 5π 3 +2πk for integers k. Using set-builder notation, the domain is _ x : x ,= π 3 + 2πk and x ,= 5π 3 + 2πk for integers k _ , or _ x : x ,= ± π 3 , ± 5π 3 , ± 7π 3 , ± 11π 3 , . . . _ , so we have − 11π 3 − 7π 3 − 5π 3 − π 3 π 3 5π 3 7π 3 11π 3 Unlike the previous example, we have two different families of points to consider, and we present two ways of dealing with this kind of situation. One way is to generalize what we did in the previous example and use the formulas we found in our domain work to describe the intervals. To that end, we let a k = π 3 + 2πk = (6k+1)π 3 and b k = 5π 3 + 2πk = (6k+5)π 3 for integers k. The goal now is to write the domain in terms of the a’s an b’s. We find a 0 = π 3 , a 1 = 7π 3 , a −1 = − 5π 3 , a 2 = 13π 3 , a −2 = − 11π 3 , b 0 = 5π 3 , b 1 = 11π 3 , b −1 = − π 3 , b 2 = 17π 3 and b −2 = − 7π 3 . Hence, in terms of the a’s and b’s, our domain is . . . (a −2 , b −2 ) ∪ (b −2 , a −1 ) ∪ (a −1 , b −1 ) ∪ (b −1 , a 0 ) ∪ (a 0 , b 0 ) ∪ (b 0 , a 1 ) ∪ (a 1 , b 1 ) ∪ . . . If we group these intervals in pairs, (a −2 , b −2 )∪(b −2 , a −1 ), (a −1 , b −1 )∪(b −1 , a 0 ), (a 0 , b 0 )∪(b 0 , a 1 ) and so forth, we see a pattern emerge of the form (a k , b k ) ∪ (b k , a k + 1 ) for integers k so that our domain can be written as ∞ _ k=−∞ (a k , b k ) ∪ (b k , a k + 1 ) = ∞ _ k=−∞ _ (6k + 1)π 3 , (6k + 5)π 3 _ ∪ _ (6k + 5)π 3 , (6k + 7)π 3 _ A second approach to the problem exploits the periodic nature of f. Since cos(x) and sin(x) have period 2π, it’s not too difficult to show the function f repeats itself every 2π units. 15 This means if we can find a formula for the domain on an interval of length 2π, we can express the entire domain by translating our answer left and right on the x-axis by adding integer multiples of 2π. One such interval that arises from our domain work is _ π 3 , 7π 3 ¸ . The portion of the domain here is _ π 3 , 5π 3 _ ∪ _ 5π 3 , 7π 3 _ . Adding integer multiples of 2π, we get the family of intervals _ π 3 + 2πk, 5π 3 + 2πk _ ∪ _ 5π 3 + 2πk, 7π 3 + 2πk _ for integers k. We leave it to the reader to show that getting common denominators leads to our previous answer. 15 This doesn’t necessarily mean the period of f is 2π. The tangent function is comprised of cos(x) and sin(x), but its period is half theirs. The reader is invited to investigate the period of f. 870 Foundations of Trigonometry 3. To find the domain of f(x) = _ 1 −cot(x), we first note that, due to the presence of the cot(x) term, x ,= πk for integers k. Next, we recall that for the square root to be defined, we need 1−cot(x) ≥ 0. Unlike the inequalities we solved in Example 10.7.3, we are not restricted here to a given interval. Our strategy is to solve this inequality over (0, π) (the same interval which generates a fundamental cycle of cotangent) and then add integer multiples of the period, in this case, π. We let g(x) = 1 −cot(x) and set about making a sign diagram for g over the interval (0, π) to find where g(x) ≥ 0. We note that g is undefined for x = πk for integers k, in particular, at the endpoints of our interval x = 0 and x = π. Next, we look for the zeros of g. Solving g(x) = 0, we get cot(x) = 1 or x = π 4 + πk for integers k and only one of these, x = π 4 , lies in (0, π). Choosing the test values x = π 6 and x = π 2 , we get g _ π 6 _ = 1 − √ 3, and g _ π 2 _ = 1. 0 ‽ (−) π 4 0 (+) π ‽ We find g(x) ≥ 0 on _ π 4 , π _ . Adding multiples of the period we get our solution to consist of the intervals _ π 4 +πk, π +πk _ = _ (4k+1)π 4 , (k + 1)π _ . Using extended interval notation, we express our final answer as ∞ _ k=−∞ _ (4k + 1)π 4 , (k + 1)π _ 10.7 Trigonometric Equations and Inequalities 871 10.7.1 Exercises In Exercises 1 - 18, find all of the exact solutions of the equation and then list those solutions which are in the interval [0, 2π). 1. sin (5x) = 0 2. cos (3x) = 1 2 3. sin (−2x) = √ 3 2 4. tan (6x) = 1 5. csc (4x) = −1 6. sec (3x) = √ 2 7. cot (2x) = − √ 3 3 8. cos (9x) = 9 9. sin _ x 3 _ = √ 2 2 10. cos _ x + 5π 6 _ = 0 11. sin _ 2x − π 3 _ = − 1 2 12. 2 cos _ x + 7π 4 _ = √ 3 13. csc(x) = 0 14. tan (2x −π) = 1 15. tan 2 (x) = 3 16. sec 2 (x) = 4 3 17. cos 2 (x) = 1 2 18. sin 2 (x) = 3 4 In Exercises 19 - 42, solve the equation, giving the exact solutions which lie in [0, 2π) 19. sin (x) = cos (x) 20. sin (2x) = sin (x) 21. sin (2x) = cos (x) 22. cos (2x) = sin (x) 23. cos (2x) = cos (x) 24. cos(2x) = 2 −5 cos(x) 25. 3 cos(2x) + cos(x) + 2 = 0 26. cos(2x) = 5 sin(x) −2 27. 3 cos(2x) = sin(x) + 2 28. 2 sec 2 (x) = 3 −tan(x) 29. tan 2 (x) = 1 −sec(x) 30. cot 2 (x) = 3 csc(x) −3 31. sec(x) = 2 csc(x) 32. cos(x) csc(x) cot(x) = 6 −cot 2 (x) 33. sin(2x) = tan(x) 34. cot 4 (x) = 4 csc 2 (x) −7 35. cos(2x) + csc 2 (x) = 0 36. tan 3 (x) = 3 tan (x) 37. tan 2 (x) = 3 2 sec (x) 38. cos 3 (x) = −cos (x) 39. tan(2x) −2 cos(x) = 0 40. csc 3 (x) + csc 2 (x) = 4 csc(x) + 4 41. 2 tan(x) = 1 −tan 2 (x) 42. tan (x) = sec (x) 872 Foundations of Trigonometry In Exercises 43 - 58, solve the equation, giving the exact solutions which lie in [0, 2π) 43. sin(6x) cos(x) = −cos(6x) sin(x) 44. sin(3x) cos(x) = cos(3x) sin(x) 45. cos(2x) cos(x) + sin(2x) sin(x) = 1 46. cos(5x) cos(3x) −sin(5x) sin(3x) = √ 3 2 47. sin(x) + cos(x) = 1 48. sin(x) + √ 3 cos(x) = 1 49. √ 2 cos(x) − √ 2 sin(x) = 1 50. √ 3 sin(2x) + cos(2x) = 1 51. cos(2x) − √ 3 sin(2x) = √ 2 52. 3 √ 3 sin(3x) −3 cos(3x) = 3 √ 3 53. cos(3x) = cos(5x) 54. cos(4x) = cos(2x) 55. sin(5x) = sin(3x) 56. cos(5x) = −cos(2x) 57. sin(6x) + sin(x) = 0 58. tan(x) = cos(x) In Exercises 59 - 70, solve the inequality. Express the exact answer in interval notation, restricting your attention to 0 ≤ x ≤ 2π. 59. sin (x) ≤ 0 60. tan (x) ≥ √ 3 61. sec 2 (x) ≤ 4 62. cos 2 (x) > 1 2 63. cos (2x) ≤ 0 64. sin _ x + π 3 _ > 1 2 65. cot 2 (x) ≥ 1 3 66. 2 cos(x) ≥ 1 67. sin(5x) ≥ 5 68. cos(3x) ≤ 1 69. sec(x) ≤ √ 2 70. cot(x) ≤ 4 In Exercises 71 - 76, solve the inequality. Express the exact answer in interval notation, restricting your attention to −π ≤ x ≤ π. 71. cos (x) > √ 3 2 72. sin(x) > 1 3 73. sec (x) ≤ 2 74. sin 2 (x) < 3 4 75. cot (x) ≥ −1 76. cos(x) ≥ sin(x) In Exercises 77 - 82, solve the inequality. Express the exact answer in interval notation, restricting your attention to −2π ≤ x ≤ 2π. 77. csc (x) > 1 78. cos(x) ≤ 5 3 79. cot(x) ≥ 5 10.7 Trigonometric Equations and Inequalities 873 80. tan 2 (x) ≥ 1 81. sin(2x) ≥ sin(x) 82. cos(2x) ≤ sin(x) In Exercises 83 - 91, express the domain of the function using the extended interval notation. (See page 756 in Section 10.3.1 for details.) 83. f(x) = 1 cos(x) −1 84. f(x) = cos(x) sin(x) + 1 85. f(x) = _ tan 2 (x) −1 86. f(x) = _ 2 −sec(x) 87. f(x) = csc(2x) 88. f(x) = sin(x) 2 + cos(x) 89. f(x) = 3 csc(x) + 4 sec(x) 90. f(x) = ln ([ cos(x)[) 91. f(x) = arcsin(tan(x)) 92. With the help of your classmates, determine the number of solutions to sin(x) = 1 2 in [0, 2π). Then find the number of solutions to sin(2x) = 1 2 , sin(3x) = 1 2 and sin(4x) = 1 2 in [0, 2π). A pattern should emerge. Explain how this pattern would help you solve equations like sin(11x) = 1 2 . Now consider sin _ x 2 _ = 1 2 , sin _ 3x 2 _ = 1 2 and sin _ 5x 2 _ = 1 2 . What do you find? Replace 1 2 with −1 and repeat the whole exploration. 874 Foundations of Trigonometry 10.7.2 Answers 1. x = πk 5 ; x = 0, π 5 , 2π 5 , 3π 5 , 4π 5 , π, 6π 5 , 7π 5 , 8π 5 , 9π 5 2. x = π 9 + 2πk 3 or x = 5π 9 + 2πk 3 ; x = π 9 , 5π 9 , 7π 9 , 11π 9 , 13π 9 , 17π 9 3. x = 2π 3 +πk or x = 5π 6 +πk; x = 2π 3 , 5π 6 , 5π 3 , 11π 6 4. x = π 24 + πk 6 ; x = π 24 , 5π 24 , 3π 8 , 13π 24 , 17π 24 , 7π 8 , 25π 24 , 29π 24 , 11π 8 , 37π 24 , 41π 24 , 15π 8 5. x = 3π 8 + πk 2 ; x = 3π 8 , 7π 8 , 11π 8 , 15π 8 6. x = π 12 + 2πk 3 or x = 7π 12 + 2πk 3 ; x = π 12 , 7π 12 , 3π 4 , 5π 4 , 17π 12 , 23π 12 7. x = π 3 + πk 2 ; x = π 3 , 5π 6 , 4π 3 , 11π 6 8. No solution 9. x = 3π 4 + 6πk or x = 9π 4 + 6πk; x = 3π 4 10. x = − π 3 +πk; x = 2π 3 , 5π 3 11. x = 3π 4 +πk or x = 13π 12 +πk; x = π 12 , 3π 4 , 13π 12 , 7π 4 12. x = − 19π 12 + 2πk or x = π 12 + 2πk; x = π 12 , 5π 12 13. No solution 14. x = 5π 8 + πk 2 ; x = π 8 , 5π 8 , 9π 8 , 13π 8 15. x = π 3 +πk or x = 2π 3 +πk; x = π 3 , 2π 3 , 4π 3 , 5π 3 16. x = π 6 +πk or x = 5π 6 +πk; x = π 6 , 5π 6 , 7π 6 , 11π 6 17. x = π 4 + πk 2 ; x = π 4 , 3π 4 , 5π 4 , 7π 4 18. x = π 3 +πk or x = 2π 3 +πk; x = π 3 , 2π 3 , 4π 3 , 5π 3 10.7 Trigonometric Equations and Inequalities 875 19. x = π 4 , 5π 4 20. x = 0, π 3 , π, 5π 3 21. x = π 6 , π 2 , 5π 6 , 3π 2 22. x = π 6 , 5π 6 , 3π 2 23. x = 0, 2π 3 , 4π 3 24. x = π 3 , 5π 3 25. x = 2π 3 , 4π 3 , arccos _ 1 3 _ , 2π −arccos _ 1 3 _ 26. x = π 6 , 5π 6 27. x = 7π 6 , 11π 6 , arcsin _ 1 3 _ , π −arcsin _ 1 3 _ 28. x = 3π 4 , 7π 4 , arctan _ 1 2 _ , π + arctan _ 1 2 _ 29. x = 0, 2π 3 , 4π 3 30. x = π 6 , 5π 6 , π 2 31. x = arctan(2), π + arctan(2) 32. x = π 6 , 7π 6 , 5π 6 , 11π 6 33. x = 0, π, π 4 , 3π 4 , 5π 4 , 7π 4 34. x = π 6 , π 4 , 3π 4 , 5π 6 , 7π 6 , 5π 4 , 7π 4 , 11π 6 35. x = π 2 , 3π 2 36. x = 0, π 3 , 2π 3 , π, 4π 3 , 5π 3 37. x = π 3 , 5π 3 38. x = π 2 , 3π 2 39. x = π 6 , π 2 , 5π 6 , 3π 2 40. x = π 6 , 5π 6 , 7π 6 , 3π 2 , 11π 6 41. x = π 8 , 5π 8 , 9π 8 , 13π 8 42. No solution 43. x = 0, π 7 , 2π 7 , 3π 7 , 4π 7 , 5π 7 , 6π 7 , π, 8π 7 , 9π 7 , 10π 7 , 11π 7 , 12π 7 , 13π 7 44. x = 0, π 2 , π, 3π 2 45. x = 0 46. x = π 48 , 11π 48 , 13π 48 , 23π 48 , 25π 48 , 35π 48 , 37π 48 , 47π 48 , 49π 48 , 59π 48 , 61π 48 , 71π 48 , 73π 48 , 83π 48 , 85π 48 , 95π 48 47. x = 0, π 2 48. x = π 2 , 11π 6 49. x = π 12 , 17π 12 50. x = 0, π, π 3 , 4π 3 51. x = 17π 24 , 41π 24 , 23π 24 , 47π 24 52. x = π 6 , 5π 18 , 5π 6 , 17π 18 , 3π 2 , 29π 18 876 Foundations of Trigonometry 53. x = 0, π 4 , π 2 , 3π 4 , π, 5π 4 , 3π 2 , 7π 4 54. x = 0, π 3 , 2π 3 , π, 4π 3 , 5π 3 55. x = 0, π 8 , 3π 8 , 5π 8 , 7π 8 , π, 9π 8 , 11π 8 , 13π 8 , 15π 8 56. x = π 7 , π 3 , 3π 7 , 5π 7 , π, 9π 7 , 11π 7 , 5π 3 , 13π 7 57. x = 2π 7 , 4π 7 , 6π 7 , 8π 7 , 10π 7 , 12π 7 , π 5 , 3π 5 , π, 7π 5 , 9π 5 58. x = arcsin _ −1 + √ 5 2 _ ≈ 0.6662, π −arcsin _ −1 + √ 5 2 _ ≈ 2.4754 59. [π, 2π] 60. _ π 3 , π 2 _ ∪ _ 4π 3 , 3π 2 _ 61. _ 0, π 3 _ ∪ _ 2π 3 , 4π 3 _ ∪ _ 5π 3 , 2π _ 62. _ 0, π 4 _ ∪ _ 3π 4 , 5π 4 _ ∪ _ 7π 4 , 2π _ 63. _ π 4 , 3π 4 _ ∪ _ 5π 4 , 7π 4 _ 64. _ 0, π 2 _ ∪ _ 11π 6 , 2π _ 65. _ 0, π 3 _ ∪ _ 2π 3 , π _ ∪ _ π, 4π 3 _ ∪ _ 5π 3 , 2π _ 66. _ 0, π 3 _ ∪ _ 5π 3 , 2π _ 67. No solution 68. [0, 2π] 69. _ 0, π 4 _ ∪ _ π 2 , 3π 2 _ ∪ _ 7π 4 , 2π _ 70. [arccot(4), π) ∪ [π + arccot(4), 2π) 71. _ − π 6 , π 6 _ 72. _ arcsin _ 1 3 _ , π −arcsin _ 1 3 __ 73. _ −π, − π 2 _ ∪ _ − π 3 , π 3 _ ∪ _ π 2 , π _ 74. _ − 2π 3 , − π 3 _ ∪ _ π 3 , 2π 3 _ 75. _ −π, − π 4 _ ∪ _ 0, 3π 4 _ 76. _ − 3π 4 , π 4 _ 77. _ −2π, − 3π 2 _ ∪ _ − 3π 2 , −π _ ∪ _ 0, π 2 _ ∪ _ π 2 , π _ 78. [−2π, 2π] 79. (−2π, arccot(5) −2π] ∪ (−π, arccot(5) −π] ∪ (0, arccot(5)] ∪ (π, π + arccot(5)] 80. _ − 7π 4 , − 3π 2 _ ∪ _ − 3π 2 , − 5π 4 _ ∪ _ − 3π 4 , − π 2 _ ∪ _ − π 2 , − π 4 _ ∪ _ π 4 , π 2 _ ∪ _ π 2 , 3π 4 _ ∪ _ 5π 4 , 3π 2 _ ∪ _ 3π 2 , 7π 4 _ 81. _ −2π, − 5π 3 _ ∪ _ −π, − π 3 _ ∪ _ 0, π 3 _ ∪ _ π, 5π 3 _ 10.7 Trigonometric Equations and Inequalities 877 82. _ − 11π 6 , − 7π 6 _ ∪ _ π 6 , 5π 6 _ ∪, _ − π 2 , 3π 2 _ 83. ∞ _ k=−∞ (2kπ, (2k + 2)π) 84. ∞ _ k=−∞ _ (4k −1)π 2 , (4k + 3)π 2 _ 85. ∞ _ k=−∞ __ (4k + 1)π 4 , (2k + 1)π 2 _ ∪ _ (2k + 1)π 2 , (4k + 3)π 4 __ 86. ∞ _ k=−∞ __ (6k −1)π 3 , (6k + 1)π 3 _ ∪ _ (4k + 1)π 2 , (4k + 3)π 2 __ 87. ∞ _ k=−∞ _ kπ 2 , (k + 1)π 2 _ 88. (−∞, ∞) 89. ∞ _ k=−∞ _ kπ 2 , (k + 1)π 2 _ 90. ∞ _ k=−∞ _ (2k −1)π 2 , (2k + 1)π 2 _ 91. ∞ _ k=−∞ _ (4k −1)π 4 , (4k + 1)π 4 _ 878 Foundations of Trigonometry Chapter 11 Applications of Trigonometry 11.1 Applications of Sinusoids In the same way exponential functions can be used to model a wide variety of phenomena in nature, 1 the cosine and sine functions can be used to model their fair share of natural behaviors. In section 10.5, we introduced the concept of a sinusoid as a function which can be written either in the form C(x) = Acos(ωx+φ)+B for ω > 0 or equivalently, in the form S(x) = Asin(ωx+φ)+B for ω > 0. At the time, we remained undecided as to which form we preferred, but the time for such indecision is over. For clarity of exposition we focus on the sine function 2 in this section and switch to the independent variable t, since the applications in this section are time-dependent. We reintroduce and summarize all of the important facts and definitions about this form of the sinusoid below. Properties of the Sinusoid S(t) = Asin(ωt +φ) +B • The amplitude is [A[ • The angular frequency is ω and the ordinary frequency is f = ω 2π • The period is T = 1 f = 2π ω • The phase is φ and the phase shift is − φ ω • The vertical shift or baseline is B Along with knowing these formulas, it is helpful to remember what these quantities mean in context. The amplitude measures the maximum displacement of the sine wave from its baseline (determined by the vertical shift), the period is the length of time it takes to complete one cycle of the sinusoid, the angular frequency tells how many cycles are completed over an interval of length 2π, and the ordinary frequency measures how many cycles occur per unit of time. The phase indicates what 1 See Section 6.5. 2 Sine haters can use the co-function identity cos π 2 −θ = sin(θ) to turn all of the sines into cosines. 880 Applications of Trigonometry angle φ corresponds to t = 0, and the phase shift represents how much of a ‘head start’ the sinusoid has over the un-shifted sine function. The figure below is repeated from Section 10.5. amplitude baseline period In Section 10.1.1, we introduced the concept of circular motion and in Section 10.2.1, we developed formulas for circular motion. Our first foray into sinusoidal motion puts these notions to good use. Example 11.1.1. Recall from Exercise 55 in Section 10.1 that The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height 136 feet. It completes two revolutions in 2 minutes and 7 seconds. Assuming that the riders are at the edge of the circle, find a sinusoid which describes the height of the passengers above the ground t seconds after they pass the point on the wheel closest to the ground. Solution. We sketch the problem situation below and assume a counter-clockwise rotation. 3 O P Q θ h 3 Otherwise, we could just observe the motion of the wheel from the other side. 11.1 Applications of Sinusoids 881 We know from the equations given on page 732 in Section 10.2.1 that the y-coordinate for counter- clockwise motion on a circle of radius r centered at the origin with constant angular velocity (frequency) ω is given by y = r sin(ωt). Here, t = 0 corresponds to the point (r, 0) so that θ, the angle measuring the amount of rotation, is in standard position. In our case, the diameter of the wheel is 128 feet, so the radius is r = 64 feet. Since the wheel completes two revolutions in 2 minutes and 7 seconds (which is 127 seconds) the period T = 1 2 (127) = 127 2 seconds. Hence, the angular frequency is ω = 2π T = 4π 127 radians per second. Putting these two pieces of information together, we have that y = 64 sin _ 4π 127 t _ describes the y-coordinate on the Giant Wheel after t seconds, assuming it is centered at (0, 0) with t = 0 corresponding to the point Q. In order to find an expression for h, we take the point O in the figure as the origin. Since the base of the Giant Wheel ride is 8 feet above the ground and the Giant Wheel itself has a radius of 64 feet, its center is 72 feet above the ground. To account for this vertical shift upward, 4 we add 72 to our formula for y to obtain the new formula h = y + 72 = 64 sin _ 4π 127 t _ + 72. Next, we need to adjust things so that t = 0 corresponds to the point P instead of the point Q. This is where the phase comes into play. Geometrically, we need to shift the angle θ in the figure back π 2 radians. From Section 10.2.1, we know θ = ωt = 4π 127 t, so we (temporarily) write the height in terms of θ as h = 64 sin (θ) + 72. Subtracting π 2 from θ gives the final answer h(t) = 64 sin _ θ − π 2 _ +72 = 64 sin _ 4π 127 t − π 2 _ +72. We can check the reasonableness of our answer by graphing y = h(t) over the interval _ 0, 127 2 ¸ . t y 127 2 8 72 136 A few remarks about Example 11.1.1 are in order. First, note that the amplitude of 64 in our answer corresponds to the radius of the Giant Wheel. This means that passengers on the Giant Wheel never stray more than 64 feet vertically from the center of the Wheel, which makes sense. Second, the phase shift of our answer works out to be π/2 4π/127 = 127 8 = 15.875. This represents the ‘time delay’ (in seconds) we introduce by starting the motion at the point P as opposed to the point Q. Said differently, passengers which ‘start’ at P take 15.875 seconds to ‘catch up’ to the point Q. Our next example revisits the daylight data first introduced in Section 2.5, Exercise 6b. 4 We are readjusting our ‘baseline’ from y = 0 to y = 72. 882 Applications of Trigonometry Example 11.1.2. According to the U.S. Naval Observatory website, the number of hours H of daylight that Fairbanks, Alaska received on the 21st day of the nth month of 2009 is given below. Here t = 1 represents January 21, 2009, t = 2 represents February 21, 2009, and so on. Month Number 1 2 3 4 5 6 7 8 9 10 11 12 Hours of Daylight 5.8 9.3 12.4 15.9 19.4 21.8 19.4 15.6 12.4 9.1 5.6 3.3 1. Find a sinusoid which models these data and use a graphing utility to graph your answer along with the data. 2. Compare your answer to part 1 to one obtained using the regression feature of a calculator. Solution. 1. To get a feel for the data, we plot it below. t H 1 2 3 4 5 6 7 8 9 10 11 12 2 4 6 8 10 12 14 16 18 20 22 The data certainly appear sinusoidal, 5 but when it comes down to it, fitting a sinusoid to data manually is not an exact science. We do our best to find the constants A, ω, φ and B so that the function H(t) = Asin(ωt + φ) + B closely matches the data. We first go after the vertical shift B whose value determines the baseline. In a typical sinusoid, the value of B is the average of the maximum and minimum values. So here we take B = 3.3+21.8 2 = 12.55. Next is the amplitude A which is the displacement from the baseline to the maximum (and minimum) values. We find A = 21.8 − 12.55 = 12.55 − 3.3 = 9.25. At this point, we have H(t) = 9.25 sin(ωt + φ) + 12.55. Next, we go after the angular frequency ω. Since the data collected is over the span of a year (12 months), we take the period T = 12 months. 6 This 5 Okay, it appears to be the ‘∧’ shape we saw in some of the graphs in Section 2.2. Just humor us. 6 Even though the data collected lies in the interval [1, 12], which has a length of 11, we need to think of the data point at t = 1 as a representative sample of the amount of daylight for every day in January. That is, it represents H(t) over the interval [0, 1]. Similarly, t = 2 is a sample of H(t) over [1, 2], and so forth. 11.1 Applications of Sinusoids 883 means ω = 2π T = 2π 12 = π 6 . The last quantity to find is the phase φ. Unlike the previous example, it is easier in this case to find the phase shift − φ ω . Since we picked A > 0, the phase shift corresponds to the first value of t with H(t) = 12.55 (the baseline value). 7 Here, we choose t = 3, since its corresponding H value of 12.4 is closer to 12.55 than the next value, 15.9, which corresponds to t = 4. Hence, − φ ω = 3, so φ = −3ω = −3 _ π 6 _ = − π 2 . We have H(t) = 9.25 sin _ π 6 t − π 2 _ + 12.55. Below is a graph of our data with the curve y = H(t). 2. Using the ‘SinReg’ command, we graph the calculator’s regression below. While both models seem to be reasonable fits to the data, the calculator model is possibly the better fit. The calculator does not give us an r 2 value like it did for linear regressions in Section 2.5, nor does it give us an R 2 value like it did for quadratic, cubic and quartic regressions as in Section 3.1. The reason for this, much like the reason for the absence of R 2 for the logistic model in Section 6.5, is beyond the scope of this course. We’ll just have to use our own good judgment when choosing the best sinusoid model. 11.1.1 Harmonic Motion One of the major applications of sinusoids in Science and Engineering is the study of harmonic motion. The equations for harmonic motion can be used to describe a wide range of phenomena, from the motion of an object on a spring, to the response of an electronic circuit. In this subsection, we restrict our attention to modeling a simple spring system. Before we jump into the Mathematics, there are some Physics terms and concepts we need to discuss. In Physics, ‘mass’ is defined as a measure of an object’s resistance to straight-line motion whereas ‘weight’ is the amount of force (pull) gravity exerts on an object. An object’s mass cannot change, 8 while its weight could change. 7 See the figure on page 880. 8 Well, assuming the object isn’t subjected to relativistic speeds . . . 884 Applications of Trigonometry An object which weighs 6 pounds on the surface of the Earth would weigh 1 pound on the surface of the Moon, but its mass is the same in both places. In the English system of units, ‘pounds’ (lbs.) is a measure of force (weight), and the corresponding unit of mass is the ‘slug’. In the SI system, the unit of force is ‘Newtons’ (N) and the associated unit of mass is the ‘kilogram’ (kg). We convert between mass and weight using the formula 9 w = mg. Here, w is the weight of the object, m is the mass and g is the acceleration due to gravity. In the English system, g = 32 feet second 2 , and in the SI system, g = 9.8 meters second 2 . Hence, on Earth a mass of 1 slug weighs 32 lbs. and a mass of 1 kg weighs 9.8 N. 10 Suppose we attach an object with mass m to a spring as depicted below. The weight of the object will stretch the spring. The system is said to be in ‘equilibrium’ when the weight of the object is perfectly balanced with the restorative force of the spring. How far the spring stretches to reach equilibrium depends on the spring’s ‘spring constant’. Usually denoted by the letter k, the spring constant relates the force F applied to the spring to the amount d the spring stretches in accordance with Hooke’s Law 11 F = kd. If the object is released above or below the equilibrium position, or if the object is released with an upward or downward velocity, the object will bounce up and down on the end of the spring until some external force stops it. If we let x(t) denote the object’s displacement from the equilibrium position at time t, then x(t) = 0 means the object is at the equilibrium position, x(t) < 0 means the object is above the equilibrium position, and x(t) > 0 means the object is below the equilibrium position. The function x(t) is called the ‘equation of motion’ of the object. 12 x(t) = 0 at the x(t) < 0 above the x(t) > 0 below the equilibrium position equilibrium position equilibrium position If we ignore all other influences on the system except gravity and the spring force, then Physics tells us that gravity and the spring force will battle each other forever and the object will oscillate indefinitely. In this case, we describe the motion as ‘free’ (meaning there is no external force causing the motion) and ‘undamped’ (meaning we ignore friction caused by surrounding medium, which in our case is air). The following theorem, which comes from Differential Equations, gives x(t) as a function of the mass m of the object, the spring constant k, the initial displacement x 0 of the 9 This is a consequence of Newton’s Second Law of Motion F = ma where F is force, m is mass and a is acceleration. In our present setting, the force involved is weight which is caused by the acceleration due to gravity. 10 Note that 1 pound = 1 slug foot second 2 and 1 Newton = 1 kg meter second 2 . 11 Look familiar? We saw Hooke’s Law in Section 4.3.1. 12 To keep units compatible, if we are using the English system, we use feet (ft.) to measure displacement. If we are in the SI system, we measure displacement in meters (m). Time is always measured in seconds (s). 11.1 Applications of Sinusoids 885 object and initial velocity v 0 of the object. As with x(t), x 0 = 0 means the object is released from the equilibrium position, x 0 < 0 means the object is released above the equilibrium position and x 0 > 0 means the object is released below the equilibrium position. As far as the initial velocity v 0 is concerned, v 0 = 0 means the object is released ‘from rest,’ v 0 < 0 means the object is heading upwards and v 0 > 0 means the object is heading downwards. 13 Theorem 11.1. Equation for Free Undamped Harmonic Motion: Suppose an object of mass m is suspended from a spring with spring constant k. If the initial displacement from the equilibrium position is x 0 and the initial velocity of the object is v 0 , then the displacement x from the equilibrium position at time t is given by x(t) = Asin(ωt +φ) where • ω = _ k m and A = _ x 2 0 + _ v 0 ω _ 2 • Asin(φ) = x 0 and Aω cos(φ) = v 0 . It is a great exercise in ‘dimensional analysis’ to verify that the formulas given in Theorem 11.1 work out so that ω has units 1 s and A has units ft. or m, depending on which system we choose. Example 11.1.3. Suppose an object weighing 64 pounds stretches a spring 8 feet. 1. If the object is attached to the spring and released 3 feet below the equilibrium position from rest, find the equation of motion of the object, x(t). When does the object first pass through the equilibrium position? Is the object heading upwards or downwards at this instant? 2. If the object is attached to the spring and released 3 feet below the equilibrium position with an upward velocity of 8 feet per second, find the equation of motion of the object, x(t). What is the longest distance the object travels above the equilibrium position? When does this first happen? Confirm your result using a graphing utility. Solution. In order to use the formulas in Theorem 11.1, we first need to determine the spring constant k and the mass of the object m. To find k, we use Hooke’s Law F = kd. We know the object weighs 64 lbs. and stretches the spring 8 ft.. Using F = 64 and d = 8, we get 64 = k 8, or k = 8 lbs. ft. . To find m, we use w = mg with w = 64 lbs. and g = 32 ft. s 2 . We get m = 2 slugs. We can now proceed to apply Theorem 11.1. 1. With k = 8 and m = 2, we get ω = _ k m = _ 8 2 = 2. We are told that the object is released 3 feet below the equilibrium position ‘from rest.’ This means x 0 = 3 and v 0 = 0. Therefore, A = _ x 2 0 + _ v 0 ω _ 2 = √ 3 2 + 0 2 = 3. To determine the phase φ, we have Asin(φ) = x 0 , which in this case gives 3 sin(φ) = 3 so sin(φ) = 1. Only φ = π 2 and angles coterminal to it 13 The sign conventions here are carried over from Physics. If not for the spring, the object would fall towards the ground, which is the ‘natural’ or ‘positive’ direction. Since the spring force acts in direct opposition to gravity, any movement upwards is considered ‘negative’. 886 Applications of Trigonometry satisfy this condition, so we pick 14 the phase to be φ = π 2 . Hence, the equation of motion is x(t) = 3 sin _ 2t + π 2 _ . To find when the object passes through the equilibrium position we solve x(t) = 3 sin _ 2t + π 2 _ = 0. Going through the usual analysis we find t = − π 4 + π 2 k for integers k. Since we are interested in the first time the object passes through the equilibrium position, we look for the smallest positive t value which in this case is t = π 4 ≈ 0.78 seconds after the start of the motion. Common sense suggests that if we release the object below the equilibrium position, the object should be traveling upwards when it first passes through it. To check this answer, we graph one cycle of x(t). Since our applied domain in this situation is t ≥ 0, and the period of x(t) is T = 2π ω = 2π 2 = π, we graph x(t) over the interval [0, π]. Remembering that x(t) > 0 means the object is below the equilibrium position and x(t) < 0 means the object is above the equilibrium position, the fact our graph is crossing through the t-axis from positive x to negative x at t = π 4 confirms our answer. 2. The only difference between this problem and the previous problem is that we now release the object with an upward velocity of 8 ft s . We still have ω = 2 and x 0 = 3, but now we have v 0 = −8, the negative indicating the velocity is directed upwards. Here, we get A = _ x 2 0 + _ v 0 ω _ 2 = _ 3 2 + (−4) 2 = 5. From Asin(φ) = x 0 , we get 5 sin(φ) = 3 which gives sin(φ) = 3 5 . From Aω cos(φ) = v 0 , we get 10 cos(φ) = −8, or cos(φ) = − 4 5 . This means that φ is a Quadrant II angle which we can describe in terms of either arcsine or arccosine. Since x(t) is expressed in terms of sine, we choose to express φ = π − arcsin _ 3 5 _ . Hence, x(t) = 5 sin _ 2t + _ π −arcsin _ 3 5 _¸_ . Since the amplitude of x(t) is 5, the object will travel at most 5 feet above the equilibrium position. To find when this happens, we solve the equation x(t) = 5 sin _ 2t + _ π −arcsin _ 3 5 _¸_ = −5, the negative once again signifying that the object is above the equilibrium position. Going through the usual machinations, we get t = 1 2 arcsin _ 3 5 _ + π 4 + πk for integers k. The smallest of these values occurs when k = 0, that is, t = 1 2 arcsin _ 3 5 _ + π 4 ≈ 1.107 seconds after the start of the motion. To check our answer using the calculator, we graph y = 5 sin _ 2x + _ π −arcsin _ 3 5 _¸_ on a graphing utility and confirm the coordinates of the first relative minimum to be approximately (1.107, −5). t x π 4 π 2 3π 4 π −3 −2 −1 1 2 3 x(t) = 3 sin _ 2t + π 2 _ y = 5 sin _ 2x + _ π −arcsin _ 3 5 _¸_ It is possible, though beyond the scope of this course, to model the effects of friction and other external forces acting on the system. 15 While we may not have the Physics and Calculus background 14 For confirmation, we note that Aω cos(φ) = v0, which in this case reduces to 6 cos(φ) = 0. 15 Take a good Differential Equations class to see this! 11.1 Applications of Sinusoids 887 to derive equations of motion for these scenarios, we can certainly analyze them. We examine three cases in the following example. Example 11.1.4. 1. Write x(t) = 5e −t/5 cos(t) + 5e −t/5 √ 3 sin(t) in the form x(t) = A(t) sin(ωt + φ). Graph x(t) using a graphing utility. 2. Write x(t) = (t +3) √ 2 cos(2t) +(t +3) √ 2 sin(2t) in the form x(t) = A(t) sin(ωt +φ). Graph x(t) using a graphing utility. 3. Find the period of x(t) = 5 sin(6t) −5 sin (8t). Graph x(t) using a graphing utility. Solution. 1. We start rewriting x(t) = 5e −t/5 cos(t) + 5e −t/5 √ 3 sin(t) by factoring out 5e −t/5 from both terms to get x(t) = 5e −t/5 _ cos(t) + √ 3 sin(t) _ . We convert what’s left in parentheses to the required form using the formulas introduced in Exercise 36 from Section 10.5. We find _ cos(t) + √ 3 sin(t) _ = 2 sin _ t + π 3 _ so that x(t) = 10e −t/5 sin _ t + π 3 _ . Graphing this on the calculator as y = 10e −x/5 sin _ x + π 3 _ reveals some interesting behavior. The sinusoidal nature continues indefinitely, but it is being attenuated. In the sinusoid Asin(ωx+φ), the coefficient A of the sine function is the amplitude. In the case of y = 10e −x/5 sin _ x + π 3 _ , we can think of the function A(x) = 10e −x/5 as the amplitude. As x →∞, 10e −x/5 →0 which means the amplitude continues to shrink towards zero. Indeed, if we graph y = ±10e −x/5 along with y = 10e −x/5 sin _ x + π 3 _ , we see this attenuation taking place. This equation corresponds to the motion of an object on a spring where there is a slight force which acts to ‘damp’, or slow the motion. An example of this kind of force would be the friction of the object against the air. In this model, the object oscillates forever, but with smaller and smaller amplitude. y = 10e −x/5 sin _ x + π 3 _ y = 10e −x/5 sin _ x + π 3 _ , y = ±10e −x/5 2. Proceeding as in the first example, we factor out (t + 3) √ 2 from each term in the function x(t) = (t +3) √ 2 cos(2t) +(t +3) √ 2 sin(2t) to get x(t) = (t +3) √ 2(cos(2t) +sin(2t)). We find (cos(2t) + sin(2t)) = √ 2 sin _ 2t + π 4 _ , so x(t) = 2(t + 3) sin _ 2t + π 4 _ . Graphing this on the calculator as y = 2(x + 3) sin _ 2x + π 4 _ , we find the sinusoid’s amplitude growing. Since our amplitude function here is A(x) = 2(x+3) = 2x+6, which continues to grow without bound 888 Applications of Trigonometry as x → ∞, this is hardly surprising. The phenomenon illustrated here is ‘forced’ motion. That is, we imagine that the entire apparatus on which the spring is attached is oscillating as well. In this case, we are witnessing a ‘resonance’ effect – the frequency of the external oscillation matches the frequency of the motion of the object on the spring. 16 y = 2(x + 3) sin _ 2x + π 4 _ y = 2(x + 3) sin _ 2x + π 4 _ y = ±2(x + 3) 3. Last, but not least, we come to x(t) = 5 sin(6t)−5 sin(8t). To find the period of this function, we need to determine the length of the smallest interval on which both f(t) = 5 sin(6t) and g(t) = 5 sin(8t) complete a whole number of cycles. To do this, we take the ratio of their frequencies and reduce to lowest terms: 6 8 = 3 4 . This tells us that for every 3 cycles f makes, g makes 4. In other words, the period of x(t) is three times the period of f(t) (which is four times the period of g(t)), or π. We graph y = 5 sin(6x) − 5 sin(8x) over [0, π] on the calculator to check this. This equation of motion also results from ‘forced’ motion, but here the frequency of the external oscillation is different than that of the object on the spring. Since the sinusoids here have different frequencies, they are ‘out of sync’ and do not amplify each other as in the previous example. Taking things a step further, we can use a sum to product identity to rewrite x(t) = 5 sin(6t) −5 sin(8t) as x(t) = −10 sin(t) cos(7t). The lower frequency factor in this expression, −10 sin(t), plays an interesting role in the graph of x(t). Below we graph y = 5 sin(6x) −5 sin(8x) and y = ±10 sin(x) over [0, 2π]. This is an example of the ‘beat’ phenomena, and the curious reader is invited to explore this concept as well. 17 y = 5 sin(6x) −5 sin(8x) over [0, π] y = 5 sin(6x) −5 sin(8x) and y = ±10 sin(x) over [0, 2π] 16 The reader is invited to investigate the destructive implications of resonance. 17 A good place to start is this article on beats. 11.1 Applications of Sinusoids 889 11.1.2 Exercises 1. The sounds we hear are made up of mechanical waves. The note ‘A’ above the note ‘middle C’ is a sound wave with ordinary frequency f = 440 Hertz = 440 cycles second . Find a sinusoid which models this note, assuming that the amplitude is 1 and the phase shift is 0. 2. The voltage V in an alternating current source has amplitude 220 √ 2 and ordinary frequency f = 60 Hertz. Find a sinusoid which models this voltage. Assume that the phase is 0. 3. The London Eye is a popular tourist attraction in London, England and is one of the largest Ferris Wheels in the world. It has a diameter of 135 meters and makes one revolution (counter- clockwise) every 30 minutes. It is constructed so that the lowest part of the Eye reaches ground level, enabling passengers to simply walk on to, and off of, the ride. Find a sinsuoid which models the height h of the passenger above the ground in meters t minutes after they board the Eye at ground level. 4. On page 732 in Section 10.2.1, we found the x-coordinate of counter-clockwise motion on a circle of radius r with angular frequency ω to be x = r cos(ωt), where t = 0 corresponds to the point (r, 0). Suppose we are in the situation of Exercise 3 above. Find a sinsusoid which models the horizontal displacement x of the passenger from the center of the Eye in meters t minutes after they board the Eye. Here we take x(t) > 0 to mean the passenger is to the right of the center, while x(t) < 0 means the passenger is to the left of the center. 5. In Exercise 52 in Section 10.1, we introduced the yo-yo trick ‘Around the World’ in which a yo-yo is thrown so it sweeps out a vertical circle. As in that exercise, suppose the yo-yo string is 28 inches and it completes one revolution in 3 seconds. If the closest the yo-yo ever gets to the ground is 2 inches, find a sinsuoid which models the height h of the yo-yo above the ground in inches t seconds after it leaves its lowest point. 6. Suppose an object weighing 10 pounds is suspended from the ceiling by a spring which stretches 2 feet to its equilibrium position when the object is attached. (a) Find the spring constant k in lbs. ft. and the mass of the object in slugs. (b) Find the equation of motion of the object if it is released from 1 foot below the equilibrium position from rest. When is the first time the object passes through the equilibrium position? In which direction is it heading? (c) Find the equation of motion of the object if it is released from 6 inches above the equilibrium position with a downward velocity of 2 feet per second. Find when the object passes through the equilibrium position heading downwards for the third time. 890 Applications of Trigonometry 7. Consider the pendulum below. Ignoring air resistance, the angular displacement of the pen- dulum from the vertical position, θ, can be modeled as a sinusoid. 18 θ The amplitude of the sinusoid is the same as the initial angular displacement, θ 0 , of the pendulum and the period of the motion is given by T = 2π ¸ l g where l is the length of the pendulum and g is the acceleration due to gravity. (a) Find a sinusoid which gives the angular displacement θ as a function of time, t. Arrange things so θ(0) = θ 0 . (b) In Exercise 40 section 5.3, you found the length of the pendulum needed in Jeff’s antique Seth-Thomas clock to ensure the period of the pendulum is 1 2 of a second. Assuming the initial displacement of the pendulum is 15 ◦ , find a sinusoid which models the displace- ment of the pendulum θ as a function of time, t, in seconds. 8. The table below lists the average temperature of Lake Erie as measured in Cleveland, Ohio on the first of the month for each month during the years 1971 – 2000. 19 For example, t = 3 represents the average of the temperatures recorded for Lake Erie on every March 1 for the years 1971 through 2000. Month Number, t 1 2 3 4 5 6 7 8 9 10 11 12 Temperature ( ◦ F), T 36 33 34 38 47 57 67 74 73 67 56 46 (a) Using the techniques discussed in Example 11.1.2, fit a sinusoid to these data. (b) Using a graphing utility, graph your model along with the data set to judge the reason- ableness of the fit. 18 Provided θ is kept ‘small.’ Carl remembers the ‘Rule of Thumb’ as being 20 ◦ or less. Check with your friendly neighborhood physicist to make sure. 19 See this website: http://www.erh.noaa.gov/cle/climate/cle/normals/laketempcle.html. 11.1 Applications of Sinusoids 891 (c) Use the model you found in part 8a to predict the average temperature recorded for Lake Erie on April 15 th and September 15 th during the years 1971–2000. 20 (d) Compare your results to those obtained using a graphing utility. 9. The fraction of the moon illuminated at midnight Eastern Standard Time on the t th day of June, 2009 is given in the table below. 21 Day of June, t 3 6 9 12 15 18 21 24 27 30 Fraction Illuminated, F 0.81 0.98 0.98 0.83 0.57 0.27 0.04 0.03 0.26 0.58 (a) Using the techniques discussed in Example 11.1.2, fit a sinusoid to these data. 22 (b) Using a graphing utility, graph your model along with the data set to judge the reason- ableness of the fit. (c) Use the model you found in part 9a to predict the fraction of the moon illuminated on June 1, 2009. 23 (d) Compare your results to those obtained using a graphing utility. 10. With the help of your classmates, research the phenomena mentioned in Example 11.1.4, namely resonance and beats. 11. With the help of your classmates, research Amplitude Modulation and Frequency Modulation. 12. What other things in the world might be roughly sinusoidal? Look to see what models you can find for them and share your results with your class. 20 The computed average is 41 ◦ F for April 15 th and 71 ◦ F for September 15 th . 21 See this website: http://www.usno.navy.mil/USNO/astronomical-applications/data-services/frac-moon-ill. 22 You may want to plot the data before you find the phase shift. 23 The listed fraction is 0.62. 892 Applications of Trigonometry 11.1.3 Answers 1. S(t) = sin (880πt) 2. V (t) = 220 √ 2 sin (120πt) 3. h(t) = 67.5 sin _ π 15 t − π 2 _ + 67.5 4. x(t) = 67.5 cos _ π 15 t − π 2 _ = 67.5 sin _ π 15 t _ 5. h(t) = 28 sin _ 2π 3 t − π 2 _ + 30 6. (a) k = 5 lbs. ft. and m = 5 16 slugs (b) x(t) = sin _ 4t + π 2 _ . The object first passes through the equilibrium point when t = π 8 ≈ 0.39 seconds after the motion starts. At this time, the object is heading upwards. (c) x(t) = √ 2 2 sin _ 4t + 7π 4 _ . The object passes through the equilibrium point heading down- wards for the third time when t = 17π 16 ≈ 3.34 seconds. 7. (a) θ(t) = θ 0 sin __ g l t + π 2 _ (b) θ(t) = π 12 sin _ 4πt + π 2 _ 8. (a) T(t) = 20.5 sin _ π 6 t −π _ + 53.5 (b) Our function and the data set are graphed below. The sinusoid seems to be shifted to the right of our data. (c) The average temperature on April 15 th is approximately T(4.5) ≈ 39.00 ◦ F and the average temperature on September 15 th is approximately T(9.5) ≈ 73.38 ◦ F. (d) Using a graphing calculator, we get the following This model predicts the average temperature for April 15 th to be approximately 42.43 ◦ F and the average temperature on September 15 th to be approximately 70.05 ◦ F. This model appears to be more accurate. 11.1 Applications of Sinusoids 893 9. (a) Based on the shape of the data, we either choose A < 0 or we find the second value of t which closely approximates the ‘baseline’ value, F = 0.505. We choose the latter to obtain F(t) = 0.475 sin _ π 15 t −2π _ + 0.505 = 0.475 sin _ π 15 t _ + 0.505 (b) Our function and the data set are graphed below. It’s a pretty good fit. (c) The fraction of the moon illuminated on June 1st, 2009 is approximately F(1) ≈ 0.60 (d) Using a graphing calculator, we get the following. This model predicts that the fraction of the moon illuminated on June 1st, 2009 is approximately 0.59. This appears to be a better fit to the data than our first model. 894 Applications of Trigonometry 11.2 The Law of Sines Trigonometry literally means ‘measuring triangles’ and with Chapter 10 under our belts, we are more than prepared to do just that. The main goal of this section and the next is to develop theorems which allow us to ‘solve’ triangles – that is, find the length of each side of a triangle and the measure of each of its angles. In Sections 10.2, 10.3 and 10.6, we’ve had some experience solving right triangles. The following example reviews what we know. Example 11.2.1. Given a right triangle with a hypotenuse of length 7 units and one leg of length 4 units, find the length of the remaining side and the measures of the remaining angles. Express the angles in decimal degrees, rounded to the nearest hundreth of a degree. Solution. For definitiveness, we label the triangle below. b = 4 a α β c = 7 To find the length of the missing side a, we use the Pythagorean Theorem to get a 2 + 4 2 = 7 2 which then yields a = √ 33 units. Now that all three sides of the triangle are known, there are several ways we can find α using the inverse trigonometric functions. To decrease the chances of propagating error, however, we stick to using the data given to us in the problem. In this case, the lengths 4 and 7 were given, so we want to relate these to α. According to Theorem 10.4, cos(α) = 4 7 . Since α is an acute angle, α = arccos _ 4 7 _ radians. Converting to degrees, we find α ≈ 55.15 ◦ . Now that we have the measure of angle α, we could find the measure of angle β using the fact that α and β are complements so α + β = 90 ◦ . Once again, we opt to use the data given to us in the problem. According to Theorem 10.4, we have that sin(β) = 4 7 so β = arcsin _ 4 7 _ radians and we have β ≈ 34.85 ◦ . A few remarks about Example 11.2.1 are in order. First, we adhere to the convention that a lower case Greek letter denotes an angle 1 and the corresponding lowercase English letter represents the side 2 opposite that angle. Thus, a is the side opposite α, b is the side opposite β and c is the side opposite γ. Taken together, the pairs (α, a), (β, b) and (γ, c) are called angle-side opposite pairs. Second, as mentioned earlier, we will strive to solve for quantities using the original data given in the problem whenever possible. While this is not always the easiest or fastest way to proceed, it 1 as well as the measure of said angle 2 as well as the length of said side 11.2 The Law of Sines 895 minimizes the chances of propagated error. 3 Third, since many of the applications which require solving triangles ‘in the wild’ rely on degree measure, we shall adopt this convention for the time being. 4 The Pythagorean Theorem along with Theorems 10.4 and 10.10 allow us to easily handle any given right triangle problem, but what if the triangle isn’t a right triangle? In certain cases, we can use the Law of Sines to help. Theorem 11.2. The Law of Sines: Given a triangle with angle-side opposite pairs (α, a), (β, b) and (γ, c), the following ratios hold sin(α) a = sin(β) b = sin(γ) c or, equivalently, a sin(α) = b sin(β) = c sin(γ) The proof of the Law of Sines can be broken into three cases. For our first case, consider the triangle ´ABC below, all of whose angles are acute, with angle-side opposite pairs (α, a), (β, b) and (γ, c). If we drop an altitude from vertex B, we divide the triangle into two right triangles: ´ABQ and ´BCQ. If we call the length of the altitude h (for height), we get from Theorem 10.4 that sin(α) = h c and sin(γ) = h a so that h = c sin(α) = a sin(γ). After some rearrangement of the last equation, we get sin(α) a = sin(γ) c . If we drop an altitude from vertex A, we can proceed as above using the triangles ´ABQ and ´ACQ to get sin(β) b = sin(γ) c , completing the proof for this case. a b c α β γ A C B a c α γ A C B Q h b c β γ A C B Q h For our next case consider the triangle ´ABC below with obtuse angle α. Extending an altitude from vertex A gives two right triangles, as in the previous case: ´ABQ and ´ACQ. Proceeding as before, we get h = b sin(γ) and h = c sin(β) so that sin(β) b = sin(γ) c . a b c α γ β A B C a b c γ β A B C Q h 3 Your Science teachers should thank us for this. 4 Don’t worry! Radians will be back before you know it! 896 Applications of Trigonometry Dropping an altitude from vertex B also generates two right triangles, ´ABQ and ´BCQ. We know that sin(α ) = h c so that h = c sin(α ). Since α = 180 ◦ − α, sin(α ) = sin(α), so in fact, we have h = c sin(α). Proceeding to ´BCQ, we get sin(γ) = h a so h = a sin(γ). Putting this together with the previous equation, we get sin(γ) c = sin(α) a , and we are finished with this case. a b c α α γ β A B C Q h The remaining case is when ´ABC is a right triangle. In this case, the Law of Sines reduces to the formulas given in Theorem 10.4 and is left to the reader. In order to use the Law of Sines to solve a triangle, we need at least one angle-side opposite pair. The next example showcases some of the power, and the pitfalls, of the Law of Sines. Example 11.2.2. Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. 1. α = 120 ◦ , a = 7 units, β = 45 ◦ 2. α = 85 ◦ , β = 30 ◦ , c = 5.25 units 3. α = 30 ◦ , a = 1 units, c = 4 units 4. α = 30 ◦ , a = 2 units, c = 4 units 5. α = 30 ◦ , a = 3 units, c = 4 units 6. α = 30 ◦ , a = 4 units, c = 4 units Solution. 1. Knowing an angle-side opposite pair, namely α and a, we may proceed in using the Law of Sines. Since β = 45 ◦ , we use b sin(45 ◦ ) = 7 sin(120 ◦ ) so b = 7 sin(45 ◦ ) sin(120 ◦ ) = 7 √ 6 3 ≈ 5.72 units. Now that we have two angle-side pairs, it is time to find the third. To find γ, we use the fact that the sum of the measures of the angles in a triangle is 180 ◦ . Hence, γ = 180 ◦ −120 ◦ −45 ◦ = 15 ◦ . To find c, we have no choice but to used the derived value γ = 15 ◦ , yet we can minimize the propagation of error here by using the given angle-side opposite pair (α, a). The Law of Sines gives us c sin(15 ◦ ) = 7 sin(120 ◦ ) so that c = 7 sin(15 ◦ ) sin(120 ◦ ) ≈ 2.09 units. 5 2. In this example, we are not immediately given an angle-side opposite pair, but as we have the measures of α and β, we can solve for γ since γ = 180 ◦ − 85 ◦ − 30 ◦ = 65 ◦ . As in the previous example, we are forced to use a derived value in our computations since the only 5 The exact value of sin(15 ◦ ) could be found using the difference identity for sine or a half-angle formula, but that becomes unnecessarily messy for the discussion at hand. Thus “exact” here means 7 sin(15 ◦ ) sin(120 ◦ ) . 11.2 The Law of Sines 897 angle-side pair available is (γ, c). The Law of Sines gives a sin(85 ◦ ) = 5.25 sin(65 ◦ ) . After the usual rearrangement, we get a = 5.25 sin(85 ◦ ) sin(65 ◦ ) ≈ 5.77 units. To find b we use the angle-side pair (γ, c) which yields b sin(30 ◦ ) = 5.25 sin(65 ◦ ) hence b = 5.25 sin(30 ◦ ) sin(65 ◦ ) ≈ 2.90 units. a = 7 b ≈ 5.72 c ≈ 2.09 α = 120 ◦ γ = 15 ◦ β = 45 ◦ a ≈ 5.77 b ≈ 2.90 c = 5.25 α = 85 ◦ γ = 65 ◦ β = 30 ◦ Triangle for number 1 Triangle for number 2 3. Since we are given (α, a) and c, we use the Law of Sines to find the measure of γ. We start with sin(γ) 4 = sin(30 ◦ ) 1 and get sin(γ) = 4 sin (30 ◦ ) = 2. Since the range of the sine function is [−1, 1], there is no real number with sin(γ) = 2. Geometrically, we see that side a is just too short to make a triangle. The next three examples keep the same values for the measure of α and the length of c while varying the length of a. We will discuss this case in more detail after we see what happens in those examples. 4. In this case, we have the measure of α = 30 ◦ , a = 2 and c = 4. Using the Law of Sines, we get sin(γ) 4 = sin(30 ◦ ) 2 so sin(γ) = 2 sin (30 ◦ ) = 1. Now γ is an angle in a triangle which also contains α = 30 ◦ . This means that γ must measure between 0 ◦ and 150 ◦ in order to fit inside the triangle with α. The only angle that satisfies this requirement and has sin(γ) = 1 is γ = 90 ◦ . In other words, we have a right triangle. We find the measure of β to be β = 180 ◦ − 30 ◦ − 90 ◦ = 60 ◦ and then determine b using the Law of Sines. We find b = 2 sin(60 ◦ ) sin(30 ◦ ) = 2 √ 3 ≈ 3.46 units. In this case, the side a is precisely long enough to form a unique right triangle. a = 1 c = 4 α = 30 ◦ a = 2 c = 4 b ≈ 3.46 α = 30 ◦ β = 60 ◦ Diagram for number 3 Triangle for number 4 5. Proceeding as we have in the previous two examples, we use the Law of Sines to find γ. In this case, we have sin(γ) 4 = sin(30 ◦ ) 3 or sin(γ) = 4 sin(30 ◦ ) 3 = 2 3 . Since γ lies in a triangle with α = 30 ◦ , 898 Applications of Trigonometry we must have that 0 ◦ < γ < 150 ◦ . There are two angles γ that fall in this range and have sin(γ) = 2 3 : γ = arcsin _ 2 3 _ radians ≈ 41.81 ◦ and γ = π − arcsin _ 2 3 _ radians ≈ 138.19 ◦ . At this point, we pause to see if it makes sense that we actually have two viable cases to consider. As we have discussed, both candidates for γ are ‘compatible’ with the given angle-side pair (α, a) = (30 ◦ , 3) in that both choices for γ can fit in a triangle with α and both have a sine of 2 3 . The only other given piece of information is that c = 4 units. Since c > a, it must be true that γ, which is opposite c, has greater measure than α which is opposite a. In both cases, γ > α, so both candidates for γ are compatible with this last piece of given information as well. Thus have two triangles on our hands. In the case γ = arcsin _ 2 3 _ radians ≈ 41.81 ◦ , we find 6 β ≈ 180 ◦ −30 ◦ −41.81 ◦ = 108.19 ◦ . Using the Law of Sines with the angle-side opposite pair (α, a) and β, we find b ≈ 3 sin(108.19 ◦ ) sin(30 ◦ ) ≈ 5.70 units. In the case γ = π −arcsin _ 2 3 _ radians ≈ 138.19 ◦ , we repeat the exact same steps and find β ≈ 11.81 ◦ and b ≈ 1.23 units. 7 Both triangles are drawn below. a = 3 c = 4 α = 30 ◦ β ≈ 108.19 ◦ γ ≈ 41.81 ◦ b ≈ 5.70 a = 3 c = 4 α = 30 ◦ β ≈ 11.81 ◦ γ ≈ 138.19 ◦ b ≈ 1.23 6. For this last problem, we repeat the usual Law of Sines routine to find that sin(γ) 4 = sin(30 ◦ ) 4 so that sin(γ) = 1 2 . Since γ must inhabit a triangle with α = 30 ◦ , we must have 0 ◦ < γ < 150 ◦ . Since the measure of γ must be strictly less than 150 ◦ , there is just one angle which satisfies both required conditions, namely γ = 30 ◦ . So β = 180 ◦ − 30 ◦ − 30 ◦ = 120 ◦ and, using the Law of Sines one last time, b = 4 sin(120 ◦ ) sin(30 ◦ ) = 4 √ 3 ≈ 6.93 units. a = 4 c = 4 α = 30 ◦ β = 120 ◦ γ = 30 ◦ b ≈ 6.93 Some remarks about Example 11.2.2 are in order. We first note that if we are given the measures of two of the angles in a triangle, say α and β, the measure of the third angle γ is uniquely 6 To find an exact expression for β, we convert everything back to radians: α = 30 ◦ = π 6 radians, γ = arcsin 2 3 radians and 180 ◦ = π radians. Hence, β = π − π 6 −arcsin 2 3 = 5π 6 −arcsin 2 3 radians ≈ 108.19 ◦ . 7 An exact answer for β in this case is β = arcsin 2 3 − π 6 radians ≈ 11.81 ◦ . 11.2 The Law of Sines 899 determined using the equation γ = 180 ◦ − α − β. Knowing the measures of all three angles of a triangle completely determines its shape. If in addition we are given the length of one of the sides of the triangle, we can then use the Law of Sines to find the lengths of the remaining two sides to determine the size of the triangle. Such is the case in numbers 1 and 2 above. In number 1, the given side is adjacent to just one of the angles – this is called the ‘Angle-Angle-Side’ (AAS) case. 8 In number 2, the given side is adjacent to both angles which means we are in the so-called ‘Angle-Side-Angle’ (ASA) case. If, on the other hand, we are given the measure of just one of the angles in the triangle along with the length of two sides, only one of which is adjacent to the given angle, we are in the ‘Angle-Side-Side’ (ASS) case. 9 In number 3, the length of the one given side a was too short to even form a triangle; in number 4, the length of a was just long enough to form a right triangle; in 5, a was long enough, but not too long, so that two triangles were possible; and in number 6, side a was long enough to form a triangle but too long to swing back and form two. These four cases exemplify all of the possibilities in the Angle-Side-Side case which are summarized in the following theorem. Theorem 11.3. Suppose (α, a) and (γ, c) are intended to be angle-side pairs in a triangle where α, a and c are given. Let h = c sin(α) • If a < h, then no triangle exists which satisfies the given criteria. • If a = h, then γ = 90 ◦ so exactly one (right) triangle exists which satisfies the criteria. • If h < a < c, then two distinct triangles exist which satisfy the given criteria. • If a ≥ c, then γ is acute and exactly one triangle exists which satisfies the given criteria Theorem 11.3 is proved on a case-by-case basis. If a < h, then a < c sin(α). If a triangle were to exist, the Law of Sines would have sin(γ) c = sin(α) a so that sin(γ) = c sin(α) a > a a = 1, which is impossible. In the figure below, we see geometrically why this is the case. c α a h = c sin(α) c α a = h = c sin(α) a < h, no triangle a = h, γ = 90 ◦ Simply put, if a < h the side a is too short to connect to form a triangle. This means if a ≥ h, we are always guaranteed to have at least one triangle, and the remaining parts of the theorem 8 If this sounds familiar, it should. From high school Geometry, we know there are four congruence conditions for triangles: Angle-Angle-Side (AAS), Angle-Side-Angle (ASA), Side-Angle-Side (SAS) and Side-Side-Side (SSS). If we are given information about a triangle that meets one of these four criteria, then we are guaranteed that exactly one triangle exists which satisfies the given criteria. 9 In more reputable books, this is called the ‘Side-Side-Angle’ or SSA case. 900 Applications of Trigonometry tell us what kind and how many triangles to expect in each case. If a = h, then a = c sin(α) and the Law of Sines gives sin(α) a = sin(γ) c so that sin(γ) = c sin(α) a = a a = 1. Here, γ = 90 ◦ as required. Moving along, now suppose h < a < c. As before, the Law of Sines 10 gives sin(γ) = c sin(α) a . Since h < a, c sin(α) < a or c sin(α) a < 1 which means there are two solutions to sin(γ) = c sin(α) a : an acute angle which we’ll call γ 0 , and its supplement, 180 ◦ − γ 0 . We need to argue that each of these angles ‘fit’ into a triangle with α. Since (α, a) and (γ 0 , c) are angle-side opposite pairs, the assumption c > a in this case gives us γ 0 > α. Since γ 0 is acute, we must have that α is acute as well. This means one triangle can contain both α and γ 0 , giving us one of the triangles promised in the theorem. If we manipulate the inequality γ 0 > α a bit, we have 180 ◦ −γ 0 < 180 ◦ −α which gives (180 ◦ −γ 0 ) + α < 180 ◦ . This proves a triangle can contain both of the angles α and (180 ◦ −γ 0 ), giving us the second triangle predicted in the theorem. To prove the last case in the theorem, we assume a ≥ c. Then α ≥ γ, which forces γ to be an acute angle. Hence, we get only one triangle in this case, completing the proof. a a c h α γ 0 γ 0 h a c α γ h < a < c, two triangles a ≥ c, one triangle One last comment before we use the Law of Sines to solve an application problem. In the Angle- Side-Side case, if you are given an obtuse angle to begin with then it is impossible to have the two triangle case. Think about this before reading further. Example 11.2.3. Sasquatch Island lies off the coast of Ippizuti Lake. Two sightings, taken 5 miles apart, are made to the island. The angle between the shore and the island at the first observation point is 30 ◦ and at the second point the angle is 45 ◦ . Assuming a straight coastline, find the distance from the second observation point to the island. What point on the shore is closest to the island? How far is the island from this point? Solution. We sketch the problem below with the first observation point labeled as P and the second as Q. In order to use the Law of Sines to find the distance d from Q to the island, we first need to find the measure of β which is the angle opposite the side of length 5 miles. To that end, we note that the angles γ and 45 ◦ are supplemental, so that γ = 180 ◦ − 45 ◦ = 135 ◦ . We can now find β = 180 ◦ −30 ◦ −γ = 180 ◦ −30 ◦ −135 ◦ = 15 ◦ . By the Law of Sines, we have d sin(30 ◦ ) = 5 sin(15 ◦ ) which gives d = 5 sin(30 ◦ ) sin(15 ◦ ) ≈ 9.66 miles. Next, to find the point on the coast closest to the island, which we’ve labeled as C, we need to find the perpendicular distance from the island to the coast. 11 10 Remember, we have already argued that a triangle exists in this case! 11 Do you see why C must lie to the right of Q? 11.2 The Law of Sines 901 Let x denote the distance from the second observation point Q to the point C and let y denote the distance from C to the island. Using Theorem 10.4, we get sin (45 ◦ ) = y d . After some rearranging, we find y = d sin (45 ◦ ) ≈ 9.66 _ √ 2 2 _ ≈ 6.83 miles. Hence, the island is approximately 6.83 miles from the coast. To find the distance from Q to C, we note that β = 180 ◦ − 90 ◦ − 45 ◦ = 45 ◦ so by symmetry, 12 we get x = y ≈ 6.83 miles. Hence, the point on the shore closest to the island is approximately 6.83 miles down the coast from the second observation point. 45 ◦ 30 ◦ γ β P Q 5 miles d ≈ 9.66 miles Shoreline Sasquatch Island 45 ◦ β d ≈ 9.66 miles x miles y miles Q C Sasquatch Island We close this section with a new formula to compute the area enclosed by a triangle. Its proof uses the same cases and diagrams as the proof of the Law of Sines and is left as an exercise. Theorem 11.4. Suppose (α, a), (β, b) and (γ, c) are the angle-side opposite pairs of a triangle. Then the area A enclosed by the triangle is given by A = 1 2 bc sin(α) = 1 2 ac sin(β) = 1 2 ab sin(γ) Example 11.2.4. Find the area of the triangle in Example 11.2.2 number 1. Solution. From our work in Example 11.2.2 number 1, we have all three angles and all three sides to work with. However, to minimize propagated error, we choose A = 1 2 ac sin(β) from Theorem 11.4 because it uses the most pieces of given information. We are given a = 7 and β = 45 ◦ , and we calculated c = 7 sin(15 ◦ ) sin(120 ◦ ) . Using these values, we find A = 1 2 (7) _ 7 sin(15 ◦ ) sin(120 ◦ ) _ sin (45 ◦ ) =≈ 5.18 square units. The reader is encouraged to check this answer against the results obtained using the other formulas in Theorem 11.4. 12 Or by Theorem 10.4 again . . . 902 Applications of Trigonometry 11.2.1 Exercises In Exercises 1 - 20, solve for the remaining side(s) and angle(s) if possible. As in the text, (α, a), (β, b) and (γ, c) are angle-side opposite pairs. 1. α = 13 ◦ , β = 17 ◦ , a = 5 2. α = 73.2 ◦ , β = 54.1 ◦ , a = 117 3. α = 95 ◦ , β = 85 ◦ , a = 33.33 4. α = 95 ◦ , β = 62 ◦ , a = 33.33 5. α = 117 ◦ , a = 35, b = 42 6. α = 117 ◦ , a = 45, b = 42 7. α = 68.7 ◦ , a = 88, b = 92 8. α = 42 ◦ , a = 17, b = 23.5 9. α = 68.7 ◦ , a = 70, b = 90 10. α = 30 ◦ , a = 7, b = 14 11. α = 42 ◦ , a = 39, b = 23.5 12. γ = 53 ◦ , α = 53 ◦ , c = 28.01 13. α = 6 ◦ , a = 57, b = 100 14. γ = 74.6 ◦ , c = 3, a = 3.05 15. β = 102 ◦ , b = 16.75, c = 13 16. β = 102 ◦ , b = 16.75, c = 18 17. β = 102 ◦ , γ = 35 ◦ , b = 16.75 18. β = 29.13 ◦ , γ = 83.95 ◦ , b = 314.15 19. γ = 120 ◦ , β = 61 ◦ , c = 4 20. α = 50 ◦ , a = 25, b = 12.5 21. Find the area of the triangles given in Exercises 1, 12 and 20 above. (Another Classic Application: Grade of a Road) The grade of a road is much like the pitch of a roof (See Example 10.6.6) in that it expresses the ratio of rise/run. In the case of a road, this ratio is always positive because it is measured going uphill and it is usually given as a percentage. For example, a road which rises 7 feet for every 100 feet of (horizontal) forward progress is said to have a 7% grade. However, if we want to apply any Trigonometry to a story problem involving roads going uphill or downhill, we need to view the grade as an angle with respect to the horizontal. In Exercises 22 - 24, we first have you change road grades into angles and then use the Law of Sines in an application. 22. Using a right triangle with a horizontal leg of length 100 and vertical leg with length 7, show that a 7% grade means that the road (hypotenuse) makes about a 4 ◦ angle with the horizontal. (It will not be exactly 4 ◦ , but it’s pretty close.) 23. What grade is given by a 9.65 ◦ angle made by the road and the horizontal? 13 13 I have friends who live in Pacifica, CA and their road is actually this steep. It’s not a nice road to drive. 11.2 The Law of Sines 903 24. Along a long, straight stretch of mountain road with a 7% grade, you see a tall tree standing perfectly plumb alongside the road. 14 From a point 500 feet downhill from the tree, the angle of inclination from the road to the top of the tree is 6 ◦ . Use the Law of Sines to find the height of the tree. (Hint: First show that the tree makes a 94 ◦ angle with the road.) (Another Classic Application: Bearings) In the next several exercises we introduce and work with the navigation tool known as bearings. Simply put, a bearing is the direction you are heading according to a compass. The classic nomenclature for bearings, however, is not given as an angle in standard position, so we must first understand the notation. A bearing is given as an acute angle of rotation (to the east or to the west) away from the north-south (up and down) line of a compass rose. For example, N40 ◦ E (read “40 ◦ east of north”) is a bearing which is rotated clockwise 40 ◦ from due north. If we imagine standing at the origin in the Cartesian Plane, this bearing would have us heading into Quadrant I along the terminal side of θ = 50 ◦ . Similarly, S50 ◦ W would point into Quadrant III along the terminal side of θ = 220 ◦ because we started out pointing due south (along θ = 270 ◦ ) and rotated clockwise 50 ◦ back to 220 ◦ . Counter-clockwise rotations would be found in the bearings N60 ◦ W (which is on the terminal side of θ = 150 ◦ ) and S27 ◦ E (which lies along the terminal side of θ = 297 ◦ ). These four bearings are drawn in the plane below. N E S W N40 ◦ E 40 ◦ N60 ◦ W 60 ◦ S50 ◦ W 50 ◦ S27 ◦ E 27 ◦ The cardinal directions north, south, east and west are usually not given as bearings in the fashion described above, but rather, one just refers to them as ‘due north’, ‘due south’, ‘due east’ and ‘due west’, respectively, and it is assumed that you know which quadrantal angle goes with each cardinal direction. (Hint: Look at the diagram above.) 25. Find the angle θ in standard position with 0 ◦ ≤ θ < 360 ◦ which corresponds to each of the bearings given below. (a) due west (b) S83 ◦ E (c) N5.5 ◦ E (d) due south 14 The word ‘plumb’ here means that the tree is perpendicular to the horizontal. 904 Applications of Trigonometry (e) N31.25 ◦ W (f) S72 ◦ 41 12 W 15 (g) N45 ◦ E (h) S45 ◦ W 26. The Colonel spots a campfire at a of bearing N42 ◦ E from his current position. Sarge, who is positioned 3000 feet due east of the Colonel, reckons the bearing to the fire to be N20 ◦ W from his current position. Determine the distance from the campfire to each man, rounded to the nearest foot. 27. A hiker starts walking due west from Sasquatch Point and gets to the Chupacabra Trailhead before she realizes that she hasn’t reset her pedometer. From the Chupacabra Trailhead she hikes for 5 miles along a bearing of N53 ◦ W which brings her to the Muffin Ridge Observatory. From there, she knows a bearing of S65 ◦ E will take her straight back to Sasquatch Point. How far will she have to walk to get from the Muffin Ridge Observatory to Sasquach Point? What is the distance between Sasquatch Point and the Chupacabra Trailhead? 28. The captain of the SS Bigfoot sees a signal flare at a bearing of N15 ◦ E from her current location. From his position, the captain of the HMS Sasquatch finds the signal flare to be at a bearing of N75 ◦ W. If the SS Bigfoot is 5 miles from the HMS Sasquatch and the bearing from the SS Bigfoot to the HMS Sasquatch is N50 ◦ E, find the distances from the flare to each vessel, rounded to the nearest tenth of a mile. 29. Carl spies a potential Sasquatch nest at a bearing of N10 ◦ E and radios Jeff, who is at a bearing of N50 ◦ E from Carl’s position. From Jeff’s position, the nest is at a bearing of S70 ◦ W. If Jeff and Carl are 500 feet apart, how far is Jeff from the Sasquatch nest? Round your answer to the nearest foot. 30. A hiker determines the bearing to a lodge from her current position is S40 ◦ W. She proceeds to hike 2 miles at a bearing of S20 ◦ E at which point she determines the bearing to the lodge is S75 ◦ W. How far is she from the lodge at this point? Round your answer to the nearest hundredth of a mile. 31. A watchtower spots a ship off shore at a bearing of N70 ◦ E. A second tower, which is 50 miles from the first at a bearing of S80 ◦ E from the first tower, determines the bearing to the ship to be N25 ◦ W. How far is the boat from the second tower? Round your answer to the nearest tenth of a mile. 32. Skippy and Sally decide to hunt UFOs. One night, they position themselves 2 miles apart on an abandoned stretch of desert runway. An hour into their investigation, Skippy spies a UFO hovering over a spot on the runway directly between him and Sally. He records the angle of inclination from the ground to the craft to be 75 ◦ and radios Sally immediately to find the angle of inclination from her position to the craft is 50 ◦ . How high off the ground is the UFO at this point? Round your answer to the nearest foot. (Recall: 1 mile is 5280 feet.) 15 See Example 10.1.1 in Section 10.1 for a review of the DMS system. 11.2 The Law of Sines 905 33. The angle of depression from an observer in an apartment complex to a gargoyle on the building next door is 55 ◦ . From a point five stories below the original observer, the angle of inclination to the gargoyle is 20 ◦ . Find the distance from each observer to the gargoyle and the distance from the gargoyle to the apartment complex. Round your answers to the nearest foot. (Use the rule of thumb that one story of a building is 9 feet.) 34. Prove that the Law of Sines holds when ´ABC is a right triangle. 35. Discuss with your classmates why knowing only the three angles of a triangle is not enough to determine any of the sides. 36. Discuss with your classmates why the Law of Sines cannot be used to find the angles in the triangle when only the three sides are given. Also discuss what happens if only two sides and the angle between them are given. (Said another way, explain why the Law of Sines cannot be used in the SSS and SAS cases.) 37. Given α = 30 ◦ and b = 10, choose four different values for a so that (a) the information yields no triangle (b) the information yields exactly one right triangle (c) the information yields two distinct triangles (d) the information yields exactly one obtuse triangle Explain why you cannot choose a in such a way as to have α = 30 ◦ , b = 10 and your choice of a yield only one triangle where that unique triangle has three acute angles. 38. Use the cases and diagrams in the proof of the Law of Sines (Theorem 11.2) to prove the area formulas given in Theorem 11.4. Why do those formulas yield square units when four quantities are being multiplied together? 906 Applications of Trigonometry 11.2.2 Answers 1. α = 13 ◦ β = 17 ◦ γ = 150 ◦ a = 5 b ≈ 6.50 c ≈ 11.11 2. α = 73.2 ◦ β = 54.1 ◦ γ = 52.7 ◦ a = 117 b ≈ 99.00 c ≈ 97.22 3. Information does not produce a triangle 4. α = 95 ◦ β = 62 ◦ γ = 23 ◦ a = 33.33 b ≈ 29.54 c ≈ 13.07 5. Information does not produce a triangle 6. α = 117 ◦ β ≈ 56.3 ◦ γ ≈ 6.7 ◦ a = 45 b = 42 c ≈ 5.89 7. α = 68.7 ◦ β ≈ 76.9 ◦ γ ≈ 34.4 ◦ a = 88 b = 92 c ≈ 53.36 α = 68.7 ◦ β ≈ 103.1 ◦ γ ≈ 8.2 ◦ a = 88 b = 92 c ≈ 13.47 8. α = 42 ◦ β ≈ 67.66 ◦ γ ≈ 70.34 ◦ a = 17 b = 23.5 c ≈ 23.93 α = 42 ◦ β ≈ 112.34 ◦ γ ≈ 25.66 ◦ a = 17 b = 23.5 c ≈ 11.00 9. Information does not produce a triangle 10. α = 30 ◦ β = 90 ◦ γ = 60 ◦ a = 7 b = 14 c = 7 √ 3 11. α = 42 ◦ β ≈ 23.78 ◦ γ ≈ 114.22 ◦ a = 39 b = 23.5 c ≈ 53.15 12. α = 53 ◦ β = 74 ◦ γ = 53 ◦ a = 28.01 b ≈ 33.71 c = 28.01 13. α = 6 ◦ β ≈ 169.43 ◦ γ ≈ 4.57 ◦ a = 57 b = 100 c ≈ 43.45 α = 6 ◦ β ≈ 10.57 ◦ γ ≈ 163.43 ◦ a = 57 b = 100 c ≈ 155.51 14. α ≈ 78.59 ◦ β ≈ 26.81 ◦ γ = 74.6 ◦ a = 3.05 b ≈ 1.40 c = 3 α ≈ 101.41 ◦ β ≈ 3.99 ◦ γ = 74.6 ◦ a = 3.05 b ≈ 0.217 c = 3 15. α ≈ 28.61 ◦ β = 102 ◦ γ ≈ 49.39 ◦ a ≈ 8.20 b = 16.75 c = 13 16. Information does not produce a triangle 17. α = 43 ◦ β = 102 ◦ γ = 35 ◦ a ≈ 11.68 b = 16.75 c ≈ 9.82 18. α = 66.92 ◦ β = 29.13 ◦ γ = 83.95 ◦ a ≈ 593.69 b = 314.15 c ≈ 641.75 19. Information does not produce a triangle 20. α = 50 ◦ β ≈ 22.52 ◦ γ ≈ 107.48 ◦ a = 25 b = 12.5 c ≈ 31.13 21. The area of the triangle from Exercise 1 is about 8.1 square units. The area of the triangle from Exercise 12 is about 377.1 square units. The area of the triangle from Exercise 20 is about 149 square units. 22. arctan _ 7 100 _ ≈ 0.699 radians, which is equivalent to 4.004 ◦ 23. About 17% 24. About 53 feet 11.2 The Law of Sines 907 25. (a) θ = 180 ◦ (b) θ = 353 ◦ (c) θ = 84.5 ◦ (d) θ = 270 ◦ (e) θ = 121.25 ◦ (f) θ = 197 ◦ 18 48 (g) θ = 45 ◦ (h) θ = 225 ◦ 26. The Colonel is about 3193 feet from the campfire. Sarge is about 2525 feet to the campfire. 27. The distance from the Muffin Ridge Observatory to Sasquach Point is about 7.12 miles. The distance from Sasquatch Point to the Chupacabra Trailhead is about 2.46 miles. 28. The SS Bigfoot is about 4.1 miles from the flare. The HMS Sasquatch is about 2.9 miles from the flare. 29. Jeff is about 342 feet from the nest. 30. She is about 3.02 miles from the lodge 31. The boat is about 25.1 miles from the second tower. 32. The UFO is hovering about 9539 feet above the ground. 33. The gargoyle is about 44 feet from the observer on the upper floor. The gargoyle is about 27 feet from the observer on the lower floor. The gargoyle is about 25 feet from the other building. 908 Applications of Trigonometry 11.3 The Law of Cosines In Section 11.2, we developed the Law of Sines (Theorem 11.2) to enable us to solve triangles in the ‘Angle-Angle-Side’ (AAS), the ‘Angle-Side-Angle’ (ASA) and the ambiguous ‘Angle-Side-Side’ (ASS) cases. In this section, we develop the Law of Cosines which handles solving triangles in the ‘Side-Angle-Side’ (SAS) and ‘Side-Side-Side’ (SSS) cases. 1 We state and prove the theorem below. Theorem 11.5. Law of Cosines: Given a triangle with angle-side opposite pairs (α, a), (β, b) and (γ, c), the following equations hold a 2 = b 2 +c 2 −2bc cos(α) b 2 = a 2 +c 2 −2ac cos(β) c 2 = a 2 +b 2 −2ab cos(γ) or, solving for the cosine in each equation, we have cos(α) = b 2 +c 2 −a 2 2bc cos(β) = a 2 +c 2 −b 2 2ac cos(γ) = a 2 +b 2 −c 2 2ab To prove the theorem, we consider a generic triangle with the vertex of angle α at the origin with side b positioned along the positive x-axis. a b c α A = (0, 0) C = (b, 0) B = (c cos(α), c sin(α)) From this set-up, we immediately find that the coordinates of A and C are A(0, 0) and C(b, 0). From Theorem 10.3, we know that since the point B(x, y) lies on a circle of radius c, the coordinates 1 Here, ‘Side-Angle-Side’ means that we are given two sides and the ‘included’ angle - that is, the given angle is adjacent to both of the given sides. 11.3 The Law of Cosines 909 of B are B(x, y) = B(c cos(α), c sin(α)). (This would be true even if α were an obtuse or right angle so although we have drawn the case when α is acute, the following computations hold for any angle α drawn in standard position where 0 < α < 180 ◦ .) We note that the distance between the points B and C is none other than the length of side a. Using the distance formula, Equation 1.1, we get a = _ (c cos(α) −b) 2 + (c sin(α) −0) 2 a 2 = _ _ (c cos(α) −b) 2 +c 2 sin 2 (α) _ 2 a 2 = (c cos(α) −b) 2 +c 2 sin 2 (α) a 2 = c 2 cos 2 (α) −2bc cos(α) +b 2 +c 2 sin 2 (α) a 2 = c 2 _ cos 2 (α) + sin 2 (α) _ +b 2 −2bc cos(α) a 2 = c 2 (1) +b 2 −2bc cos(α) Since cos 2 (α) + sin 2 (α) = 1 a 2 = c 2 +b 2 −2bc cos(α) The remaining formulas given in Theorem 11.5 can be shown by simply reorienting the triangle to place a different vertex at the origin. We leave these details to the reader. What’s important about a and α in the above proof is that (α, a) is an angle-side opposite pair and b and c are the sides adjacent to α – the same can be said of any other angle-side opposite pair in the triangle. Notice that the proof of the Law of Cosines relies on the distance formula which has its roots in the Pythagorean Theorem. That being said, the Law of Cosines can be thought of as a generalization of the Pythagorean Theorem. If we have a triangle in which γ = 90 ◦ , then cos(γ) = cos (90 ◦ ) = 0 so we get the familiar relationship c 2 = a 2 +b 2 . What this means is that in the larger mathematical sense, the Law of Cosines and the Pythagorean Theorem amount to pretty much the same thing. 2 Example 11.3.1. Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. 1. β = 50 ◦ , a = 7 units, c = 2 units 2. a = 4 units, b = 7 units, c = 5 units Solution. 1. We are given the lengths of two sides, a = 7 and c = 2, and the measure of the included angle, β = 50 ◦ . With no angle-side opposite pair to use, we apply the Law of Cosines. We get b 2 = 7 2 + 2 2 − 2(7)(2) cos (50 ◦ ) which yields b = _ 53 −28 cos (50 ◦ ) ≈ 5.92 units. In order to determine the measures of the remaining angles α and γ, we are forced to used the derived value for b. There are two ways to proceed at this point. We could use the Law of Cosines again, or, since we have the angle-side opposite pair (β, b) we could use the Law of Sines. The advantage to using the Law of Cosines over the Law of Sines in cases like this is that unlike the sine function, the cosine function distinguishes between acute and obtuse angles. The cosine of an acute is positive, whereas the cosine of an obtuse angle is negative. Since the sine of both acute and obtuse angles are positive, the sine of an angle alone is not 2 This shouldn’t come as too much of a shock. All of the theorems in Trigonometry can ultimately be traced back to the definition of the circular functions along with the distance formula and hence, the Pythagorean Theorem. 910 Applications of Trigonometry enough to determine if the angle in question is acute or obtuse. Since both authors of the textbook prefer the Law of Cosines, we proceed with this method first. When using the Law of Cosines, it’s always best to find the measure of the largest unknown angle first, since this will give us the obtuse angle of the triangle if there is one. Since the largest angle is opposite the longest side, we choose to find α first. To that end, we use the formula cos(α) = b 2 +c 2 −a 2 2bc and substitute a = 7, b = _ 53 −28 cos (50 ◦ ) and c = 2. We get 3 cos(α) = 2 −7 cos (50 ◦ ) _ 53 −28 cos (50 ◦ ) Since α is an angle in a triangle, we know the radian measure of α must lie between 0 and π radians. This matches the range of the arccosine function, so we have α = arccos _ 2 −7 cos (50 ◦ ) _ 53 −28 cos (50 ◦ ) _ radians ≈ 114.99 ◦ At this point, we could find γ using γ = 180 ◦ − α − β ≈ 180 ◦ − 114.99 ◦ − 50 ◦ = 15.01 ◦ , that is if we trust our approximation for α. To minimize propagation of error, however, we could use the Law of Cosines again, 4 in this case using cos(γ) = a 2 +b 2 −c 2 2ab . Plugging in a = 7, b = _ 53 −28 cos (50 ◦ ) and c = 2, we get γ = arccos _ 7−2 cos(50 ◦ ) √ 53−28 cos(50 ◦ ) _ radians ≈ 15.01 ◦ . We sketch the triangle below. a = 7 b ≈ 5.92 c = 2 α ≈ 114.99 ◦ γ ≈ 15.01 ◦ β = 50 ◦ As we mentioned earlier, once we’ve determined b it is possible to use the Law of Sines to find the remaining angles. Here, however, we must proceed with caution as we are in the ambiguous (ASS) case. It is advisable to first find the smallest of the unknown angles, since we are guaranteed it will be acute. 5 In this case, we would find γ since the side opposite γ is smaller than the side opposite the other unknown angle, α. Using the angle-side opposite pair (β, b), we get sin(γ) 2 = sin(50 ◦ ) √ 53−28 cos(50 ◦ ) . The usual calculations produces γ ≈ 15.01 ◦ and α = 180 ◦ −β −γ ≈ 180 ◦ −50 ◦ −15.01 ◦ = 114.99 ◦ . 2. Since all three sides and no angles are given, we are forced to use the Law of Cosines. Following our discussion in the previous problem, we find β first, since it is opposite the longest side, b. We get cos(β) = a 2 +c 2 −b 2 2ac = − 1 5 , so we get β = arccos _ − 1 5 _ radians ≈ 101.54 ◦ . As in 3 after simplifying . . . 4 Your instructor will let you know which procedure to use. It all boils down to how much you trust your calculator. 5 There can only be one obtuse angle in the triangle, and if there is one, it must be the largest. 11.3 The Law of Cosines 911 the previous problem, now that we have obtained an angle-side opposite pair (β, b), we could proceed using the Law of Sines. The Law of Cosines, however, offers us a rare opportunity to find the remaining angles using only the data given to us in the statement of the problem. Using this, we get γ = arccos _ 5 7 _ radians ≈ 44.42 ◦ and α = arccos _ 29 35 _ radians ≈ 34.05 ◦ . a = 4 c = 5 α ≈ 34.05 ◦ β ≈ 101.54 ◦ γ ≈ 44.42 ◦ b = 7 We note that, depending on how many decimal places are carried through successive calculations, and depending on which approach is used to solve the problem, the approximate answers you obtain may differ slightly from those the authors obtain in the Examples and the Exercises. A great example of this is number 2 in Example 11.3.1, where the approximate values we record for the measures of the angles sum to 180.01 ◦ , which is geometrically impossible. Next, we have an application of the Law of Cosines. Example 11.3.2. A researcher wishes to determine the width of a vernal pond as drawn below. From a point P, he finds the distance to the eastern-most point of the pond to be 950 feet, while the distance to the western-most point of the pond from P is 1000 feet. If the angle between the two lines of sight is 60 ◦ , find the width of the pond. 950 feet 1000 feet 60 ◦ P Solution. We are given the lengths of two sides and the measure of an included angle, so we may apply the Law of Cosines to find the length of the missing side opposite the given angle. Calling this length w (for width), we get w 2 = 950 2 + 1000 2 −2(950)(1000) cos (60 ◦ ) = 952500 from which we get w = √ 952500 ≈ 976 feet. 912 Applications of Trigonometry In Section 11.2, we used the proof of the Law of Sines to develop Theorem 11.4 as an alternate formula for the area enclosed by a triangle. In this section, we use the Law of Cosines to derive another such formula - Heron’s Formula. Theorem 11.6. Heron’s Formula: Suppose a, b and c denote the lengths of the three sides of a triangle. Let s be the semiperimeter of the triangle, that is, let s = 1 2 (a +b +c). Then the area A enclosed by the triangle is given by A = _ s(s −a)(s −b)(s −c) We prove Theorem 11.6 using Theorem 11.4. Using the convention that the angle γ is opposite the side c, we have A = 1 2 ab sin(γ) from Theorem 11.4. In order to simplify computations, we start by manipulating the expression for A 2 . A 2 = _ 1 2 ab sin(γ) _ 2 = 1 4 a 2 b 2 sin 2 (γ) = a 2 b 2 4 _ 1 −cos 2 (γ) _ since sin 2 (γ) = 1 −cos 2 (γ). The Law of Cosines tells us cos(γ) = a 2 +b 2 −c 2 2ab , so substituting this into our equation for A 2 gives A 2 = a 2 b 2 4 _ 1 −cos 2 (γ) _ = a 2 b 2 4 _ 1 − _ a 2 +b 2 −c 2 2ab _ 2 _ = a 2 b 2 4 _ 1 − _ a 2 +b 2 −c 2 _ 2 4a 2 b 2 _ = a 2 b 2 4 _ 4a 2 b 2 − _ a 2 +b 2 −c 2 _ 2 4a 2 b 2 _ = 4a 2 b 2 − _ a 2 +b 2 −c 2 _ 2 16 = (2ab) 2 − _ a 2 +b 2 −c 2 _ 2 16 = _ 2ab − _ a 2 +b 2 −c 2 ¸_ _ 2ab + _ a 2 +b 2 −c 2 ¸_ 16 difference of squares. = _ c 2 −a 2 + 2ab −b 2 _ _ a 2 + 2ab +b 2 −c 2 _ 16 11.3 The Law of Cosines 913 A 2 = _ c 2 − _ a 2 −2ab +b 2 ¸_ __ a 2 + 2ab +b 2 ¸ −c 2 _ 16 = _ c 2 −(a −b) 2 _ _ (a +b) 2 −c 2 _ 16 perfect square trinomials. = (c −(a −b))(c + (a −b))((a +b) −c)((a +b) +c) 16 difference of squares. = (b +c −a)(a +c −b)(a +b −c)(a +b +c) 16 = (b +c −a) 2 (a +c −b) 2 (a +b −c) 2 (a +b +c) 2 At this stage, we recognize the last factor as the semiperimeter, s = 1 2 (a + b + c) = a+b+c 2 . To complete the proof, we note that (s −a) = a +b +c 2 −a = a +b +c −2a 2 = b +c −a 2 Similarly, we find (s −b) = a+c−b 2 and (s −c) = a+b−c 2 . Hence, we get A 2 = (b +c −a) 2 (a +c −b) 2 (a +b −c) 2 (a +b +c) 2 = (s −a)(s −b)(s −c)s so that A = _ s(s −a)(s −b)(s −c) as required. We close with an example of Heron’s Formula. Example 11.3.3. Find the area enclosed of the triangle in Example 11.3.1 number 2. Solution. We are given a = 4, b = 7 and c = 5. Using these values, we find s = 1 2 (4 + 7 + 5) = 8, (s − a) = 8 − 4 = 4, (s − b) = 8 − 7 = 1 and (s − c) = 8 − 5 = 3. Using Heron’s Formula, we get A = _ s(s −a)(s −b)(s −c) = _ (8)(4)(1)(3) = √ 96 = 4 √ 6 ≈ 9.80 square units. 914 Applications of Trigonometry 11.3.1 Exercises In Exercises 1 - 10, use the Law of Cosines to find the remaining side(s) and angle(s) if possible. 1. a = 7, b = 12, γ = 59.3 ◦ 2. α = 104 ◦ , b = 25, c = 37 3. a = 153, β = 8.2 ◦ , c = 153 4. a = 3, b = 4, γ = 90 ◦ 5. α = 120 ◦ , b = 3, c = 4 6. a = 7, b = 10, c = 13 7. a = 1, b = 2, c = 5 8. a = 300, b = 302, c = 48 9. a = 5, b = 5, c = 5 10. a = 5, b = 12, ; c = 13 In Exercises 11 - 16, solve for the remaining side(s) and angle(s), if possible, using any appropriate technique. 11. a = 18, α = 63 ◦ , b = 20 12. a = 37, b = 45, c = 26 13. a = 16, α = 63 ◦ , b = 20 14. a = 22, α = 63 ◦ , b = 20 15. α = 42 ◦ , b = 117, c = 88 16. β = 7 ◦ , γ = 170 ◦ , c = 98.6 17. Find the area of the triangles given in Exercises 6, 8 and 10 above. 18. The hour hand on my antique Seth Thomas schoolhouse clock in 4 inches long and the minute hand is 5.5 inches long. Find the distance between the ends of the hands when the clock reads four o’clock. Round your answer to the nearest hundredth of an inch. 19. A geologist wants to measure the diameter of a crater. From her camp, it is 4 miles to the northern-most point of the crater and 2 miles to the southern-most point. If the angle between the two lines of sight is 117 ◦ , what is the diameter of the crater? Round your answer to the nearest hundredth of a mile. 20. From the Pedimaxus International Airport a tour helicopter can fly to Cliffs of Insanity Point by following a bearing of N8.2 ◦ E for 192 miles and it can fly to Bigfoot Falls by following a bearing of S68.5 ◦ E for 207 miles. 6 Find the distance between Cliffs of Insanity Point and Bigfoot Falls. Round your answer to the nearest mile. 21. Cliffs of Insanity Point and Bigfoot Falls from Exericse 20 above both lie on a straight stretch of the Great Sasquatch Canyon. What bearing would the tour helicopter need to follow to go directly from Bigfoot Falls to Cliffs of Insanity Point? Round your angle to the nearest tenth of a degree. 6 Please refer to Page 903 in Section 11.2 for an introduction to bearings. 11.3 The Law of Cosines 915 22. A naturalist sets off on a hike from a lodge on a bearing of S80 ◦ W. After 1.5 miles, she changes her bearing to S17 ◦ W and continues hiking for 3 miles. Find her distance from the lodge at this point. Round your answer to the nearest hundredth of a mile. What bearing should she follow to return to the lodge? Round your angle to the nearest degree. 23. The HMS Sasquatch leaves port on a bearing of N23 ◦ E and travels for 5 miles. It then changes course and follows a heading of S41 ◦ E for 2 miles. How far is it from port? Round your answer to the nearest hundredth of a mile. What is its bearing to port? Round your angle to the nearest degree. 24. The SS Bigfoot leaves a harbor bound for Nessie Island which is 300 miles away at a bearing of N32 ◦ E. A storm moves in and after 100 miles, the captain of the Bigfoot finds he has drifted off course. If his bearing to the harbor is now S70 ◦ W, how far is the SS Bigfoot from Nessie Island? Round your answer to the nearest hundredth of a mile. What course should the captain set to head to the island? Round your angle to the nearest tenth of a degree. 25. From a point 300 feet above level ground in a firetower, a ranger spots two fires in the Yeti National Forest. The angle of depression 7 made by the line of sight from the ranger to the first fire is 2.5 ◦ and the angle of depression made by line of sight from the ranger to the second fire is 1.3 ◦ . The angle formed by the two lines of sight is 117 ◦ . Find the distance between the two fires. Round your answer to the nearest foot. (Hint: In order to use the 117 ◦ angle between the lines of sight, you will first need to use right angle Trigonometry to find the lengths of the lines of sight. This will give you a Side-Angle-Side case in which to apply the Law of Cosines.) firetower fire fire 117 ◦ 26. If you apply the Law of Cosines to the ambiguous Angle-Side-Side (ASS) case, the result is a quadratic equation whose variable is that of the missing side. If the equation has no positive real zeros then the information given does not yield a triangle. If the equation has only one positive real zero then exactly one triangle is formed and if the equation has two distinct positive real zeros then two distinct triangles are formed. Apply the Law of Cosines to Exercises 11, 13 and 14 above in order to demonstrate this result. 27. Discuss with your classmates why Heron’s Formula yields an area in square units even though four lengths are being multiplied together. 7 See Exercise 78 in Section 10.3 for the definition of this angle. 916 Applications of Trigonometry 11.3.2 Answers 1. α ≈ 35.54 ◦ β ≈ 85.16 ◦ γ = 59.3 ◦ a = 7 b = 12 c ≈ 10.36 2. α = 104 ◦ β ≈ 29.40 ◦ γ ≈ 46.60 ◦ a ≈ 49.41 b = 25 c = 37 3. α ≈ 85.90 ◦ β = 8.2 ◦ γ ≈ 85.90 ◦ a = 153 b ≈ 21.88 c = 153 4. α ≈ 36.87 ◦ β ≈ 53.13 ◦ γ = 90 ◦ a = 3 b = 4 c = 5 5. α = 120 ◦ β ≈ 25.28 ◦ γ ≈ 34.72 ◦ a = √ 37 b = 3 c = 4 6. α ≈ 32.31 ◦ β ≈ 49.58 ◦ γ ≈ 98.21 ◦ a = 7 b = 10 c = 13 7. Information does not produce a triangle 8. α ≈ 83.05 ◦ β ≈ 87.81 ◦ γ ≈ 9.14 ◦ a = 300 b = 302 c = 48 9. α = 60 ◦ β = 60 ◦ γ = 60 ◦ a = 5 b = 5 c = 5 10. α ≈ 22.62 ◦ β ≈ 67.38 ◦ γ = 90 ◦ a = 5 b = 12 c = 13 11. α = 63 ◦ β ≈ 98.11 ◦ γ ≈ 18.89 ◦ a = 18 b = 20 c ≈ 6.54 α = 63 ◦ β ≈ 81.89 ◦ γ ≈ 35.11 ◦ a = 18 b = 20 c ≈ 11.62 12. α ≈ 55.30 ◦ β ≈ 89.40 ◦ γ ≈ 35.30 ◦ a = 37 b = 45 c = 26 13. Information does not produce a triangle 14. α = 63 ◦ β ≈ 54.1 ◦ γ ≈ 62.9 ◦ a = 22 b = 20 c ≈ 21.98 15. α = 42 ◦ β ≈ 89.23 ◦ γ ≈ 48.77 ◦ a ≈ 78.30 b = 117 c = 88 16. α ≈ 3 ◦ β = 7 ◦ γ = 170 ◦ a ≈ 29.72 b ≈ 69.2 c = 98.6 17. The area of the triangle given in Exercise 6 is √ 1200 = 20 √ 3 ≈ 34.64 square units. The area of the triangle given in Exercise 8 is √ 51764375 ≈ 7194.75 square units. The area of the triangle given in Exercise 10 is exactly 30 square units. 18. The distance between the ends of the hands at four o’clock is about 8.26 inches. 19. The diameter of the crater is about 5.22 miles. 20. About 313 miles 21. N31.8 ◦ W 22. She is about 3.92 miles from the lodge and her bearing to the lodge is N37 ◦ E. 23. It is about 4.50 miles from port and its heading to port is S47 ◦ W. 24. It is about 229.61 miles from the island and the captain should set a course of N16.4 ◦ E to reach the island. 25. The fires are about 17456 feet apart. (Try to avoid rounding errors.) 11.4 Polar Coordinates 917 11.4 Polar Coordinates In Section 1.1, we introduced the Cartesian coordinates of a point in the plane as a means of assigning ordered pairs of numbers to points in the plane. We defined the Cartesian coordinate plane using two number lines – one horizontal and one vertical – which intersect at right angles at a point we called the ‘origin’. To plot a point, say P(−3, 4), we start at the origin, travel horizontally to the left 3 units, then up 4 units. Alternatively, we could start at the origin, travel up 4 units, then to the left 3 units and arrive at the same location. For the most part, the ‘motions’ of the Cartesian system (over and up) describe a rectangle, and most points can be thought of as the corner diagonally across the rectangle from the origin. 1 For this reason, the Cartesian coordinates of a point are often called ‘rectangular’ coordinates. In this section, we introduce a new system for assigning coordinates to points in the plane – polar coordinates. We start with an origin point, called the pole, and a ray called the polar axis. We then locate a point P using two coordinates, (r, θ), where r represents a directed distance from the pole 2 and θ is a measure of rotation from the polar axis. Roughly speaking, the polar coordinates (r, θ) of a point measure ‘how far out’ the point is from the pole (that’s r), and ‘how far to rotate’ from the polar axis, (that’s θ). x y P(−3, 4) −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 3 2 r r θ Pole Polar Axis P(r, θ) For example, if we wished to plot the point P with polar coordinates _ 4, 5π 6 _ , we’d start at the pole, move out along the polar axis 4 units, then rotate 5π 6 radians counter-clockwise. Pole r = 4 Pole θ = 5π 6 Pole P _ 4, 5π 6 _ We may also visualize this process by thinking of the rotation first. 3 To plot P _ 4, 5π 6 _ this way, we rotate 5π 6 counter-clockwise from the polar axis, then move outwards from the pole 4 units. 1 Excluding, of course, the points in which one or both coordinates are 0. 2 We will explain more about this momentarily. 3 As with anything in Mathematics, the more ways you have to look at something, the better. The authors encourage the reader to take time to think about both approaches to plotting points given in polar coordinates. 918 Applications of Trigonometry Essentially we are locating a point on the terminal side of 5π 6 which is 4 units away from the pole. Pole θ = 5π 6 Pole θ = 5π 6 Pole P _ 4, 5π 6 _ If r < 0, we begin by moving in the opposite direction on the polar axis from the pole. For example, to plot Q _ −3.5, π 4 _ we have Pole r = −3.5 Pole θ = π 4 Pole Q _ −3.5, π 4 _ If we interpret the angle first, we rotate π 4 radians, then move back through the pole 3.5 units. Here we are locating a point 3.5 units away from the pole on the terminal side of 5π 4 , not π 4 . Pole θ = π 4 Pole θ = π 4 Pole Q _ −3.5, π 4 _ As you may have guessed, θ < 0 means the rotation away from the polar axis is clockwise instead of counter-clockwise. Hence, to plot R _ 3.5, − 3π 4 _ we have the following. Pole r = 3.5 Pole θ = − 3π 4 Pole R _ 3.5, − 3π 4 _ From an ‘angles first’ approach, we rotate − 3π 4 then move out 3.5 units from the pole. We see that R is the point on the terminal side of θ = − 3π 4 which is 3.5 units from the pole. Pole θ = − 3π 4 Pole θ = − 3π 4 Pole R _ 3.5, − 3π 4 _ 11.4 Polar Coordinates 919 The points Q and R above are, in fact, the same point despite the fact that their polar coordinate representations are different. Unlike Cartesian coordinates where (a, b) and (c, d) represent the same point if and only if a = c and b = d, a point can be represented by infinitely many polar coordinate pairs. We explore this notion more in the following example. Example 11.4.1. For each point in polar coordinates given below plot the point and then give two additional expressions for the point, one of which has r > 0 and the other with r < 0. 1. P (2, 240 ◦ ) 2. P _ −4, 7π 6 _ 3. P _ 117, − 5π 2 _ 4. P _ −3, − π 4 _ Solution. 1. Whether we move 2 units along the polar axis and then rotate 240 ◦ or rotate 240 ◦ then move out 2 units from the pole, we plot P (2, 240 ◦ ) below. Pole θ = 240 ◦ Pole P (2, 240 ◦ ) We now set about finding alternate descriptions (r, θ) for the point P. Since P is 2 units from the pole, r = ±2. Next, we choose angles θ for each of the r values. The given representation for P is (2, 240 ◦ ) so the angle θ we choose for the r = 2 case must be coterminal with 240 ◦ . (Can you see why?) One such angle is θ = −120 ◦ so one answer for this case is (2, −120 ◦ ). For the case r = −2, we visualize our rotation starting 2 units to the left of the pole. From this position, we need only to rotate θ = 60 ◦ to arrive at location coterminal with 240 ◦ . Hence, our answer here is (−2, 60 ◦ ). We check our answers by plotting them. P (2, −120 ◦ ) Pole θ = −120 ◦ P (−2, 60 ◦ ) Pole θ = 60 ◦ 2. We plot _ −4, 7π 6 _ by first moving 4 units to the left of the pole and then rotating 7π 6 radians. Since r = −4 < 0, we find our point lies 4 units from the pole on the terminal side of π 6 . Pole θ = 7π 6 Pole P _ −4, 7π 6 _ 920 Applications of Trigonometry To find alternate descriptions for P, we note that the distance from P to the pole is 4 units, so any representation (r, θ) for P must have r = ±4. As we noted above, P lies on the terminal side of π 6 , so this, coupled with r = 4, gives us _ 4, π 6 _ as one of our answers. To find a different representation for P with r = −4, we may choose any angle coterminal with the angle in the original representation of P _ −4, 7π 6 _ . We pick − 5π 6 and get _ −4, − 5π 6 _ as our second answer. θ = π 6 P _ 4, π 6 _ Pole θ = − 5π 6 P _ −4, − 5π 6 _ Pole 3. To plot P _ 117, − 5π 2 _ , we move along the polar axis 117 units from the pole and rotate clockwise 5π 2 radians as illustrated below. Pole θ = − 5π 2 Pole P _ 117, − 5π 2 _ Since P is 117 units from the pole, any representation (r, θ) for P satisfies r = ±117. For the r = 117 case, we can take θ to be any angle coterminal with − 5π 2 . In this case, we choose θ = 3π 2 , and get _ 117, 3π 2 _ as one answer. For the r = −117 case, we visualize moving left 117 units from the pole and then rotating through an angle θ to reach P. We find that θ = π 2 satisfies this requirement, so our second answer is _ −117, π 2 _ . Pole θ = 3π 2 P _ 117, 3π 2 _ Pole θ = π 2 P _ −117, π 2 _ 11.4 Polar Coordinates 921 4. We move three units to the left of the pole and follow up with a clockwise rotation of π 4 radians to plot P _ −3, − π 4 _ . We see that P lies on the terminal side of 3π 4 . Pole θ = − π 4 Pole P _ −3, − π 4 _ Since P lies on the terminal side of 3π 4 , one alternative representation for P is _ 3, 3π 4 _ . To find a different representation for P with r = −3, we may choose any angle coterminal with − π 4 . We choose θ = 7π 4 for our final answer _ −3, 7π 4 _ . Pole P _ 3, 3π 4 _ θ = 3π 4 Pole P _ −3, 7π 4 _ θ = 7π 4 Now that we have had some practice with plotting points in polar coordinates, it should come as no surprise that any given point expressed in polar coordinates has infinitely many other represen- tations in polar coordinates. The following result characterizes when two sets of polar coordinates determine the same point in the plane. It could be considered as a definition or a theorem, depend- ing on your point of view. We state it as a property of the polar coordinate system. Equivalent Representations of Points in Polar Coordinates Suppose (r, θ) and (r , θ ) are polar coordinates where r ,= 0, r ,= 0 and the angles are measured in radians. Then (r, θ) and (r , θ ) determine the same point P if and only if one of the following is true: • r = r and θ = θ + 2πk for some integer k • r = −r and θ = θ + (2k + 1)π for some integer k All polar coordinates of the form (0, θ) represent the pole regardless of the value of θ. The key to understanding this result, and indeed the whole polar coordinate system, is to keep in mind that (r, θ) means (directed distance from pole, angle of rotation). If r = 0, then no matter how much rotation is performed, the point never leaves the pole. Thus (0, θ) is the pole for all 922 Applications of Trigonometry values of θ. Now let’s assume that neither r nor r is zero. If (r, θ) and (r , θ ) determine the same point P then the (non-zero) distance from P to the pole in each case must be the same. Since this distance is controlled by the first coordinate, we have that either r = r or r = −r. If r = r, then when plotting (r, θ) and (r , θ ), the angles θ and θ have the same initial side. Hence, if (r, θ) and (r , θ ) determine the same point, we must have that θ is coterminal with θ. We know that this means θ = θ + 2πk for some integer k, as required. If, on the other hand, r = −r, then when plotting (r, θ) and (r , θ ), the initial side of θ is rotated π radians away from the initial side of θ. In this case, θ must be coterminal with π + θ. Hence, θ = π + θ + 2πk which we rewrite as θ = θ +(2k+1)π for some integer k. Conversely, if r = r and θ = θ +2πk for some integer k, then the points P (r, θ) and P (r , θ ) lie the same (directed) distance from the pole on the terminal sides of coterminal angles, and hence are the same point. Now suppose r = −r and θ = θ + (2k + 1)π for some integer k. To plot P, we first move a directed distance r from the pole; to plot P , our first step is to move the same distance from the pole as P, but in the opposite direction. At this intermediate stage, we have two points equidistant from the pole rotated exactly π radians apart. Since θ = θ +(2k +1)π = (θ +π) +2πk for some integer k, we see that θ is coterminal to (θ +π) and it is this extra π radians of rotation which aligns the points P and P . Next, we marry the polar coordinate system with the Cartesian (rectangular) coordinate system. To do so, we identify the pole and polar axis in the polar system to the origin and positive x-axis, respectively, in the rectangular system. We get the following result. Theorem 11.7. Conversion Between Rectangular and Polar Coordinates: Suppose P is represented in rectangular coordinates as (x, y) and in polar coordinates as (r, θ). Then • x = r cos(θ) and y = r sin(θ) • x 2 +y 2 = r 2 and tan(θ) = y x (provided x ,= 0) In the case r > 0, Theorem 11.7 is an immediate consequence of Theorem 10.3 along with the quotient identity tan(θ) = sin(θ) cos(θ) . If r < 0, then we know an alternate representation for (r, θ) is (−r, θ + π). Since cos(θ + π) = −cos(θ) and sin(θ + π) = −sin(θ), applying the theorem to (−r, θ + π) gives x = (−r) cos(θ + π) = (−r)(−cos(θ)) = r cos(θ) and y = (−r) sin(θ + π) = (−r)(−sin(θ)) = r sin(θ). Moreover, x 2 + y 2 = (−r) 2 = r 2 , and y x = tan(θ + π) = tan(θ), so the theorem is true in this case, too. The remaining case is r = 0, in which case (r, θ) = (0, θ) is the pole. Since the pole is identified with the origin (0, 0) in rectangular coordinates, the theorem in this case amounts to checking ‘0 = 0.’ The following example puts Theorem 11.7 to good use. Example 11.4.2. Convert each point in rectangular coordinates given below into polar coordinates with r ≥ 0 and 0 ≤ θ < 2π. Use exact values if possible and round any approximate values to two decimal places. Check your answer by converting them back to rectangular coordinates. 1. P _ 2, −2 √ 3 _ 2. Q(−3, −3) 3. R(0, −3) 4. S(−3, 4) 11.4 Polar Coordinates 923 Solution. 1. Even though we are not explicitly told to do so, we can avoid many common mistakes by taking the time to plot the points before we do any calculations. Plotting P _ 2, −2 √ 3 _ shows that it lies in Quadrant IV. With x = 2 and y = −2 √ 3, we get r 2 = x 2 +y 2 = (2) 2 + _ −2 √ 3 _ 2 = 4 + 12 = 16 so r = ±4. Since we are asked for r ≥ 0, we choose r = 4. To find θ, we have that tan(θ) = y x = −2 √ 3 2 = − √ 3. This tells us θ has a reference angle of π 3 , and since P lies in Quadrant IV, we know θ is a Quadrant IV angle. We are asked to have 0 ≤ θ < 2π, so we choose θ = 5π 3 . Hence, our answer is _ 4, 5π 3 _ . To check, we convert (r, θ) = _ 4, 5π 3 _ back to rectangular coordinates and we find x = r cos(θ) = 4 cos _ 5π 3 _ = 4 _ 1 2 _ = 2 and y = r sin(θ) = 4 sin _ 5π 3 _ = 4 _ − √ 3 2 _ = −2 √ 3, as required. 2. The point Q(−3, −3) lies in Quadrant III. Using x = y = −3, we get r 2 = (−3) 2 +(−3) 2 = 18 so r = ± √ 18 = ±3 √ 2. Since we are asked for r ≥ 0, we choose r = 3 √ 2. We find tan(θ) = −3 −3 = 1, which means θ has a reference angle of π 4 . Since Q lies in Quadrant III, we choose θ = 5π 4 , which satisfies the requirement that 0 ≤ θ < 2π. Our final answer is (r, θ) = _ 3 √ 2, 5π 4 _ . To check, we find x = r cos(θ) = (3 √ 2) cos _ 5π 4 _ = (3 √ 2) _ − √ 2 2 _ = −3 and y = r sin(θ) = (3 √ 2) sin _ 5π 4 _ = (3 √ 2) _ − √ 2 2 _ = −3, so we are done. x y P θ = 5π 3 x y Q θ = 5π 4 P has rectangular coordinates (2, −2 √ 3) Q has rectangular coordinates (−3, −3) P has polar coordinates _ 4, 5π 3 _ Q has polar coordinates _ 3 √ 2, 5π 4 _ 3. The point R(0, −3) lies along the negative y-axis. While we could go through the usual computations 4 to find the polar form of R, in this case we can find the polar coordinates of R using the definition. Since the pole is identified with the origin, we can easily tell the point R is 3 units from the pole, which means in the polar representation (r, θ) of R we know r = ±3. Since we require r ≥ 0, we choose r = 3. Concerning θ, the angle θ = 3π 2 satisfies 0 ≤ θ < 2π 4 Since x = 0, we would have to determine θ geometrically. 924 Applications of Trigonometry with its terminal side along the negative y-axis, so our answer is _ 3, 3π 2 _ . To check, we note x = r cos(θ) = 3 cos _ 3π 2 _ = (3)(0) = 0 and y = r sin(θ) = 3 sin _ 3π 2 _ = 3(−1) = −3. 4. The point S(−3, 4) lies in Quadrant II. With x = −3 and y = 4, we get r 2 = (−3) 2 +(4) 2 = 25 so r = ±5. As usual, we choose r = 5 ≥ 0 and proceed to determine θ. We have tan(θ) = y x = 4 −3 = − 4 3 , and since this isn’t the tangent of one the common angles, we resort to using the arctangent function. Since θ lies in Quadrant II and must satisfy 0 ≤ θ < 2π, we choose θ = π − arctan _ 4 3 _ radians. Hence, our answer is (r, θ) = _ 5, π −arctan _ 4 3 __ ≈ (5, 2.21). To check our answers requires a bit of tenacity since we need to simplify expressions of the form: cos _ π −arctan _ 4 3 __ and sin _ π −arctan _ 4 3 __ . These are good review exercises and are hence left to the reader. We find cos _ π −arctan _ 4 3 __ = − 3 5 and sin _ π −arctan _ 4 3 __ = 4 5 , so that x = r cos(θ) = (5) _ − 3 5 _ = −3 and y = r sin(θ) = (5) _ 4 5 _ = 4 which confirms our answer. x y R θ = 3π 2 x y S θ = π −arctan _ 4 3 _ R has rectangular coordinates (0, −3) S has rectangular coordinates (−3, 4) R has polar coordinates _ 3, 3π 2 _ S has polar coordinates _ 5, π −arctan _ 4 3 __ Now that we’ve had practice converting representations of points between the rectangular and polar coordinate systems, we now set about converting equations from one system to another. Just as we’ve used equations in x and y to represent relations in rectangular coordinates, equations in the variables r and θ represent relations in polar coordinates. We convert equations between the two systems using Theorem 11.7 as the next example illustrates. Example 11.4.3. 1. Convert each equation in rectangular coordinates into an equation in polar coordinates. (a) (x −3) 2 +y 2 = 9 (b) y = −x (c) y = x 2 2. Convert each equation in polar coordinates into an equation in rectangular coordinates. (a) r = −3 (b) θ = 4π 3 (c) r = 1 −cos(θ) 11.4 Polar Coordinates 925 Solution. 1. One strategy to convert an equation from rectangular to polar coordinates is to replace every occurrence of x with r cos(θ) and every occurrence of y with r sin(θ) and use identities to simplify. This is the technique we employ below. (a) We start by substituting x = r cos(θ) and y = sin(θ) into (x−3) 2 +y 2 = 9 and simplifying. With no real direction in which to proceed, we follow our mathematical instincts and see where they take us. 5 (r cos(θ) −3) 2 + (r sin(θ)) 2 = 9 r 2 cos 2 (θ) −6r cos(θ) + 9 +r 2 sin 2 (θ) = 9 r 2 _ cos 2 (θ) + sin 2 (θ) _ −6r cos(θ) = 0 Subtract 9 from both sides. r 2 −6r cos(θ) = 0 Since cos 2 (θ) + sin 2 (θ) = 1 r(r −6 cos(θ)) = 0 Factor. We get r = 0 or r = 6 cos(θ). From Section 7.2 we know the equation (x −3) 2 +y 2 = 9 describes a circle, and since r = 0 describes just a point (namely the pole/origin), we choose r = 6 cos(θ) for our final answer. 6 (b) Substituting x = r cos(θ) and y = r sin(θ) into y = −x gives r cos(θ) = −r sin(θ). Rearranging, we get r cos(θ) +r sin(θ) = 0 or r(cos(θ) +sin(θ)) = 0. This gives r = 0 or cos(θ) + sin(θ) = 0. Solving the latter equation for θ, we get θ = − π 4 + πk for integers k. As we did in the previous example, we take a step back and think geometrically. We know y = −x describes a line through the origin. As before, r = 0 describes the origin, but nothing else. Consider the equation θ = − π 4 . In this equation, the variable r is free, 7 meaning it can assume any and all values including r = 0. If we imagine plotting points (r, − π 4 ) for all conceivable values of r (positive, negative and zero), we are essentially drawing the line containing the terminal side of θ = − π 4 which is none other than y = −x. Hence, we can take as our final answer θ = − π 4 here. 8 (c) We substitute x = r cos(θ) and y = r sin(θ) into y = x 2 and get r sin(θ) = (r cos(θ)) 2 , or r 2 cos 2 (θ) −r sin(θ) = 0. Factoring, we get r(r cos 2 (θ) −sin(θ)) = 0 so that either r = 0 or r cos 2 (θ) = sin(θ). We can solve the latter equation for r by dividing both sides of the equation by cos 2 (θ), but as a general rule, we never divide through by a quantity that may be 0. In this particular case, we are safe since if cos 2 (θ) = 0, then cos(θ) = 0, and for the equation r cos 2 (θ) = sin(θ) to hold, then sin(θ) would also have to be 0. Since there are no angles with both cos(θ) = 0 and sin(θ) = 0, we are not losing any 5 Experience is the mother of all instinct, and necessity is the mother of invention. Study this example and see what techniques are employed, then try your best to get your answers in the homework to match Jeff’s. 6 Note that when we substitute θ = π 2 into r = 6 cos(θ), we recover the point r = 0, so we aren’t losing anything by disregarding r = 0. 7 See Section 8.1. 8 We could take it to be any of θ = − π 4 +πk for integers k. 926 Applications of Trigonometry information by dividing both sides of r cos 2 (θ) = sin(θ) by cos 2 (θ). Doing so, we get r = sin(θ) cos 2 (θ) , or r = sec(θ) tan(θ). As before, the r = 0 case is recovered in the solution r = sec(θ) tan(θ) (let θ = 0), so we state the latter as our final answer. 2. As a general rule, converting equations from polar to rectangular coordinates isn’t as straight forward as the reverse process. We could solve r 2 = x 2 + y 2 for r to get r = ± _ x 2 +y 2 and solving tan(θ) = y x requires the arctangent function to get θ = arctan _ y x _ + πk for integers k. Neither of these expressions for r and θ are especially user-friendly, so we opt for a second strategy – rearrange the given polar equation so that the expressions r 2 = x 2 + y 2 , r cos(θ) = x, r sin(θ) = y and/or tan(θ) = y x present themselves. (a) Starting with r = −3, we can square both sides to get r 2 = (−3) 2 or r 2 = 9. We may now substitute r 2 = x 2 +y 2 to get the equation x 2 +y 2 = 9. As we have seen, 9 squaring an equation does not, in general, produce an equivalent equation. The concern here is that the equation r 2 = 9 might be satisfied by more points than r = −3. On the surface, this appears to be the case since r 2 = 9 is equivalent to r = ±3, not just r = −3. However, any point with polar coordinates (3, θ) can be represented as (−3, θ + π), which means any point (r, θ) whose polar coordinates satisfy the relation r = ±3 has an equivalent 10 representation which satisfies r = −3. (b) We take the tangent of both sides the equation θ = 4π 3 to get tan(θ) = tan _ 4π 3 _ = √ 3. Since tan(θ) = y x , we get y x = √ 3 or y = x √ 3. Of course, we pause a moment to wonder if, geometrically, the equations θ = 4π 3 and y = x √ 3 generate the same set of points. 11 The same argument presented in number 1b applies equally well here so we are done. (c) Once again, we need to manipulate r = 1 − cos(θ) a bit before using the conversion formulas given in Theorem 11.7. We could square both sides of this equation like we did in part 2a above to obtain an r 2 on the left hand side, but that does nothing helpful for the right hand side. Instead, we multiply both sides by r to obtain r 2 = r − r cos(θ). We now have an r 2 and an r cos(θ) in the equation, which we can easily handle, but we also have another r to deal with. Rewriting the equation as r = r 2 + r cos(θ) and squaring both sides yields r 2 = _ r 2 +r cos(θ) _ 2 . Substituting r 2 = x 2 + y 2 and r cos(θ) = x gives x 2 + y 2 = _ x 2 +y 2 +x _ 2 . Once again, we have performed some 9 Exercise 5.3.1 in Section 5.3, for instance . . . 10 Here, ‘equivalent’ means they represent the same point in the plane. As ordered pairs, (3, 0) and (−3, π) are different, but when interpreted as polar coordinates, they correspond to the same point in the plane. Mathematically speaking, relations are sets of ordered pairs, so the equations r 2 = 9 and r = −3 represent different relations since they correspond to different sets of ordered pairs. Since polar coordinates were defined geometrically to describe the location of points in the plane, however, we concern ourselves only with ensuring that the sets of points in the plane generated by two equations are the same. This was not an issue, by the way, when we first defined relations as sets of points in the plane in Section 1.2. Back then, a point in the plane was identified with a unique ordered pair given by its Cartesian coordinates. 11 In addition to taking the tangent of both sides of an equation (There are infinitely many solutions to tan(θ) = √ 3, and θ = 4π 3 is only one of them!), we also went from y x = √ 3, in which x cannot be 0, to y = x √ 3 in which we assume x can be 0. 11.4 Polar Coordinates 927 algebraic maneuvers which may have altered the set of points described by the original equation. First, we multiplied both sides by r. This means that now r = 0 is a viable solution to the equation. In the original equation, r = 1−cos(θ), we see that θ = 0 gives r = 0, so the multiplication by r doesn’t introduce any new points. The squaring of both sides of this equation is also a reason to pause. Are there points with coordinates (r, θ) which satisfy r 2 = _ r 2 +r cos(θ) _ 2 but do not satisfy r = r 2 + r cos(θ)? Suppose (r , θ ) satisfies r 2 = _ r 2 +r cos(θ) _ 2 . Then r = ± _ (r ) 2 +r cos(θ ) _ . If we have that r = (r ) 2 +r cos(θ ), we are done. What if r = − _ (r ) 2 +r cos(θ ) _ = −(r ) 2 −r cos(θ )? We claim that the coordinates (−r , θ +π), which determine the same point as (r , θ ), satisfy r = r 2 +r cos(θ). We substitute r = −r and θ = θ +π into r = r 2 +r cos(θ) to see if we get a true statement. −r ? = (−r ) 2 + (−r cos(θ +π)) − _ −(r ) 2 −r cos(θ ) _ ? = (r ) 2 −r cos(θ +π) Since r = −(r ) 2 −r cos(θ ) (r ) 2 +r cos(θ ) ? = (r ) 2 −r (−cos(θ )) Since cos(θ +π) = −cos(θ ) (r ) 2 +r cos(θ ) = (r ) 2 +r cos(θ ) Since both sides worked out to be equal, (−r , θ + π) satisfies r = r 2 + r cos(θ) which means that any point (r, θ) which satisfies r 2 = _ r 2 +r cos(θ) _ 2 has a representation which satisfies r = r 2 +r cos(θ), and we are done. In practice, much of the pedantic verification of the equivalence of equations in Example 11.4.3 is left unsaid. Indeed, in most textbooks, squaring equations like r = −3 to arrive at r 2 = 9 happens without a second thought. Your instructor will ultimately decide how much, if any, justification is warranted. If you take anything away from Example 11.4.3, it should be that relatively nice things in rectangular coordinates, such as y = x 2 , can turn ugly in polar coordinates, and vice-versa. In the next section, we devote our attention to graphing equations like the ones given in Example 11.4.3 number 2 on the Cartesian coordinate plane without converting back to rectangular coordinates. If nothing else, number 2c above shows the price we pay if we insist on always converting to back to the more familiar rectangular coordinate system. 928 Applications of Trigonometry 11.4.1 Exercises In Exercises 1 - 16, plot the point given in polar coordinates and then give three different expressions for the point such that (a) r < 0 and 0 ≤ θ ≤ 2π, (b) r > 0 and θ ≤ 0 (c) r > 0 and θ ≥ 2π 1. _ 2, π 3 _ 2. _ 5, 7π 4 _ 3. _ 1 3 , 3π 2 _ 4. _ 5 2 , 5π 6 _ 5. _ 12, − 7π 6 _ 6. _ 3, − 5π 4 _ 7. _ 2 √ 2, −π _ 8. _ 7 2 , − 13π 6 _ 9. (−20, 3π) 10. _ −4, 5π 4 _ 11. _ −1, 2π 3 _ 12. _ −3, π 2 _ 13. _ −3, − 11π 6 _ 14. _ −2.5, − π 4 _ 15. _ − √ 5, − 4π 3 _ 16. (−π, −π) In Exercises 17 - 36, convert the point from polar coordinates into rectangular coordinates. 17. _ 5, 7π 4 _ 18. _ 2, π 3 _ 19. _ 11, − 7π 6 _ 20. (−20, 3π) 21. _ 3 5 , π 2 _ 22. _ −4, 5π 6 _ 23. _ 9, 7π 2 _ 24. _ −5, − 9π 4 _ 25. _ 42, 13π 6 _ 26. (−117, 117π) 27. (6, arctan(2)) 28. (10, arctan(3)) 29. _ −3, arctan _ 4 3 __ 30. _ 5, arctan _ − 4 3 __ 31. _ 2, π −arctan _ 1 2 __ 32. _ − 1 2 , π −arctan (5) _ 33. _ −1, π + arctan _ 3 4 __ 34. _ 2 3 , π + arctan _ 2 √ 2 _ _ 35. (π, arctan(π)) 36. _ 13, arctan _ 12 5 __ In Exercises 37 - 56, convert the point from rectangular coordinates into polar coordinates with r ≥ 0 and 0 ≤ θ < 2π. 37. (0, 5) 38. (3, √ 3) 39. (7, −7) 40. (−3, − √ 3) 41. (−3, 0) 42. _ − √ 2, √ 2 _ 43. _ −4, −4 √ 3 _ 44. _ √ 3 4 , − 1 4 _ 11.4 Polar Coordinates 929 45. _ − 3 10 , − 3 √ 3 10 _ 46. _ − √ 5, − √ 5 _ 47. (6, 8) 48. ( √ 5, 2 √ 5) 49. (−8, 1) 50. (−2 √ 10, 6 √ 10) 51. (−5, −12) 52. _ − √ 5 15 , − 2 √ 5 15 _ 53. (24, −7) 54. (12, −9) 55. _ √ 2 4 , √ 6 4 _ 56. _ − √ 65 5 , 2 √ 65 5 _ In Exercises 57 - 76, convert the equation from rectangular coordinates into polar coordinates. Solve for r in all but #60 through #63. In Exercises 60 - 63, you need to solve for θ 57. x = 6 58. x = −3 59. y = 7 60. y = 0 61. y = −x 62. y = x √ 3 63. y = 2x 64. x 2 +y 2 = 25 65. x 2 +y 2 = 117 66. y = 4x −19 67. x = 3y + 1 68. y = −3x 2 69. 4x = y 2 70. x 2 +y 2 −2y = 0 71. x 2 −4x +y 2 = 0 72. x 2 +y 2 = x 73. y 2 = 7y −x 2 74. (x + 2) 2 +y 2 = 4 75. x 2 + (y −3) 2 = 9 76. 4x 2 + 4 _ y − 1 2 _ 2 = 1 In Exercises 77 - 96, convert the equation from polar coordinates into rectangular coordinates. 77. r = 7 78. r = −3 79. r = √ 2 80. θ = π 4 81. θ = 2π 3 82. θ = π 83. θ = 3π 2 84. r = 4 cos(θ) 85. 5r = cos(θ) 86. r = 3 sin(θ) 87. r = −2 sin(θ) 88. r = 7 sec(θ) 89. 12r = csc(θ) 90. r = −2 sec(θ) 91. r = − √ 5 csc(θ) 92. r = 2 sec(θ) tan(θ) 93. r = −csc(θ) cot(θ) 94. r 2 = sin(2θ) 95. r = 1 −2 cos(θ) 96. r = 1 + sin(θ) 97. Convert the origin (0, 0) into polar coordinates in four different ways. 98. With the help of your classmates, use the Law of Cosines to develop a formula for the distance between two points in polar coordinates. 930 Applications of Trigonometry 11.4.2 Answers 1. _ 2, π 3 _ , _ −2, 4π 3 _ _ 2, − 5π 3 _ , _ 2, 7π 3 _ x y −1 1 2 1 2 2. _ 5, 7π 4 _ , _ −5, 3π 4 _ _ 5, − π 4 _ , _ 5, 15π 4 _ x y −1 1 2 3 −3 −2 −1 1 3. _ 1 3 , 3π 2 _ , _ − 1 3 , π 2 _ _ 1 3 , − π 2 _ , _ 1 3 , 7π 2 _ x y 1 1 4. _ 5 2 , 5π 6 _ , _ − 5 2 , 11π 6 _ _ 5 2 , − 7π 6 _ , _ 5 2 , 17π 6 _ x y −3 −2 −1 1 2 3 −1 −2 −3 1 2 3 11.4 Polar Coordinates 931 5. _ 12, − 7π 6 _ , _ −12, 11π 6 _ _ 12, − 13π 6 _ , _ 12, 17π 6 _ x y −12 −9 −6 −3 3 6 6. _ 3, − 5π 4 _ , _ −3, 7π 4 _ _ 3, − 13π 4 _ , _ 3, 11π 4 _ x y −3 −2 −1 1 2 3 −3 −2 −1 1 2 3 7. _ 2 √ 2, −π _ , _ −2 √ 2, 0 _ _ 2 √ 2, −3π _ , _ 2 √ 2, 3π _ x y −3 −2 −1 1 2 3 −3 −2 −1 1 2 3 8. _ 7 2 , − 13π 6 _ , _ − 7 2 , 5π 6 _ _ 7 2 , − π 6 _ , _ 7 2 , 23π 6 _ x y −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 932 Applications of Trigonometry 9. (−20, 3π), (−20, π) (20, −2π), (20, 4π) x y −20 −10 10 20 −1 1 10. _ −4, 5π 4 _ , _ −4, 13π 4 _ _ 4, − 7π 4 _ , _ 4, 9π 4 _ x y −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 11. _ −1, 2π 3 _ , _ −1, 8π 3 _ _ 1, − π 3 _ , _ 1, 11π 3 _ x y −2 −1 1 2 −2 −1 1 2 12. _ −3, π 2 _ , _ −3, 5π 2 _ _ 3, − π 2 _ , _ 3, 7π 2 _ x y −3 −2 −1 1 2 3 −3 −2 −1 1 2 3 11.4 Polar Coordinates 933 13. _ −3, − 11π 6 _ , _ −3, π 6 _ _ 3, − 5π 6 _ , _ 3, 19π 6 _ x y −3 −2 −1 1 2 3 −3 −2 −1 1 2 3 14. _ −2.5, − π 4 _ , _ −2.5, 7π 4 _ _ 2.5, − 5π 4 _ , _ 2.5, 11π 4 _ x y −2 −1 1 2 −2 −1 1 2 15. _ − √ 5, − 4π 3 _ , _ − √ 5, 2π 3 _ _ √ 5, − π 3 _ , _ √ 5, 11π 3 _ x y −2 −1 1 2 −2 −1 1 2 16. (−π, −π) , (−π, π) (π, −2π) , (π, 2π) x y −3 −2 −1 1 2 3 −3 −2 −1 1 2 3 934 Applications of Trigonometry 17. _ 5 √ 2 2 , − 5 √ 2 2 _ 18. _ 1, √ 3 _ 19. _ − 11 √ 3 2 , 11 2 _ 20. (20, 0) 21. _ 0, 3 5 _ 22. _ 2 √ 3, −2 _ 23. (0, −9) 24. _ − 5 √ 2 2 , 5 √ 2 2 _ 25. _ 21 √ 3, 21 _ 26. (117, 0) 27. _ 6 √ 5 5 , 12 √ 5 5 _ 28. _√ 10, 3 √ 10 _ 29. _ − 9 5 , − 12 5 _ 30. (3, −4) 31. _ − 4 √ 5 5 , 2 √ 5 5 _ 32. _ √ 26 52 , − 5 √ 26 52 _ 33. _ 4 5 , 3 5 _ 34. _ − 2 9 , − 4 √ 2 9 _ 35. _ π √ 1+π 2 , π 2 √ 1+π 2 _ 36. (5, 12) 37. _ 5, π 2 _ 38. _ 2 √ 3, π 6 _ 39. _ 7 √ 2, 7π 4 _ 40. _ 2 √ 3, 7π 6 _ 41. (3, π) 42. _ 2, 3π 4 _ 43. _ 8, 4π 3 _ 44. _ 1 2 , 11π 6 _ 45. _ 3 5 , 2π 3 _ 46. _ √ 10, 5π 4 _ 47. _ 10, arctan _ 4 3 __ 48. (5, arctan (2)) 49. _ √ 65, π −arctan _ 1 8 __ 50. (20, π −arctan(3)) 51. _ 13, π + arctan _ 12 5 __ 52. _ 1 3 , π + arctan (2) _ 53. _ 25, 2π −arctan _ 7 24 __ 54. _ 15, 2π −arctan _ 3 4 __ 55. _ √ 2 2 , π 3 _ 56. _√ 13, π −arctan(2) _ 57. r = 6 sec(θ) 58. r = −3 sec(θ) 59. r = 7 csc(θ) 60. θ = 0 61. θ = 3π 4 62. θ = π 3 63. θ = arctan(2) 64. r = 5 65. r = √ 117 66. r = 19 4 cos(θ)−sin(θ) 67. x = 1 cos(θ)−3 sin(θ) 68. r = −sec(θ) tan(θ) 3 69. r = 4 csc(θ) cot(θ) 70. r = 2 sin(θ) 71. r = 4 cos(θ) 72. r = cos(θ) 11.4 Polar Coordinates 935 73. r = 7 sin(θ) 74. r = −4 cos(θ) 75. r = 6 sin(θ) 76. r = 1 4 sin(θ) 77. x 2 +y 2 = 49 78. x 2 +y 2 = 9 79. x 2 +y 2 = 2 80. y = x 81. y = − √ 3x 82. y = 0 83. x = 0 84. x 2 +y 2 = 4x or (x −2) 2 +y 2 = 4 85. 5x 2 + 5y 2 = x or _ x − 1 10 _ 2 +y 2 = 1 100 86. x 2 +y 2 = 3y or x 2 + _ y − 3 2 _ 2 = 9 4 87. x 2 +y 2 = −2y or x 2 + (y + 1) 2 = 1 88. x = 7 89. y = 1 12 90. x = −2 91. y = − √ 5 92. x 2 = 2y 93. y 2 = −x 94. _ x 2 +y 2 _ 2 = 2xy 95. _ x 2 + 2x +y 2 _ 2 = x 2 +y 2 96. _ x 2 +y 2 +y _ 2 = x 2 +y 2 97. Any point of the form (0, θ) will work, e.g. (0, π), (0, −117), _ 0, 23π 4 _ and (0, 0). 936 Applications of Trigonometry 11.5 Graphs of Polar Equations In this section, we discuss how to graph equations in polar coordinates on the rectangular coordinate plane. Since any given point in the plane has infinitely many different representations in polar coordinates, our ‘Fundamental Graphing Principle’ in this section is not as clean as it was for graphs of rectangular equations on page 23. We state it below for completeness. The Fundamental Graphing Principle for Polar Equations The graph of an equation in polar coordinates is the set of points which satisfy the equation. That is, a point P(r, θ) is on the graph of an equation if and only if there is a representation of P, say (r , θ ), such that r and θ satisfy the equation. Our first example focuses on the some of the more structurally simple polar equations. Example 11.5.1. Graph the following polar equations. 1. r = 4 2. r = −3 √ 2 3. θ = 5π 4 4. θ = − 3π 2 Solution. In each of these equations, only one of the variables r and θ is present making the other variable free. 1 This makes these graphs easier to visualize than others. 1. In the equation r = 4, θ is free. The graph of this equation is, therefore, all points which have a polar coordinate representation (4, θ), for any choice of θ. Graphically this translates into tracing out all of the points 4 units away from the origin. This is exactly the definition of circle, centered at the origin, with a radius of 4. x y θ > 0 θ < 0 x y −4 4 −4 4 In r = 4, θ is free The graph of r = 4 2. Once again we have θ being free in the equation r = −3 √ 2. Plotting all of the points of the form (−3 √ 2, θ) gives us a circle of radius 3 √ 2 centered at the origin. 1 See the discussion in Example 11.4.3 number 2a. 11.5 Graphs of Polar Equations 937 x y θ < 0 θ > 0 x y −4 4 −4 4 In r = −3 √ 2, θ is free The graph of r = −3 √ 2 3. In the equation θ = 5π 4 , r is free, so we plot all of the points with polar representation _ r, 5π 4 _ . What we find is that we are tracing out the line which contains the terminal side of θ = 5π 4 when plotted in standard position. x y r < 0 r > 0 r = 0 θ = 5π 4 x y −4 4 −4 4 In θ = 5π 4 , r is free The graph of θ = 5π 4 4. As in the previous example, the variable r is free in the equation θ = − 3π 2 . Plotting _ r, − 3π 2 _ for various values of r shows us that we are tracing out the y-axis. 938 Applications of Trigonometry x y r > 0 r < 0 r = 0 θ = − 3π 2 x y −4 4 −4 4 In θ = − 3π 2 , r is free The graph of θ = − 3π 2 Hopefully, our experience in Example 11.5.1 makes the following result clear. Theorem 11.8. Graphs of Constant r and θ: Suppose a and α are constants, a ,= 0. • The graph of the polar equation r = a on the Cartesian plane is a circle centered at the origin of radius [a[. • The graph of the polar equation θ = α on the Cartesian plane is the line containing the terminal side of α when plotted in standard position. Suppose we wish to graph r = 6 cos(θ). A reasonable way to start is to treat θ as the independent variable, r as the dependent variable, evaluate r = f(θ) at some ‘friendly’ values of θ and plot the resulting points. 2 We generate the table below. θ r = 6 cos(θ) (r, θ) 0 6 (6, 0) π 4 3 √ 2 _ 3 √ 2, π 4 _ π 2 0 _ 0, π 2 _ 3π 4 −3 √ 2 _ −3 √ 2, 3π 4 _ π −6 (−6, π) 5π 4 −3 √ 2 _ −3 √ 2, 5π 4 _ 3π 2 0 _ 0, 3π 2 _ 7π 4 3 √ 2 _ 3 √ 2, 7π 4 _ 2π 6 (6, 2π) x y 3 6 −3 3 2 For a review of these concepts and this process, see Sections 1.4 and 1.6. 11.5 Graphs of Polar Equations 939 Despite having nine ordered pairs, we get only four distinct points on the graph. For this reason, we employ a slightly different strategy. We graph one cycle of r = 6 cos(θ) on the θr-plane 3 and use it to help graph the equation on the xy-plane. We see that as θ ranges from 0 to π 2 , r ranges from 6 to 0. In the xy-plane, this means that the curve starts 6 units from the origin on the positive x-axis (θ = 0) and gradually returns to the origin by the time the curve reaches the y-axis (θ = π 2 ). The arrows drawn in the figure below are meant to help you visualize this process. In the θr-plane, the arrows are drawn from the θ-axis to the curve r = 6 cos(θ). In the xy-plane, each of these arrows starts at the origin and is rotated through the corresponding angle θ, in accordance with how we plot polar coordinates. It is a less-precise way to generate the graph than computing the actual function values, but it is markedly faster. π 2 π 3π 2 2π −6 −3 3 6 θ r x y θ runs from 0 to π 2 Next, we repeat the process as θ ranges from π 2 to π. Here, the r values are all negative. This means that in the xy-plane, instead of graphing in Quadrant II, we graph in Quadrant IV, with all of the angle rotations starting from the negative x-axis. π 2 π 3π 2 2π −6 −3 3 6 θ r x y θ runs from π 2 to π r < 0 so we plot here As θ ranges from π to 3π 2 , the r values are still negative, which means the graph is traced out in Quadrant I instead of Quadrant III. Since the [r[ for these values of θ match the r values for θ in 3 The graph looks exactly like y = 6 cos(x) in the xy-plane, and for good reason. At this stage, we are just graphing the relationship between r and θ before we interpret them as polar coordinates (r, θ) on the xy-plane. 940 Applications of Trigonometry _ 0, π 2 ¸ , we have that the curve begins to retrace itself at this point. Proceeding further, we find that when 3π 2 ≤ θ ≤ 2π, we retrace the portion of the curve in Quadrant IV that we first traced out as π 2 ≤ θ ≤ π. The reader is invited to verify that plotting any range of θ outside the interval [0, π] results in retracting some portion of the curve. 4 We present the final graph below. π 2 π −6 −3 3 6 θ r x y 3 6 −3 3 r = 6 cos(θ) in the θr-plane r = 6 cos(θ) in the xy-plane Example 11.5.2. Graph the following polar equations. 1. r = 4 −2 sin(θ) 2. r = 2 + 4 cos(θ) 3. r = 5 sin(2θ) 4. r 2 = 16 cos(2θ) Solution. 1. We first plot the fundamental cycle of r = 4 − 2 sin(θ) on the θr-axes. To help us visualize what is going on graphically, we divide up [0, 2π] into the usual four subintervals _ 0, π 2 ¸ , _ π 2 , π ¸ , _ π, 3π 2 ¸ and _ 3π 2 , 2π ¸ , and proceed as we did above. As θ ranges from 0 to π 2 , r decreases from 4 to 2. This means that the curve in the xy-plane starts 4 units from the origin on the positive x-axis and gradually pulls in towards the origin as it moves towards the positive y-axis. π 2 π 3π 2 2π 2 4 6 θ r x y θ runs from 0 to π 2 4 The graph of r = 6 cos(θ) looks suspiciously like a circle, for good reason. See number 1a in Example 11.4.3. 11.5 Graphs of Polar Equations 941 Next, as θ runs from π 2 to π, we see that r increases from 2 to 4. Picking up where we left off, we gradually pull the graph away from the origin until we reach the negative x-axis. π 2 π 3π 2 2π 2 4 6 θ r x y θ runs from π 2 to π Over the interval _ π, 3π 2 ¸ , we see that r increases from 4 to 6. On the xy-plane, the curve sweeps out away from the origin as it travels from the negative x-axis to the negative y-axis. π 2 π 3π 2 2π 2 4 6 θ r x y θ runs from π to 3π 2 Finally, as θ takes on values from 3π 2 to 2π, r decreases from 6 back to 4. The graph on the xy-plane pulls in from the negative y-axis to finish where we started. π 2 π 3π 2 2π 2 4 6 θ r x y θ runs from 3π 2 to 2π We leave it to the reader to verify that plotting points corresponding to values of θ outside the interval [0, 2π] results in retracing portions of the curve, so we are finished. 942 Applications of Trigonometry π 2 π 3π 2 2π 2 4 6 θ r x y −4 4 −6 2 r = 4 −2 sin(θ) in the θr-plane r = 4 −2 sin(θ) in the xy-plane. 2. The first thing to note when graphing r = 2 + 4 cos(θ) on the θr-plane over the interval [0, 2π] is that the graph crosses through the θ-axis. This corresponds to the graph of the curve passing through the origin in the xy-plane, and our first task is to determine when this happens. Setting r = 0 we get 2 + 4 cos(θ) = 0, or cos(θ) = − 1 2 . Solving for θ in [0, 2π] gives θ = 2π 3 and θ = 4π 3 . Since these values of θ are important geometrically, we break the interval [0, 2π] into six subintervals: _ 0, π 2 ¸ , _ π 2 , 2π 3 ¸ , _ 2π 3 , π ¸ , _ π, 4π 3 ¸ , _ 4π 3 , 3π 2 ¸ and _ 3π 2 , 2π ¸ . As θ ranges from 0 to π 2 , r decreases from 6 to 2. Plotting this on the xy-plane, we start 6 units out from the origin on the positive x-axis and slowly pull in towards the positive y-axis. π 2 π 3π 2 2π −2 2 4 6 2π 3 4π 3 θ r x y θ runs from 0 to π 2 On the interval _ π 2 , 2π 3 ¸ , r decreases from 2 to 0, which means the graph is heading into (and will eventually cross through) the origin. Not only do we reach the origin when θ = 2π 3 , a theorem from Calculus 5 states that the curve hugs the line θ = 2π 3 as it approaches the origin. 5 The ‘tangents at the pole’ theorem from second semester Calculus. 11.5 Graphs of Polar Equations 943 π 2 π 3π 2 2π −2 2 4 6 2π 3 4π 3 θ r x y θ = 2π 3 On the interval _ 2π 3 , π ¸ , r ranges from 0 to −2. Since r ≤ 0, the curve passes through the origin in the xy-plane, following the line θ = 2π 3 and continues upwards through Quadrant IV towards the positive x-axis. 6 Since [r[ is increasing from 0 to 2, the curve pulls away from the origin to finish at a point on the positive x-axis. π 2 π 3π 2 2π −2 2 4 6 2π 3 4π 3 θ r x y θ = 2π 3 Next, as θ progresses from π to 4π 3 , r ranges from −2 to 0. Since r ≤ 0, we continue our graph in the first quadrant, heading into the origin along the line θ = 4π 3 . 6 Recall that one way to visualize plotting polar coordinates (r, θ) with r < 0 is to start the rotation from the left side of the pole - in this case, the negative x-axis. Rotating between 2π 3 and π radians from the negative x-axis in this case determines the region between the line θ = 2π 3 and the x-axis in Quadrant IV. 944 Applications of Trigonometry π 2 π 3π 2 2π −2 2 4 6 2π 3 4π 3 θ r x y θ = 4π 3 On the interval _ 4π 3 , 3π 2 ¸ , r returns to positive values and increases from 0 to 2. We hug the line θ = 4π 3 as we move through the origin and head towards the negative y-axis. π 2 π 3π 2 2π −2 2 4 6 2π 3 4π 3 θ r x y θ = 4π 3 As we round out the interval, we find that as θ runs through 3π 2 to 2π, r increases from 2 out to 6, and we end up back where we started, 6 units from the origin on the positive x-axis. π 2 π 3π 2 2π −2 2 4 6 2π 3 4π 3 θ r x y θ runs from 3π 2 to 2π 11.5 Graphs of Polar Equations 945 Again, we invite the reader to show that plotting the curve for values of θ outside [0, 2π] results in retracing a portion of the curve already traced. Our final graph is below. π 2 π 3π 2 2π −2 2 4 6 2π 3 4π 3 θ r x y 2 6 −2 2 θ = 4π 3 θ = 2π 3 r = 2 + 4 cos(θ) in the θr-plane r = 2 + 4 cos(θ) in the xy-plane 3. As usual, we start by graphing a fundamental cycle of r = 5 sin(2θ) in the θr-plane, which in this case, occurs as θ ranges from 0 to π. We partition our interval into subintervals to help us with the graphing, namely _ 0, π 4 ¸ , _ π 4 , π 2 ¸ , _ π 2 , 3π 4 ¸ and _ 3π 4 , π ¸ . As θ ranges from 0 to π 4 , r increases from 0 to 5. This means that the graph of r = 5 sin(2θ) in the xy-plane starts at the origin and gradually sweeps out so it is 5 units away from the origin on the line θ = π 4 . π 4 π 2 3π 4 π −5 5 θ r x y Next, we see that r decreases from 5 to 0 as θ runs through _ π 4 , π 2 ¸ , and furthermore, r is heading negative as θ crosses π 2 . Hence, we draw the curve hugging the line θ = π 2 (the y-axis) as the curve heads to the origin. 946 Applications of Trigonometry π 4 π 2 3π 4 π −5 5 θ r x y As θ runs from π 2 to 3π 4 , r becomes negative and ranges from 0 to −5. Since r ≤ 0, the curve pulls away from the negative y-axis into Quadrant IV. π 4 π 2 3π 4 π −5 5 θ r x y For 3π 4 ≤ θ ≤ π, r increases from −5 to 0, so the curve pulls back to the origin. π 4 π 2 3π 4 π −5 5 θ r x y 11.5 Graphs of Polar Equations 947 Even though we have finished with one complete cycle of r = 5 sin(2θ), if we continue plotting beyond θ = π, we find that the curve continues into the third quadrant! Below we present a graph of a second cycle of r = 5 sin(2θ) which continues on from the first. The boxed labels on the θ-axis correspond to the portions with matching labels on the curve in the xy-plane. π 5π 4 3π 2 7π 4 2π −5 5 θ r 1 2 3 4 x y 1 2 3 4 We have the final graph below. π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π −5 5 θ r x y −5 5 −5 5 r = 5 sin(2θ) in the θr-plane r = 5 sin(2θ) in the xy-plane 4. Graphing r 2 = 16 cos(2θ) is complicated by the r 2 , so we solve to get r = ± _ 16 cos(2θ) = ±4 _ cos(2θ). How do we sketch such a curve? First off, we sketch a fundamental period of r = cos(2θ) which we have dotted in the figure below. When cos(2θ) < 0, _ cos(2θ) is undefined, so we don’t have any values on the interval _ π 4 , 3π 4 _ . On the intervals which remain, cos(2θ) ranges from 0 to 1, inclusive. Hence, _ cos(2θ) ranges from 0 to 1 as well. 7 From this, we know r = ±4 _ cos(2θ) ranges continuously from 0 to ±4, respectively. Below we graph both r = 4 _ cos(2θ) and r = −4 _ cos(2θ) on the θr plane and use them to sketch the corresponding pieces of the curve r 2 = 16 cos(2θ) in the xy-plane. As we have seen in earlier 7 Owing to the relationship between y = x and y = √ x over [0, 1], we also know cos(2θ) ≥ cos(2θ) wherever the former is defined. 948 Applications of Trigonometry examples, the lines θ = π 4 and θ = 3π 4 , which are the zeros of the functions r = ±4 _ cos(2θ), serve as guides for us to draw the curve as is passes through the origin. π 4 π 2 3π 4 π −4 4 θ r 1 2 3 4 x y 1 2 3 4 θ = π 4 θ = 3π 4 r = 4 _ cos(2θ) and r = −4 _ cos(2θ) As we plot points corresponding to values of θ outside of the interval [0, π], we find ourselves retracing parts of the curve, 8 so our final answer is below. π 4 π 2 3π 4 π −4 4 θ r x y −4 4 −4 4 θ = π 4 θ = 3π 4 r = ±4 _ cos(2θ) r 2 = 16 cos(2θ) in the θr-plane in the xy-plane A few remarks are in order. First, there is no relation, in general, between the period of the function f(θ) and the length of the interval required to sketch the complete graph of r = f(θ) in the xy- plane. As we saw on page 939, despite the fact that the period of f(θ) = 6 cos(θ) is 2π, we sketched the complete graph of r = 6 cos(θ) in the xy-plane just using the values of θ as θ ranged from 0 to π. In Example 11.5.2, number 3, the period of f(θ) = 5 sin(2θ) is π, but in order to obtain the complete graph of r = 5 sin(2θ), we needed to run θ from 0 to 2π. While many of the ‘common’ polar graphs can be grouped into families, 9 the authors truly feel that taking the time to work through each graph in the manner presented here is the best way to not only understand the polar 8 In this case, we could have generated the entire graph by using just the plot r = 4 cos(2θ), but graphed over the interval [0, 2π] in the θr-plane. We leave the details to the reader. 9 Numbers 1 and 2 in Example 11.5.2 are examples of ‘lima¸ cons,’ number 3 is an example of a ‘polar rose,’ and number 4 is the famous ‘Lemniscate of Bernoulli.’ 11.5 Graphs of Polar Equations 949 coordinate system, but also prepare you for what is needed in Calculus. Second, the symmetry seen in the examples is also a common occurrence when graphing polar equations. In addition to the usual kinds of symmetry discussed up to this point in the text (symmetry about each axis and the origin), it is possible to talk about rotational symmetry. We leave the discussion of symmetry to the Exercises. In our next example, we are given the task of finding the intersection points of polar curves. According to the Fundamental Graphing Principle for Polar Equations on page 936, in order for a point P to be on the graph of a polar equation, it must have a representation P(r, θ) which satisfies the equation. What complicates matters in polar coordinates is that any given point has infinitely many representations. As a result, if a point P is on the graph of two different polar equations, it is entirely possible that the representation P(r, θ) which satisfies one of the equations does not satisfy the other equation. Here, more than ever, we need to rely on the Geometry as much as the Algebra to find our solutions. Example 11.5.3. Find the points of intersection of the graphs of the following polar equations. 1. r = 2 sin(θ) and r = 2 −2 sin(θ) 2. r = 2 and r = 3 cos(θ) 3. r = 3 and r = 6 cos(2θ) 4. r = 3 sin _ θ 2 _ and r = 3 cos _ θ 2 _ Solution. 1. Following the procedure in Example 11.5.2, we graph r = 2 sin(θ) and find it to be a circle centered at the point with rectangular coordinates (0, 1) with a radius of 1. The graph of r = 2 −2 sin(θ) is a special kind of lima¸ con called a ‘cardioid.’ 10 x y −2 2 −4 2 r = 2 −2 sin(θ) and r = 2 sin(θ) It appears as if there are three intersection points: one in the first quadrant, one in the second quadrant, and the origin. Our next task is to find polar representations of these points. In 10 Presumably, the name is derived from its resemblance to a stylized human heart. 950 Applications of Trigonometry order for a point P to be on the graph of r = 2 sin(θ), it must have a representation P(r, θ) which satisfies r = 2 sin(θ). If P is also on the graph of r = 2−2 sin(θ), then P has a (possibly different) representation P(r , θ ) which satisfies r = 2 sin(θ ). We first try to see if we can find any points which have a single representation P(r, θ) that satisfies both r = 2 sin(θ) and r = 2 − 2 sin(θ). Assuming such a pair (r, θ) exists, then equating 11 the expressions for r gives 2 sin(θ) = 2 − 2 sin(θ) or sin(θ) = 1 2 . From this, we get θ = π 6 + 2πk or θ = 5π 6 + 2πk for integers k. Plugging θ = π 6 into r = 2 sin(θ), we get r = 2 sin _ π 6 _ = 2 _ 1 2 _ = 1, which is also the value we obtain when we substitute it into r = 2 − 2 sin(θ). Hence, _ 1, π 6 _ is one representation for the point of intersection in the first quadrant. For the point of intersection in the second quadrant, we try θ = 5π 6 . Both equations give us the point _ 1, 5π 6 _ , so this is our answer here. What about the origin? We know from Section 11.4 that the pole may be represented as (0, θ) for any angle θ. On the graph of r = 2 sin(θ), we start at the origin when θ = 0 and return to it at θ = π, and as the reader can verify, we are at the origin exactly when θ = πk for integers k. On the curve r = 2 −2 sin(θ), however, we reach the origin when θ = π 2 , and more generally, when θ = π 2 + 2πk for integers k. There is no integer value of k for which πk = π 2 + 2πk which means while the origin is on both graphs, the point is never reached simultaneously. In any case, we have determined the three points of intersection to be _ 1, π 6 _ , _ 1, 5π 6 _ and the origin. 2. As before, we make a quick sketch of r = 2 and r = 3 cos(θ) to get feel for the number and location of the intersection points. The graph of r = 2 is a circle, centered at the origin, with a radius of 2. The graph of r = 3 cos(θ) is also a circle - but this one is centered at the point with rectangular coordinates _ 3 2 , 0 _ and has a radius of 3 2 . x y −2 2 3 −2 2 r = 2 and r = 3 cos(θ) We have two intersection points to find, one in Quadrant I and one in Quadrant IV. Pro- ceeding as above, we first determine if any of the intersection points P have a represen- tation (r, θ) which satisfies both r = 2 and r = 3 cos(θ). Equating these two expressions for r, we get cos(θ) = 2 3 . To solve this equation, we need the arccosine function. We get 11 We are really using the technique of substitution to solve the system of equations r = 2 sin(θ) r = 2 −2 sin(θ) 11.5 Graphs of Polar Equations 951 θ = arccos _ 2 3 _ +2πk or θ = 2π−arccos _ 2 3 _ +2πk for integers k. From these solutions, we get _ 2, arccos _ 2 3 __ as one representation for our answer in Quadrant I, and _ 2, 2π −arccos _ 2 3 __ as one representation for our answer in Quadrant IV. The reader is encouraged to check these results algebraically and geometrically. 3. Proceeding as above, we first graph r = 3 and r = 6 cos(2θ) to get an idea of how many intersection points to expect and where they lie. The graph of r = 3 is a circle centered at the origin with a radius of 3 and the graph of r = 6 cos(2θ) is another four-leafed rose. 12 x y −6 −3 3 6 −6 −3 3 6 r = 3 and r = 6 cos(2θ) It appears as if there are eight points of intersection - two in each quadrant. We first look to see if there any points P(r, θ) with a representation that satisfies both r = 3 and r = 6 cos(2θ). For these points, 6 cos(2θ) = 3 or cos(2θ) = 1 2 . Solving, we get θ = π 6 + πk or θ = 5π 6 + πk for integers k. Out of all of these solutions, we obtain just four distinct points represented by _ 3, π 6 _ , _ 3, 5π 6 _ , _ 3, 7π 6 _ and _ 3, 11π 6 _ . To determine the coordinates of the remaining four points, we have to consider how the representations of the points of intersection can differ. We know from Section 11.4 that if (r, θ) and (r , θ ) represent the same point and r ,= 0, then either r = r or r = −r . If r = r , then θ = θ+2πk, so one possibility is that an intersection point P has a representation (r, θ) which satisfies r = 3 and another representation (r, θ+2πk) for some integer, k which satisfies r = 6 cos(2θ). At this point, 13 f we replace every occurrence of θ in the equation r = 6 cos(2θ) with (θ+2πk) and then see if, by equating the resulting expressions for r, we get any more solutions for θ. Since cos(2(θ + 2πk)) = cos(2θ + 4πk) = cos(2θ) for every integer k, however, the equation r = 6 cos(2(θ + 2πk)) reduces to the same equation we had before, r = 6 cos(2θ), which means we get no additional solutions. Moving on to the case where r = −r , we have that θ = θ + (2k + 1)π for integers k. We look to see if we can find points P which have a representation (r, θ) that satisfies r = 3 and another, 12 See Example 11.5.2 number 3. 13 The authors have chosen to replace θ with θ+2πk in the equation r = 6 cos(2θ) for illustration purposes only. We could have just as easily chosen to do this substitution in the equation r = 3. Since there is no θ in r = 3, however, this case would reduce to the previous case instantly. The reader is encouraged to follow this latter procedure in the interests of efficiency. 952 Applications of Trigonometry (−r, θ + (2k + 1)π), that satisfies r = 6 cos(2θ). To do this, we substitute 14 (−r) for r and (θ + (2k + 1)π) for θ in the equation r = 6 cos(2θ) and get −r = 6 cos(2(θ + (2k + 1)π)). Since cos(2(θ +(2k +1)π)) = cos(2θ +(2k +1)(2π)) = cos(2θ) for all integers k, the equation −r = 6 cos(2(θ + (2k + 1)π)) reduces to −r = 6 cos(2θ), or r = −6 cos(2θ). Coupling this equation with r = 3 gives −6 cos(2θ) = 3 or cos(2θ) = − 1 2 . We get θ = π 3 +πk or θ = 2π 3 +πk. From these solutions, we obtain 15 the remaining four intersection points with representations _ −3, π 3 _ , _ −3, 2π 3 _ , _ −3, 4π 3 _ and _ −3, 5π 3 _ , which we can readily check graphically. 4. As usual, we begin by graphing r = 3 sin _ θ 2 _ and r = 3 cos _ θ 2 _ . Using the techniques presented in Example 11.5.2, we find that we need to plot both functions as θ ranges from 0 to 4π to obtain the complete graph. To our surprise and/or delight, it appears as if these two equations describe the same curve! x y −3 3 −3 3 r = 3 sin _ θ 2 _ and r = 3 cos _ θ 2 _ appear to determine the same curve in the xy-plane To verify this incredible claim, 16 we need to show that, in fact, the graphs of these two equations intersect at all points on the plane. Suppose P has a representation (r, θ) which satisfies both r = 3 sin _ θ 2 _ and r = 3 cos _ θ 2 _ . Equating these two expressions for r gives the equation 3 sin _ θ 2 _ = 3 cos _ θ 2 _ . While normally we discourage dividing by a variable expression (in case it could be 0), we note here that if 3 cos _ θ 2 _ = 0, then for our equation to hold, 3 sin _ θ 2 _ = 0 as well. Since no angles have both cosine and sine equal to zero, we are safe to divide both sides of the equation 3 sin _ θ 2 _ = 3 cos _ θ 2 _ by 3 cos _ θ 2 _ to get tan _ θ 2 _ = 1 which gives θ = π 2 + 2πk for integers k. From these solutions, however, we 14 Again, we could have easily chosen to substitute these into r = 3 which would give −r = 3, or r = −3. 15 We obtain these representations by substituting the values for θ into r = 6 cos(2θ), once again, for illustration purposes. Again, in the interests of efficiency, we could ‘plug’ these values for θ into r = 3 (where there is no θ) and get the list of points: 3, π 3 , 3, 2π 3 , 3, 4π 3 and 3, 5π 3 . While it is not true that 3, π 3 represents the same point as −3, π 3 , we still get the same set of solutions. 16 A quick sketch of r = 3 sin θ 2 and r = 3 cos θ 2 in the θr-plane will convince you that, viewed as functions of r, these are two different animals. 11.5 Graphs of Polar Equations 953 get only one intersection point which can be represented by _ 3 √ 2 2 , π 2 _ . We now investigate other representations for the intersection points. Suppose P is an intersection point with a representation (r, θ) which satisfies r = 3 sin _ θ 2 _ and the same point P has a different representation (r, θ + 2πk) for some integer k which satisfies r = 3 cos _ θ 2 _ . Substituting into the latter, we get r = 3 cos _ 1 2 [θ + 2πk] _ = 3 cos _ θ 2 +πk _ . Using the sum formula for cosine, we expand 3 cos _ θ 2 +πk _ = 3 cos _ θ 2 _ cos(πk) − 3 sin _ θ 2 _ sin (πk) = ±3 cos _ θ 2 _ , since sin(πk) = 0 for all integers k, and cos (πk) = ±1 for all integers k. If k is an even integer, we get the same equation r = 3 cos _ θ 2 _ as before. If k is odd, we get r = −3 cos _ θ 2 _ . This latter expression for r leads to the equation 3 sin _ θ 2 _ = −3 cos _ θ 2 _ , or tan _ θ 2 _ = −1. Solving, we get θ = − π 2 + 2πk for integers k, which gives the intersection point _ 3 √ 2 2 , − π 2 _ . Next, we assume P has a representation (r, θ) which satisfies r = 3 sin _ θ 2 _ and a representation (−r, θ + (2k + 1)π) which satisfies r = 3 cos _ θ 2 _ for some integer k. Substituting (−r) for r and (θ + (2k + 1)π) in for θ into r = 3 cos _ θ 2 _ gives −r = 3 cos _ 1 2 [θ + (2k + 1)π] _ . Once again, we use the sum formula for cosine to get cos _ 1 2 [θ + (2k + 1)π] _ = cos _ θ 2 + (2k+1)π 2 _ = cos _ θ 2 _ cos _ (2k+1)π 2 _ −sin _ θ 2 _ sin _ (2k+1)π 2 _ = ±sin _ θ 2 _ where the last equality is true since cos _ (2k+1)π 2 _ = 0 and sin _ (2k+1)π 2 _ = ±1 for integers k. Hence, −r = 3 cos _ 1 2 [θ + (2k + 1)π] _ can be rewritten as r = ±3 sin _ θ 2 _ . If we choose k = 0, then sin _ (2k+1)π 2 _ = sin _ π 2 _ = 1, and the equation −r = 3 cos _ 1 2 [θ + (2k + 1)π] _ in this case reduces to −r = −3 sin _ θ 2 _ , or r = 3 sin _ θ 2 _ which is the other equation under consideration! What this means is that if a polar representation (r, θ) for the point P satisfies r = 3 sin( θ 2 ), then the representation (−r, θ + π) for P automatically satisfies r = 3 cos _ θ 2 _ . Hence the equations r = 3 sin( θ 2 ) and r = 3 cos( θ 2 ) determine the same set of points in the plane. Our work in Example 11.5.3 justifies the following. Guidelines for Finding Points of Intersection of Graphs of Polar Equations To find the points of intersection of the graphs of two polar equations E 1 and E 2 : • Sketch the graphs of E 1 and E 2 . Check to see if the curves intersect at the origin (pole). • Solve for pairs (r, θ) which satisfy both E 1 and E 2 . • Substitute (θ +2πk) for θ in either one of E 1 or E 2 (but not both) and solve for pairs (r, θ) which satisfy both equations. Keep in mind that k is an integer. • Substitute (−r) for r and (θ + (2k + 1)π) for θ in either one of E 1 or E 2 (but not both) and solve for pairs (r, θ) which satisfy both equations. Keep in mind that k is an integer. 954 Applications of Trigonometry Our last example ties together graphing and points of intersection to describe regions in the plane. Example 11.5.4. Sketch the region in the xy-plane described by the following sets. 1. _ (r, θ) [ 0 ≤ r ≤ 5 sin(2θ), 0 ≤ θ ≤ π 2 _ 2. _ (r, θ) [ 3 ≤ r ≤ 6 cos(2θ), 0 ≤ θ ≤ π 6 _ 3. _ (r, θ) [ 2 + 4 cos(θ) ≤ r ≤ 0, 2π 3 ≤ θ ≤ 4π 3 _ 4. _ (r, θ) [ 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π 6 _ ∪ _ (r, θ) [ 0 ≤ r ≤ 2 −2 sin(θ), π 6 ≤ θ ≤ π 2 _ Solution. Our first step in these problems is to sketch the graphs of the polar equations involved to get a sense of the geometric situation. Since all of the equations in this example are found in either Example 11.5.2 or Example 11.5.3, most of the work is done for us. 1. We know from Example 11.5.2 number 3 that the graph of r = 5 sin(2θ) is a rose. Moreover, we know from our work there that as 0 ≤ θ ≤ π 2 , we are tracing out the ‘leaf’ of the rose which lies in the first quadrant. The inequality 0 ≤ r ≤ 5 sin(2θ) means we want to capture all of the points between the origin (r = 0) and the curve r = 5 sin(2θ) as θ runs through _ 0, π 2 ¸ . Hence, the region we seek is the leaf itself. π 4 π 2 3π 4 π −5 5 θ r x y _ (r, θ) [ 0 ≤ r ≤ 5 sin(2θ), 0 ≤ θ ≤ π 2 _ 2. We know from Example 11.5.3 number 3 that r = 3 and r = 6 cos(2θ) intersect at θ = π 6 , so the region that is being described here is the set of points whose directed distance r from the origin is at least 3 but no more than 6 cos(2θ) as θ runs from 0 to π 6 . In other words, we are looking at the points outside or on the circle (since r ≥ 3) but inside or on the rose (since r ≤ 6 cos(2θ)). We shade the region below. x y θ = π 6 x y r = 3 and r = 6 cos(2θ) _ (r, θ) [ 3 ≤ r ≤ 6 cos(2θ), 0 ≤ θ ≤ π 6 _ 11.5 Graphs of Polar Equations 955 3. From Example 11.5.2 number 2, we know that the graph of r = 2 + 4 cos(θ) is a lima¸con whose ‘inner loop’ is traced out as θ runs through the given values 2π 3 to 4π 3 . Since the values r takes on in this interval are non-positive, the inequality 2 + 4 cos(θ) ≤ r ≤ 0 makes sense, and we are looking for all of the points between the pole r = 0 and the lima¸con as θ ranges over the interval _ 2π 3 , 4π 3 ¸ . In other words, we shade in the inner loop of the lima¸con. π 2 π 3π 2 2π −2 2 4 6 2π 3 4π 3 θ r x y θ = 4π 3 θ = 2π 3 _ (r, θ) [ 2 + 4 cos(θ) ≤ r ≤ 0, 2π 3 ≤ θ ≤ 4π 3 _ 4. We have two regions described here connected with the union symbol ‘∪.’ We shade each in turn and find our final answer by combining the two. In Example 11.5.3, number 1, we found that the curves r = 2 sin(θ) and r = 2 −2 sin(θ) intersect when θ = π 6 . Hence, for the first region, _ (r, θ) [ 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π 6 _ , we are shading the region between the origin (r = 0) out to the circle (r = 2 sin(θ)) as θ ranges from 0 to π 6 , which is the angle of intersection of the two curves. For the second region, _ (r, θ) [ 0 ≤ r ≤ 2 −2 sin(θ), π 6 ≤ θ ≤ π 2 _ , θ picks up where it left off at π 6 and continues to π 2 . In this case, however, we are shading from the origin (r = 0) out to the cardioid r = 2 − 2 sin(θ) which pulls into the origin at θ = π 2 . Putting these two regions together gives us our final answer. x y 1 1 θ = π 6 x y 1 1 r = 2 −2 sin(θ) and r = 2 sin(θ) _ (r, θ) [ 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π 6 _ ∪ _ (r, θ) [ 0 ≤ r ≤ 2 −2 sin(θ), π 6 ≤ θ ≤ π 2 _ 956 Applications of Trigonometry 11.5.1 Exercises In Exercises 1 - 20, plot the graph of the polar equation by hand. Carefully label your graphs. 1. Circle: r = 6 sin(θ) 2. Circle: r = 2 cos(θ) 3. Rose: r = 2 sin(2θ) 4. Rose: r = 4 cos(2θ) 5. Rose: r = 5 sin(3θ) 6. Rose: r = cos(5θ) 7. Rose: r = sin(4θ) 8. Rose: r = 3 cos(4θ) 9. Cardioid: r = 3 −3 cos(θ) 10. Cardioid: r = 5 + 5 sin(θ) 11. Cardioid: r = 2 + 2 cos(θ) 12. Cardioid: r = 1 −sin(θ) 13. Lima¸con: r = 1 −2 cos(θ) 14. Lima¸con: r = 1 −2 sin(θ) 15. Lima¸con: r = 2 √ 3 + 4 cos(θ) 16. Lima¸con: r = 3 −5 cos(θ) 17. Lima¸con: r = 3 −5 sin(θ) 18. Lima¸ con: r = 2 + 7 sin(θ) 19. Lemniscate: r 2 = sin(2θ) 20. Lemniscate: r 2 = 4 cos(2θ) In Exercises 21 - 30, find the exact polar coordinates of the points of intersection of graphs of the polar equations. Remember to check for intersection at the pole (origin). 21. r = 3 cos(θ) and r = 1 + cos(θ) 22. r = 1 + sin(θ) and r = 1 −cos(θ) 23. r = 1 −2 sin(θ) and r = 2 24. r = 1 −2 cos(θ) and r = 1 25. r = 2 cos(θ) and r = 2 √ 3 sin(θ) 26. r = 3 cos(θ) and r = sin(θ) 27. r 2 = 4 cos(2θ) and r = √ 2 28. r 2 = 2 sin(2θ) and r = 1 29. r = 4 cos(2θ) and r = 2 30. r = 2 sin(2θ) and r = 1 In Exercises 31 - 40, sketch the region in the xy-plane described by the given set. 31. ¦(r, θ) [ 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π¦ 32. ¦(r, θ) [ 0 ≤ r ≤ 4 sin(θ), 0 ≤ θ ≤ π¦ 33. _ (r, θ) [ 0 ≤ r ≤ 3 cos(θ), − π 2 ≤ θ ≤ π 2 _ 34. _ (r, θ) [ 0 ≤ r ≤ 2 sin(2θ), 0 ≤ θ ≤ π 2 _ 11.5 Graphs of Polar Equations 957 35. _ (r, θ) [ 0 ≤ r ≤ 4 cos(2θ), − π 4 ≤ θ ≤ π 4 _ 36. _ (r, θ) [ 1 ≤ r ≤ 1 −2 cos(θ), π 2 ≤ θ ≤ 3π 2 _ 37. _ (r, θ) [ 1 + cos(θ) ≤ r ≤ 3 cos(θ), − π 3 ≤ θ ≤ π 3 _ 38. _ (r, θ) [ 1 ≤ r ≤ _ 2 sin(2θ), 13π 12 ≤ θ ≤ 17π 12 _ 39. _ (r, θ) [ 0 ≤ r ≤ 2 √ 3 sin(θ), 0 ≤ θ ≤ π 6 _ ∪ _ (r, θ) [ 0 ≤ r ≤ 2 cos(θ), π 6 ≤ θ ≤ π 2 _ 40. _ (r, θ) [ 0 ≤ r ≤ 2 sin(2θ), 0 ≤ θ ≤ π 12 _ ∪ _ (r, θ) [ 0 ≤ r ≤ 1, π 12 ≤ θ ≤ π 4 _ In Exercises 41 - 50, use set-builder notation to describe the polar region. Assume that the region contains its bounding curves. 41. The region inside the circle r = 5. 42. The region inside the circle r = 5 which lies in Quadrant III. 43. The region inside the left half of the circle r = 6 sin(θ). 44. The region inside the circle r = 4 cos(θ) which lies in Quadrant IV. 45. The region inside the top half of the cardioid r = 3 −3 cos(θ) 46. The region inside the cardioid r = 2 −2 sin(θ) which lies in Quadrants I and IV. 47. The inside of the petal of the rose r = 3 cos(4θ) which lies on the positive x-axis 48. The region inside the circle r = 5 but outside the circle r = 3. 49. The region which lies inside of the circle r = 3 cos(θ) but outside of the circle r = sin(θ) 50. The region in Quadrant I which lies inside both the circle r = 3 as well as the rose r = 6 sin(2θ) While the authors truly believe that graphing polar curves by hand is fundamental to your under- standing of the polar coordinate system, we would be derelict in our duties if we totally ignored the graphing calculator. Indeed, there are some important polar curves which are simply too dif- ficult to graph by hand and that makes the calculator an important tool for your further studies in Mathematics, Science and Engineering. We now give a brief demonstration of how to use the graphing calculator to plot polar curves. The first thing you must do is switch the MODE of your calculator to POL, which stands for “polar”. 958 Applications of Trigonometry This changes the “Y=” menu as seen above in the middle. Let’s plot the polar rose given by r = 3 cos(4θ) from Exercise 8 above. We type the function into the “r=” menu as seen above on the right. We need to set the viewing window so that the curve displays properly, but when we look at the WINDOW menu, we find three extra lines. In order for the calculator to be able to plot r = 3 cos(4θ) in the xy-plane, we need to tell it not only the dimensions which x and y will assume, but we also what values of θ to use. From our previous work, we know that we need 0 ≤ θ ≤ 2π, so we enter the data you see above. (I’ll say more about the θ-step in just a moment.) Hitting GRAPH yields the curve below on the left which doesn’t look quite right. The issue here is that the calculator screen is 96 pixels wide but only 64 pixels tall. To get a true geometric perspective, we need to hit ZOOM SQUARE (seen below in the middle) to produce a more accurate graph which we present below on the right. In function mode, the calculator automatically divided the interval [Xmin, Xmax] into 96 equal subintervals. In polar mode, however, we must specify how to split up the interval [θmin, θmax] using the θstep. For most graphs, a θstep of 0.1 is fine. If you make it too small then the calculator takes a long time to graph. It you make it too big, you get chunky garbage like this. You will need to experiment with the settings in order to get a nice graph. Exercises 51 - 60 give you some curves to graph using your calculator. Notice that some of them have explicit bounds on θ and others do not. 11.5 Graphs of Polar Equations 959 51. r = θ, 0 ≤ θ ≤ 12π 52. r = ln(θ), 1 ≤ θ ≤ 12π 53. r = e .1θ , 0 ≤ θ ≤ 12π 54. r = θ 3 −θ, −1.2 ≤ θ ≤ 1.2 55. r = sin(5θ) −3 cos(θ) 56. r = sin 3 _ θ 2 _ + cos 2 _ θ 3 _ 57. r = arctan(θ), −π ≤ θ ≤ π 58. r = 1 1 −cos(θ) 59. r = 1 2 −cos(θ) 60. r = 1 2 −3 cos(θ) 61. How many petals does the polar rose r = sin(2θ) have? What about r = sin(3θ), r = sin(4θ) and r = sin(5θ)? With the help of your classmates, make a conjecture as to how many petals the polar rose r = sin(nθ) has for any natural number n. Replace sine with cosine and repeat the investigation. How many petals does r = cos(nθ) have for each natural number n? Looking back through the graphs in the section, it’s clear that many polar curves enjoy various forms of symmetry. However, classifying symmetry for polar curves is not as straight-forward as it was for equations back on page 26. In Exercises 62 - 64, we have you and your classmates explore some of the more basic forms of symmetry seen in common polar curves. 62. Show that if f is even 17 then the graph of r = f(θ) is symmetric about the x-axis. (a) Show that f(θ) = 2 + 4 cos(θ) is even and verify that the graph of r = 2 + 4 cos(θ) is indeed symmetric about the x-axis. (See Example 11.5.2 number 2.) (b) Show that f(θ) = 3 sin _ θ 2 _ is not even, yet the graph of r = 3 sin _ θ 2 _ is symmetric about the x-axis. (See Example 11.5.3 number 4.) 63. Show that if f is odd 18 then the graph of r = f(θ) is symmetric about the origin. (a) Show that f(θ) = 5 sin(2θ) is odd and verify that the graph of r = 5 sin(2θ) is indeed symmetric about the origin. (See Example 11.5.2 number 3.) (b) Show that f(θ) = 3 cos _ θ 2 _ is not odd, yet the graph of r = 3 cos _ θ 2 _ is symmetric about the origin. (See Example 11.5.3 number 4.) 64. Show that if f(π − θ) = f(θ) for all θ in the domain of f then the graph of r = f(θ) is symmetric about the y-axis. (a) For f(θ) = 4 − 2 sin(θ), show that f(π − θ) = f(θ) and the graph of r = 4 − 2 sin(θ) is symmetric about the y-axis, as required. (See Example 11.5.2 number 1.) 17 Recall that this means f(−θ) = f(θ) for θ in the domain of f. 18 Recall that this means f(−θ) = −f(θ) for θ in the domain of f. 960 Applications of Trigonometry (b) For f(θ) = 5 sin(2θ), show that f _ π − π 4 _ ,= f _ π 4 _ , yet the graph of r = 5 sin(2θ) is symmetric about the y-axis. (See Example 11.5.2 number 3.) In Section 1.7, we discussed transformations of graphs. In Exercise 65 we have you and your classmates explore transformations of polar graphs. 65. For Exercises 65a and 65b below, let f(θ) = cos(θ) and g(θ) = 2 −sin(θ). (a) Using your graphing calculator, compare the graph of r = f(θ) to each of the graphs of r = f _ θ + π 4 _ , r = f _ θ + 3π 4 _ , r = f _ θ − π 4 _ and r = f _ θ − 3π 4 _ . Repeat this process for g(θ). In general, how do you think the graph of r = f(θ + α) compares with the graph of r = f(θ)? (b) Using your graphing calculator, compare the graph of r = f(θ) to each of the graphs of r = 2f (θ), r = 1 2 f (θ), r = −f (θ) and r = −3f(θ). Repeat this process for g(θ). In general, how do you think the graph of r = k f(θ) compares with the graph of r = f(θ)? (Does it matter if k > 0 or k < 0?) 66. In light of Exercises 62 - 64, how would the graph of r = f(−θ) compare with the graph of r = f(θ) for a generic function f? What about the graphs of r = −f(θ) and r = f(θ)? What about r = f(θ) and r = f(π−θ)? Test out your conjectures using a variety of polar functions found in this section with the help of a graphing utility. 67. With the help of your classmates, research cardioid microphones. 68. Back in Section 1.2, in the paragraph before Exercise 53, we gave you this link to a fascinating list of curves. Some of these curves have polar representations which we invite you and your classmates to research. 11.5 Graphs of Polar Equations 961 11.5.2 Answers 1. Circle: r = 6 sin(θ) x y −6 6 −6 6 2. Circle: r = 2 cos(θ) x y −2 2 −2 2 3. Rose: r = 2 sin(2θ) x y −2 2 −2 2 4. Rose: r = 4 cos(2θ) x y −4 4 −4 4 θ = π 4 θ = 3π 4 5. Rose: r = 5 sin(3θ) x y −5 5 −5 5 θ = π 3 θ = 2π 3 6. Rose: r = cos(5θ) x y −1 1 −1 1 θ = π 10 θ = 3π 10 θ = 7π 10 θ = 9π 10 962 Applications of Trigonometry 7. Rose: r = sin(4θ) x y −1 1 −1 1 θ = π 4 θ = 3π 4 8. Rose: r = 3 cos(4θ) x y −3 3 −3 3 θ = π 8 θ = 3π 8 θ = 5π 8 θ = 7π 8 9. Cardioid: r = 3 −3 cos(θ) x y −6 −3 3 6 −6 −3 3 6 10. Cardioid: r = 5 + 5 sin(θ) x y −10 −5 5 10 −10 −5 5 10 11. Cardioid: r = 2 + 2 cos(θ) x y −4 −2 2 4 −4 −2 2 4 12. Cardioid: r = 1 −sin(θ) x y −2 −1 1 2 −2 −1 1 2 11.5 Graphs of Polar Equations 963 13. Lima¸con: r = 1 −2 cos(θ) x y −3 −1 1 3 −3 −1 1 3 θ = π 3 θ = 5π 3 14. Lima¸con: r = 1 −2 sin(θ) x y −3 −1 1 3 −3 −1 1 3 θ = π 6 θ = 5π 6 15. Lima¸con: r = 2 √ 3 + 4 cos(θ) x y −2 √ 3 −4 2 √ 3 + 4 −2 √ 3 −4 −2 √ 3 2 √ 3 2 √ 3 + 4 θ = 7π 6 θ = 5π 6 16. Lima¸con: r = 3 −5 cos(θ) x y −8 −3 3 8 −8 −2 8 θ = arccos 3 5 θ = 2π − arccos 3 5 17. Lima¸con: r = 3 −5 sin(θ) x y −8 −3 3 8 −8 −2 8 θ = arcsin 3 5 θ = π − arcsin 3 5 18. Lima¸con: r = 2 + 7 sin(θ) x y −9 −2 2 9 −9 5 9 θ = π + arcsin 2 7 θ = 2π − arcsin 2 7 964 Applications of Trigonometry 19. Lemniscate: r 2 = sin(2θ) x y −1 1 −1 1 20. Lemniscate: r 2 = 4 cos(2θ) x y −2 2 −2 2 θ = π 4 θ = 3π 4 21. r = 3 cos(θ) and r = 1 + cos(θ) x y −3 −2 −1 1 2 3 −3 −2 −1 1 2 3 _ 3 2 , π 3 _ , _ 3 2 , 5π 3 _ , pole 22. r = 1 + sin(θ) and r = 1 −cos(θ) x y −2 −1 1 2 −2 −1 1 2 _ 2 + √ 2 2 , 3π 4 _ , _ 2 − √ 2 2 , 7π 4 _ , pole 11.5 Graphs of Polar Equations 965 23. r = 1 −2 sin(θ) and r = 2 x y −3 −1 1 3 −3 −1 1 3 _ 2, 7π 6 _ , _ 2, 11π 6 _ 24. r = 1 −2 cos(θ) and r = 1 x y −3 −1 1 3 −3 −1 1 3 _ 1, π 2 _ , _ 1, 3π 2 _ , (−1, 0) 25. r = 2 cos(θ) and r = 2 √ 3 sin(θ) x y −3 −2 −1 1 2 3 −4 −3 −2 −1 1 2 3 4 _ √ 3, π 6 _ , pole 966 Applications of Trigonometry 26. r = 3 cos(θ) and r = sin(θ) x y −3 −2 −1 1 2 3 −3 −2 −1 1 2 3 _ 3 √ 10 10 , arctan(3) _ , pole 27. r 2 = 4 cos(2θ) and r = √ 2 x y −2 2 −2 2 _ √ 2, π 6 _ , _ √ 2, 5π 6 _ , _ √ 2, 7π 6 _ , _ √ 2, 11π 6 _ 28. r 2 = 2 sin(2θ) and r = 1 x y − √ 2 −1 1 √ 2 − √ 2 −1 1 √ 2 _ 1, π 12 _ , _ 1, 5π 12 _ , _ 1, 13π 12 _ , _ 1, 17π 12 _ 11.5 Graphs of Polar Equations 967 29. r = 4 cos(2θ) and r = 2 x y −4 4 −4 4 _ 2, π 6 _ , _ 2, 5π 6 _ , _ 2, 7π 6 _ , _ 2, 11π 6 _ , _ −2, π 3 _ , _ −2, 2π 3 _ , _ −2, 4π 3 _ , _ −2, 5π 3 _ 30. r = 2 sin(2θ) and r = 1 x y −2 2 −2 2 _ 1, π 12 _ , _ 1, 5π 12 _ , _ 1, 13π 12 _ , _ 1, 17π 12 _ , _ −1, 7π 12 _ , _ −1, 11π 12 _ , _ −1, 19π 12 _ , _ −1, 23π 12 _ 968 Applications of Trigonometry 31. ¦(r, θ) [ 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π¦ x y −3 −2 −1 1 2 3 −3 −2 −1 1 2 3 32. ¦(r, θ) [ 0 ≤ r ≤ 4 sin(θ), 0 ≤ θ ≤ π¦ x y −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 33. _ (r, θ) [ 0 ≤ r ≤ 3 cos(θ), − π 2 ≤ θ ≤ π 2 _ x y −3 −2 −1 1 2 3 −3 −2 −1 1 2 3 34. _ (r, θ) [ 0 ≤ r ≤ 2 sin(2θ), 0 ≤ θ ≤ π 2 _ x y −2 2 −2 2 35. _ (r, θ) [ 0 ≤ r ≤ 4 cos(2θ), − π 4 ≤ θ ≤ π 4 _ x y −4 4 −4 4 36. _ (r, θ) [ 1 ≤ r ≤ 1 −2 cos(θ), π 2 ≤ θ ≤ 3π 2 _ x y −3 −1 1 3 −3 −1 1 3 11.5 Graphs of Polar Equations 969 37. _ (r, θ) [ 1 + cos(θ) ≤ r ≤ 3 cos(θ), − π 3 ≤ θ ≤ π 3 _ x y −3 −2 −1 1 2 3 −3 −2 −1 1 2 3 38. _ (r, θ) [ 1 ≤ r ≤ _ 2 sin(2θ), 13π 12 ≤ θ ≤ 17π 12 _ x y − √ 2 −1 1 √ 2 − √ 2 −1 1 √ 2 39. _ (r, θ) [ 0 ≤ r ≤ 2 √ 3 sin(θ), 0 ≤ θ ≤ π 6 _ ∪ _ (r, θ) [ 0 ≤ r ≤ 2 cos(θ), π 6 ≤ θ ≤ π 2 _ x y −3 −2 −1 1 2 3 −4 −3 −2 −1 1 2 3 4 970 Applications of Trigonometry 40. _ (r, θ) [ 0 ≤ r ≤ 2 sin(2θ), 0 ≤ θ ≤ π 12 _ ∪ _ (r, θ) [ 0 ≤ r ≤ 1, π 12 ≤ θ ≤ π 4 _ x y −2 2 −2 2 41. ¦(r, θ) [ 0 ≤ r ≤ 5, 0 ≤ θ ≤ 2π¦ 42. _ (r, θ) [ 0 ≤ r ≤ 5, π ≤ θ ≤ 3π 2 _ 43. _ (r, θ) [ 0 ≤ r ≤ 6 sin(θ), π 2 ≤ θ ≤ π _ 44. _ (r, θ) [ 4 cos(θ) ≤ r ≤ 0, π 2 ≤ θ ≤ π _ 45. ¦(r, θ) [ 0 ≤ r ≤ 3 −3 cos(θ), 0 ≤ θ ≤ π¦ 46. _ (r, θ) [ 0 ≤ r ≤ 2 −2 sin(θ), 0 ≤ θ ≤ π 2 _ ∪ _ (r, θ) [ 0 ≤ r ≤ 2 −2 sin(θ), 3π 2 ≤ θ ≤ 2π _ or _ (r, θ) [ 0 ≤ r ≤ 2 −2 sin(θ), 3π 2 ≤ θ ≤ 5π 2 _ 47. _ (r, θ) [ 0 ≤ r ≤ 3 cos(4θ), 0 ≤ θ ≤ π 8 _ ∪ _ (r, θ) [ 0 ≤ r ≤ 3 cos(4θ), 15π 8 ≤ θ ≤ 2π _ or _ (r, θ) [ 0 ≤ r ≤ 3 cos(4θ), − π 8 ≤ θ ≤ π 8 _ 48. ¦(r, θ) [ 3 ≤ r ≤ 5, 0 ≤ θ ≤ 2π¦ 49. _ (r, θ) [ 0 ≤ r ≤ 3 cos(θ), − π 2 ≤ θ ≤ 0 _ ∪ ¦(r, θ) [ sin(θ) ≤ r ≤ 3 cos(θ), 0 ≤ θ ≤ arctan(3)¦ 50. _ (r, θ) [ 0 ≤ r ≤ 6 sin(2θ), 0 ≤ θ ≤ π 12 _ ∪ _ (r, θ) [ 0 ≤ r ≤ 3, π 12 ≤ θ ≤ 5π 12 _ ∪ _ (r, θ) [ 0 ≤ r ≤ 6 sin(2θ), 5π 12 ≤ θ ≤ π 2 _ 11.6 Hooked on Conics Again 971 11.6 Hooked on Conics Again In this section, we revisit our friends the Conic Sections which we began studying in Chapter 7. Our first task is to formalize the notion of rotating axes so this subsection is actually a follow-up to Example 8.3.3 in Section 8.3. In that example, we saw that the graph of y = 2 x is actually a hyperbola. More specifically, it is the hyperbola obtained by rotating the graph of x 2 − y 2 = 4 counter-clockwise through a 45 ◦ angle. Armed with polar coordinates, we can generalize the process of rotating axes as shown below. 11.6.1 Rotation of Axes Consider the x- and y-axes below along with the dashed x - and y -axes obtained by rotating the x- and y-axes counter-clockwise through an angle θ and consider the point P(x, y). The coordinates (x, y) are rectangular coordinates and are based on the x- and y-axes. Suppose we wished to find rectangular coordinates based on the x - and y -axes. That is, we wish to determine P(x , y ). While this seems like a formidable challenge, it is nearly trivial if we use polar coordinates. Consider the angle φ whose initial side is the positive x -axis and whose terminal side contains the point P. x y x y P(x, y) = P(x , y ) θ θ φ We relate P(x, y) and P(x , y ) by converting them to polar coordinates. Converting P(x, y) to polar coordinates with r > 0 yields x = r cos(θ + φ) and y = r sin(θ + φ). To convert the point P(x , y ) into polar coordinates, we first match the polar axis with the positive x -axis, choose the same r > 0 (since the origin is the same in both systems) and get x = r cos(φ) and y = r sin(φ). Using the sum formulas for sine and cosine, we have x = r cos(θ +φ) = r cos(θ) cos(φ) −r sin(θ) sin(φ) Sum formula for cosine = (r cos(φ)) cos(θ) −(r sin(φ)) sin(θ) = x cos(θ) −y sin(θ) Since x = r cos(φ) and y = r sin(φ) 972 Applications of Trigonometry Similarly, using the sum formula for sine we get y = x sin(θ) +y cos(θ). These equations enable us to easily convert points with x y -coordinates back into xy-coordinates. They also enable us to easily convert equations in the variables x and y into equations in the variables in terms of x and y . 1 If we want equations which enable us to convert points with xy-coordinates into x y -coordinates, we need to solve the system _ x cos(θ) −y sin(θ) = x x sin(θ) +y cos(θ) = y for x and y . Perhaps the cleanest way 2 to solve this system is to write it as a matrix equation. Using the machinery developed in Section 8.4, we write the above system as the matrix equation AX = X where A = _ cos(θ) −sin(θ) sin(θ) cos(θ) _ , X = _ x y _ , X = _ x y _ Since det(A) = (cos(θ))(cos(θ)) − (−sin(θ))(sin(θ)) = cos 2 (θ) + sin 2 (θ) = 1, the determinant of A is not zero so A is invertible and X = A −1 X. Using the formula given in Equation 8.2 with det(A) = 1, we find A −1 = _ cos(θ) sin(θ) −sin(θ) cos(θ) _ so that X = A −1 X _ x y _ = _ cos(θ) sin(θ) −sin(θ) cos(θ) _ _ x y _ _ x y _ = _ xcos(θ) +y sin(θ) −xsin(θ) +y cos(θ) _ From which we get x = xcos(θ) +y sin(θ) and y = −xsin(θ) +y cos(θ). To summarize, Theorem 11.9. Rotation of Axes: Suppose the positive x and y axes are rotated counter- clockwise through an angle θ to produce the axes x and y , respectively. Then the coordinates P(x, y) and P(x , y ) are related by the following systems of equations _ x = x cos(θ) −y sin(θ) y = x sin(θ) +y cos(θ) and _ x = xcos(θ) +y sin(θ) y = −xsin(θ) +y cos(θ) We put the formulas in Theorem 11.9 to good use in the following example. 1 Sound familiar? In Section 11.4, the equations x = r cos(θ) and y = r sin(θ) make it easy to convert points from polar coordinates into rectangular coordinates, and they make it easy to convert equations from rectangular coordinates into polar coordinates. 2 We could, of course, interchange the roles of x and x , y and y and replace φ with −φ to get x and y in terms of x and y, but that seems like cheating. The matrix A introduced here is revisited in the Exercises. 11.6 Hooked on Conics Again 973 Example 11.6.1. Suppose the x- and y- axes are both rotated counter-clockwise through the angle θ = π 3 to produce the x - and y - axes, respectively. 1. Let P(x, y) = (2, −4) and find P(x , y ). Check your answer algebraically and graphically. 2. Convert the equation 21x 2 + 10xy √ 3 + 31y 2 = 144 to an equation in x and y and graph. Solution. 1. If P(x, y) = (2, −4) then x = 2 and y = −4. Using these values for x and y along with θ = π 3 , Theorem 11.9 gives x = xcos(θ) +y sin(θ) = 2 cos _ π 3 _ +(−4) sin _ π 3 _ which simplifies to x = 1 − 2 √ 3. Similarly, y = −xsin(θ) + y cos(θ) = (−2) sin _ π 3 _ + (−4) cos _ π 3 _ which gives y = − √ 3 −2 = −2 − √ 3. Hence P(x , y ) = _ 1 −2 √ 3, −2 − √ 3 _ . To check our answer algebraically, we use the formulas in Theorem 11.9 to convert P(x , y ) = _ 1 −2 √ 3, −2 − √ 3 _ back into x and y coordinates. We get x = x cos(θ) −y sin(θ) = (1 −2 √ 3) cos _ π 3 _ −(−2 − √ 3) sin _ π 3 _ = _ 1 2 − √ 3 _ − _ − √ 3 − 3 2 _ = 2 Similarly, using y = x sin(θ) +y cos(θ), we obtain y = −4 as required. To check our answer graphically, we sketch in the x -axis and y -axis to see if the new coordinates P(x , y ) = _ 1 −2 √ 3, −2 − √ 3 _ ≈ (−2.46, −3.73) seem reasonable. Our graph is below. x y x y P(x, y) = (2, −4) P(x , y ) ≈ (−2.46, −3.73) π 3 π 3 2. To convert the equation 21x 2 +10xy √ 3+31y 2 = 144 to an equation in the variables x and y , we substitute x = x cos _ π 3 _ −y sin _ π 3 _ = x 2 − y √ 3 2 and y = x sin _ π 3 _ +y cos _ π 3 _ = x √ 3 2 + y 2 974 Applications of Trigonometry and simplify. While this is by no means a trivial task, it is nothing more than a hefty dose of Beginning Algebra. We will not go through the entire computation, but rather, the reader should take the time to do it. Start by verifying that x 2 = (x ) 2 4 − x y √ 3 2 + 3(y ) 2 4 , xy = (x ) 2 √ 3 4 − x y 2 − (y ) 2 √ 3 4 , y 2 = 3(x ) 2 4 + x y √ 3 2 + (y ) 2 4 To our surprise and delight, the equation 21x 2 + 10xy √ 3 + 31y 2 = 144 in xy-coordinates reduces to 36(x ) 2 + 16(y ) 2 = 144, or (x ) 2 4 + (y ) 2 9 = 1 in x y -coordinates. The latter is an ellipse centered at (0, 0) with vertices along the y -axis with (x y -coordinates) (0, ±3) and whose minor axis has endpoints with (x y -coordinates) (±2, 0). We graph it below. x y x y π 3 π 3 21x 2 + 10xy √ 3 + 31y 2 = 144 The elimination of the troublesome ‘xy’ term from the equation 21x 2 + 10xy √ 3 + 31y 2 = 144 in Example 11.6.1 number 2 allowed us to graph the equation by hand using what we learned in Chapter 7. It is natural to wonder if we can always do this. That is, given an equation of the form Ax 2 +Bxy+Cy 2 +Dx+Ey+F = 0, with B ,= 0, is there an angle θ so that if we rotate the x and y- axes counter-clockwise through that angle θ, the equation in the rotated variables x and y contains no x y term? To explore this conjecture, we make the usual substitutions x = x cos(θ) −y sin(θ) and y = x sin(θ) + y cos(θ) into the equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 and set the coefficient of the x y term equal to 0. Terms containing x y in this expression will come from the first three terms of the equation: Ax 2 , Bxy and Cy 2 . We leave it to the reader to verify that x 2 = (x ) 2 cos 2 (θ) −2x y cos(θ) sin(θ) + (y ) 2 sin(θ) xy = (x ) 2 cos(θ) sin(θ) +x y _ cos 2 (θ) −sin 2 (θ) _ −(y ) 2 cos(θ) sin(θ) y 2 = (x ) 2 sin 2 (θ) + 2x y cos(θ) sin(θ) + (y ) 2 cos 2 (θ) 11.6 Hooked on Conics Again 975 The contribution to the x y -term fromAx 2 is −2Acos(θ) sin(θ), fromBxy it is B _ cos 2 (θ) −sin 2 (θ) _ , and from Cy 2 it is 2C cos(θ) sin(θ). Equating the x y -term to 0, we get −2Acos(θ) sin(θ) +B _ cos 2 (θ) −sin 2 (θ) _ + 2C cos(θ) sin(θ) = 0 −Asin(2θ) +Bcos(2θ) +C sin(2θ) = 0 Double Angle Identities From this, we get Bcos(2θ) = (A − C) sin(2θ), and our goal is to solve for θ in terms of the coefficients A, B and C. Since we are assuming B ,= 0, we can divide both sides of this equation by B. To solve for θ we would like to divide both sides of the equation by sin(2θ), provided of course that we have assurances that sin(2θ) ,= 0. If sin(2θ) = 0, then we would have Bcos(2θ) = 0, and since B ,= 0, this would force cos(2θ) = 0. Since no angle θ can have both sin(2θ) = 0 and cos(2θ) = 0, we can safely assume 3 sin(2θ) ,= 0. We get cos(2θ) sin(2θ) = A−C B , or cot(2θ) = A−C B . We have just proved the following theorem. Theorem 11.10. The equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 with B ,= 0 can be transformed into an equation in variables x and y without any x y terms by rotating the x- and y- axes counter-clockwise through an angle θ which satisfies cot(2θ) = A−C B . We put Theorem 11.10 to good use in the following example. Example 11.6.2. Graph the following equations. 1. 5x 2 + 26xy + 5y 2 −16x √ 2 + 16y √ 2 −104 = 0 2. 16x 2 + 24xy + 9y 2 + 15x −20y = 0 Solution. 1. Since the equation 5x 2 + 26xy + 5y 2 − 16x √ 2 + 16y √ 2 − 104 = 0 is already given to us in the form required by Theorem 11.10, we identify A = 5, B = 26 and C = 5 so that cot(2θ) = A−C B = 5−5 26 = 0. This means cot(2θ) = 0 which gives θ = π 4 + π 2 k for integers k. We choose θ = π 4 so that our rotation equations are x = x √ 2 2 − y √ 2 2 and y = x √ 2 2 + y √ 2 2 . The reader should verify that x 2 = (x ) 2 2 −x y + (y ) 2 2 , xy = (x ) 2 2 − (y ) 2 2 , y 2 = (x ) 2 2 +x y + (y ) 2 2 Making the other substitutions, we get that 5x 2 + 26xy + 5y 2 − 16x √ 2 + 16y √ 2 − 104 = 0 reduces to 18(x ) 2 −8(y ) 2 +32y −104 = 0, or (x ) 2 4 − (y −2) 2 9 = 1. The latter is the equation of a hyperbola centered at the x y -coordinates (0, 2) opening in the x direction with vertices (±2, 2) (in x y -coordinates) and asymptotes y = ± 3 2 x + 2. We graph it below. 3 The reader is invited to think about the case sin(2θ) = 0 geometrically. What happens to the axes in this case? 976 Applications of Trigonometry 2. From 16x 2 + 24xy + 9y 2 + 15x − 20y = 0, we get A = 16, B = 24 and C = 9 so that cot(2θ) = 7 24 . Since this isn’t one of the values of the common angles, we will need to use inverse functions. Ultimately, we need to find cos(θ) and sin(θ), which means we have two options. If we use the arccotangent function immediately, after the usual calculations we get θ = 1 2 arccot _ 7 24 _ . To get cos(θ) and sin(θ) from this, we would need to use half angle identities. Alternatively, we can start with cot(2θ) = 7 24 , use a double angle identity, and then go after cos(θ) and sin(θ). We adopt the second approach. From cot(2θ) = 7 24 , we have tan(2θ) = 24 7 . Using the double angle identity for tangent, we have 2 tan(θ) 1−tan 2 (θ) = 24 7 , which gives 24 tan 2 (θ) +14 tan(θ) −24 = 0. Factoring, we get 2(3 tan(θ) +4)(4 tan(θ) −3) = 0 which gives tan(θ) = − 4 3 or tan(θ) = 3 4 . While either of these values of tan(θ) satisfies the equation cot(2θ) = 7 24 , we choose tan(θ) = 3 4 , since this produces an acute angle, 4 θ = arctan _ 3 4 _ . To find the rotation equations, we need cos(θ) = cos _ arctan _ 3 4 __ and sin(θ) = sin _ arctan _ 3 4 __ . Using the techniques developed in Section 10.6 we get cos(θ) = 4 5 and sin(θ) = 3 5 . Our rotation equations are x = x cos(θ) − y sin(θ) = 4x 5 − 3y 5 and y = x sin(θ) + y cos(θ) = 3x 5 + 4y 5 . As usual, we now substitute these quantities into 16x 2 + 24xy + 9y 2 + 15x − 20y = 0 and simplify. As a first step, the reader can verify x 2 = 16(x ) 2 25 − 24x y 25 + 9(y ) 2 25 , xy = 12(x ) 2 25 + 7x y 25 − 12(y ) 2 25 , y 2 = 9(x ) 2 25 + 24x y 25 + 16(y ) 2 25 Once the dust settles, we get 25(x ) 2 − 25y = 0, or y = (x ) 2 , whose graph is a parabola opening along the positive y -axis with vertex (0, 0). We graph this equation below. x y x y θ = π 4 5x 2 + 26xy + 5y 2 −16x √ 2 + 16y √ 2 −104 = 0 x y x y θ = arctan _ 3 4 _ 16x 2 + 24xy + 9y 2 + 15x −20y = 0 4 As usual, there are infinitely many solutions to tan(θ) = 3 4 . We choose the acute angle θ = arctan 3 4 . The reader is encouraged to think about why there is always at least one acute answer to cot(2θ) = A−C B and what this means geometrically in terms of what we are trying to accomplish by rotating the axes. The reader is also encouraged to keep a sharp lookout for the angles which satisfy tan(θ) = − 4 3 in our final graph. (Hint: 3 4 − 4 3 = −1.) 11.6 Hooked on Conics Again 977 We note that even though the coefficients of x 2 and y 2 were both positive numbers in parts 1 and 2 of Example 11.6.2, the graph in part 1 turned out to be a hyperbola and the graph in part 2 worked out to be a parabola. Whereas in Chapter 7, we could easily pick out which conic section we were dealing with based on the presence (or absence) of quadratic terms and their coefficients, Example 11.6.2 demonstrates that all bets are off when it comes to conics with an xy term which require rotation of axes to put them into a more standard form. Nevertheless, it is possible to determine which conic section we have by looking at a special, familiar combination of the coefficients of the quadratic terms. We have the following theorem. Theorem 11.11. Suppose the equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 describes a non-degenerate conic section. a • If B 2 −4AC > 0 then the graph of the equation is a hyperbola. • If B 2 −4AC = 0 then the graph of the equation is a parabola. • If B 2 −4AC < 0 then the graph of the equation is an ellipse or circle. a Recall that this means its graph is either a circle, parabola, ellipse or hyperbola. See page 497. As you may expect, the quantity B 2 −4AC mentioned in Theorem 11.11 is called the discriminant of the conic section. While we will not attempt to explain the deep Mathematics which produces this ‘coincidence’, we will at least work through the proof of Theorem 11.11 mechanically to show that it is true. 5 First note that if the coefficient B = 0 in the equation Ax 2 +Bxy+Cy 2 +Dx+Ey+F = 0, Theorem 11.11 reduces to the result presented in Exercise 34 in Section 7.5, so we proceed here under the assumption that B ,= 0. We rotate the xy-axes counter-clockwise through an angle θ which satisfies cot(2θ) = A−C B to produce an equation with no x y -term in accordance with Theorem 11.10: A (x ) 2 + C(y ) 2 + Dx + Ey + F = 0. In this form, we can invoke Exercise 34 in Section 7.5 once more using the product A C . Our goal is to find the product A C in terms of the coefficients A, B and C in the original equation. To that end, we make the usual substitutions x = x cos(θ) − y sin(θ) y = x sin(θ) + y cos(θ) into Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. We leave it to the reader to show that, after gathering like terms, the coefficient A on (x ) 2 and the coefficient C on (y ) 2 are A = Acos 2 (θ) +Bcos(θ) sin(θ) +C sin 2 (θ) C = Asin 2 (θ) −Bcos(θ) sin(θ) +C cos 2 (θ) In order to make use of the condition cot(2θ) = A−C B , we rewrite our formulas for A and C using the power reduction formulas. After some regrouping, we get 2A = [(A+C) + (A−C) cos(2θ)] +Bsin(2θ) 2C = [(A+C) −(A−C) cos(2θ)] −Bsin(2θ) Next, we try to make sense of the product (2A )(2C ) = ¦[(A+C) + (A−C) cos(2θ)] +Bsin(2θ)¦ ¦[(A+C) −(A−C) cos(2θ)] −Bsin(2θ)¦ 5 We hope that someday you get to see why this works the way it does. 978 Applications of Trigonometry We break this product into pieces. First, we use the difference of squares to multiply the ‘first’ quantities in each factor to get [(A+C) + (A−C) cos(2θ)] [(A+C) −(A−C) cos(2θ)] = (A+C) 2 −(A−C) 2 cos 2 (2θ) Next, we add the product of the ‘outer’ and ‘inner’ quantities in each factor to get −Bsin(2θ) [(A+C) + (A−C) cos(2θ)] +Bsin(2θ) [(A+C) −(A−C) cos(2θ)] = −2B(A−C) cos(2θ) sin(2θ) The product of the ‘last’ quantity in each factor is (Bsin(2θ))(−Bsin(2θ)) = −B 2 sin 2 (2θ). Putting all of this together yields 4A C = (A+C) 2 −(A−C) 2 cos 2 (2θ) −2B(A−C) cos(2θ) sin(2θ) −B 2 sin 2 (2θ) From cot(2θ) = A−C B , we get cos(2θ) sin(2θ) = A−C B , or (A−C) sin(2θ) = Bcos(2θ). We use this substitution twice along with the Pythagorean Identity cos 2 (2θ) = 1 −sin 2 (2θ) to get 4A C = (A+C) 2 −(A−C) 2 cos 2 (2θ) −2B(A−C) cos(2θ) sin(2θ) −B 2 sin 2 (2θ) = (A+C) 2 −(A−C) 2 _ 1 −sin 2 (2θ) ¸ −2Bcos(2θ)Bcos(2θ) −B 2 sin 2 (2θ) = (A+C) 2 −(A−C) 2 + (A−C) 2 sin 2 (2θ) −2B 2 cos 2 (2θ) −B 2 sin 2 (2θ) = (A+C) 2 −(A−C) 2 + [(A−C) sin(2θ)] 2 −2B 2 cos 2 (2θ) −B 2 sin 2 (2θ) = (A+C) 2 −(A−C) 2 + [Bcos(2θ)] 2 −2B 2 cos 2 (2θ) −B 2 sin 2 (2θ) = (A+C) 2 −(A−C) 2 +B 2 cos 2 (2θ) −2B 2 cos 2 (2θ) −B 2 sin 2 (2θ) = (A+C) 2 −(A−C) 2 −B 2 cos 2 (2θ) −B 2 sin 2 (2θ) = (A+C) 2 −(A−C) 2 −B 2 _ cos 2 (2θ) + sin 2 (2θ) ¸ = (A+C) 2 −(A−C) 2 −B 2 = _ A 2 + 2AC +C 2 _ − _ A 2 −2AC +C 2 _ −B 2 = 4AC −B 2 Hence, B 2 −4AC = −4A C , so the quantity B 2 −4AC has the opposite sign of A C . The result now follows by applying Exercise 34 in Section 7.5. Example 11.6.3. Use Theorem 11.11 to classify the graphs of the following non-degenerate conics. 1. 21x 2 + 10xy √ 3 + 31y 2 = 144 2. 5x 2 + 26xy + 5y 2 −16x √ 2 + 16y √ 2 −104 = 0 3. 16x 2 + 24xy + 9y 2 + 15x −20y = 0 Solution. This is a straightforward application of Theorem 11.11. 11.6 Hooked on Conics Again 979 1. We have A = 21, B = 10 √ 3 and C = 31 so B 2 −4AC = (10 √ 3) 2 −4(21)(31) = −2304 < 0. Theorem 11.11 predicts the graph is an ellipse, which checks with our work from Example 11.6.1 number 2. 2. Here, A = 5, B = 26 and C = 5, so B 2 − 4AC = 26 2 − 4(5)(5) = 576 > 0. Theorem 11.11 classifies the graph as a hyperbola, which matches our answer to Example 11.6.2 number 1. 3. Finally, we have A = 16, B = 24 and C = 9 which gives 24 2 −4(16)(9) = 0. Theorem 11.11 tells us that the graph is a parabola, matching our result from Example 11.6.2 number 2. 11.6.2 The Polar Form of Conics In this subsection, we start from scratch to reintroduce the conic sections from a more unified perspective. We have our ‘new’ definition below. Definition 11.1. Given a fixed line L, a point F not on L, and a positive number e, a conic section is the set of all points P such that the distance from P to F the distance from P to L = e The line L is called the directrix of the conic section, the point F is called a focus of the conic section, and the constant e is called the eccentricity of the conic section. We have seen the notions of focus and directrix before in the definition of a parabola, Definition 7.3. There, a parabola is defined as the set of points equidistant from the focus and directrix, giving an eccentricity e = 1 according to Definition 11.1. We have also seen the concept of eccentricity before. It was introduced for ellipses in Definition 7.5 in Section 7.4, and later extended to hyperbolas in Exercise 31 in Section 7.5. There, e was also defined as a ratio of distances, though in these cases the distances involved were measurements from the center to a focus and from the center to a vertex. One way to reconcile the ‘old’ ideas of focus, directrix and eccentricity with the ‘new’ ones presented in Definition 11.1 is to derive equations for the conic sections using Definition 11.1 and compare these parameters with what we know from Chapter 7. We begin by assuming the conic section has eccentricity e, a focus F at the origin and that the directrix is the vertical line x = −d as in the figure below. y x P(r, θ) x = −d r cos(θ) d O = F θ r 980 Applications of Trigonometry Using a polar coordinate representation P(r, θ) for a point on the conic with r > 0, we get e = the distance from P to F the distance from P to L = r d +r cos(θ) so that r = e(d +r cos(θ)). Solving this equation for r, yields r = ed 1 −e cos(θ) At this point, we convert the equation r = e(d + r cos(θ)) back into a rectangular equation in the variables x and y. If e > 0, but e ,= 1, the usual conversion process outlined in Section 11.4 gives 6 _ _ 1 −e 2 _ 2 e 2 d 2 _ _ x − e 2 d 1 −e 2 _ 2 + _ 1 −e 2 e 2 d 2 _ y 2 = 1 We leave it to the reader to show if 0 < e < 1, this is the equation of an ellipse centered at _ e 2 d 1−e 2 , 0 _ with major axis along the x-axis. Using the notation from Section 7.4, we have a 2 = e 2 d 2 (1−e 2 ) 2 and b 2 = e 2 d 2 1−e 2 , so the major axis has length 2ed 1−e 2 and the minor axis has length 2ed √ 1−e 2 . Moreover, we find that one focus is (0, 0) and working through the formula given in Definition 7.5 gives the eccentricity to be e, as required. If e > 1, then the equation generates a hyperbola with center _ e 2 d 1−e 2 , 0 _ whose transverse axis lies along the x-axis. Since such hyperbolas have the form (x−h) 2 a 2 − y 2 b 2 = 1, we need to take the opposite reciprocal of the coefficient of y 2 to find b 2 . We get 7 a 2 = e 2 d 2 (1−e 2 ) 2 = e 2 d 2 (e 2 −1) 2 and b 2 = − e 2 d 2 1−e 2 = e 2 d 2 e 2 −1 , so the transverse axis has length 2ed e 2 −1 and the conjugate axis has length 2ed √ e 2 −1 . Additionally, we verify that one focus is at (0, 0), and the formula given in Exercise 31 in Section 7.5 gives the eccentricity is e in this case as well. If e = 1, the equation r = ed 1−e cos(θ) reduces to r = d 1−cos(θ) which gives the rectangular equation y 2 = 2d _ x + d 2 _ . This is a parabola with vertex _ − d 2 , 0 _ opening to the right. In the language of Section 7.3, 4p = 2d so p = d 2 , the focus is (0, 0), the focal diameter is 2d and the directrix is x = −d, as required. Hence, we have shown that in all cases, our ‘new’ understanding of ‘conic section’, ‘focus’, ‘eccentricity’ and ‘directrix’ as presented in Definition 11.1 correspond with the ‘old’ definitions given in Chapter 7. Before we summarize our findings, we note that in order to arrive at our general equation of a conic r = ed 1−e cos(θ) , we assumed that the directrix was the line x = −d for d > 0. We could have just as easily chosen the directrix to be x = d, y = −d or y = d. As the reader can verify, in these cases we obtain the forms r = ed 1+e cos(θ) , r = ed 1−e sin(θ) and r = ed 1+e sin(θ) , respectively. The key thing to remember is that in any of these cases, the directrix is always perpendicular to the major axis of an ellipse and it is always perpendicular to the transverse axis of the hyperbola. For parabolas, knowing the focus is (0, 0) and the directrix also tells us which way the parabola opens. We have established the following theorem. 6 Turn r = e(d +r cos(θ)) into r = e(d +x) and square both sides to get r 2 = e 2 (d +x) 2 . Replace r 2 with x 2 +y 2 , expand (d +x) 2 , combine like terms, complete the square on x and clean things up. 7 Since e > 1 in this case, 1 −e 2 < 0. Hence, we rewrite 1 −e 2 2 = e 2 −1 2 to help simplify things later on. 11.6 Hooked on Conics Again 981 Theorem 11.12. Suppose e and d are positive numbers. Then • the graph of r = ed 1−e cos(θ) is the graph of a conic section with directrix x = −d. • the graph of r = ed 1+e cos(θ) is the graph of a conic section with directrix x = d. • the graph of r = ed 1−e sin(θ) is the graph of a conic section with directrix y = −d. • the graph of r = ed 1+e sin(θ) is the graph of a conic section with directrix y = d. In each case above, (0, 0) is a focus of the conic and the number e is the eccentricity of the conic. • If 0 < e < 1, the graph is an ellipse whose major axis has length 2ed 1−e 2 and whose minor axis has length 2ed √ 1−e 2 • If e = 1, the graph is a parabola whose focal diameter is 2d. • If e > 1, the graph is a hyperbola whose transverse axis has length 2ed e 2 −1 and whose conjugate axis has length 2ed √ e 2 −1 . We test out Theorem 11.12 in the next example. Example 11.6.4. Sketch the graphs of the following equations. 1. r = 4 1 −sin(θ) 2. r = 12 3 −cos(θ) 3. r = 6 1 + 2 sin(θ) Solution. 1. From r = 4 1−sin(θ) , we first note e = 1 which means we have a parabola on our hands. Since ed = 4, we have d = 4 and considering the form of the equation, this puts the directrix at y = −4. Since the focus is at (0, 0), we know that the vertex is located at the point (in rectangular coordinates) (0, −2) and must open upwards. With d = 4, we have a focal diameter of 2d = 8, so the parabola contains the points (±4, 0). We graph r = 4 1−sin(θ) below. 2. We first rewrite r = 12 3−cos(θ) in the form found in Theorem 11.12, namely r = 4 1−(1/3) cos(θ) . Since e = 1 3 satisfies 0 < e < 1, we know that the graph of this equation is an ellipse. Since ed = 4, we have d = 12 and, based on the form of the equation, the directrix is x = −12. This means that the ellipse has its major axis along the x-axis. We can find the vertices of the ellipse by finding the points of the ellipse which lie on the x-axis. We find r(0) = 6 and r(π) = 3 which correspond to the rectangular points (−3, 0) and (6, 0), so these are our vertices. The center of the ellipse is the midpoint of the vertices, which in this case is _ 3 2 , 0 _ . 8 We know one focus is (0, 0), which is 3 2 from the center _ 3 2 , 0 _ and this allows us to find the 8 As a quick check, we have from Theorem 11.12 the major axis should have length 2ed 1−e 2 = (2)(4) 1−(1/3) 2 = 9. 982 Applications of Trigonometry other focus (3, 0), even though we are not asked to do so. Finally, we know from Theorem 11.12 that the length of the minor axis is 2ed √ 1−e 2 = 4 √ 1−(1/3) 2 = 6 √ 3 which means the endpoints of the minor axis are _ 3 2 , ±3 √ 2 _ . We now have everything we need to graph r = 12 3−cos(θ) . y = −4 −4 −3 −2 −1 1 2 3 4 −3 −2 −1 1 2 3 r = 4 1−sin(θ) x y x = −12 −3 −2 −1 1 2 3 4 5 6 −4 −3 −2 −1 1 2 3 4 r = 12 3−cos(θ) 3. From r = 6 1+2 sin(θ) we get e = 2 > 1 so the graph is a hyperbola. Since ed = 6, we get d = 3, and from the form of the equation, we know the directrix is y = 3. This means the transverse axis of the hyperbola lies along the y-axis, so we can find the vertices by looking where the hyperbola intersects the y-axis. We find r _ π 2 _ = 2 and r _ 3π 2 _ = −6. These two points correspond to the rectangular points (0, 2) and (0, 6) which puts the center of the hyperbola at (0, 4). Since one focus is at (0, 0), which is 4 units away from the center, we know the other focus is at (0, 8). According to Theorem 11.12, the conjugate axis has a length of 2ed √ e 2 −1 = (2)(6) √ 2 2 −1 = 4 √ 3. Putting this together with the location of the vertices, we get that the asymptotes of the hyperbola have slopes ± 2 2 √ 3 = ± √ 3 3 . Since the center of the hyperbola is (0, 4), the asymptotes are y = ± √ 3 3 x + 4. We graph the hyperbola below. x y y = 3 −5 −4 −3 −2 −1 1 2 3 4 5 1 2 4 5 6 7 8 r = 6 1+2 sin(θ) 11.6 Hooked on Conics Again 983 In light of Section 11.6.1, the reader may wonder what the rotated form of the conic sections would look like in polar form. We know from Exercise 65 in Section 11.5 that replacing θ with (θ −φ) in an expression r = f(θ) rotates the graph of r = f(θ) counter-clockwise by an angle φ. For instance, to graph r = 4 1−sin(θ− π 4 ) all we need to do is rotate the graph of r = 4 1−sin(θ) , which we obtained in Example 11.6.4 number 1, counter-clockwise by π 4 radians, as shown below. −4−3−2−1 1 2 3 4 −3 −2 −1 1 2 3 r = 4 1−sin(θ− π 4 ) Using rotations, we can greatly simplify the form of the conic sections presented in Theorem 11.12, since any three of the forms given there can be obtained from the fourth by rotating through some multiple of π 2 . Since rotations do not affect lengths, all of the formulas for lengths Theorem 11.12 remain intact. In the theorem below, we also generalize our formula for conic sections to include circles centered at the origin by extending the concept of eccentricity to include e = 0. We conclude this section with the statement of the following theorem. Theorem 11.13. Given constants > 0, e ≥ 0 and φ, the graph of the equation r = 1 −e cos(θ −φ) is a conic section with eccentricity e and one focus at (0, 0). • If e = 0, the graph is a circle centered at (0, 0) with radius . • If e ,= 0, then the conic has a focus at (0, 0) and the directrix contains the point with polar coordinates (−d, φ) where d = e . – If 0 < e < 1, the graph is an ellipse whose major axis has length 2ed 1−e 2 and whose minor axis has length 2ed √ 1−e 2 – If e = 1, the graph is a parabola whose focal diameter is 2d. – If e > 1, the graph is a hyperbola whose transverse axis has length 2ed e 2 −1 and whose conjugate axis has length 2ed √ e 2 −1 . 984 Applications of Trigonometry 11.6.3 Exercises Graph the following equations. 1. x 2 + 2xy +y 2 −x √ 2 +y √ 2 −6 = 0 2. 7x 2 −4xy √ 3 + 3y 2 −2x −2y √ 3 −5 = 0 3. 5x 2 + 6xy + 5y 2 −4 √ 2x + 4 √ 2y = 0 4. x 2 + 2 √ 3xy + 3y 2 + 2 √ 3x −2y −16 = 0 5. 13x 2 −34xy √ 3 + 47y 2 −64 = 0 6. x 2 −2 √ 3xy −y 2 + 8 = 0 7. x 2 −4xy + 4y 2 −2x √ 5 −y √ 5 = 0 8. 8x 2 + 12xy + 17y 2 −20 = 0 Graph the following equations. 9. r = 2 1 −cos(θ) 10. r = 3 2 + sin(θ) 11. r = 3 2 −cos(θ) 12. r = 2 1 + sin(θ) 13. r = 4 1 + 3 cos(θ) 14. r = 2 1 −2 sin(θ) 15. r = 2 1 + sin(θ − π 3 ) 16. r = 6 3 −cos _ θ + π 4 _ The matrix A(θ) = _ cos(θ) −sin(θ) sin(θ) cos(θ) _ is called a rotation matrix. We’ve seen this matrix most recently in the proof of used in the proof of Theorem 11.9. 17. Show the matrix from Example 8.3.3 in Section 8.3 is none other than A _ π 4 _ . 18. Discuss with your classmates how to use A(θ) to rotate points in the plane. 19. Using the even / odd identities for cosine and sine, show A(θ) −1 = A(−θ). Interpret this geometrically. 11.6 Hooked on Conics Again 985 11.6.4 Answers 1. x 2 + 2xy +y 2 −x √ 2 +y √ 2 −6 = 0 becomes (x ) 2 = −(y −3) after rotating counter-clockwise through θ = π 4 . x y x y θ = π 4 x 2 + 2xy +y 2 −x √ 2 +y √ 2 −6 = 0 2. 7x 2 −4xy √ 3 + 3y 2 −2x −2y √ 3 −5 = 0 becomes (x −2) 2 9 + (y ) 2 = 1 after rotating counter-clockwise through θ = π 3 x y x y θ = π 3 7x 2 −4xy √ 3 + 3y 2 −2x −2y √ 3 −5 = 0 3. 5x 2 + 6xy + 5y 2 −4 √ 2x + 4 √ 2y = 0 becomes (x ) 2 + (y +2) 2 4 = 1 after rotating counter-clockwise through θ = π 4 . x y x y θ = π 4 5x 2 + 6xy + 5y 2 −4 √ 2x + 4 √ 2y = 0 4. x 2 + 2 √ 3xy + 3y 2 + 2 √ 3x −2y −16 = 0 becomes(x ) 2 = y + 4 after rotating counter-clockwise through θ = π 3 x y x y θ = π 3 x 2 + 2 √ 3xy + 3y 2 + 2 √ 3x −2y −16 = 0 986 Applications of Trigonometry 5. 13x 2 −34xy √ 3 + 47y 2 −64 = 0 becomes (y ) 2 − (x ) 2 16 = 1 after rotating counter-clockwise through θ = π 6 . x y x y θ = π 6 13x 2 −34xy √ 3 + 47y 2 −64 = 0 6. x 2 −2 √ 3xy −y 2 + 8 = 0 becomes (x ) 2 4 − (y ) 2 4 = 1 after rotating counter-clockwise through θ = π 3 x y x y θ = π 3 x 2 −2 √ 3xy −y 2 + 8 = 0 7. x 2 −4xy + 4y 2 −2x √ 5 −y √ 5 = 0 becomes (y ) 2 = x after rotating counter-clockwise through θ = arctan _ 1 2 _ . x y x y θ = arctan _ 1 2 _ x 2 −4xy + 4y 2 −2x √ 5 −y √ 5 = 0 8. 8x 2 + 12xy + 17y 2 −20 = 0 becomes (x ) 2 + (y ) 2 4 = 1 after rotating counter-clockwise through θ = arctan(2) x y x y θ = arctan(2) 8x 2 + 12xy + 17y 2 −20 = 0 11.6 Hooked on Conics Again 987 9. r = 2 1−cos(θ) is a parabola directrix x = −2 , vertex (−1, 0) focus (0, 0), focal diameter 4 x y −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 10. r = 3 2+sin(θ) = 3 2 1+ 1 2 sin(θ) is an ellipse directrix y = 3 , vertices (0, 1), (0, −3) center (0, −2) , foci (0, 0), (0, −2) minor axis length 2 √ 3 x y −4 −3 −2 −1 1 2 3 4 −4 −2 −1 1 2 3 4 11. r = 3 2−cos(θ) = 3 2 1− 1 2 cos(θ) is an ellipse directrix x = −3 , vertices (−1, 0), (3, 0) center (1, 0) , foci (0, 0), (2, 0) minor axis length 2 √ 3 x y −4 −3 −2 −1 1 2 3 4 −4 −2 −1 1 2 3 4 12. r = 2 1+sin(θ) is a parabola directrix y = 2 , vertex (0, 1) focus (0, 0), focal diameter 4 x y −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 988 Applications of Trigonometry 13. r = 4 1+3 cos(θ) is a hyperbola directrix x = 4 3 , vertices (1, 0), (2, 0) center _ 3 2 , 0 _ , foci (0, 0), (3, 0) conjugate axis length 2 √ 2 x y −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 14. r = 2 1−2 sin(θ) is a hyperbola directrix y = −1, vertices _ 0, − 2 3 _ , (0, −2) center _ 0, − 4 3 _ , foci (0, 0), _ 0, − 8 3 _ conjugate axis length 2 √ 3 3 x y −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 15. r = 2 1+sin(θ− π 3 ) is the parabola r = 2 1+sin(θ) rotated through φ = π 3 x y x y φ = π 3 16. r = 6 3−cos(θ+ π 4 ) is the ellipse r = 6 3−cos(θ) = 2 1− 1 3 cos(θ) rotated through φ = − π 4 x y x y −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 φ = − π 4 11.7 Polar Form of Complex Numbers 989 11.7 Polar Form of Complex Numbers In this section, we return to our study of complex numbers which were first introduced in Section 3.4. Recall that a complex number is a number of the form z = a + bi where a and b are real numbers and i is the imaginary unit defined by i = √ −1. The number a is called the real part of z, denoted Re(z), while the real number b is called the imaginary part of z, denoted Im(z). From Intermediate Algebra, we know that if z = a + bi = c + di where a, b, c and d are real numbers, then a = c and b = d, which means Re(z) and Im(z) are well-defined. 1 To start off this section, we associate each complex number z = a + bi with the point (a, b) on the coordinate plane. In this case, the x-axis is relabeled as the real axis, which corresponds to the real number line as usual, and the y-axis is relabeled as the imaginary axis, which is demarcated in increments of the imaginary unit i. The plane determined by these two axes is called the complex plane. Real Axis Imaginary Axis (−4, 2) ←→ z = −4 + 2i (0, −3) ←→ z = −3i (3, 0) ←→ z = 3 0 −4 −3 −2 −1 1 2 3 4 −4i −3i −2i −i i 2i 3i 4i The Complex Plane Since the ordered pair (a, b) gives the rectangular coordinates associated with the complex number z = a +bi, the expression z = a +bi is called the rectangular form of z. Of course, we could just as easily associate z with a pair of polar coordinates (r, θ). Although it is not a straightforward as the definitions of Re(z) and Im(z), we can still give r and θ special names in relation to z. Definition 11.2. The Modulus and Argument of Complex Numbers: Let z = a +bi be a complex number with a = Re(z) and b = Im(z). Let (r, θ) be a polar representation of the point with rectangular coordinates (a, b) where r ≥ 0. • The modulus of z, denoted [z[, is defined by [z[ = r. • The angle θ is an argument of z. The set of all arguments of z is denoted arg(z). • If z ,= 0 and −π < θ ≤ π, then θ is the principal argument of z, written θ = Arg(z). 1 ‘Well-defined’ means that no matter how we express z, the number Re(z) is always the same, and the number Im(z) is always the same. In other words, Re and Im are functions of complex numbers. 990 Applications of Trigonometry Some remarks about Definition 11.2 are in order. We know from Section 11.4 that every point in the plane has infinitely many polar coordinate representations (r, θ) which means it’s worth our time to make sure the quantities ‘modulus’, ‘argument’ and ‘principal argument’ are well-defined. Concerning the modulus, if z = 0 then the point associated with z is the origin. In this case, the only r-value which can be used here is r = 0. Hence for z = 0, [z[ = 0 is well-defined. If z ,= 0, then the point associated with z is not the origin, and there are two possibilities for r: one positive and one negative. However, we stipulated r ≥ 0 in our definition so this pins down the value of [z[ to one and only one number. Thus the modulus is well-defined in this case, too. 2 Even with the requirement r ≥ 0, there are infinitely many angles θ which can be used in a polar representation of a point (r, θ). If z ,= 0 then the point in question is not the origin, so all of these angles θ are coterminal. Since coterminal angles are exactly 2π radians apart, we are guaranteed that only one of them lies in the interval (−π, π], and this angle is what we call the principal argument of z, Arg(z). In fact, the set arg(z) of all arguments of z can be described using set-builder notation as arg(z) = ¦Arg(z) + 2πk [ k is an integer¦. Note that since arg(z) is a set, we will write ‘θ ∈ arg(z)’ to mean ‘θ is in 3 the set of arguments of z’. If z = 0 then the point in question is the origin, which we know can be represented in polar coordinates as (0, θ) for any angle θ. In this case, we have arg(0) = (−∞, ∞) and since there is no one value of θ which lies (−π, π], we leave Arg(0) undefined. 4 It is time for an example. Example 11.7.1. For each of the following complex numbers find Re(z), Im(z), [z[, arg(z) and Arg(z). Plot z in the complex plane. 1. z = √ 3 −i 2. z = −2 + 4i 3. z = 3i 4. z = −117 Solution. 1. For z = √ 3 − i = √ 3 + (−1)i, we have Re(z) = √ 3 and Im(z) = −1. To find [z[, arg(z) and Arg(z), we need to find a polar representation (r, θ) with r ≥ 0 for the point P( √ 3, −1) associated with z. We know r 2 = ( √ 3) 2 + (−1) 2 = 4, so r = ±2. Since we require r ≥ 0, we choose r = 2, so [z[ = 2. Next, we find a corresponding angle θ. Since r > 0 and P lies in Quadrant IV, θ is a Quadrant IV angle. We know tan(θ) = −1 √ 3 = − √ 3 3 , so θ = − π 6 + 2πk for integers k. Hence, arg(z) = _ − π 6 + 2πk [ k is an integer _ . Of these values, only θ = − π 6 satisfies the requirement that −π < θ ≤ π, hence Arg(z) = − π 6 . 2. The complex number z = −2 + 4i has Re(z) = −2, Im(z) = 4, and is associated with the point P(−2, 4). Our next task is to find a polar representation (r, θ) for P where r ≥ 0. Running through the usual calculations gives r = 2 √ 5, so [z[ = 2 √ 5. To find θ, we get tan(θ) = −2, and since r > 0 and P lies in Quadrant II, we know θ is a Quadrant II angle. We find θ = π + arctan(−2) + 2πk, or, more succinctly θ = π −arctan(2) + 2πk for integers k. Hence arg(z) = ¦π −arctan(2) + 2πk [ k is an integer¦. Only θ = π − arctan(2) satisfies the requirement −π < θ ≤ π, so Arg(z) = π −arctan(2). 2 In case you’re wondering, the use of the absolute value notation |z| for modulus will be explained shortly. 3 Recall the symbol being used here, ‘∈,’ is the mathematical symbol which denotes membership in a set. 4 If we had Calculus, we would regard Arg(0) as an ‘indeterminate form.’ But we don’t, so we won’t. 11.7 Polar Form of Complex Numbers 991 3. We rewrite z = 3i as z = 0 + 3i to find Re(z) = 0 and Im(z) = 3. The point in the plane which corresponds to z is (0, 3) and while we could go through the usual calculations to find the required polar form of this point, we can almost ‘see’ the answer. The point (0, 3) lies 3 units away from the origin on the positive y-axis. Hence, r = [z[ = 3 and θ = π 2 + 2πk for integers k. We get arg(z) = _ π 2 + 2πk [ k is an integer _ and Arg(z) = π 2 . 4. As in the previous problem, we write z = −117 = −117 +0i so Re(z) = −117 and Im(z) = 0. The number z = −117 corresponds to the point (−117, 0), and this is another instance where we can determine the polar form ‘by eye’. The point (−117, 0) is 117 units away from the origin along the negative x-axis. Hence, r = [z[ = 117 and θ = π + 2π = (2k + 1)πk for integers k. We have arg(z) = ¦(2k + 1)π [ k is an integer¦. Only one of these values, θ = π, just barely lies in the interval (−π, π] which means and Arg(z) = π. We plot z along with the other numbers in this example below. Real Axis Imaginary Axis z = √ 3 −i z = −2 + 4i z = 3i z = −117 −117 −2 −1 1 2 3 4 −i i 2i 3i 4i Now that we’ve had some practice computing the modulus and argument of some complex numbers, it is time to explore their properties. We have the following theorem. Theorem 11.14. Properties of the Modulus: Let z and w be complex numbers. • [z[ is the distance from z to 0 in the complex plane • [z[ ≥ 0 and [z[ = 0 if and only if z = 0 • [z[ = _ Re(z) 2 + Im(z) 2 • Product Rule: [zw[ = [z[[w[ • Power Rule: [z n [ = [z[ n for all natural numbers, n • Quotient Rule: ¸ ¸ ¸ z w ¸ ¸ ¸ = [z[ [w[ , provided w ,= 0 To prove the first three properties in Theorem 11.14, suppose z = a + bi where a and b are real numbers. To determine [z[, we find a polar representation (r, θ) with r ≥ 0 for the point (a, b). From Section 11.4, we know r 2 = a 2 +b 2 so that r = ± √ a 2 +b 2 . Since we require r ≥ 0, then it must be that r = √ a 2 +b 2 , which means [z[ = √ a 2 +b 2 . Using the distance formula, we find the distance 992 Applications of Trigonometry from (0, 0) to (a, b) is also √ a 2 +b 2 , establishing the first property. 5 For the second property, note that since [z[ is a distance, [z[ ≥ 0. Furthermore, [z[ = 0 if and only if the distance from z to 0 is 0, and the latter happens if and only if z = 0, which is what we were asked to show. 6 For the third property, we note that since a = Re(z) and b = Im(z), z = √ a 2 +b 2 = _ Re(z) 2 + Im(z) 2 . To prove the product rule, suppose z = a +bi and w = c +di for real numbers a, b, c and d. Then zw = (a +bi)(c +di). After the usual arithmetic 7 we get zw = (ac −bd) + (ad +bc)i. Therefore, [zw[ = _ (ac −bd) 2 + (ad +bc) 2 = √ a 2 c 2 −2abcd +b 2 d 2 +a 2 d 2 + 2abcd +b 2 c 2 Expand = √ a 2 c 2 +a 2 d 2 +b 2 c 2 +b 2 d 2 Rearrange terms = _ a 2 (c 2 +d 2 ) +b 2 (c 2 +d 2 ) Factor = _ (a 2 +b 2 ) (c 2 +d 2 ) Factor = √ a 2 +b 2 √ c 2 +d 2 Product Rule for Radicals = [z[[w[ Definition of [z[ and [w[ Hence [zw[ = [z[[w[ as required. Now that the Product Rule has been established, we use it and the Principle of Mathematical Induction 8 to prove the power rule. Let P(n) be the statement [z n [ = [z[ n . Then P(1) is true since ¸ ¸ z 1 ¸ ¸ = [z[ = [z[ 1 . Next, assume P(k) is true. That is, assume ¸ ¸ z k ¸ ¸ = [z[ k for some k ≥ 1. Our job is to show that P(k + 1) is true, namely ¸ ¸ z k+1 ¸ ¸ = [z[ k+1 . As is customary with induction proofs, we first try to reduce the problem in such a way as to use the Induction Hypothesis. ¸ ¸ z k+1 ¸ ¸ = ¸ ¸ z k z ¸ ¸ Properties of Exponents = ¸ ¸ z k ¸ ¸ [z[ Product Rule = [z[ k [z[ Induction Hypothesis = [z[ k+1 Properties of Exponents Hence, P(k + 1) is true, which means [z n [ = [z[ n is true for all natural numbers n. Like the Power Rule, the Quotient Rule can also be established with the help of the Product Rule. We assume w ,= 0 (so [w[ , = 0) and we get ¸ ¸ ¸ z w ¸ ¸ ¸ = ¸ ¸ ¸ ¸ (z) _ 1 w _¸ ¸ ¸ ¸ = [z[ ¸ ¸ ¸ ¸ 1 w ¸ ¸ ¸ ¸ Product Rule. 5 Since the absolute value |x| of a real number x can be viewed as the distance from x to 0 on the number line, this first property justifies the notation |z| for modulus. We leave it to the reader to show that if z is real, then the definition of modulus coincides with absolute value so the notation |z| is unambiguous. 6 This may be considered by some to be a bit of a cheat, so we work through the underlying Algebra to see this is true. We know |z| = 0 if and only if √ a 2 +b 2 = 0 if and only if a 2 + b 2 = 0, which is true if and only if a = b = 0. The latter happens if and only if z = a +bi = 0. There. 7 See Example 3.4.1 in Section 3.4 for a review of complex number arithmetic. 8 See Section 9.3 for a review of this technique. 11.7 Polar Form of Complex Numbers 993 Hence, the proof really boils down to showing ¸ ¸ 1 w ¸ ¸ = 1 |w| . This is left as an exercise. Next, we characterize the argument of a complex number in terms of its real and imaginary parts. Theorem 11.15. Properties of the Argument: Let z be a complex number. • If Re(z) ,= 0 and θ ∈ arg(z), then tan(θ) = Im(z) Re(z) . • If Re(z) = 0 and Im(z) > 0, then arg(z) = _ π 2 + 2πk [ k is an integer _ . • If Re(z) = 0 and Im(z) < 0, then arg(z) = _ − π 2 + 2πk [ k is an integer _ . • If Re(z) = Im(z) = 0, then z = 0 and arg(z) = (−∞, ∞). To prove Theorem 11.15, suppose z = a+bi for real numbers a and b. By definition, a = Re(z) and b = Im(z), so the point associated with z is (a, b) = (Re(z), Im(z)). From Section 11.4, we know that if (r, θ) is a polar representation for (Re(z), Im(z)), then tan(θ) = Im(z) Re(z) , provided Re(z) ,= 0. If Re(z) = 0 and Im(z) > 0, then z lies on the positive imaginary axis. Since we take r > 0, we have that θ is coterminal with π 2 , and the result follows. If Re(z) = 0 and Im(z) < 0, then z lies on the negative imaginary axis, and a similar argument shows θ is coterminal with − π 2 . The last property in the theorem was already discussed in the remarks following Definition 11.2. Our next goal is to completely marry the Geometry and the Algebra of the complex numbers. To that end, consider the figure below. Real Axis Imaginary Axis (a, b) ←→ z = a + bi ←→ (r, θ) 0 θ ∈ arg(z) a bi | z | = √ a 2 + b 2 = r Polar coordinates, (r, θ) associated with z = a + bi with r ≥ 0. We know from Theorem 11.7 that a = r cos(θ) and b = r sin(θ). Making these substitutions for a and b gives z = a +bi = r cos(θ) +r sin(θ)i = r [cos(θ) +i sin(θ)]. The expression ‘cos(θ) +i sin(θ)’ is abbreviated cis(θ) so we can write z = rcis(θ). Since r = [z[ and θ ∈ arg(z), we get Definition 11.3. A Polar Form of a Complex Number: Suppose z is a complex number and θ ∈ arg(z). The expression: [z[cis(θ) = [z[ [cos(θ) +i sin(θ)] is called a polar form for z. 994 Applications of Trigonometry Since there are infinitely many choices for θ ∈ arg(z), there infinitely many polar forms for z, so we used the indefinite article ‘a’ in Definition 11.3. It is time for an example. Example 11.7.2. 1. Find the rectangular form of the following complex numbers. Find Re(z) and Im(z). (a) z = 4cis _ 2π 3 _ (b) z = 2cis _ − 3π 4 _ (c) z = 3cis(0) (d) z = cis _ π 2 _ 2. Use the results from Example 11.7.1 to find a polar form of the following complex numbers. (a) z = √ 3 −i (b) z = −2 + 4i (c) z = 3i (d) z = −117 Solution. 1. The key to this problem is to write out cis(θ) as cos(θ) +i sin(θ). (a) By definition, z = 4cis _ 2π 3 _ = 4 _ cos _ 2π 3 _ +i sin _ 2π 3 _¸ . After some simplifying, we get z = −2 + 2i √ 3, so that Re(z) = −2 and Im(z) = 2 √ 3. (b) Expanding, we get z = 2cis _ − 3π 4 _ = 2 _ cos _ − 3π 4 _ +i sin _ − 3π 4 _¸ . From this, we find z = − √ 2 −i √ 2, so Re(z) = − √ 2 = Im(z). (c) We get z = 3cis(0) = 3 [cos(0) +i sin(0)] = 3. Writing 3 = 3 + 0i, we get Re(z) = 3 and Im(z) = 0, which makes sense seeing as 3 is a real number. (d) Lastly, we have z = cis _ π 2 _ = cos _ π 2 _ +i sin _ π 2 _ = i. Since i = 0 + 1i, we get Re(z) = 0 and Im(z) = 1. Since i is called the ‘imaginary unit,’ these answers make perfect sense. 2. To write a polar form of a complex number z, we need two pieces of information: the modulus [z[ and an argument (not necessarily the principal argument) of z. We shamelessly mine our solution to Example 11.7.1 to find what we need. (a) For z = √ 3 − i, [z[ = 2 and θ = − π 6 , so z = 2cis _ − π 6 _ . We can check our answer by converting it back to rectangular form to see that it simplifies to z = √ 3 −i. (b) For z = −2 + 4i, [z[ = 2 √ 5 and θ = π − arctan(2). Hence, z = 2 √ 5cis(π − arctan(2)). It is a good exercise to actually show that this polar form reduces to z = −2 + 4i. (c) For z = 3i, [z[ = 3 and θ = π 2 . In this case, z = 3cis _ π 2 _ . This can be checked geometrically. Head out 3 units from 0 along the positive real axis. Rotating π 2 radians counter-clockwise lands you exactly 3 units above 0 on the imaginary axis at z = 3i. (d) Last but not least, for z = −117, [z[ = 117 and θ = π. We get z = 117cis(π). As with the previous problem, our answer is easily checked geometrically. 11.7 Polar Form of Complex Numbers 995 The following theorem summarizes the advantages of working with complex numbers in polar form. Theorem 11.16. Products, Powers and Quotients Complex Numbers in Polar Form: Suppose z and w are complex numbers with polar forms z = [z[cis(α) and w = [w[cis(β). Then • Product Rule: zw = [z[[w[cis(α +β) • Power Rule (DeMoivre’s Theorem) : z n = [z[ n cis(nθ) for every natural number n • Quotient Rule: z w = [z[ [w[ cis(α −β), provided [w[ , = 0 The proof of Theorem 11.16 requires a healthy mix of definition, arithmetic and identities. We first start with the product rule. zw = [[z[cis(α)] [[w[cis(β)] = [z[[w[ [cos(α) +i sin(α)] [cos(β) +i sin(β)] We now focus on the quantity in brackets on the right hand side of the equation. [cos(α) +i sin(α)] [cos(β) +i sin(β)] = cos(α) cos(β) +i cos(α) sin(β) +i sin(α) cos(β) +i 2 sin(α) sin(β) = cos(α) cos(β) +i 2 sin(α) sin(β) Rearranging terms +i sin(α) cos(β) +i cos(α) sin(β) = (cos(α) cos(β) −sin(α) sin(β)) Since i 2 = −1 +i (sin(α) cos(β) + cos(α) sin(β)) Factor out i = cos(α +β) +i sin(α +β) Sum identities = cis(α +β) Definition of ‘cis’ Putting this together with our earlier work, we get zw = [z[[w[cis(α +β), as required. Moving right along, we next take aim at the Power Rule, better known as DeMoivre’s Theorem. 9 We proceed by induction on n. Let P(n) be the sentence z n = [z[ n cis(nθ). Then P(1) is true, since z 1 = z = [z[cis(θ) = [z[ 1 cis(1 θ). We now assume P(k) is true, that is, we assume z k = [z[ k cis(kθ) for some k ≥ 1. Our goal is to show that P(k + 1) is true, or that z k+1 = [z[ k+1 cis((k + 1)θ). We have z k+1 = z k z Properties of Exponents = _ [z[ k cis(kθ) _ ([z[cis(θ)) Induction Hypothesis = _ [z[ k [z[ _ cis(kθ +θ) Product Rule = [z[ k+1 cis((k + 1)θ) 9 Compare this proof with the proof of the Power Rule in Theorem 11.14. 996 Applications of Trigonometry Hence, assuming P(k) is true, we have that P(k + 1) is true, so by the Principle of Mathematical Induction, z n = [z[ n cis(nθ) for all natural numbers n. The last property in Theorem 11.16 to prove is the quotient rule. Assuming [w[ , = 0 we have z w = [z[cis(α) [w[cis(β) = _ [z[ [w[ _ cos(α) +i sin(α) cos(β) +i sin(β) Next, we multiply both the numerator and denominator of the right hand side by (cos(β)−i sin(β)) which is the complex conjugate of (cos(β) +i sin(β)) to get z w = _ [z[ [w[ _ cos(α) +i sin(α) cos(β) +i sin(β) cos(β) −i sin(β) cos(β) −i sin(β) If we let the numerator be N = [cos(α) +i sin(α)] [cos(β) −i sin(β)] and simplify we get N = [cos(α) +i sin(α)] [cos(β) −i sin(β)] = cos(α) cos(β) −i cos(α) sin(β) +i sin(α) cos(β) −i 2 sin(α) sin(β) Expand = [cos(α) cos(β) + sin(α) sin(β)] +i [sin(α) cos(β) −cos(α) sin(β)] Rearrange and Factor = cos(α −β) +i sin(α −β) Difference Identities = cis(α −β) Definition of ‘cis’ If we call the denominator D then we get D = [cos(β) +i sin(β)] [cos(β) −i sin(β)] = cos 2 (β) −i cos(β) sin(β) +i cos(β) sin(β) −i 2 sin 2 (β) Expand = cos 2 (β) −i 2 sin 2 (β) Simplify = cos 2 (β) + sin 2 (β) Again, i 2 = −1 = 1 Pythagorean Identity Putting it all together, we get z w = _ [z[ [w[ _ cos(α) +i sin(α) cos(β) +i sin(β) cos(β) −i sin(β) cos(β) −i sin(β) = _ [z[ [w[ _ cis(α −β) 1 = [z[ [w[ cis(α −β) and we are done. The next example makes good use of Theorem 11.16. 11.7 Polar Form of Complex Numbers 997 Example 11.7.3. Let z = 2 √ 3 +2i and w = −1 +i √ 3. Use Theorem 11.16 to find the following. 1. zw 2. w 5 3. z w Write your final answers in rectangular form. Solution. In order to use Theorem 11.16, we need to write z and w in polar form. For z = 2 √ 3+2i, we find [z[ = _ (2 √ 3) 2 + (2) 2 = √ 16 = 4. If θ ∈ arg(z), we know tan(θ) = Im(z) Re(z) = 2 2 √ 3 = √ 3 3 . Since z lies in Quadrant I, we have θ = π 6 + 2πk for integers k. Hence, z = 4cis _ π 6 _ . For w = −1 +i √ 3, we have [w[ = _ (−1) 2 + ( √ 3) 2 = 2. For an argument θ of w, we have tan(θ) = √ 3 −1 = − √ 3. Since w lies in Quadrant II, θ = 2π 3 + 2πk for integers k and w = 2cis _ 2π 3 _ . We can now proceed. 1. We get zw = _ 4cis _ π 6 __ _ 2cis _ 2π 3 __ = 8cis _ π 6 + 2π 3 _ = 8cis _ 5π 6 _ = 8 _ cos _ 5π 6 _ +i sin _ 5π 6 _¸ . After simplifying, we get zw = −4 √ 3 + 4i. 2. We use DeMoivre’s Theorem which yields w 5 = _ 2cis _ 2π 3 _¸ 5 = 2 5 cis _ 5 2π 3 _ = 32cis _ 10π 3 _ . Since 10π 3 is coterminal with 4π 3 , we get w 5 = 32 _ cos _ 4π 3 _ +i sin _ 4π 3 _¸ = −16 −16i √ 3. 3. Last, but not least, we have z w = 4cis( π 6 ) 2cis( 2π 3 ) = 4 2 cis _ π 6 − 2π 3 _ = 2cis _ − π 2 _ . Since − π 2 is a quadrantal angle, we can ‘see’ the rectangular form by moving out 2 units along the positive real axis, then rotating π 2 radians clockwise to arrive at the point 2 units below 0 on the imaginary axis. The long and short of it is that z w = −2i. Some remarks are in order. First, the reader may not be sold on using the polar form of complex numbers to multiply complex numbers – especially if they aren’t given in polar form to begin with. Indeed, a lot of work was needed to convert the numbers z and w in Example 11.7.3 into polar form, compute their product, and convert back to rectangular form – certainly more work than is required to multiply out zw = (2 √ 3 + 2i)(−1 +i √ 3) the old-fashioned way. However, Theorem 11.16 pays huge dividends when computing powers of complex numbers. Consider how we computed w 5 above and compare that to using the Binomial Theorem, Theorem 9.4, to accomplish the same feat by expanding (−1 + i √ 3) 5 . Division is tricky in the best of times, and we saved ourselves a lot of time and effort using Theorem 11.16 to find and simplify z w using their polar forms as opposed to starting with 2 √ 3+2i −1+i √ 3 , rationalizing the denominator, and so forth. There is geometric reason for studying these polar forms and we would be derelict in our duties if we did not mention the Geometry hidden in Theorem 11.16. Take the product rule, for instance. If z = [z[cis(α) and w = [w[cis(β), the formula zw = [z[[w[cis(α +β) can be viewed geometrically as a two step process. The multiplication of [z[ by [w[ can be interpreted as magnifying 10 the distance [z[ from z to 0, by the factor [w[. Adding the argument of w to the argument of z can be interpreted geometrically as a rotation of β radians counter-clockwise. 11 Focusing on z and w from Example 10 Assuming |w| > 1. 11 Assuming β > 0. 998 Applications of Trigonometry 11.7.3, we can arrive at the product zw by plotting z, doubling its distance from 0 (since [w[ = 2), and rotating 2π 3 radians counter-clockwise. The sequence of diagrams below attempts to describe this process geometrically. Real Axis Imaginary Axis 0 z = 4cis _ π 6 _ z|w| = 8cis _ π 6 _ 1 2 3 4 5 6 7 i 2i 3i 4i 5i 6i Real Axis Imaginary Axis 0 zw = 8cis _ π 6 + 2π 3 _ z|w| = 8cis _ π 6 _ −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 i 2i 3i 4i 5i 6i Multiplying z by |w| = 2. Rotating counter-clockwise by Arg(w) = 2π 3 radians. Visualizing zw for z = 4cis _ π 6 _ and w = 2cis _ 2π 3 _ . We may also visualize division similarly. Here, the formula z w = |z| |w| cis(α − β) may be interpreted as shrinking 12 the distance from 0 to z by the factor [w[, followed up by a clockwise 13 rotation of β radians. In the case of z and w from Example 11.7.3, we arrive at z w by first halving the distance from 0 to z, then rotating clockwise 2π 3 radians. Real Axis Imaginary Axis 0 _ 1 |w| _ z = 2cis _ π 6 _ z = 4cis _ π 6 _ 1 2 3 i 2i 3i Real Axis Imaginary Axis 0 zw = 2cis _ π 6 2π 3 _ _ 1 |w| _ z = 2cis _ π 6 _ 1 2 3 −2i −i i Dividing z by |w| = 2. Rotating clockwise by Arg(w) = 2π 3 radians. Visualizing z w for z = 4cis _ π 6 _ and w = 2cis _ 2π 3 _ . Our last goal of the section is to reverse DeMoivre’s Theorem to extract roots of complex numbers. Definition 11.4. Let z and w be complex numbers. If there is a natural number n such that w n = z, then w is an n th root of z. Unlike Definition 5.4 in Section 5.3, we do not specify one particular prinicpal n th root, hence the use of the indefinite article ‘an’ as in ‘an n th root of z’. Using this definition, both 4 and −4 are 12 Again, assuming |w| > 1. 13 Again, assuming β > 0. 11.7 Polar Form of Complex Numbers 999 square roots of 16, while √ 16 means the principal square root of 16 as in √ 16 = 4. Suppose we wish to find all complex third (cube) roots of 8. Algebraically, we are trying to solve w 3 = 8. We know that there is only one real solution to this equation, namely w = 3 √ 8 = 2, but if we take the time to rewrite this equation as w 3 − 8 = 0 and factor, we get (w − 2) _ w 2 + 2w + 4 _ = 0. The quadratic factor gives two more cube roots w = −1 ±i √ 3, for a total of three cube roots of 8. In accordance with Theorem 3.14, since the degree of p(w) = w 3 −8 is three, there are three complex zeros, counting multiplicity. Since we have found three distinct zeros, we know these are all of the zeros, so there are exactly three distinct cube roots of 8. Let us now solve this same problem using the machinery developed in this section. To do so, we express z = 8 in polar form. Since z = 8 lies 8 units away on the positive real axis, we get z = 8cis(0). If we let w = [w[cis(α) be a polar form of w, the equation w 3 = 8 becomes w 3 = 8 ([w[cis(α)) 3 = 8cis(0) [w[ 3 cis(3α) = 8cis(0) DeMoivre’s Theorem The complex number on the left hand side of the equation corresponds to the point with polar coordinates _ [w[ 3 , 3α _ , while the complex number on the right hand side corresponds to the point with polar coordinates (8, 0). Since [w[ ≥ 0, so is [w[ 3 , which means _ [w[ 3 , 3α _ and (8, 0) are two polar representations corresponding to the same complex number, both with positive r values. From Section 11.4, we know [w[ 3 = 8 and 3α = 0 + 2πk for integers k. Since [w[ is a real number, we solve [w[ 3 = 8 by extracting the principal cube root to get [w[ = 3 √ 8 = 2. As for α, we get α = 2πk 3 for integers k. This produces three distinct points with polar coordinates corresponding to k = 0, 1 and 2: specifically (2, 0), _ 2, 2π 3 _ and _ 2, 4π 3 _ . These correspond to the complex numbers w 0 = 2cis(0), w 1 = 2cis _ 2π 3 _ and w 2 = 2cis _ 4π 3 _ , respectively. Writing these out in rectangular form yields w 0 = 2, w 1 = −1 + i √ 3 and w 2 = −1 − i √ 3. While this process seems a tad more involved than our previous factoring approach, this procedure can be generalized to find, for example, all of the fifth roots of 32. (Try using Chapter 3 techniques on that!) If we start with a generic complex number in polar form z = [z[cis(θ) and solve w n = z in the same manner as above, we arrive at the following theorem. Theorem 11.17. The n th roots of a Complex Number: Let z ,= 0 be a complex number with polar form z = rcis(θ). For each natural number n, z has n distinct n th roots, which we denote by w 0 , w 1 , . . . , w n − 1 , and they are given by the formula w k = n √ rcis _ θ n + 2π n k _ The proof of Theorem 11.17 breaks into to two parts: first, showing that each w k is an n th root, and second, showing that the set ¦w k [ k = 0, 1, . . . , (n −1)¦ consists of n different complex numbers. To show w k is an n th root of z, we use DeMoivre’s Theorem to show (w k ) n = z. 1000 Applications of Trigonometry (w k ) n = _ n √ rcis _ θ n + 2π n k __ n = ( n √ r) n cis _ n _ θ n + 2π n k ¸_ DeMoivre’s Theorem = rcis (θ + 2πk) Since k is a whole number, cos(θ +2πk) = cos(θ) and sin(θ +2πk) = sin(θ). Hence, it follows that cis(θ + 2πk) = cis(θ), so (w k ) n = rcis(θ) = z, as required. To show that the formula in Theorem 11.17 generates n distinct numbers, we assume n ≥ 2 (or else there is nothing to prove) and note that the modulus of each of the w k is the same, namely n √ r. Therefore, the only way any two of these polar forms correspond to the same number is if their arguments are coterminal – that is, if the arguments differ by an integer multiple of 2π. Suppose k and j are whole numbers between 0 and (n−1), inclusive, with k ,= j. Since k and j are different, let’s assume for the sake of argument that k > j. Then _ θ n + 2π n k _ − _ θ n + 2π n j _ = 2π _ k−j n _ . For this to be an integer multiple of 2π, (k − j) must be a multiple of n. But because of the restrictions on k and j, 0 < k − j ≤ n − 1. (Think this through.) Hence, (k − j) is a positive number less than n, so it cannot be a multiple of n. As a result, w k and w j are different complex numbers, and we are done. By Theorem 3.14, we know there at most n distinct solutions to w n = z, and we have just found all of them. We illustrate Theorem 11.17 in the next example. Example 11.7.4. Use Theorem 11.17 to find the following: 1. both square roots of z = −2 + 2i √ 3 2. the four fourth roots of z = −16 3. the three cube roots of z = √ 2 +i √ 2 4. the five fifth roots of z = 1. Solution. 1. We start by writing z = −2 + 2i √ 3 = 4cis _ 2π 3 _ . To use Theorem 11.17, we identify r = 4, θ = 2π 3 and n = 2. We know that z has two square roots, and in keeping with the notation in Theorem 11.17, we’ll call them w 0 and w 1 . We get w 0 = √ 4cis _ (2π/3) 2 + 2π 2 (0) _ = 2cis _ π 3 _ and w 1 = √ 4cis _ (2π/3) 2 + 2π 2 (1) _ = 2cis _ 4π 3 _ . In rectangular form, the two square roots of z are w 0 = 1 + i √ 3 and w 1 = −1 − i √ 3. We can check our answers by squaring them and showing that we get z = −2 + 2i √ 3. 2. Proceeding as above, we get z = −16 = 16cis(π). With r = 16, θ = π and n = 4, we get the four fourth roots of z to be w 0 = 4 √ 16cis _ π 4 + 2π 4 (0) _ = 2cis _ π 4 _ , w 1 = 4 √ 16cis _ π 4 + 2π 4 (1) _ = 2cis _ 3π 4 _ , w 2 = 4 √ 16cis _ π 4 + 2π 4 (2) _ = 2cis _ 5π 4 _ and w 3 = 4 √ 16cis _ π 4 + 2π 4 (3) _ = 2cis _ 7π 4 _ . Converting these to rectangular form gives w 0 = √ 2+i √ 2, w 1 = − √ 2+i √ 2, w 2 = − √ 2−i √ 2 and w 3 = √ 2 −i √ 2. 11.7 Polar Form of Complex Numbers 1001 3. For z = √ 2+i √ 2, we have z = 2cis _ π 4 _ . With r = 2, θ = π 4 and n = 3 the usual computations yield w 0 = 3 √ 2cis _ π 12 _ , w 1 = 3 √ 2cis _ 9π 12 _ = 3 √ 2cis _ 3π 4 _ and w 2 = 3 √ 2cis _ 17π 12 _ . If we were to convert these to rectangular form, we would need to use either the Sum and Difference Identities in Theorem 10.16 or the Half-Angle Identities in Theorem 10.19 to evaluate w 0 and w 2 . Since we are not explicitly told to do so, we leave this as a good, but messy, exercise. 4. To find the five fifth roots of 1, we write 1 = 1cis(0). We have r = 1, θ = 0 and n = 5. Since 5 √ 1 = 1, the roots are w 0 = cis(0) = 1, w 1 = cis _ 2π 5 _ , w 2 = cis _ 4π 5 _ , w 3 = cis _ 6π 5 _ and w 4 = cis _ 8π 5 _ . The situation here is even graver than in the previous example, since we have not developed any identities to help us determine the cosine or sine of 2π 5 . At this stage, we could approximate our answers using a calculator, and we leave this as an exercise. Now that we have done some computations using Theorem 11.17, we take a step back to look at things geometrically. Essentially, Theorem 11.17 says that to find the n th roots of a complex number, we first take the n th root of the modulus and divide the argument by n. This gives the first root w 0 . Each succeessive root is found by adding 2π n to the argument, which amounts to rotating w 0 by 2π n radians. This results in n roots, spaced equally around the complex plane. As an example of this, we plot our answers to number 2 in Example 11.7.4 below. Real Axis Imaginary Axis 0 w 0 w 1 w 2 w 3 −2 −1 1 2 −2i −i i 2i The four fourth roots of z = −16 equally spaced 2π 4 = π 2 around the plane. We have only glimpsed at the beauty of the complex numbers in this section. The complex plane is without a doubt one of the most important mathematical constructs ever devised. Coupled with Calculus, it is the venue for incredibly important Science and Engineering applications. 14 For now, the following exercises will have to suffice. 14 For more on this, see the beautifully written epilogue to Section 3.4 found on page 293. 1002 Applications of Trigonometry 11.7.1 Exercises In Exercises 1 - 20, find a polar representation for the complex number z and then identify Re(z), Im(z), [z[, arg(z) and Arg(z). 1. z = 9 + 9i 2. z = 5 + 5i √ 3 3. z = 6i 4. z = −3 √ 2+3i √ 2 5. z = −6 √ 3 + 6i 6. z = −2 7. z = − √ 3 2 − 1 2 i 8. z = −3 −3i 9. z = −5i 10. z = 2 √ 2 −2i √ 2 11. z = 6 12. z = i 3 √ 7 13. z = 3 + 4i 14. z = √ 2 +i 15. z = −7 + 24i 16. z = −2 + 6i 17. z = −12 −5i 18. z = −5 −2i 19. z = 4 −2i 20. z = 1 −3i In Exercises 21 - 40, find the rectangular form of the given complex number. Use whatever identities are necessary to find the exact values. 21. z = 6cis(0) 22. z = 2cis _ π 6 _ 23. z = 7 √ 2cis _ π 4 _ 24. z = 3cis _ π 2 _ 25. z = 4cis _ 2π 3 _ 26. z = √ 6cis _ 3π 4 _ 27. z = 9cis (π) 28. z = 3cis _ 4π 3 _ 29. z = 7cis _ − 3π 4 _ 30. z = √ 13cis _ 3π 2 _ 31. z = 1 2 cis _ 7π 4 _ 32. z = 12cis _ − π 3 _ 33. z = 8cis _ π 12 _ 34. z = 2cis _ 7π 8 _ 35. z = 5cis _ arctan _ 4 3 __ 36. z = √ 10cis _ arctan _ 1 3 __ 37. z = 15cis (arctan (−2)) 38. z = √ 3 _ arctan _ − √ 2 __ 39. z = 50cis _ π −arctan _ 7 24 __ 40. z = 1 2 cis _ π + arctan _ 5 12 __ For Exercises 41 - 52, use z = − 3 √ 3 2 + 3 2 i and w = 3 √ 2 −3i √ 2 to compute the quantity. Express your answers in polar form using the principal argument. 41. zw 42. z w 43. w z 44. z 4 45. w 3 46. z 5 w 2 47. z 3 w 2 48. z 2 w 11.7 Polar Form of Complex Numbers 1003 49. w z 2 50. z 3 w 2 51. w 2 z 3 52. _ w z _ 6 In Exercises 53 - 64, use DeMoivre’s Theorem to find the indicated power of the given complex number. Express your final answers in rectangular form. 53. _ −2 + 2i √ 3 _ 3 54. (− √ 3 −i) 3 55. (−3 + 3i) 4 56. ( √ 3 +i) 4 57. _ 5 2 + 5 2 i _ 3 58. _ − 1 2 − √ 3 2 i _ 6 59. _ 3 2 − 3 2 i _ 3 60. _ √ 3 3 − 1 3 i _ 4 61. _ √ 2 2 + √ 2 2 i _ 4 62. (2 + 2i) 5 63. ( √ 3 −i) 5 64. (1 −i) 8 In Exercises 65 - 76, find the indicated complex roots. Express your answers in polar form and then convert them into rectangular form. 65. the two square roots of z = 4i 66. the two square roots of z = −25i 67. the two square roots of z = 1 +i √ 3 68. the two square roots of 5 2 − 5 √ 3 2 i 69. the three cube roots of z = 64 70. the three cube roots of z = −125 71. the three cube roots of z = i 72. the three cube roots of z = −8i 73. the four fourth roots of z = 16 74. the four fourth roots of z = −81 75. the six sixth roots of z = 64 76. the six sixth roots of z = −729 77. Use the Sum and Difference Identities in Theorem 10.16 or the Half Angle Identities in Theorem 10.19 to express the three cube roots of z = √ 2 + i √ 2 in rectangular form. (See Example 11.7.4, number 3.) 78. Use a calculator to approximate the five fifth roots of 1. (See Example 11.7.4, number 4.) 79. According to Theorem 3.16 in Section 3.4, the polynomial p(x) = x 4 + 4 can be factored into the product linear and irreducible quadratic factors. In Exercise 28 in Section 8.7, we showed you how to factor this polynomial into the product of two irreducible quadratic factors using a system of non-linear equations. Now that we can compute the complex fourth roots of −4 directly, we can simply apply the Complex Factorization Theorem, Theorem 3.14, to obtain the linear factorization p(x) = (x − (1 + i))(x − (1 − i))(x − (−1 + i))(x − (−1 − i)). By multiplying the first two factors together and then the second two factors together, thus pairing up the complex conjugate pairs of zeros Theorem 3.15 told us we’d get, we have that p(x) = (x 2 − 2x + 2)(x 2 + 2x + 2). Use the 12 complex 12 th roots of 4096 to factor p(x) = x 12 −4096 into a product of linear and irreducible quadratic factors. 1004 Applications of Trigonometry 80. Complete the proof of Theorem 11.14 by showing that if w ,= 0 than ¸ ¸ 1 w ¸ ¸ = 1 |w| . 81. Recall from Section 3.4 that given a complex number z = a+bi its complex conjugate, denoted z, is given by z = a −bi. (a) Prove that [z[ = [z[. (b) Prove that [z[ = √ zz (c) Show that Re(z) = z +z 2 and Im(z) = z −z 2i (d) Show that if θ ∈ arg(z) then −θ ∈ arg (z). Interpret this result geometrically. (e) Is it always true that Arg (z) = −Arg(z)? 82. Given any natural number n ≥ 2, the n complex n th roots of the number z = 1 are called the n th Roots of Unity. In the following exercises, assume that n is a fixed, but arbitrary, natural number such that n ≥ 2. (a) Show that w = 1 is an n th root of unity. (b) Show that if both w j and w k are n th roots of unity then so is their product w j w k . (c) Show that if w j is an n th root of unity then there exists another n th root of unity w j such that w j w j = 1. Hint: If w j = cis(θ) let w j = cis(2π − θ). You’ll need to verify that w j = cis(2π −θ) is indeed an n th root of unity. 83. Another way to express the polar form of a complex number is to use the exponential function. For real numbers t, Euler’s Formula defines e it = cos(t) +i sin(t). (a) Use Theorem 11.16 to show that e ix e iy = e i(x+y) for all real numbers x and y. (b) Use Theorem 11.16 to show that _ e ix _ n = e i(nx) for any real number x and any natural number n. (c) Use Theorem 11.16 to show that e ix e iy = e i(x−y) for all real numbers x and y. (d) If z = rcis(θ) is the polar form of z, show that z = re it where θ = t radians. (e) Show that e iπ +1 = 0. (This famous equation relates the five most important constants in all of Mathematics with the three most fundamental operations in Mathematics.) (f) Show that cos(t) = e it +e −it 2 and that sin(t) = e it −e −it 2i for all real numbers t. 11.7 Polar Form of Complex Numbers 1005 11.7.2 Answers 1. z = 9 + 9i = 9 √ 2cis _ π 4 _ , Re(z) = 9, Im(z) = 9, [z[ = 9 √ 2 arg(z) = _ π 4 + 2πk [ k is an integer _ and Arg(z) = π 4 . 2. z = 5 + 5i √ 3 = 10cis _ π 3 _ , Re(z) = 5, Im(z) = 5 √ 3, [z[ = 10 arg(z) = _ π 3 + 2πk [ k is an integer _ and Arg(z) = π 3 . 3. z = 6i = 6cis _ π 2 _ , Re(z) = 0, Im(z) = 6, [z[ = 6 arg(z) = _ π 2 + 2πk [ k is an integer _ and Arg(z) = π 2 . 4. z = −3 √ 2 + 3i √ 2 = 6cis _ 3π 4 _ , Re(z) = −3 √ 2, Im(z) = 3 √ 2, [z[ = 6 arg(z) = _ 3π 4 + 2πk [ k is an integer _ and Arg(z) = 3π 4 . 5. z = −6 √ 3 + 6i = 12cis _ 5π 6 _ , Re(z) = −6 √ 3, Im(z) = 6, [z[ = 12 arg(z) = _ 5π 6 + 2πk [ k is an integer _ and Arg(z) = 5π 6 . 6. z = −2 = 2cis (π), Re(z) = −2, Im(z) = 0, [z[ = 2 arg(z) = ¦(2k + 1)π [ k is an integer¦ and Arg(z) = π. 7. z = − √ 3 2 − 1 2 i = cis _ 7π 6 _ , Re(z) = − √ 3 2 , Im(z) = − 1 2 , [z[ = 1 arg(z) = _ 7π 6 + 2πk [ k is an integer _ and Arg(z) = − 5π 6 . 8. z = −3 −3i = 3 √ 2cis _ 5π 4 _ , Re(z) = −3, Im(z) = −3, [z[ = 3 √ 2 arg(z) = _ 5π 4 + 2πk [ k is an integer _ and Arg(z) = − 3π 4 . 9. z = −5i = 5cis _ 3π 2 _ , Re(z) = 0, Im(z) = −5, [z[ = 5 arg(z) = _ 3π 2 + 2πk [ k is an integer _ and Arg(z) = − π 2 . 10. z = 2 √ 2 −2i √ 2 = 4cis _ 7π 4 _ , Re(z) = 2 √ 2, Im(z) = −2 √ 2, [z[ = 4 arg(z) = _ 7π 4 + 2πk [ k is an integer _ and Arg(z) = − π 4 . 11. z = 6 = 6cis (0), Re(z) = 6, Im(z) = 0, [z[ = 6 arg(z) = ¦2πk [ k is an integer¦ and Arg(z) = 0. 12. z = i 3 √ 7 = 3 √ 7cis _ π 2 _ , Re(z) = 0, Im(z) = 3 √ 7, [z[ = 3 √ 7 arg(z) = _ π 2 + 2πk [ k is an integer _ and Arg(z) = π 2 . 13. z = 3 + 4i = 5cis _ arctan _ 4 3 __ , Re(z) = 3, Im(z) = 4, [z[ = 5 arg(z) = _ arctan _ 4 3 _ + 2πk [ k is an integer _ and Arg(z) = arctan _ 4 3 _ . 1006 Applications of Trigonometry 14. z = √ 2 +i = √ 3cis _ arctan _ √ 2 2 __ , Re(z) = √ 2, Im(z) = 1, [z[ = √ 3 arg(z) = _ arctan _ √ 2 2 _ + 2πk [ k is an integer _ and Arg(z) = arctan _ √ 2 2 _ . 15. z = −7 + 24i = 25cis _ π −arctan _ 24 7 __ , Re(z) = −7, Im(z) = 24, [z[ = 25 arg(z) = _ π −arctan _ 24 7 _ + 2πk [ k is an integer _ and Arg(z) = π −arctan _ 24 7 _ . 16. z = −2 + 6i = 2 √ 10cis (π −arctan (3)), Re(z) = −2, Im(z) = 6, [z[ = 2 √ 10 arg(z) = ¦π −arctan (3) + 2πk [ k is an integer¦ and Arg(z) = π −arctan (3). 17. z = −12 −5i = 13cis _ π + arctan _ 5 12 __ , Re(z) = −12, Im(z) = −5, [z[ = 13 arg(z) = _ π + arctan _ 5 12 _ + 2πk [ k is an integer _ and Arg(z) = arctan _ 5 12 _ −π. 18. z = −5 −2i = √ 29cis _ π + arctan _ 2 5 __ , Re(z) = −5, Im(z) = −2, [z[ = √ 29 arg(z) = _ π + arctan _ 2 5 _ + 2πk [ k is an integer _ and Arg(z) = arctan _ 2 5 _ −π. 19. z = 4 −2i = 2 √ 5cis _ arctan _ − 1 2 __ , Re(z) = 4, Im(z) = −2, [z[ = 2 √ 5 arg(z) = _ arctan _ − 1 2 _ + 2πk [ k is an integer _ and Arg(z) = arctan _ − 1 2 _ = −arctan _ 1 2 _ . 20. z = 1 −3i = √ 10cis (arctan (−3)), Re(z) = 1, Im(z) = −3, [z[ = √ 10 arg(z) = ¦arctan (−3) + 2πk [ k is an integer¦ and Arg(z) = arctan (−3) = −arctan(3). 21. z = 6cis(0) = 6 22. z = 2cis _ π 6 _ = √ 3 +i 23. z = 7 √ 2cis _ π 4 _ = 7 + 7i 24. z = 3cis _ π 2 _ = 3i 25. z = 4cis _ 2π 3 _ = −2 + 2i √ 3 26. z = √ 6cis _ 3π 4 _ = − √ 3 +i √ 3 27. z = 9cis (π) = −9 28. z = 3cis _ 4π 3 _ = − 3 2 − 3i √ 3 2 29. z = 7cis _ − 3π 4 _ = − 7 √ 2 2 − 7 √ 2 2 i 30. z = √ 13cis _ 3π 2 _ = −i √ 13 31. z = 1 2 cis _ 7π 4 _ = √ 2 4 −i √ 2 4 32. z = 12cis _ − π 3 _ = 6 −6i √ 3 33. z = 8cis _ π 12 _ = 4 _ 2 + √ 3 + 4i _ 2 − √ 3 34. z = 2cis _ 7π 8 _ = − _ 2 + √ 2 +i _ 2 − √ 2 35. z = 5cis _ arctan _ 4 3 __ = 3 + 4i 36. z = √ 10cis _ arctan _ 1 3 __ = 3 +i 37. z = 15cis (arctan (−2)) = 3 √ 5 −6i √ 5 38. z = √ 3cis _ arctan _ − √ 2 __ = 1 −i √ 2 39. z = 50cis _ π −arctan _ 7 24 __ = −48 + 14i 40. z = 1 2 cis _ π + arctan _ 5 12 __ = − 6 13 − 5i 26 11.7 Polar Form of Complex Numbers 1007 In Exercises 41 - 52, we have that z = − 3 √ 3 2 + 3 2 i = 3cis _ 5π 6 _ and w = 3 √ 2 −3i √ 2 = 6cis _ − π 4 _ so we get the following. 41. zw = 18cis _ 7π 12 _ 42. z w = 1 2 cis _ − 11π 12 _ 43. w z = 2cis _ 11π 12 _ 44. z 4 = 81cis _ − 2π 3 _ 45. w 3 = 216cis _ − 3π 4 _ 46. z 5 w 2 = 8748cis _ − π 3 _ 47. z 3 w 2 = 972cis(0) 48. z 2 w = 3 2 cis _ − π 12 _ 49. w z 2 = 2 3 cis _ π 12 _ 50. z 3 w 2 = 3 4 cis(π) 51. w 2 z 3 = 4 3 cis(π) 52. _ w z _ 6 = 64cis _ − π 2 _ 53. _ −2 + 2i √ 3 _ 3 = 64 54. (− √ 3 −i) 3 = −8i 55. (−3 + 3i) 4 = −324 56. ( √ 3 +i) 4 = −8 + 8i √ 3 57. _ 5 2 + 5 2 i _ 3 = − 125 4 + 125 4 i 58. _ − 1 2 − i √ 3 2 _ 6 = 1 59. _ 3 2 − 3 2 i _ 3 = − 27 4 − 27 4 i 60. _ √ 3 3 − 1 3 i _ 4 = − 8 81 − 8i √ 3 81 61. _ √ 2 2 + √ 2 2 i _ 4 = −1 62. (2 + 2i) 5 −128 −128i 63. ( √ 3 −i) 5 = −16 √ 3 −16i 64. (1 −i) 8 = 16 65. Since z = 4i = 4cis _ π 2 _ we have w 0 = 2cis _ π 4 _ = √ 2 +i √ 2 w 1 = 2cis _ 5π 4 _ = − √ 2 −i √ 2 66. Since z = −25i = 25cis _ 3π 2 _ we have w 0 = 5cis _ 3π 4 _ = − 5 √ 2 2 + 5 √ 2 2 i w 1 = 5cis _ 7π 4 _ = 5 √ 2 2 − 5 √ 2 2 i 67. Since z = 1 +i √ 3 = 2cis _ π 3 _ we have w 0 = √ 2cis _ π 6 _ = √ 6 2 + √ 2 2 i w 1 = √ 2cis _ 7π 6 _ = − √ 6 2 − √ 2 2 i 68. Since z = 5 2 − 5 √ 3 2 i = 5cis _ 5π 3 _ we have w 0 = √ 5cis _ 5π 6 _ = − √ 15 2 + √ 5 2 i w 1 = √ 5cis _ 11π 6 _ = √ 15 2 − √ 5 2 i 69. Since z = 64 = 64cis (0) we have w 0 = 4cis (0) = 4 w 1 = 4cis _ 2π 3 _ = −2 + 2i √ 3 w 2 = 4cis _ 4π 3 _ = −2 −2i √ 3 1008 Applications of Trigonometry 70. Since z = −125 = 125cis (π) we have w 0 = 5cis _ π 3 _ = 5 2 + 5 √ 3 2 i w 1 = 5cis (π) = −5 w 2 = 5cis _ 5π 3 _ = 5 2 − 5 √ 3 2 i 71. Since z = i = cis _ π 2 _ we have w 0 = cis _ π 6 _ = √ 3 2 + 1 2 i w 1 = cis _ 5π 6 _ = − √ 3 2 + 1 2 i w 2 = cis _ 3π 2 _ = −i 72. Since z = −8i = 8cis _ 3π 2 _ we have w 0 = 2cis _ π 2 _ = 2i w 1 = 2cis _ 7π 6 _ = − √ 3 −i w 2 = cis _ 11π 6 _ = √ 3 −i 73. Since z = 16 = 16cis (0) we have w 0 = 2cis (0) = 2 w 1 = 2cis _ π 2 _ = 2i w 2 = 2cis (π) = −2 w 3 = 2cis _ 3π 2 _ = −2i 74. Since z = −81 = 81cis (π) we have w 0 = 3cis _ π 4 _ = 3 √ 2 2 + 3 √ 2 2 i w 1 = 3cis _ 3π 4 _ = − 3 √ 2 2 + 3 √ 2 2 i w 2 = 3cis _ 5π 4 _ = − 3 √ 2 2 − 3 √ 2 2 i w 3 = 3cis _ 7π 4 _ = 3 √ 2 2 − 3 √ 2 2 i 75. Since z = 64 = 64cis(0) we have w 0 = 2cis(0) = 2 w 1 = 2cis _ π 3 _ = 1 + √ 3i w 2 = 2cis _ 2π 3 _ = −1 + √ 3i w 3 = 2cis (π) = −2 w 4 = 2cis _ − 2π 3 _ = −1 − √ 3i w 5 = 2cis _ − π 3 _ = 1 − √ 3i 76. Since z = −729 = 729cis(π) we have w 0 = 3cis _ π 6 _ = 3 √ 3 2 + 3 2 i w 1 = 3cis _ π 2 _ = 3i w 2 = 3cis _ 5π 6 _ = − 3 √ 3 2 + 3 2 i w 3 = 3cis _ 7π 6 _ = − 3 √ 3 2 − 3 2 i w 4 = 3cis _ − 3π 2 _ = −3i w 5 = 3cis _ − 11π 6 _ = 3 √ 3 2 − 3 2 i 77. Note: In the answers for w 0 and w 2 the first rectangular form comes from applying the appropriate Sum or Difference Identity ( π 12 = π 3 − π 4 and 17π 12 = 2π 3 + 3π 4 , respectively) and the second comes from using the Half-Angle Identities. w 0 = 3 √ 2cis _ π 12 _ = 3 √ 2 _ √ 6+ √ 2 4 +i _ √ 6− √ 2 4 __ = 3 √ 2 _ √ 2+ √ 3 2 +i √ 2− √ 3 2 _ w 1 = 3 √ 2cis _ 3π 4 _ = 3 √ 2 _ − √ 2 2 + √ 2 2 i _ w 2 = 3 √ 2cis _ 17π 12 _ = 3 √ 2 _ √ 2− √ 6 4 +i _ − √ 2− √ 6 4 __ = 3 √ 2 _ √ 2− √ 3 2 +i √ 2+ √ 3 2 _ 11.7 Polar Form of Complex Numbers 1009 78. w 0 = cis(0) = 1 w 1 = cis _ 2π 5 _ ≈ 0.309 + 0.951i w 2 = cis _ 4π 5 _ ≈ −0.809 + 0.588i w 3 = cis _ 6π 5 _ ≈ −0.809 −0.588i w 4 = cis _ 8π 5 _ ≈ 0.309 −0.951i 79. p(x) = x 12 −4096 = (x−2)(x+2)(x 2 +4)(x 2 −2x+4)(x 2 +2x+4)(x 2 −2 √ 3x+4)(x 2 +2 √ 3+4) 1010 Applications of Trigonometry 11.8 Vectors As we have seen numerous times in this book, Mathematics can be used to model and solve real-world problems. For many applications, real numbers suffice; that is, real numbers with the appropriate units attached can be used to answer questions like “How close is the nearest Sasquatch nest?” There are other times though, when these kinds of quantities do not suffice. Perhaps it is important to know, for instance, how close the nearest Sasquatch nest is as well as the direction in which it lies. (Foreshadowing the use of bearings in the exercises, perhaps?) To answer questions like these which involve both a quantitative answer, or magnitude, along with a direction, we use the mathematical objects called vectors. 1 A vector is represented geometrically as a directed line segment where the magnitude of the vector is taken to be the length of the line segment and the direction is made clear with the use of an arrow at one endpoint of the segment. When referring to vectors in this text, we shall adopt 2 the ‘arrow’ notation, so the symbol v is read as ‘the vector v’. Below is a typical vector v with endpoints P (1, 2) and Q(4, 6). The point P is called the initial point or tail of v and the point Q is called the terminal point or head of v. Since we can reconstruct v completely from P and Q, we write v = −−→ PQ, where the order of points P (initial point) and Q (terminal point) is important. (Think about this before moving on.) P (1, 2) Q(4, 6) v = −−→ PQ While it is true that P and Q completely determine v, it is important to note that since vectors are defined in terms of their two characteristics, magnitude and direction, any directed line segment with the same length and direction as v is considered to be the same vector as v, regardless of its initial point. In the case of our vector v above, any vector which moves three units to the right and four up 3 from its initial point to arrive at its terminal point is considered the same vector as v. The notation we use to capture this idea is the component form of the vector, v = ¸3, 4¸, where the first number, 3, is called the x-component of v and the second number, 4, is called the y-component of v. If we wanted to reconstruct v = ¸3, 4¸ with initial point P (−2, 3), then we would find the terminal point of v by adding 3 to the x-coordinate and adding 4 to the y-coordinate to obtain the terminal point Q (1, 7), as seen below. 1 The word ‘vector’ comes from the Latin vehere meaning ‘to convey’ or ‘to carry.’ 2 Other textbook authors use bold vectors such as v. We find that writing in bold font on the chalkboard is inconvenient at best, so we have chosen the ‘arrow’ notation. 3 If this idea of ‘over’ and ‘up’ seems familiar, it should. The slope of the line segment containing v is 4 3 . 11.8 Vectors 1011 P (−2, 3) Q (1, 7) over 3 up 4 v = 3, 4 with initial point P (−2, 3). The component form of a vector is what ties these very geometric objects back to Algebra and ultimately Trigonometry. We generalize our example in our definition below. Definition 11.5. Suppose v is represented by a directed line segment with initial point P (x 0 , y 0 ) and terminal point Q(x 1 , y 1 ). The component form of v is given by v = −−→ PQ = ¸x 1 −x 0 , y 1 −y 0 ¸ Using the language of components, we have that two vectors are equal if and only if their corre- sponding components are equal. That is, ¸v 1 , v 2 ¸ = ¸v 1 , v 2 ¸ if and only if v 1 = v 1 and v 2 = v 2 . (Again, think about this before reading on.) We now set about defining operations on vectors. Suppose we are given two vectors v and w. The sum, or resultant vector v + w is obtained as follows. First, plot v. Next, plot w so that its initial point is the terminal point of v. To plot the vector v + w we begin at the initial point of v and end at the terminal point of w. It is helpful to think of the vector v + w as the ‘net result’ of moving along v then moving along w. v w v + w v, w, and v + w Our next example makes good use of resultant vectors and reviews bearings and the Law of Cosines. 4 Example 11.8.1. A plane leaves an airport with an airspeed 5 of 175 miles per hour at a bearing of N40 ◦ E. A 35 mile per hour wind is blowing at a bearing of S60 ◦ E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree. 4 If necessary, review page 903 and Section 11.3. 5 That is, the speed of the plane relative to the air around it. If there were no wind, plane’s airspeed would be the same as its speed as observed from the ground. How does wind affect this? Keep reading! 1012 Applications of Trigonometry Solution: For both the plane and the wind, we are given their speeds and their directions. Coupling speed (as a magnitude) with direction is the concept of velocity which we’ve seen a few times before in this textbook. 6 We let v denote the plane’s velocity and w denote the wind’s velocity in the diagram below. The ‘true’ speed and bearing is found by analyzing the resultant vector, v + w. From the vector diagram, we get a triangle, the lengths of whose sides are the magnitude of v, which is 175, the magnitude of w, which is 35, and the magnitude of v + w, which we’ll call c. From the given bearing information, we go through the usual geometry to determine that the angle between the sides of length 35 and 175 measures 100 ◦ . E N 40 ◦ 60 ◦ v w v + w E N 175 c 35 40 ◦ 60 ◦ 100 ◦ α From the Law of Cosines, we determine c = _ 31850 −12250 cos(100 ◦ ) ≈ 184, which means the true speed of the plane is (approximately) 184 miles per hour. To determine the true bearing of the plane, we need to determine the angle α. Using the Law of Cosines once more, 7 we find cos(α) = c 2 +29400 350c so that α ≈ 11 ◦ . Given the geometry of the situation, we add α to the given 40 ◦ and find the true bearing of the plane to be (approximately) N51 ◦ E. Our next step is to define addition of vectors component-wise to match the geometric action. 8 Definition 11.6. Suppose v = ¸v 1 , v 2 ¸ and w = ¸w 1 , w 2 ¸. The vector v + w is defined by v + w = ¸v 1 +w 1 , v 2 +w 2 ¸ Example 11.8.2. Let v = ¸3, 4¸ and suppose w = −−→ PQ where P(−3, 7) and Q(−2, 5). Find v + w and interpret this sum geometrically. Solution. Before can add the vectors using Definition 11.6, we need to write w in component form. Using Definition 11.5, we get w = ¸−2 −(−3), 5 −7¸ = ¸1, −2¸. Thus 6 See Section 10.1.1, for instance. 7 Or, since our given angle, 100 ◦ , is obtuse, we could use the Law of Sines without any ambiguity here. 8 Adding vectors ‘component-wise’ should seem hauntingly familiar. Compare this with how matrix addition was defined in section 8.3. In fact, in more advanced courses such as Linear Algebra, vectors are defined as 1 ×n or n×1 matrices, depending on the situation. 11.8 Vectors 1013 v + w = ¸3, 4¸ +¸1, −2¸ = ¸3 + 1, 4 + (−2)¸ = ¸4, 2¸ To visualize this sum, we draw v with its initial point at (0, 0) (for convenience) so that its terminal point is (3, 4). Next, we graph w with its initial point at (3, 4). Moving one to the right and two down, we find the terminal point of w to be (4, 2). We see that the vector v + w has initial point (0, 0) and terminal point (4, 2) so its component form is ¸4, 2¸, as required. x y v w v + w 1 2 3 4 1 2 3 4 In order for vector addition to enjoy the same kinds of properties as real number addition, it is necessary to extend our definition of vectors to include a ‘zero vector’, 0 = ¸0, 0¸. Geometrically, 0 represents a point, which we can think of as a directed line segment with the same initial and terminal points. The reader may well object to the inclusion of 0, since after all, vectors are supposed to have both a magnitude (length) and a direction. While it seems clear that the magnitude of 0 should be 0, it is not clear what its direction is. As we shall see, the direction of 0 is in fact undefined, but this minor hiccup in the natural flow of things is worth the benefits we reap by including 0 in our discussions. We have the following theorem. Theorem 11.18. Properties of Vector Addition • Commutative Property: For all vectors v and w, v + w = w +v. • Associative Property: For all vectors u, v and w, (u +v) + w = u + (v + w). • Identity Property: The vector 0 acts as the additive identity for vector addition. That is, for all vectors v, v + 0 = 0 +v = v. • Inverse Property: Every vector v has a unique additive inverse, denoted −v. That is, for every vector v, there is a vector −v so that v + (−v) = (−v) +v = 0. 1014 Applications of Trigonometry The properties in Theorem 11.18 are easily verified using the definition of vector addition. 9 For the commutative property, we note that if v = ¸v 1 , v 2 ¸ and w = ¸w 1 , w 2 ¸ then v + w = ¸v 1 , v 2 ¸ +¸w 1 , w 2 ¸ = ¸v 1 +w 1 , v 2 +w 2 ¸ = ¸w 1 +v 1 , w 2 +v 2 ¸ = w +v Geometrically, we can ‘see’ the commutative property by realizing that the sums v + w and w +v are the same directed diagonal determined by the parallelogram below. v w v w w + v v + w Demonstrating the commutative property of vector addition. The proofs of the associative and identity properties proceed similarly, and the reader is encouraged to verify them and provide accompanying diagrams. The existence and uniqueness of the additive inverse is yet another property inherited from the real numbers. Given a vector v = ¸v 1 , v 2 ¸, suppose we wish to find a vector w = ¸w 1 , w 2 ¸ so that v + w = 0. By the definition of vector addition, we have ¸v 1 +w 1 , v 2 +w 2 ¸ = ¸0, 0¸, and hence, v 1 + w 1 = 0 and v 2 + w 2 = 0. We get w 1 = −v 1 and w 2 = −v 2 so that w = ¸−v 1 , −v 2 ¸. Hence, v has an additive inverse, and moreover, it is unique and can be obtained by the formula −v = ¸−v 1 , −v 2 ¸. Geometrically, the vectors v = ¸v 1 , v 2 ¸ and −v = ¸−v 1 , −v 2 ¸ have the same length, but opposite directions. As a result, when adding the vectors geometrically, the sum v +(−v) results in starting at the initial point of v and ending back at the initial point of v, or in other words, the net result of moving v then −v is not moving at all. v −v Using the additive inverse of a vector, we can define the difference of two vectors, v − w = v +(− w). If v = ¸v 1 , v 2 ¸ and w = ¸w 1 , w 2 ¸ then 9 The interested reader is encouraged to compare Theorem 11.18 and the ensuing discussion with Theorem 8.3 in Section 8.3 and the discussion there. 11.8 Vectors 1015 v − w = v + (− w) = ¸v 1 , v 2 ¸ +¸−w 1 , −w 2 ¸ = ¸v 1 + (−w 1 ) , v 2 + (−w 2 )¸ = ¸v 1 −w 1 , v 2 −w 2 ¸ In other words, like vector addition, vector subtraction works component-wise. To interpret the vector v − w geometrically, we note w + (v − w) = w + (v + (− w)) Definition of Vector Subtraction = w + ((− w) +v) Commutativity of Vector Addition = ( w + (− w)) +v Associativity of Vector Addition = 0 +v Definition of Additive Inverse = v Definition of Additive Identity This means that the ‘net result’ of moving along w then moving along v − w is just v itself. From the diagram below, we see that v − w may be interpreted as the vector whose initial point is the terminal point of w and whose terminal point is the terminal point of v as depicted below. It is also worth mentioning that in the parallelogram determined by the vectors v and w, the vector v − w is one of the diagonals – the other being v + w. v w v − w v w v w v − w Next, we discuss scalar multiplication – that is, taking a real number times a vector. We define scalar multiplication for vectors in the same way we defined it for matrices in Section 8.3. Definition 11.7. If k is a real number and v = ¸v 1 , v 2 ¸, we define kv by kv = k ¸v 1 , v 2 ¸ = ¸kv 1 , kv 2 ¸ Scalar multiplication by k in vectors can be understood geometrically as scaling the vector (if k > 0) or scaling the vector and reversing its direction (if k < 0) as demonstrated below. 1016 Applications of Trigonometry v 2v 1 2 v −2v Note that, by definition 11.7, (−1)v = (−1) ¸v 1 , v 2 ¸ = ¸(−1)v 1 , (−1)v 2 ¸ = ¸−v 1 , −v 2 ¸ = −v. This, and other properties of scalar multiplication are summarized below. Theorem 11.19. Properties of Scalar Multiplication • Associative Property: For every vector v and scalars k and r, (kr)v = k(rv). • Identity Property: For all vectors v, 1v = v. • Additive Inverse Property: For all vectors v, −v = (−1)v. • Distributive Property of Scalar Multiplication over Scalar Addition: For every vector v and scalars k and r, (k +r)v = kv +rv • Distributive Property of Scalar Multiplication over Vector Addition: For all vectors v and w and scalars k, k(v + w) = kv +k w • Zero Product Property: If v is vector and k is a scalar, then kv = 0 if and only if k = 0 or v = 0 The proof of Theorem 11.19, like the proof of Theorem 11.18, ultimately boils down to the definition of scalar multiplication and properties of real numbers. For example, to prove the associative property, we let v = ¸v 1 , v 2 ¸. If k and r are scalars then (kr)v = (kr) ¸v 1 , v 2 ¸ = ¸(kr)v 1 , (kr)v 2 ¸ Definition of Scalar Multiplication = ¸k(rv 1 ), k(rv 2 )¸ Associative Property of Real Number Multiplication = k ¸rv 1 , rv 2 ¸ Definition of Scalar Multiplication = k (r ¸v 1 , v 2 ¸) Definition of Scalar Multiplication = k(rv) 11.8 Vectors 1017 The remaining properties are proved similarly and are left as exercises. Our next example demonstrates how Theorem 11.19 allows us to do the same kind of algebraic manipulations with vectors as we do with variables – multiplication and division of vectors notwith- standing. If the pedantry seems familiar, it should. This is the same treatment we gave Example 8.3.1 in Section 8.3. As in that example, we spell out the solution in excruciating detail to encourage the reader to think carefully about why each step is justified. Example 11.8.3. Solve 5v −2 (v +¸1, −2¸) = 0 for v. Solution. 5v −2 (v + ¸1, −2¸) = 0 5v + (−1) [2 (v +¸1, −2¸)] = 0 5v + [(−1)(2)] (v +¸1, −2¸) = 0 5v + (−2) (v +¸1, −2¸) = 0 5v + [(−2)v + (−2) ¸1, −2¸] = 0 5v + [(−2)v +¸(−2)(1), (−2)(−2)¸] = 0 [5v + (−2)v] +¸−2, 4¸ = 0 (5 + (−2))v +¸−2, 4¸ = 0 3v +¸−2, 4¸ = 0 (3v +¸−2, 4¸) + (−¸−2, 4¸) = 0 + (−¸−2, 4¸) 3v + [¸−2, 4¸ + (−¸−2, 4¸)] = 0 + (−1) ¸−2, 4¸ 3v + 0 = 0 +¸(−1)(−2), (−1)(4)¸ 3v = ¸2, −4¸ 1 3 (3v) = 1 3 (¸2, −4¸) __ 1 3 _ (3) ¸ v = ¸_ 1 3 _ (2), _ 1 3 _ (−4) _ 1v = ¸ 2 3 , − 4 3 _ v = ¸ 2 3 , − 4 3 _ A vector whose initial point is (0, 0) is said to be in standard position. If v = ¸v 1 , v 2 ¸ is plotted in standard position, then its terminal point is necessarily (v 1 , v 2 ). (Once more, think about this before reading on.) x y (v 1 , v 2 ) v = ¸v 1 , v 2 ¸ in standard position. 1018 Applications of Trigonometry Plotting a vector in standard position enables us to more easily quantify the concepts of magnitude and direction of the vector. We can convert the point (v 1 , v 2 ) in rectangular coordinates to a pair (r, θ) in polar coordinates where r ≥ 0. The magnitude of v, which we said earlier was length of the directed line segment, is r = _ v 2 1 +v 2 2 and is denoted by |v|. From Section 11.4, we know v 1 = r cos(θ) = |v| cos(θ) and v 2 = r sin(θ) = |v| sin(θ). From the definition of scalar multiplication and vector equality, we get v = ¸v 1 , v 2 ¸ = ¸|v| cos(θ), |v| sin(θ)¸ = |v| ¸cos(θ), sin(θ)¸ This motivates the following definition. Definition 11.8. Suppose v is a vector with component form v = ¸v 1 , v 2 ¸. Let (r, θ) be a polar representation of the point with rectangular coordinates (v 1 , v 2 ) with r ≥ 0. • The magnitude of v, denoted |v|, is given by |v| = r = _ v 2 1 +v 2 2 • If v ,= 0, the (vector) direction of v, denoted ˆ v is given by ˆ v = ¸cos(θ), sin(θ)¸ Taken together, we get v = ¸|v| cos(θ), |v| sin(θ)¸. A few remarks are in order. First, we note that if v ,= 0 then even though there are infinitely many angles θ which satisfy Definition 11.8, the stipulation r > 0 means that all of the angles are coterminal. Hence, if θ and θ both satisfy the conditions of Definition 11.8, then cos(θ) = cos(θ ) and sin(θ) = sin(θ ), and as such, ¸cos(θ), sin(θ)¸ = ¸cos(θ ), sin(θ )¸ making ˆ v is well-defined. 10 If v = 0, then v = ¸0, 0¸, and we know from Section 11.4 that (0, θ) is a polar representation for the origin for any angle θ. For this reason, ˆ 0 is undefined. The following theorem summarizes the important facts about the magnitude and direction of a vector. Theorem 11.20. Properties of Magnitude and Direction: Suppose v is a vector. • |v| ≥ 0 and |v| = 0 if and only if v = 0 • For all scalars k, |k v| = [k[|v|. • If v ,= 0 then v = |v|ˆ v, so that ˆ v = _ 1 v _ v. The proof of the first property in Theorem 11.20 is a direct consequence of the definition of |v|. If v = ¸v 1 , v 2 ¸, then |v| = _ v 2 1 +v 2 2 which is by definition greater than or equal to 0. Moreover, _ v 2 1 +v 2 2 = 0 if and only of v 2 1 + v 2 2 = 0 if and only if v 1 = v 2 = 0. Hence, |v| = 0 if and only if v = ¸0, 0¸ = 0, as required. The second property is a result of the definition of magnitude and scalar multiplication along with a propery of radicals. If v = ¸v 1 , v 2 ¸ and k is a scalar then 10 If this all looks familiar, it should. The interested reader is invited to compare Definition 11.8 to Definition 11.2 in Section 11.7. 11.8 Vectors 1019 |k v| = |k ¸v 1 , v 2 ¸ | = | ¸kv 1 , kv 2 ¸ | Definition of scalar multiplication = _ (kv 1 ) 2 + (kv 2 ) 2 Definition of magnitude = _ k 2 v 2 1 +k 2 v 2 2 = _ k 2 (v 2 1 +v 2 2 ) = √ k 2 _ v 2 1 +v 2 2 Product Rule for Radicals = [k[ _ v 2 1 +v 2 2 Since √ k 2 = [k[ = [k[|v| The equation v = |v|ˆ v in Theorem 11.20 is a consequence of the definitions of |v| and ˆ v and was worked out in the discussion just prior to Definition 11.8 on page 1018. In words, the equation v = |v|ˆ v says that any given vector is the product of its magnitude and its direction – an important concept to keep in mind when studying and using vectors. The equation ˆ v = _ 1 v _ v is a result of solving v = |v|ˆ v for ˆ v by multiplying 11 both sides of the equation by 1 v and using the properties of Theorem 11.19. We are overdue for an example. Example 11.8.4. 1. Find the component form of the vector v with |v| = 5 so that when v is plotted in standard position, it lies in Quadrant II and makes a 60 ◦ angle 12 with the negative x-axis. 2. For v = ¸ 3, −3 √ 3 _ , find |v| and θ, 0 ≤ θ < 2π so that v = |v| ¸cos(θ), sin(θ)¸. 3. For the vectors v = ¸3, 4¸ and w = ¸1, −2¸, find the following. (a) ˆ v (b) |v| −2| w| (c) |v −2 w| (d) | ˆ w| Solution. 1. We are told that |v| = 5 and are given information about its direction, so we can use the formula v = |v|ˆ v to get the component form of v. To determine ˆ v, we appeal to Definition 11.8. We are told that v lies in Quadrant II and makes a 60 ◦ angle with the negative x-axis, so the polar form of the terminal point of v, when plotted in standard position is (5, 120 ◦ ). (See the diagram below.) Thus ˆ v = ¸cos (120 ◦ ) , sin (120 ◦ )¸ = _ − 1 2 , √ 3 2 _ , so v = |v|ˆ v = 5 _ − 1 2 , √ 3 2 _ = _ − 5 2 , 5 √ 3 2 _ . 11 Of course, to go from v = vˆ v to ˆ v = 1 v v, we are essentially ‘dividing both sides’ of the equation by the scalar v. The authors encourage the reader, however, to work out the details carefully to gain an appreciation of the properties in play. 12 Due to the utility of vectors in ‘real-world’ applications, we will usually use degree measure for the angle when giving the vector’s direction. However, since Carl doesn’t want you to forget about radians, he’s made sure there are examples and exercises which use them. 1020 Applications of Trigonometry x y θ = 120 ◦ 60 ◦ v −3 −2 −1 1 2 3 1 2 3 4 5 2. For v = ¸ 3, −3 √ 3 _ , we get |v| = _ (3) 2 + (−3 √ 3) 2 = 6. In light of Definition 11.8, we can find the θ we’re after by converting the point with rectangular coordinates (3, −3 √ 3) to polar form (r, θ) where r = |v| > 0. From Section 11.4, we have tan(θ) = −3 √ 3 3 = − √ 3. Since (3, −3 √ 3) is a point in Quadrant IV, θ is a Quadrant IV angle. Hence, we pick θ = 5π 3 . We may check our answer by verifying v = ¸ 3, −3 √ 3 _ = 6 ¸ cos _ 5π 3 _ , sin _ 5π 3 __ . 3. (a) Since we are given the component form of v, we’ll use the formula ˆ v = _ 1 v _ v. For v = ¸3, 4¸, we have |v| = √ 3 2 + 4 2 = √ 25 = 5. Hence, ˆ v = 1 5 ¸3, 4¸ = ¸ 3 5 , 4 5 _ . (b) We know from our work above that |v| = 5, so to find |v|−2| w|, we need only find | w|. Since w = ¸1, −2¸, we get | w| = _ 1 2 + (−2) 2 = √ 5. Hence, |v| −2| w| = 5 −2 √ 5. (c) In the expression |v−2 w|, notice that the arithmetic on the vectors comes first, then the magnitude. Hence, our first step is to find the component form of the vector v −2 w. We get v −2 w = ¸3, 4¸ −2 ¸1, −2¸ = ¸1, 8¸. Hence, |v −2 w| = | ¸1, 8¸ | = √ 1 2 + 8 2 = √ 65. (d) To find | ˆ w|, we first need ˆ w. Using the formula ˆ w = _ 1 w _ w along with | w| = √ 5, which we found the in the previous problem, we get ˆ w = 1 √ 5 ¸1, −2¸ = _ 1 √ 5 , − 2 √ 5 _ = _ √ 5 5 , − 2 √ 5 5 _ . Hence, | ˆ w| = _ _ √ 5 5 _ 2 + _ − 2 √ 5 5 _ 2 = _ 5 25 + 20 25 = √ 1 = 1. The process exemplified by number 1 in Example 11.8.4 above by which we take information about the magnitude and direction of a vector and find the component form of a vector is called resolving a vector into its components. As an application of this process, we revisit Example 11.8.1 below. Example 11.8.5. A plane leaves an airport with an airspeed of 175 miles per hour with bearing N40 ◦ E. A 35 mile per hour wind is blowing at a bearing of S60 ◦ E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree. Solution: We proceed as we did in Example 11.8.1 and let v denote the plane’s velocity and w denote the wind’s velocity, and set about determining v + w. If we regard the airport as being 11.8 Vectors 1021 at the origin, the positive y-axis acting as due north and the positive x-axis acting as due east, we see that the vectors v and w are in standard position and their directions correspond to the angles 50 ◦ and −30 ◦ , respectively. Hence, the component form of v = 175 ¸cos(50 ◦ ), sin(50 ◦ )¸ = ¸175 cos(50 ◦ ), 175 sin(50 ◦ )¸ and the component form of w = ¸35 cos(−30 ◦ ), 35 sin(−30 ◦ )¸. Since we have no convenient way to express the exact values of cosine and sine of 50 ◦ , we leave both vectors in terms of cosines and sines. 13 Adding corresponding components, we find the resultant vector v + w = ¸175 cos(50 ◦ ) + 35 cos(−30 ◦ ), 175 sin(50 ◦ ) + 35 sin(−30 ◦ )¸. To find the ‘true’ speed of the plane, we compute the magnitude of this resultant vector |v + w| = _ (175 cos(50 ◦ ) + 35 cos(−30 ◦ )) 2 + (175 sin(50 ◦ ) + 35 sin(−30 ◦ )) 2 ≈ 184 Hence, the ‘true’ speed of the plane is approximately 184 miles per hour. To find the true bear- ing, we need to find the angle θ which corresponds to the polar form (r, θ), r > 0, of the point (x, y) = (175 cos(50 ◦ ) + 35 cos(−30 ◦ ), 175 sin(50 ◦ ) + 35 sin(−30 ◦ )). Since both of these coordinates are positive, 14 we know θ is a Quadrant I angle, as depicted below. Furthermore, tan(θ) = y x = 175 sin(50 ◦ ) + 35 sin(−30 ◦ ) 75 cos(50 ◦ ) + 35 cos(−30 ◦ ) , so using the arctangent function, we get θ ≈ 39 ◦ . Since, for the purposes of bearing, we need the angle between v+ w and the positive y-axis, we take the complement of θ and find the ‘true’ bearing of the plane to be approximately N51 ◦ E. x (E) y (N) 40 ◦ 50 ◦ 60 ◦ −30 ◦ v w x (E) y (N) v w v + w θ In part 3d of Example 11.8.4, we saw that | ˆ w| = 1. Vectors with length 1 have a special name and are important in our further study of vectors. Definition 11.9. Unit Vectors: Let v be a vector. If |v| = 1, we say that v is a unit vector. 13 Keeping things ‘calculator’ friendly, for once! 14 Yes, a calculator approximation is the quickest way to see this, but you can also use good old-fashioned inequalities and the fact that 45 ◦ ≤ 50 ◦ ≤ 60 ◦ . 1022 Applications of Trigonometry If v is a unit vector, then necessarily, v = |v|ˆ v = 1 ˆ v = ˆ v. Conversely, we leave it as an exercise 15 to show that ˆ v = _ 1 v _ v is a unit vector for any nonzero vector v. In practice, if v is a unit vector we write it as ˆ v as opposed to v because we have reserved the ‘ˆ’ notation for unit vectors. The process of multiplying a nonzero vector by the factor 1 v to produce a unit vector is called ‘normalizing the vector,’ and the resulting vector ˆ v is called the ‘unit vector in the direction of v’. The terminal points of unit vectors, when plotted in standard position, lie on the Unit Circle. (You should take the time to show this.) As a result, we visualize normalizing a nonzero vector v as shrinking 16 its terminal point, when plotted in standard position, back to the Unit Circle. x y v ˆ v −1 1 −1 1 Visualizing vector normalization ˆ v = _ 1 v _ v Of all of the unit vectors, two deserve special mention. Definition 11.10. The Principal Unit Vectors: • The vector ˆı is defined by ˆı = ¸1, 0¸ • The vector ˆ  is defined by ˆı = ¸0, 1¸ We can think of the vector ˆı as representing the positive x-direction, while ˆ  represents the positive y-direction. We have the following ‘decomposition’ theorem. 17 Theorem 11.21. Principal Vector Decomposition Theorem: Let v be a vector with component form v = ¸v 1 , v 2 ¸. Then v = v 1 ˆı +v 2 ˆ . The proof of Theorem 11.21 is straightforward. Since ˆı = ¸1, 0¸ and ˆ  = ¸0, 1¸, we have from the definition of scalar multiplication and vector addition that v 1 ˆı +v 2 ˆ  = v 1 ¸1, 0¸ +v 2 ¸0, 1¸ = ¸v 1 , 0¸ +¸0, v 2 ¸ = ¸v 1 , v 2 ¸ = v 15 One proof uses the properties of scalar multiplication and magnitude. If v = 0, consider ˆ v = 1 v v . Use the fact that v ≥ 0 is a scalar and consider factoring. 16 . . . if v > 1 . . . 17 We will see a generalization of Theorem 11.21 in Section 11.9. Stay tuned! 11.8 Vectors 1023 Geometrically, the situation looks like this: x y v = v 1 , v 2 v 1 ˆı v 2 ˆ  ˆı ˆ  v = ¸v 1 , v 2 ¸ = v 1 ˆı +v 2 ˆ . We conclude this section with a classic example which demonstrates how vectors are used to model forces. A ‘force’ is defined as a ‘push’ or a ‘pull.’ The intensity of the push or pull is the magnitude of the force, and is measured in Netwons (N) in the SI system or pounds (lbs.) in the English system. 18 The following example uses all of the concepts in this section, and should be studied in great detail. Example 11.8.6. A 50 pound speaker is suspended from the ceiling by two support braces. If one of them makes a 60 ◦ angle with the ceiling and the other makes a 30 ◦ angle with the ceiling, what are the tensions on each of the supports? Solution. We represent the problem schematically below and then provide the corresponding vector diagram. 30 ◦ 60 ◦ 50 lbs. 30 ◦ 60 ◦ 30 ◦ 60 ◦ w T 1 T 2 We have three forces acting on the speaker: the weight of the speaker, which we’ll call w, pulling the speaker directly downward, and the forces on the support rods, which we’ll call T 1 and T 2 (for ‘tensions’) acting upward at angles 60 ◦ and 30 ◦ , respectively. We are looking for the tensions on the support, which are the magnitudes | T 1 | and | T 2 |. In order for the speaker to remain stationary, 19 we require w + T 1 + T 2 = 0. Viewing the common initial point of these vectors as the 18 See also Section 11.1.1. 19 This is the criteria for ‘static equilbrium’. 1024 Applications of Trigonometry origin and the dashed line as the x-axis, we use Theorem 11.20 to get component representations for the three vectors involved. We can model the weight of the speaker as a vector pointing directly downwards with a magnitude of 50 pounds. That is, | w| = 50 and ˆ w = −ˆ  = ¸0, −1¸. Hence, w = 50 ¸0, −1¸ = ¸0, −50¸. For the force in the first support, we get T 1 = | T 1 | ¸cos (60 ◦ ) , sin (60 ◦ )¸ = _ | T 1 | 2 , | T 1 | √ 3 2 _ For the second support, we note that the angle 30 ◦ is measured from the negative x-axis, so the angle needed to write T 2 in component form is 150 ◦ . Hence T 2 = | T 2 | ¸cos (150 ◦ ) , sin (150 ◦ )¸ = _ − | T 2 | √ 3 2 , | T 2 | 2 _ The requirement w + T 1 + T 2 = 0 gives us this vector equation. w + T 1 + T 2 = 0 ¸0, −50¸ + _ | T 1 | 2 , | T 1 | √ 3 2 _ + _ − | T 2 | √ 3 2 , | T 2 | 2 _ = ¸0, 0¸ _ | T 1 | 2 − | T 2 | √ 3 2 , | T 1 | √ 3 2 + | T 2 | 2 −50 _ = ¸0, 0¸ Equating the corresponding components of the vectors on each side, we get a system of linear equations in the variables | T 1 | and | T 2 |. _ ¸ ¸ _ ¸ ¸ _ (E1) | T 1 | 2 − | T 2 | √ 3 2 = 0 (E2) | T 1 | √ 3 2 + | T 2 | 2 −50 = 0 From (E1), we get | T 1 | = | T 2 | √ 3. Substituting that into (E2) gives ( T 2 √ 3) √ 3 2 + T 2 2 − 50 = 0 which yields 2| T 2 | −50 = 0. Hence, | T 2 | = 25 pounds and | T 1 | = | T 2 | √ 3 = 25 √ 3 pounds. 11.8 Vectors 1025 11.8.1 Exercises In Exercises 1 - 10, use the given pair of vectors v and w to find the following quantities. State whether the result is a vector or a scalar. •v + w • w −2v • |v + w| • |v| +| w| • |v| w −| w|v • |w|ˆ v Finally, verify that the vectors satisfy the Parallelogram Law |v| 2 +| w| 2 = 1 2 _ |v + w| 2 +|v − w| 2 ¸ 1. v = ¸12, −5¸, w = ¸3, 4¸ 2. v = ¸−7, 24¸, w = ¸−5, −12¸ 3. v = ¸2, −1¸, w = ¸−2, 4¸ 4. v = ¸10, 4¸, w = ¸−2, 5¸ 5. v = ¸ − √ 3, 1 _ , w = ¸ 2 √ 3, 2 _ 6. v = ¸ 3 5 , 4 5 _ , w = ¸ − 4 5 , 3 5 _ 7. v = _ √ 2 2 , − √ 2 2 _ , w = _ − √ 2 2 , √ 2 2 _ 8. v = _ 1 2 , √ 3 2 _ , w = ¸ −1, − √ 3 _ 9. v = 3ˆı + 4ˆ , w = −2ˆ  10. v = 1 2 (ˆı + ˆ ), w = 1 2 (ˆı − ˆ ) In Exercises 11 - 25, find the component form of the vector v using the information given about its magnitude and direction. Give exact values. 11. |v| = 6; when drawn in standard position v lies in Quadrant I and makes a 60 ◦ angle with the positive x-axis 12. |v| = 3; when drawn in standard position v lies in Quadrant I and makes a 45 ◦ angle with the positive x-axis 13. |v| = 2 3 ; when drawn in standard position v lies in Quadrant I and makes a 60 ◦ angle with the positive y-axis 14. |v| = 12; when drawn in standard position v lies along the positive y-axis 15. |v| = 4; when drawn in standard position v lies in Quadrant II and makes a 30 ◦ angle with the negative x-axis 16. |v| = 2 √ 3; when drawn in standard position v lies in Quadrant II and makes a 30 ◦ angle with the positive y-axis 17. |v| = 7 2 ; when drawn in standard position v lies along the negative x-axis 18. |v| = 5 √ 6; when drawn in standard position v lies in Quadrant III and makes a 45 ◦ angle with the negative x-axis 19. |v| = 6.25; when drawn in standard position v lies along the negative y-axis 1026 Applications of Trigonometry 20. |v| = 4 √ 3; when drawn in standard position v lies in Quadrant IV and makes a 30 ◦ angle with the positive x-axis 21. |v| = 5 √ 2; when drawn in standard position v lies in Quadrant IV and makes a 45 ◦ angle with the negative y-axis 22. |v| = 2 √ 5; when drawn in standard position v lies in Quadrant I and makes an angle measuring arctan(2) with the positive x-axis 23. |v| = √ 10; when drawn in standard position v lies in Quadrant II and makes an angle measuring arctan(3) with the negative x-axis 24. |v| = 5; when drawn in standard position v lies in Quadrant III and makes an angle measuring arctan _ 4 3 _ with the negative x-axis 25. |v| = 26; when drawn in standard position v lies in Quadrant IV and makes an angle measuring arctan _ 5 12 _ with the positive x-axis In Exercises 26 - 31, approximate the component form of the vector v using the information given about its magnitude and direction. Round your approximations to two decimal places. 26. |v| = 392; when drawn in standard position v makes a 117 ◦ angle with the positive x-axis 27. |v| = 63.92; when drawn in standard position v makes a 78.3 ◦ angle with the positive x-axis 28. |v| = 5280; when drawn in standard position v makes a 12 ◦ angle with the positive x-axis 29. |v| = 450; when drawn in standard position v makes a 210.75 ◦ angle with the positive x-axis 30. |v| = 168.7; when drawn in standard position v makes a 252 ◦ angle with the positive x-axis 31. |v| = 26; when drawn in standard position v makes a 304.5 ◦ angle with the positive x-axis In Exercises 32 - 52, for the given vector v, find the magnitude |v| and an angle θ with 0 ≤ θ < 360 ◦ so that v = |v| ¸cos(θ), sin(θ)¸ (See Definition 11.8.) Round approximations to two decimal places. 32. v = ¸ 1, √ 3 _ 33. v = ¸5, 5¸ 34. v = ¸ −2 √ 3, 2 _ 35. v = ¸ − √ 2, √ 2 _ 36. v = _ − √ 2 2 , − √ 2 2 _ 37. v = _ − 1 2 , − √ 3 2 _ 38. v = ¸6, 0¸ 39. v = ¸−2.5, 0¸ 40. v = ¸ 0, √ 7 _ 41. v = −10ˆ  42. v = ¸3, 4¸ 43. v = ¸12, 5¸ 11.8 Vectors 1027 44. v = ¸−4, 3¸ 45. v = ¸−7, 24¸ 46. v = ¸−2, −1¸ 47. v = ¸−2, −6¸ 48. v = ˆı + ˆ  49. v = ˆı −4ˆ  50. v = ¸123.4, −77.05¸ 51. v = ¸965.15, 831.6¸ 52. v = ¸−114.1, 42.3¸ 53. A small boat leaves the dock at Camp DuNuthin and heads across the Nessie River at 17 miles per hour (that is, with respect to the water) at a bearing of S68 ◦ W. The river is flowing due east at 8 miles per hour. What is the boat’s true speed and heading? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 54. The HMS Sasquatch leaves port with bearing S20 ◦ E maintaining a speed of 42 miles per hour (that is, with respect to the water). If the ocean current is 5 miles per hour with a bearing of N60 ◦ E, find the HMS Sasquatch’s true speed and bearing. Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 55. If the captain of the HMS Sasquatch in Exercise 54 wishes to reach Chupacabra Cove, an island 100 miles away at a bearing of S20 ◦ E from port, in three hours, what speed and heading should she set to take into account the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. HINT: If v denotes the velocity of the HMS Sasquatch and w denotes the velocity of the current, what does v + w need to be to reach Chupacabra Cove in three hours? 56. In calm air, a plane flying from the Pedimaxus International Airport can reach Cliffs of Insanity Point in two hours by following a bearing of N8.2 ◦ E at 96 miles an hour. (The distance between the airport and the cliffs is 192 miles.) If the wind is blowing from the southeast at 25 miles per hour, what speed and bearing should the pilot take so that she makes the trip in two hours along the original heading? Round the speed to the nearest hundredth of a mile per hour and your angle to the nearest tenth of a degree. 57. The SS Bigfoot leaves Yeti Bay on a course of N37 ◦ W at a speed of 50 miles per hour. After traveling half an hour, the captain determines he is 30 miles from the bay and his bearing back to the bay is S40 ◦ E. What is the speed and bearing of the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 58. A 600 pound Sasquatch statue is suspended by two cables from a gymnasium ceiling. If each cable makes a 60 ◦ angle with the ceiling, find the tension on each cable. Round your answer to the nearest pound. 59. Two cables are to support an object hanging from a ceiling. If the cables are each to make a 42 ◦ angle with the ceiling, and each cable is rated to withstand a maximum tension of 100 pounds, what is the heaviest object that can be supported? Round your answer down to the nearest pound. 1028 Applications of Trigonometry 60. A 300 pound metal star is hanging on two cables which are attached to the ceiling. The left hand cable makes a 72 ◦ angle with the ceiling while the right hand cable makes a 18 ◦ angle with the ceiling. What is the tension on each of the cables? Round your answers to three decimal places. 61. Two drunken college students have filled an empty beer keg with rocks and tied ropes to it in order to drag it down the street in the middle of the night. The stronger of the two students pulls with a force of 100 pounds at a heading of N77 ◦ E and the other pulls at a heading of S68 ◦ E. What force should the weaker student apply to his rope so that the keg of rocks heads due east? What resultant force is applied to the keg? Round your answer to the nearest pound. 62. Emboldened by the success of their late night keg pull in Exercise 61 above, our intrepid young scholars have decided to pay homage to the chariot race scene from the movie ‘Ben-Hur’ by tying three ropes to a couch, loading the couch with all but one of their friends and pulling it due west down the street. The first rope points N80 ◦ W, the second points due west and the third points S80 ◦ W. The force applied to the first rope is 100 pounds, the force applied to the second rope is 40 pounds and the force applied (by the non-riding friend) to the third rope is 160 pounds. They need the resultant force to be at least 300 pounds otherwise the couch won’t move. Does it move? If so, is it heading due west? 63. Let v = ¸v 1 , v 2 ¸ be any non-zero vector. Show that 1 |v| v has length 1. 64. We say that two non-zero vectors v and w are parallel if they have same or opposite directions. That is, v ,= 0 and w ,= 0 are parallel if either ˆ v = ˆ w or ˆ v = −ˆ w. Show that this means v = k w for some non-zero scalar k and that k > 0 if the vectors have the same direction and k < 0 if they point in opposite directions. 65. The goal of this exercise is to use vectors to describe non-vertical lines in the plane. To that end, consider the line y = 2x − 4. Let v 0 = ¸0, −4¸ and let s = ¸1, 2¸. Let t be any real number. Show that the vector defined by v = v 0 +ts, when drawn in standard position, has its terminal point on the line y = 2x −4. (Hint: Show that v 0 + ts = ¸t, 2t −4¸ for any real number t.) Now consider the non-vertical line y = mx+b. Repeat the previous analysis with v 0 = ¸0, b¸ and let s = ¸1, m¸. Thus any non-vertical line can be thought of as a collection of terminal points of the vector sum of ¸0, b¸ (the position vector of the y-intercept) and a scalar multiple of the slope vector s = ¸1, m¸. 66. Prove the associative and identity properties of vector addition in Theorem 11.18. 67. Prove the properties of scalar multiplication in Theorem 11.19. 11.8 Vectors 1029 11.8.2 Answers 1. • v + w = ¸15, −1¸, vector • w −2v = ¸−21, 14¸, vector • |v + w| = √ 226, scalar • |v| +| w| = 18, scalar • |v| w −| w|v = ¸−21, 77¸, vector • |w|ˆ v = ¸ 60 13 , − 25 13 _ , vector 2. • v + w = ¸−12, 12¸, vector • w −2v = ¸9, −60¸, vector • |v + w| = 12 √ 2, scalar • |v| +| w| = 38, scalar • |v| w −| w|v = ¸−34, −612¸, vector • |w|ˆ v = ¸ − 91 25 , 312 25 _ , vector 3. • v + w = ¸0, 3¸, vector • w −2v = ¸−6, 6¸, vector • |v + w| = 3, scalar • |v| +| w| = 3 √ 5, scalar • |v| w −| w|v = ¸ −6 √ 5, 6 √ 5 _ , vector • |w|ˆ v = ¸4, −2¸, vector 4. • v + w = ¸8, 9¸, vector • w −2v = ¸−22, −3¸, vector • |v + w| = √ 145, scalar • |v| +| w| = 3 √ 29, scalar • |v| w−| w|v = ¸ −14 √ 29, 6 √ 29 _ , vector • |w|ˆ v = ¸5, 2¸, vector 5. • v + w = ¸√ 3, 3 _ , vector • w −2v = ¸ 4 √ 3, 0 _ , vector • |v + w| = 2 √ 3, scalar • |v| +| w| = 6, scalar • |v| w −| w|v = ¸ 8 √ 3, 0 _ , vector • |w|ˆ v = ¸ −2 √ 3, 2 _ , vector 6. • v + w = ¸ − 1 5 , 7 5 _ , vector • w −2v = ¸−2, −1¸, vector • |v + w| = √ 2, scalar • |v| +| w| = 2, scalar • |v| w −| w|v = ¸ − 7 5 , − 1 5 _ , vector • |w|ˆ v = ¸ 3 5 , 4 5 _ , vector 7. • v + w = ¸0, 0¸, vector • w −2v = _ − 3 √ 2 2 , 3 √ 2 2 _ , vector • |v + w| = 0, scalar • |v| +| w| = 2, scalar • |v| w −| w|v = ¸ − √ 2, √ 2 _ , vector • |w|ˆ v = _ √ 2 2 , − √ 2 2 _ , vector 1030 Applications of Trigonometry 8. • v + w = _ − 1 2 , − √ 3 2 _ , vector • w −2v = ¸ −2, −2 √ 3 _ , vector • |v + w| = 1, scalar • |v| +| w| = 3, scalar • |v| w −| w|v = ¸ −2, −2 √ 3 _ , vector • |w|ˆ v = ¸ 1, √ 3 _ , vector 9. • v + w = ¸3, 2¸, vector • w −2v = ¸−6, −10¸, vector • |v + w| = √ 13, scalar • |v| +| w| = 7, scalar • |v| w −| w|v = ¸−6, −18¸, vector • |w|ˆ v = ¸ 6 5 , 8 5 _ , vector 10. • v + w = ¸1, 0¸, vector • w −2v = ¸ − 1 2 , − 3 2 _ , vector • |v + w| = 1, scalar • |v| +| w| = √ 2, scalar • |v| w −| w|v = _ 0, − √ 2 2 _ , vector • |w|ˆ v = ¸ 1 2 , 1 2 _ , vector 11. v = ¸ 3, 3 √ 3 _ 12. v = _ 3 √ 2 2 , 3 √ 2 2 _ 13. v = _ √ 3 3 , 1 3 _ 14. v = ¸0, 12¸ 15. v = ¸ −2 √ 3, 2 _ 16. v = ¸ − √ 3, 3 _ 17. v = ¸ − 7 2 , 0 _ 18. v = ¸ −5 √ 3, −5 √ 3 _ 19. v = ¸0, −6.25¸ 20. v = ¸ 6, −2 √ 3 _ 21. v = ¸5, −5¸ 22. v = ¸2, 4¸ 23. v = ¸−1, 3¸ 24. v = ¸−3, −4¸ 25. v = ¸24, −10¸ 26. v ≈ ¸−177.96, 349.27¸ 27. v ≈ ¸12.96, 62.59¸ 28. v ≈ ¸5164.62, 1097.77¸ 29. v ≈ ¸−386.73, −230.08¸ 30. v ≈ ¸−52.13, −160.44¸ 31. v ≈ ¸14.73, −21.43¸ 32. |v| = 2, θ = 60 ◦ 33. |v| = 5 √ 2, θ = 45 ◦ 34. |v| = 4, θ = 150 ◦ 35. |v| = 2, θ = 135 ◦ 36. |v| = 1, θ = 225 ◦ 37. |v| = 1, θ = 240 ◦ 38. |v| = 6, θ = 0 ◦ 39. |v| = 2.5, θ = 180 ◦ 40. |v| = √ 7, θ = 90 ◦ 41. |v| = 10, θ = 270 ◦ 42. |v| = 5, θ ≈ 53.13 ◦ 43. |v| = 13, θ ≈ 22.62 ◦ 44. |v| = 5, θ ≈ 143.13 ◦ 45. |v| = 25, θ ≈ 106.26 ◦ 46. |v| = √ 5, θ ≈ 206.57 ◦ 11.8 Vectors 1031 47. |v| = 2 √ 10, θ ≈ 251.57 ◦ 48. |v| = √ 2, θ ≈ 45 ◦ 49. |v| = √ 17, θ ≈ 284.04 ◦ 50. |v| ≈ 145.48, θ ≈ 328.02 ◦ 51. |v| ≈ 1274.00, θ ≈ 40.75 ◦ 52. |v| ≈ 121.69, θ ≈ 159.66 ◦ 53. The boat’s true speed is about 10 miles per hour at a heading of S50.6 ◦ W. 54. The HMS Sasquatch’s true speed is about 41 miles per hour at a heading of S26.8 ◦ E. 55. She should maintain a speed of about 35 miles per hour at a heading of S11.8 ◦ E. 56. She should fly at 83.46 miles per hour with a heading of N22.1 ◦ E 57. The current is moving at about 10 miles per hour bearing N54.6 ◦ W. 58. The tension on each of the cables is about 346 pounds. 59. The maximum weight that can be held by the cables in that configuration is about 133 pounds. 60. The tension on the left hand cable is 285.317 lbs. and on the right hand cable is 92.705 lbs. 61. The weaker student should pull about 60 pounds. The net force on the keg is about 153 pounds. 62. The resultant force is only about 296 pounds so the couch doesn’t budge. Even if it did move, the stronger force on the third rope would have made the couch drift slightly to the south as it traveled down the street. 1032 Applications of Trigonometry 11.9 The Dot Product and Projection In Section 11.8, we learned how add and subtract vectors and how to multiply vectors by scalars. In this section, we define a product of vectors. We begin with the following definition. Definition 11.11. Suppose v and w are vectors whose component forms are v = ¸v 1 , v 2 ¸ and w = ¸w 1 , w 2 ¸. The dot product of v and w is given by v w = ¸v 1 , v 2 ¸ ¸w 1 , w 2 ¸ = v 1 w 1 +v 2 w 2 For example, let v = ¸3, 4¸ and w = ¸1, −2¸. Then v w = ¸3, 4¸ ¸1, −2¸ = (3)(1) + (4)(−2) = −5. Note that the dot product takes two vectors and produces a scalar. For that reason, the quantity v w is often called the scalar product of v and w. The dot product enjoys the following properties. Theorem 11.22. Properties of the Dot Product • Commutative Property: For all vectors v and w, v w = w v. • Distributive Property: For all vectors u, v and w, u (v + w) = u v +u w. • Scalar Property: For all vectors v and w and scalars k, (kv) w = k(v w) = v (k w). • Relation to Magnitude: For all vectors v, v v = |v| 2 . Like most of the theorems involving vectors, the proof of Theorem 11.22 amounts to using the definition of the dot product and properties of real number arithmetic. To show the commutative property for instance, let v = ¸v 1 , v 2 ¸ and w = ¸w 1 , w 2 ¸. Then v w = ¸v 1 , v 2 ¸ ¸w 1 , w 2 ¸ = v 1 w 1 +v 2 w 2 Definition of Dot Product = w 1 v 1 +w 2 v 2 Commutativity of Real Number Multiplication = ¸w 1 , w 2 ¸ ¸v 1 , v 2 ¸ Definition of Dot Product = w v The distributive property is proved similarly and is left as an exercise. For the scalar property, assume that v = ¸v 1 , v 2 ¸ and w = ¸w 1 , w 2 ¸ and k is a scalar. Then (kv) w = (k ¸v 1 , v 2 ¸) ¸w 1 , w 2 ¸ = ¸kv 1 , kv 2 ¸ ¸w 1 , w 2 ¸ Definition of Scalar Multiplication = (kv 1 )(w 1 ) + (kv 2 )(w 2 ) Definition of Dot Product = k(v 1 w 1 ) +k(v 2 w 2 ) Associativity of Real Number Multiplication = k(v 1 w 1 +v 2 w 2 ) Distributive Law of Real Numbers = k ¸v 1 , v 2 ¸ ¸w 1 , w 2 ¸ Definition of Dot Product = k(v w) We leave the proof of k(v w) = v (k w) as an exercise. 11.9 The Dot Product and Projection 1033 For the last property, we note that if v = ¸v 1 , v 2 ¸, then v v = ¸v 1 , v 2 ¸ ¸v 1 , v 2 ¸ = v 2 1 + v 2 2 = |v| 2 , where the last equality comes courtesy of Definition 11.8. The following example puts Theorem 11.22 to good use. As in Example 11.8.3, we work out the problem in great detail and encourage the reader to supply the justification for each step. Example 11.9.1. Prove the identity: |v − w| 2 = |v| 2 −2(v w) +| w| 2 . Solution. We begin by rewriting |v − w| 2 in terms of the dot product using Theorem 11.22. |v − w| 2 = (v − w) (v − w) = (v + [− w]) (v + [− w]) = (v + [− w]) v + (v + [− w]) [− w] = v (v + [− w]) + [− w] (v + [− w]) = v v +v [− w] + [− w] v + [− w] [− w] = v v +v [(−1) w] + [(−1) w] v + [(−1) w] [(−1) w] = v v + (−1)(v w) + (−1)( w v) + [(−1)(−1)]( w w) = v v + (−1)(v w) + (−1)(v w) + w w = v v −2(v w) + w w = |v| 2 −2(v w) +| w| 2 Hence, |v − w| 2 = |v| 2 −2(v w) +| w| 2 as required. If we take a step back from the pedantry in Example 11.9.1, we see that the bulk of the work is needed to show that (v− w)(v− w) = vv−2(v w)+ w w. If this looks familiar, it should. Since the dot product enjoys many of the same properties enjoyed by real numbers, the machinations required to expand (v − w) (v − w) for vectors v and w match those required to expand (v −w)(v −w) for real numbers v and w, and hence we get similar looking results. The identity verified in Example 11.9.1 plays a large role in the development of the geometric properties of the dot product, which we now explore. Suppose v and w are two nonzero vectors. If we draw v and w with the same initial point, we define the angle between v and w to be the angle θ determined by the rays containing the vectors v and w, as illustrated below. We require 0 ≤ θ ≤ π. (Think about why this is needed in the definition.) v w θ v w v w θ = 0 0 < θ < π θ = π The following theorem gives us some insight into the geometric role the dot product plays. Theorem 11.23. Geometric Interpretation of Dot Product: If v and w are nonzero vectors then v w = |v|| w| cos(θ), where θ is the angle between v and w. 1034 Applications of Trigonometry We prove Theorem 11.23 in cases. If θ = 0, then v and w have the same direction. It follows 1 that there is a real number k > 0 so that w = kv. Hence, v w = v (kv) = k(v v) = k|v| 2 = k|v||v|. Since k > 0, k = [k[, so k|v| = [k[|v| = |kv| by Theorem 11.20. Hence, k|v||v| = |v|(k|v|) = |v||kv| = |v|| w|. Since cos(0) = 1, we get v w = k|v||v| = |v|| w| = |v|| w| cos(0), proving that the formula holds for θ = 0. If θ = π, we repeat the argument with the difference being w = kv where k < 0. In this case, [k[ = −k, so k|v| = −[k[|v| = −|kv| = −| w|. Since cos(π) = −1, we get v w = −|v|| w| = |v|| w| cos(π), as required. Next, if 0 < θ < π, the vectors v, w and v − w determine a triangle with side lengths |v|, | w| and |v − w|, respectively, as seen below. θ v w v − w θ |v| | w| |v − w| The Law of Cosines yields |v − w| 2 = |v| 2 + | w| 2 − 2|v|| w| cos(θ). From Example 11.9.1, we know |v − w| 2 = |v| 2 − 2(v w) + | w| 2 . Equating these two expressions for |v − w| 2 gives |v| 2 +| w| 2 −2|v|| w| cos(θ) = |v| 2 −2(v w)+| w| 2 which reduces to −2|v|| w| cos(θ) = −2(v w), or v w = |v|| w| cos(θ), as required. An immediate consequence of Theorem 11.23 is the following. Theorem 11.24. Let v and w be nonzero vectors and let θ the angle between v and w. Then θ = arccos _ v w |v|| w| _ = arccos(ˆ v ˆ w) We obtain the formula in Theorem 11.24 by solving the equation given in Theorem 11.23 for θ. Since v and w are nonzero, so are |v| and | w|. Hence, we may divide both sides of v w = |v|| w| cos(θ) by |v|| w| to get cos(θ) = v· w v w . Since 0 ≤ θ ≤ π by definition, the values of θ exactly match the range of the arccosine function. Hence, θ = arccos _ v· w v w _ . Using Theorem 11.22, we can rewrite v· w v w = _ 1 v v _ _ 1 w w _ = ˆ v ˆ w, giving us the alternative formula θ = arccos(ˆ v ˆ w). We are overdue for an example. Example 11.9.2. Find the angle between the following pairs of vectors. 1. v = ¸ 3, −3 √ 3 _ , and w = ¸ − √ 3, 1 _ 2. v = ¸2, 2¸, and w = ¸5, −5¸ 3. v = ¸3, −4¸, and w = ¸2, 1¸ Solution. We use the formula θ = arccos _ v· w v w _ from Theorem 11.24 in each case below. 1 Since v = vˆ v and w = w ˆ w, if ˆ v = ˆ w then w = wˆ v = w v (vˆ v) = w v v. In this case, k = w v > 0. 11.9 The Dot Product and Projection 1035 1. We have v w = ¸ 3, −3 √ 3 _ ¸ − √ 3, 1 _ = −3 √ 3−3 √ 3 = −6 √ 3. Since |v| = _ 3 2 + (−3 √ 3) 2 = √ 36 = 6 and | w| = _ (− √ 3) 2 + 1 2 = √ 4 = 2, θ = arccos _ −6 √ 3 12 _ = arccos _ − √ 3 2 _ = 5π 6 . 2. For v = ¸2, 2¸ and w = ¸5, −5¸, we find v w = ¸2, 2¸ ¸5, −5¸ = 10 −10 = 0. Hence, it doesn’t matter what |v| and | w| are, 2 θ = arccos _ v· w v w _ = arccos(0) = π 2 . 3. We find v w = ¸3, −4¸ ¸2, 1¸ = 6 − 4 = 2. Also |v| = _ 3 2 + (−4) 2 = √ 25 = 5 and w = √ 2 2 + 1 2 = √ 5, so θ = arccos _ 2 5 √ 5 _ = arccos _ 2 √ 5 25 _ . Since 2 √ 5 25 isn’t the cosine of one of the common angles, we leave our answer as θ = arccos _ 2 √ 5 25 _ . The vectors v = ¸2, 2¸, and w = ¸5, −5¸ in Example 11.9.2 are called orthogonal and we write v ⊥ w, because the angle between them is π 2 radians = 90 ◦ . Geometrically, when orthogonal vectors are sketched with the same initial point, the lines containing the vectors are perpendicular. v w v and w are orthogonal, v ⊥ w We state the relationship between orthogonal vectors and their dot product in the following theorem. Theorem 11.25. The Dot Product Detects Orthogonality: Let v and w be nonzero vectors. Then v ⊥ w if and only if v w = 0. To prove Theorem 11.25, we first assume v and w are nonzero vectors with v ⊥ w. By definition, the angle between v and w is π 2 . By Theorem 11.23, v w = |v|| w| cos _ π 2 _ = 0. Conversely, if v and w are nonzero vectors and v w = 0, then Theorem 11.24 gives θ = arccos _ v· w v w _ = arccos _ 0 v w _ = arccos(0) = π 2 , so v ⊥ w. We can use Theorem 11.25 in the following example to provide a different proof about the relationship between the slopes of perpendicular lines. 3 Example 11.9.3. Let L 1 be the line y = m 1 x+b 1 and let L 2 be the line y = m 2 x+b 2 . Prove that L 1 is perpendicular to L 2 if and only if m 1 m 2 = −1. Solution. Our strategy is to find two vectors: v 1 , which has the same direction as L 1 , and v 2 , which has the same direction as L 2 and show v 1 ⊥ v 2 if and only if m 1 m 2 = −1. To that end, we substitute x = 0 and x = 1 into y = m 1 x + b 1 to find two points which lie on L 1 , namely P(0, b 1 ) 2 Note that there is no ‘zero product property’ for the dot product since neither v nor w is 0, yet v · w = 0. 3 See Exercise 2.1.1 in Section 2.1. 1036 Applications of Trigonometry and Q(1, m 1 + b 1 ). We let v 1 = −−→ PQ = ¸1 −0, (m 1 +b 1 ) −b 1 ¸ = ¸1, m 1 ¸, and note that since v 1 is determined by two points on L 1 , it may be viewed as lying on L 1 . Hence it has the same direction as L 1 . Similarly, we get the vector v 2 = ¸1, m 2 ¸ which has the same direction as the line L 2 . Hence, L 1 and L 2 are perpendicular if and only if v 1 ⊥ v 2 . According to Theorem 11.25, v 1 ⊥ v 2 if and only if v 1 v 2 = 0. Notice that v 1 v 2 = ¸1, m 1 ¸ ¸1, m 2 ¸ = 1 +m 1 m 2 . Hence, v 1 v 2 = 0 if and only if 1 +m 1 m 2 = 0, which is true if and only if m 1 m 2 = −1, as required. While Theorem 11.25 certainly gives us some insight into what the dot product means geometrically, there is more to the story of the dot product. Consider the two nonzero vectors v and w drawn with a common initial point O below. For the moment, assume that the angle between v and w, which we’ll denote θ, is acute. We wish to develop a formula for the vector p, indicated below, which is called the orthogonal projection of v onto w. The vector p is obtained geometrically as follows: drop a perpendicular from the terminal point T of v to the vector w and call the point of intersection R. The vector p is then defined as p = −−→ OR. Like any vector, p is determined by its magnitude | p| and its direction ˆ p according to the formula p = | p|ˆ p. Since we want ˆ p to have the same direction as w, we have ˆ p = ˆ w. To determine | p|, we make use of Theorem 10.4 as applied to the right triangle ´ORT. We find cos(θ) = p v , or | p| = |v| cos(θ). To get things in terms of just v and w, we use Theorem 11.23 to get | p| = |v| cos(θ) = v w cos(θ) w = v· w w . Using Theorem 11.22, we rewrite v· w w = v _ 1 w w _ = v ˆ w. Hence, | p| = v ˆ w, and since ˆ p = ˆ w, we now have a formula for p completely in terms of v and w, namely p = | p|ˆ p = (v ˆ w) ˆ w. O w v θ O w v R T p = −−→ OR θ O R T |v| | p| θ Now suppose that the angle θ between v and w is obtuse, and consider the diagram below. In this case, we see that ˆ p = −ˆ w and using the triangle ´ORT, we find | p| = |v| cos(θ ). Since θ+θ = π, it follows that cos(θ ) = −cos(θ), which means | p| = |v| cos(θ ) = −|v| cos(θ). Rewriting this last equation in terms of v and w as before, we get | p| = −(v ˆ w). Putting this together with ˆ p = −ˆ w, we get p = | p|ˆ p = −(v ˆ w)(−ˆ w) = (v ˆ w) ˆ w in this case as well. 11.9 The Dot Product and Projection 1037 O w v R T p = −−→ OR θ θ If the angle between v and w is π 2 then it is easy to show 4 that p = 0. Since v ⊥ w in this case, v w = 0. It follows that v ˆ w = 0 and p = 0 = 0 ˆ w = (v ˆ w) ˆ w in this case, too. This gives us Definition 11.12. Let v and w be nonzero vectors. The orthogonal projection of v onto w, denoted proj w (v) is given by proj w (v) = (v ˆ w) ˆ w. Definition 11.12 gives us a good idea what the dot product does. The scalar v ˆ w is a measure of how much of the vector v is in the direction of the vector w and is thus called the scalar projection of v onto w. While the formula given in Definition 11.12 is theoretically appealing, because of the presence of the normalized unit vector ˆ w, computing the projection using the formula proj w (v) = (v ˆ w) ˆ w can be messy. We present two other formulas that are often used in practice. Theorem 11.26. Alternate Formulas for Vector Projections: If v and w are nonzero vectors then proj w (v) = (v ˆ w) ˆ w = _ v w | w| 2 _ w = _ v w w w _ w The proof of Theorem 11.26, which we leave to the reader as an exercise, amounts to using the formula ˆ w = _ 1 w _ w and properties of the dot product. It is time for an example. Example 11.9.4. Let v = ¸1, 8¸ and w = ¸−1, 2¸. Find p = proj w (v), and plot v, w and p in standard position. Solution. We find v w = ¸1, 8¸ ¸−1, 2¸ = (−1) +16 = 15 and w w = ¸−1, 2¸ ¸−1, 2¸ = 1+4 = 5. Hence, p = v· w w· w w = 15 5 ¸−1, 2¸ = ¸−3, 6¸. We plot v, w and p below. 4 In this case, the point R coincides with the point O, so p = −→ OR = −−→ OO = 0. 1038 Applications of Trigonometry v w p −3 −2 −1 1 2 3 4 5 6 7 8 Suppose we wanted to verify that our answer p in Example 11.9.4 is indeed the orthogonal projection of v onto w. We first note that since p is a scalar multiple of w, it has the correct direction, so what remains to check is the orthogonality condition. Consider the vector q whose initial point is the terminal point of p and whose terminal point is the terminal point of v. v w p q −3 −2 −1 1 2 3 4 5 6 7 8 From the definition of vector arithmetic, p+q = v, so that q = v − p. In the case of Example 11.9.4, v = ¸1, 8¸ and p = ¸−3, 6¸, so q = ¸1, 8¸−¸−3, 6¸ = ¸4, 2¸. Then q w = ¸4, 2¸¸−1, 2¸ = (−4)+4 = 0, which shows q ⊥ w, as required. This result is generalized in the following theorem. Theorem 11.27. Generalized Decomposition Theorem: Let v and w be nonzero vectors. There are unique vectors p and q such that v = p + q where p = k w for some scalar k, and q w = 0. Note that if the vectors p and q in Theorem 11.27 are nonzero, then we can say p is parallel 5 to w and q is orthogonal to w. In this case, the vector p is sometimes called the ‘vector component of v parallel to w’ and q is called the ‘vector component of v orthogonal to w.’ To prove Theorem 11.27, we take p = proj w (v) and q = v − p. Then p is, by definition, a scalar multiple of w. Next, we compute q w. 5 See Exercise 64 in Section 11.8. 11.9 The Dot Product and Projection 1039 q w = (v − p) w Definition of q. = v w − p w Properties of Dot Product = v w − _ v w w w w _ w Since p = proj w (v). = v w − _ v w w w _ ( w w) Properties of Dot Product. = v w −v w = 0 Hence, q w = 0, as required. At this point, we have shown that the vectors p and q guaranteed by Theorem 11.27 exist. Now we need to show that they are unique. Suppose v = p +q = p +q where the vectors p and q satisfy the same properties described in Theorem 11.27 as p and q. Then p − p = q − q, so w ( p − p ) = w (q − q) = w q − w q = 0 − 0 = 0. Hence, w ( p − p ) = 0. Now there are scalars k and k so that p = k w and p = k w. This means w ( p − p ) = w (k w − k w) = w ([k − k ] w) = (k − k )( w w) = (k − k )| w| 2 . Since w ,= 0, | w| 2 ,= 0, which means the only way w ( p − p ) = (k −k )| w| 2 = 0 is for k −k = 0, or k = k . This means p = k w = k w = p . With q − q = p − p = p − p = 0, it must be that q = q as well. Hence, we have shown there is only one way to write v as a sum of vectors as described in Theorem 11.27. We close this section with an application of the dot product. In Physics, if a constant force F is exerted over a distance d, the work W done by the force is given by W = Fd. Here, we assume the force is being applied in the direction of the motion. If the force applied is not in the direction of the motion, we can use the dot product to find the work done. Consider the scenario below where the constant force F is applied to move an object from the point P to the point Q. P Q F F θ θ To find the work W done in this scenario, we need to find how much of the force F is in the direction of the motion −−→ PQ. This is precisely what the dot product F ¯ PQ represents. Since the distance the object travels is | −−→ PQ|, we get W = ( F ¯ PQ)| −−→ PQ|. Since −−→ PQ = | −−→ PQ| ¯ PQ, W = ( F ¯ PQ)| −−→ PQ| = F (| −−→ PQ| ¯ PQ) = F −−→ PQ = | F|| −−→ PQ| cos(θ), where θ is the angle between the applied force F and the trajectory of the motion −−→ PQ. We have proved the following. 1040 Applications of Trigonometry Theorem 11.28. Work as a Dot Product: Suppose a constant force F is applied along the vector −−→ PQ. The work W done by F is given by W = F −−→ PQ = | F|| −−→ PQ| cos(θ), where θ is the angle between F and −−→ PQ. Example 11.9.5. Taylor exerts a force of 10 pounds to pull her wagon a distance of 50 feet over level ground. If the handle of the wagon makes a 30 ◦ angle with the horizontal, how much work did Taylor do pulling the wagon? Assume Taylor exerts the force of 10 pounds at a 30 ◦ angle for the duration of the 50 feet. 30 ◦ Solution. There are two ways to attack this problem. One way is to find the vectors F and −−→ PQ mentioned in Theorem 11.9.5 and compute W = F −−→ PQ. To do this, we assume the origin is at the point where the handle of the wagon meets the wagon and the positive x-axis lies along the dashed line in the figure above. Since the force applied is a constant 10 pounds, we have | F| = 10. Since it is being applied at a constant angle of θ = 30 ◦ with respect to the positive x-axis, Definition 11.8 gives us F = 10 ¸cos(30 ◦ , sin(30 ◦ )¸ = ¸ 5 √ 3, 5 _ . Since the wagon is being pulled along 50 feet in the positive direction, the displacement vector is −−→ PQ = 50ˆı = 50 ¸1, 0¸ = ¸50, 0¸. We get W = F −−→ PQ = ¸ 5 √ 3, 5 _ ¸50, 0¸ = 250 √ 3. Since force is measured in pounds and distance is measured in feet, we get W = 250 √ 3 foot-pounds. Alternatively, we can use the formulation W = | F|| −−→ PQ| cos(θ) to get W = (10 pounds)(50 feet) cos (30 ◦ ) = 250 √ 3 foot-pounds of work. 11.9 The Dot Product and Projection 1041 11.9.1 Exercises In Exercises 1 - 20, use the pair of vectors v and w to find the following quantities. • v w • The angle θ (in degrees) between v and w • proj w (v) • q = v −proj w (v) (Show that q w = 0.) 1. v = ¸−2, −7¸ and w = ¸5, −9¸ 2. v = ¸−6, −5¸ and w = ¸10, −12¸ 3. v = ¸ 1, √ 3 _ and w = ¸ 1, − √ 3 _ 4. v = ¸3, 4¸ and w = ¸−6, −8¸ 5. v = ¸−2, 1¸ and w = ¸3, 6¸ 6. v = ¸ −3 √ 3, 3 _ and w = ¸ − √ 3, −1 _ 7. v = ¸1, 17¸ and w = ¸−1, 0¸ 8. v = ¸3, 4¸ and w = ¸5, 12¸ 9. v = ¸−4, −2¸ and w = ¸1, −5¸ 10. v = ¸−5, 6¸ and w = ¸4, −7¸ 11. v = ¸−8, 3¸ and w = ¸2, 6¸ 12. v = ¸34, −91¸ and w = ¸0, 1¸ 13. v = 3ˆı − ˆ  and w = 4ˆ  14. v = −24ˆı + 7ˆ  and w = 2ˆı 15. v = 3 2 ˆı + 3 2 ˆ  and w = ˆı − ˆ  16. v = 5ˆı + 12ˆ  and w = −3ˆı + 4ˆ  17. v = _ 1 2 , √ 3 2 _ and w = _ − √ 2 2 , √ 2 2 _ 18. v = _ √ 2 2 , √ 2 2 _ and w = _ 1 2 , − √ 3 2 _ 19. v = _ √ 3 2 , 1 2 _ and w = _ − √ 2 2 , − √ 2 2 _ 20. v = _ 1 2 , − √ 3 2 _ and w = _ √ 2 2 , − √ 2 2 _ 21. A force of 1500 pounds is required to tow a trailer. Find the work done towing the trailer along a flat stretch of road 300 feet. Assume the force is applied in the direction of the motion. 22. Find the work done lifting a 10 pound book 3 feet straight up into the air. Assume the force of gravity is acting straight downwards. 23. Suppose Taylor fills her wagon with rocks and must exert a force of 13 pounds to pull her wagon across the yard. If she maintains a 15 ◦ angle between the handle of the wagon and the horizontal, compute how much work Taylor does pulling her wagon 25 feet. Round your answer to two decimal places. 24. In Exercise 61 in Section 11.8, two drunken college students have filled an empty beer keg with rocks which they drag down the street by pulling on two attached ropes. The stronger of the two students pulls with a force of 100 pounds on a rope which makes a 13 ◦ angle with the direction of motion. (In this case, the keg was being pulled due east and the student’s heading was N77 ◦ E.) Find the work done by this student if the keg is dragged 42 feet. 1042 Applications of Trigonometry 25. Find the work done pushing a 200 pound barrel 10 feet up a 12.5 ◦ incline. Ignore all forces acting on the barrel except gravity, which acts downwards. Round your answer to two decimal places. HINT: Since you are working to overcome gravity only, the force being applied acts directly upwards. This means that the angle between the applied force in this case and the motion of the object is not the 12.5 ◦ of the incline! 26. Prove the distributive property of the dot product in Theorem 11.22. 27. Finish the proof of the scalar property of the dot product in Theorem 11.22. 28. Use the identity in Example 11.9.1 to prove the Parallelogram Law |v| 2 +| w| 2 = 1 2 _ |v + w| 2 +|v − w| 2 ¸ 29. We know that [x + y[ ≤ [x[ + [y[ for all real numbers x and y by the Triangle Inequality established in Exercise 36 in Section 2.2. We can now establish a Triangle Inequality for vectors. In this exercise, we prove that |u +v| ≤ |u| +|v| for all pairs of vectors u and v. (a) (Step 1) Show that |u +v| 2 = |u| 2 + 2u v +|v| 2 . (b) (Step 2) Show that [u v[ ≤ |u||v|. This is the celebrated Cauchy-Schwarz Inequality. 6 (Hint: To show this inequality, start with the fact that [u v[ = [ |u||v| cos(θ) [ and use the fact that [ cos(θ)[ ≤ 1 for all θ.) (c) (Step 3) Show that |u +v| 2 = |u| 2 + 2u v + |v| 2 ≤ |u| 2 + 2[u v[ + |v| 2 ≤ |u| 2 + 2|u||v| +|v| 2 = (|u| +|v|) 2 . (d) (Step 4) Use Step 3 to show that |u +v| ≤ |u| +|v| for all pairs of vectors u and v. (e) As an added bonus, we can now show that the Triangle Inequality [z + w[ ≤ [z[ + [w[ holds for all complex numbers z and w as well. Identify the complex number z = a +bi with the vector u = ¸a, b¸ and identify the complex number w = c + di with the vector v = ¸c, d¸ and just follow your nose! 6 It is also known by other names. Check out this site for details. 11.9 The Dot Product and Projection 1043 11.9.2 Answers 1. v = ¸−2, −7¸ and w = ¸5, −9¸ v w = 53 θ = 45 ◦ proj w (v) = ¸ 5 2 , − 9 2 _ q = ¸ − 9 2 , − 5 2 _ 2. v = ¸−6, −5¸ and w = ¸10, −12¸ v w = 0 θ = 90 ◦ proj w (v) = ¸0, 0¸ q = ¸−6, −5¸ 3. v = ¸ 1, √ 3 _ and w = ¸ 1, − √ 3 _ v w = −2 θ = 120 ◦ proj w (v) = _ − 1 2 , √ 3 2 _ q = _ 3 2 , √ 3 2 _ 4. v = ¸3, 4¸ and w = ¸−6, −8¸ v w = −50 θ = 180 ◦ proj w (v) = ¸3, 4¸ q = ¸0, 0¸ 5. v = ¸−2, 1¸ and w = ¸3, 6¸ v w = 0 θ = 90 ◦ proj w (v) = ¸0, 0¸ q = ¸−2, 1¸ 6. v = ¸ −3 √ 3, 3 _ and w = ¸ − √ 3, −1 _ v w = 6 θ = 60 ◦ proj w (v) = _ − 3 √ 3 2 , − 3 2 _ q = _ − 3 √ 3 2 , 9 2 _ 7. v = ¸1, 17¸ and w = ¸−1, 0¸ v w = −1 θ ≈ 93.37 ◦ proj w (v) = ¸1, 0¸ q = ¸0, 17¸ 8. v = ¸3, 4¸ and w = ¸5, 12¸ v w = 63 θ ≈ 14.25 ◦ proj w (v) = ¸ 315 169 , 756 169 _ q = ¸ 192 169 , − 80 169 _ 9. v = ¸−4, −2¸ and w = ¸1, −5¸ v w = 6 θ ≈ 74.74 ◦ proj w (v) = ¸ 3 13 , − 15 13 _ q = ¸ − 55 13 , − 11 13 _ 10. v = ¸−5, 6¸ and w = ¸4, −7¸ v w = −62 θ ≈ 169.94 ◦ proj w (v) = ¸ − 248 65 , 434 65 _ q = ¸ − 77 65 , − 44 65 _ 1044 Applications of Trigonometry 11. v = ¸−8, 3¸ and w = ¸2, 6¸ v w = 2 θ ≈ 87.88 ◦ proj w (v) = ¸ 1 10 , 3 10 _ q = ¸ − 81 10 , 27 10 _ 12. v = ¸34, −91¸ and w = ¸0, 1¸ v w = −91 θ ≈ 159.51 ◦ proj w (v) = ¸0, −91¸ q = ¸34, 0¸ 13. v = 3ˆı − ˆ  and w = 4ˆ  v w = −4 θ ≈ 108.43 ◦ proj w (v) = ¸0, −1¸ q = ¸3, 0¸ 14. v = −24ˆı + 7ˆ  and w = 2ˆı v w = −48 θ ≈ 163.74 ◦ proj w (v) = ¸−24, 0¸ q = ¸0, 7¸ 15. v = 3 2 ˆı + 3 2 ˆ  and w = ˆı − ˆ  v w = 0 θ = 90 ◦ proj w (v) = ¸0, 0¸ q = ¸ 3 2 , 3 2 _ 16. v = 5ˆı + 12ˆ  and w = −3ˆı + 4ˆ  v w = 33 θ ≈ 59.49 ◦ proj w (v) = ¸ − 99 25 , 132 25 _ q = ¸ 224 25 , 168 25 _ 17. v = _ 1 2 , √ 3 2 _ and w = _ − √ 2 2 , √ 2 2 _ v w = √ 6− √ 2 4 θ = 75 ◦ proj w (v) = _ 1− √ 3 4 , √ 3−1 4 _ q = _ 1+ √ 3 4 , 1+ √ 3 4 _ 18. v = _ √ 2 2 , √ 2 2 _ and w = _ 1 2 , − √ 3 2 _ v w = √ 2− √ 6 4 θ = 105 ◦ proj w (v) = _ √ 2− √ 6 8 , 3 √ 2− √ 6 8 _ q = _ 3 √ 2+ √ 6 8 , √ 2+ √ 6 8 _ 19. v = _ √ 3 2 , 1 2 _ and w = _ − √ 2 2 , − √ 2 2 _ v w = − √ 6+ √ 2 4 θ = 165 ◦ proj w (v) = _ √ 3+1 4 , √ 3+1 4 _ q = _ √ 3−1 4 , 1− √ 3 4 _ 20. v = _ 1 2 , − √ 3 2 _ and w = _ √ 2 2 , − √ 2 2 _ v w = √ 6+ √ 2 4 θ = 15 ◦ proj w (v) = _ √ 3+1 4 , − √ 3+1 4 _ q = _ 1− √ 3 4 , 1− √ 3 4 _ 21. (1500 pounds)(300 feet) cos (0 ◦ ) = 450, 000 foot-pounds 22. (10 pounds)(3 feet) cos (0 ◦ ) = 30 foot-pounds 11.9 The Dot Product and Projection 1045 23. (13 pounds)(25 feet) cos (15 ◦ ) ≈ 313.92 foot-pounds 24. (100 pounds)(42 feet) cos (13 ◦ ) ≈ 4092.35 foot-pounds 25. (200 pounds)(10 feet) cos (77.5 ◦ ) ≈ 432.88 foot-pounds 1046 Applications of Trigonometry 11.10 Parametric Equations As we have seen in Exercises 53 - 56 in Section 1.2, Chapter 7 and most recently in Section 11.5, there are scores of interesting curves which, when plotted in the xy-plane, neither represent y as a function of x nor x as a function of y. In this section, we present a new concept which allows us to use functions to study these kinds of curves. To motivate the idea, we imagine a bug crawling across a table top starting at the point O and tracing out a curve C in the plane, as shown below. x y O Q P(x, y) = (f(t), g(t)) 1 2 3 4 5 1 2 3 4 5 The curve C does not represent y as a function of x because it fails the Vertical Line Test and it does not represent x as a function of y because it fails the Horizontal Line Test. However, since the bug can be in only one place P(x, y) at any given time t, we can define the x-coordinate of P as a function of t and the y-coordinate of P as a (usually, but not necessarily) different function of t. (Traditionally, f(t) is used for x and g(t) is used for y.) The independent variable t in this case is called a parameter and the system of equations _ x = f(t) y = g(t) is called a system of parametric equations or a parametrization of the curve C. 1 The parametrization of C endows it with an orientation and the arrows on C indicate motion in the direction of increasing values of t. In this case, our bug starts at the point O, travels upwards to the left, then loops back around to cross its path 2 at the point Q and finally heads off into the first quadrant. It is important to note that the curve itself is a set of points and as such is devoid of any orientation. The parametrization determines the orientation and as we shall see, different parametrizations can determine different orientations. If all of this seems hauntingly familiar, it should. By definition, the system of equations ¦x = cos(t), y = sin(t) parametrizes the Unit Circle, giving it a counter-clockwise orientation. More generally, the equations of circular motion ¦x = r cos(ωt), y = r sin(ωt) developed on page 732 in Section 10.2.1 are parametric equations which trace out a circle of radius r centered at the origin. If ω > 0, the orientation is counter- clockwise; if ω < 0, the orientation is clockwise. The angular frequency ω determines ‘how fast’ the 1 Note the use of the indefinite article ‘a’. As we shall see, there are infinitely many different parametric represen- tations for any given curve. 2 Here, the bug reaches the point Q at two different times. While this does not contradict our claim that f(t) and g(t) are functions of t, it shows that neither f nor g can be one-to-one. (Think about this before reading on.) 11.10 Parametric Equations 1047 object moves around the circle. In particular, the equations _ x = 2960 cos _ π 12 t _ , y = 2960 sin _ π 12 t _ that model the motion of Lakeland Community College as the earth rotates (see Example 10.2.7 in Section 10.2) parameterize a circle of radius 2960 with a counter-clockwise rotation which completes one revolution as t runs through the interval [0, 24). It is time for another example. Example 11.10.1. Sketch the curve described by _ x = t 2 −3 y = 2t −1 for t ≥ −2. Solution. We follow the same procedure here as we have time and time again when asked to graph anything new – choose friendly values of t, plot the corresponding points and connect the results in a pleasing fashion. Since we are told t ≥ −2, we start there and as we plot successive points, we draw an arrow to indicate the direction of the path for increasing values of t. t x(t) y(t) (x(t), y(t)) −2 1 −5 (1, −5) −1 −2 −3 (−2, −3) 0 −3 −1 (−3, −1) 1 −2 1 (−2, 1) 2 1 3 (1, 3) 3 6 5 (6, 5) x y −2−1 1 2 3 4 5 6 −5 −3 −2 −1 1 2 3 4 5 The curve sketched out in Example 11.10.1 certainly looks like a parabola, and the presence of the t 2 term in the equation x = t 2 − 3 reinforces this hunch. Since the parametric equations _ x = t 2 −3, y = 2t −1 given to describe this curve are a system of equations, we can use the technique of substitution as described in Section 8.7 to eliminate the parameter t and get an equation involving just x and y. To do so, we choose to solve the equation y = 2t − 1 for t to get t = y+1 2 . Substituting this into the equation x = t 2 − 3 yields x = _ y+1 2 _ 2 − 3 or, after some rearrangement, (y + 1) 2 = 4(x + 3). Thinking back to Section 7.3, we see that the graph of this equation is a parabola with vertex (−3, −1) which opens to the right, as required. Technically speaking, the equation (y + 1) 2 = 4(x + 3) describes the entire parabola, while the parametric equations _ x = t 2 −3, y = 2t −1 for t ≥ −2 describe only a portion of the parabola. In this case, 3 we can remedy this situation by restricting the bounds on y. Since the portion of the parabola we want is exactly the part where y ≥ −5, the equation (y +1) 2 = 4(x+3) coupled with the restriction y ≥ −5 describes the same curve as the given parametric equations. The one piece of information we can never recover after eliminating the parameter is the orientation of the curve. Eliminating the parameter and obtaining an equation in terms of x and y, whenever possible, can be a great help in graphing curves determined by parametric equations. If the system of parametric equations contains algebraic functions, as was the case in Example 11.10.1, then the usual techniques of substitution and elimination as learned in Section 8.7 can be applied to the 3 We will have an example shortly where no matter how we restrict x and y, we can never accurately describe the curve once we’ve eliminated the parameter. 1048 Applications of Trigonometry system ¦x = f(t), y = g(t) to eliminate the parameter. If, on the other hand, the parametrization involves the trigonometric functions, the strategy changes slightly. In this case, it is often best to solve for the trigonometric functions and relate them using an identity. We demonstrate these techniques in the following example. Example 11.10.2. Sketch the curves described by the following parametric equations. 1. _ x = t 3 y = 2t 2 for −1 ≤ t ≤ 1 2. _ x = e −t y = e −2t for t ≥ 0 3. _ x = sin(t) y = csc(t) for 0 < t < π 4. _ x = 1 + 3 cos(t) y = 2 sin(t) for 0 ≤ t ≤ 3π 2 Solution. 1. To get a feel for the curve described by the system _ x = t 3 , y = 2t 2 we first sketch the graphs of x = t 3 and y = 2t 2 over the interval [−1, 1]. We note that as t takes on values in the interval [−1, 1], x = t 3 ranges between −1 and 1, and y = 2t 2 ranges between 0 and 2. This means that all of the action is happening on a portion of the plane, namely ¦(x, y) [ −1 ≤ x ≤ 1, 0 ≤ y ≤ 2¦. Next, we plot a few points to get a sense of the position and orientation of the curve. Certainly, t = −1 and t = 1 are good values to pick since these are the extreme values of t. We also choose t = 0, since that corresponds to a relative minimum 4 on the graph of y = 2t 2 . Plugging in t = −1 gives the point (−1, 2), t = 0 gives (0, 0) and t = 1 gives (1, 2). More generally, we see that x = t 3 is increasing over the entire interval [−1, 1] whereas y = 2t 2 is decreasing over the interval [−1, 0] and then increasing over [0, 1]. Geometrically, this means that in order to trace out the path described by the parametric equations, we start at (−1, 2) (where t = −1), then move to the right (since x is increasing) and down (since y is decreasing) to (0, 0) (where t = 0). We continue to move to the right (since x is still increasing) but now move upwards (since y is now increasing) until we reach (1, 2) (where t = 1). Finally, to get a good sense of the shape of the curve, we eliminate the parameter. Solving x = t 3 for t, we get t = 3 √ x. Substituting this into y = 2t 2 gives y = 2( 3 √ x) 2 = 2x 2/3 . Our experience in Section 5.3 yields the graph of our final answer below. t x −1 1 −1 1 t y −1 1 1 2 x y −1 1 1 2 x = t 3 , −1 ≤ t ≤ 1 y = 2t 2 , −1 ≤ t ≤ 1 _ x = t 3 , y = 2t 2 , −1 ≤ t ≤ 1 4 You should review Section 1.6.1 if you’ve forgotten what ‘increasing’, ‘decreasing’ and ‘relative minimum’ mean. 11.10 Parametric Equations 1049 2. For the system _ x = 2e −t , y = e −2t for t ≥ 0, we proceed as in the previous example and graph x = 2e −t and y = e −2t over the interval [0, ∞). We find that the range of x in this case is (0, 2] and the range of y is (0, 1]. Next, we plug in some friendly values of t to get a sense of the orientation of the curve. Since t lies in the exponent here, ‘friendly’ values of t involve natural logarithms. Starting with t = ln(1) = 0 we get 5 (2, 1), for t = ln(2) we get _ 1, 1 4 _ and for t = ln(3) we get _ 2 3 , 1 9 _ . Since t is ranging over the unbounded interval [0, ∞), we take the time to analyze the end behavior of both x and y. As t → ∞, x = 2e −t → 0 + and y = e −2t → 0 + as well. This means the graph of _ x = 2e −t , y = e −2t approaches the point (0, 0). Since both x = 2e −t and y = e −2t are always decreasing for t ≥ 0, we know that our final graph will start at (2, 1) (where t = 0), and move consistently to the left (since x is decreasing) and down (since y is decreasing) to approach the origin. To eliminate the parameter, one way to proceed is to solve x = 2e −t for t to get t = −ln _ x 2 _ . Substituting this for t in y = e −2t gives y = e −2(−ln(x/2)) = e 2 ln(x/2) = e ln(x/2) 2 = _ x 2 _ 2 = x 2 4 . Or, we could recognize that y = e −2t = _ e −t _ 2 , and since x = 2e −t means e −t = x 2 , we get y = _ x 2 _ 2 = x 2 4 this way as well. Either way, the graph of _ x = 2e −t , y = e −2t for t ≥ 0 is a portion of the parabola y = x 2 4 which starts at the point (2, 1) and heads towards, but never reaches, 6 (0, 0). t x 1 2 1 2 t y 1 2 1 x y 1 2 1 x = 2e −t , t ≥ 0 y = e −2t , t ≥ 0 _ x = 2e −t , y = e −2t , t ≥ 0 3. For the system ¦x = sin(t), y = csc(t) for 0 < t < π, we start by graphing x = sin(t) and y = csc(t) over the interval (0, π). We find that the range of x is (0, 1] while the range of y is [1, ∞). Plotting a few friendly points, we see that t = π 6 gives the point _ 1 2 , 2 _ , t = π 2 gives (1, 1) and t = 5π 6 returns us to _ 1 2 , 2 _ . Since t = 0 and t = π aren’t included in the domain for t, (because y = csc(t) is undefined at these t-values), we analyze the behavior of the system as t approaches 0 and π. We find that as t →0 + as well as when t →π − , we get x = sin(t) →0 + and y = csc(t) →∞. Piecing all of this information together, we get that for t near 0, we have points with very small positive x-values, but very large positive y-values. As t ranges through the interval _ 0, π 2 ¸ , x = sin(t) is increasing and y = csc(t) is decreasing. This means that we are moving to the right and downwards, through _ 1 2 , 2 _ when t = π 6 to (1, 1) when t = π 2 . Once t = π 2 , the orientation reverses, and we start to head to the left, since x = sin(t) is now decreasing, and up, since y = csc(t) is now increasing. We pass back through _ 1 2 , 2 _ when t = 5π 6 back to the points with small positive x-coordinates and large 5 The reader is encouraged to review Sections 6.1 and 6.2 as needed. 6 Note the open circle at the origin. See the solution to part 3 in Example 1.2.1 on page 22 and Theorem 4.1 in Section 4.1 for a review of this concept. 1050 Applications of Trigonometry positive y-coordinates. To better explain this behavior, we eliminate the parameter. Using a reciprocal identity, we write y = csc(t) = 1 sin(t) . Since x = sin(t), the curve traced out by this parametrization is a portion of the graph of y = 1 x . We now can explain the unusual behavior as t →0 + and t →π − – for these values of t, we are hugging the vertical asymptote x = 0 of the graph of y = 1 x . We see that the parametrization given above traces out the portion of y = 1 x for 0 < x ≤ 1 twice as t runs through the interval (0, π). t x π 2 π 1 t y π 2 π 1 x y 1 1 2 3 x = sin(t), 0 < t < π y = csc(t), 0 < t < π {x = sin(t), y = csc(t) , 0 < t < π 4. Proceeding as above, we set about graphing ¦x = 1 + 3 cos(t), y = 2 sin(t) for 0 ≤ t ≤ 3π 2 by first graphing x = 1 + 3 cos(t) and y = 2 sin(t) on the interval _ 0, 3π 2 ¸ . We see that x ranges from −2 to 4 and y ranges from −2 to 2. Plugging in t = 0, π 2 , π and 3π 2 gives the points (4, 0), (1, 2), (−2, 0) and (1, −2), respectively. As t ranges from 0 to π 2 , x = 1+3 cos(t) is decreasing, while y = 2 sin(t) is increasing. This means that we start tracing out our answer at (4, 0) and continue moving to the left and upwards towards (1, 2). For π 2 ≤ t ≤ π, x is decreasing, as is y, so the motion is still right to left, but now is downwards from (1, 2) to (−2, 0). On the interval _ π, 3π 2 ¸ , x begins to increase, while y continues to decrease. Hence, the motion becomes left to right but continues downwards, connecting (−2, 0) to (1, −2). To eliminate the parameter here, we note that the trigonometric functions involved, namely cos(t) and sin(t), are related by the Pythagorean Identity cos 2 (t) +sin 2 (t) = 1. Hence, we solve x = 1 +3 cos(t) for cos(t) to get cos(t) = x−1 3 , and we solve y = 2 sin(t) for sin(t) to get sin(t) = y 2 . Substituting these expressions into cos 2 (t)+sin 2 (t) = 1 gives _ x−1 3 _ 2 + _ y 2 _ 2 = 1, or (x−1) 2 9 + y 2 4 = 1. From Section 7.4, we know that the graph of this equation is an ellipse centered at (1, 0) with vertices at (−2, 0) and (4, 0) with a minor axis of length 4. Our parametric equations here are tracing out three-quarters of this ellipse, in a counter-clockwise direction. t x π 2 π 3π 2 −2 −1 1 2 3 4 t y π 2 π 3π 2 −2 −1 1 2 x y −1 1 2 3 4 −2 −1 1 2 x = 1 + 3 cos(t), 0 ≤ t ≤ 3π 2 y = 2 sin(t), 0 ≤ t ≤ 3π 2 {x = 1 + 3 cos(t), y = 2 sin(t) , 0 ≤ t ≤ 3π 2 11.10 Parametric Equations 1051 Now that we have had some good practice sketching the graphs of parametric equations, we turn to the problem of finding parametric representations of curves. We start with the following. Parametrizations of Common Curves • To parametrize y = f(x) as x runs through some interval I, let x = t and y = f(t) and let t run through I. • To parametrize x = g(y) as y runs through some interval I, let x = g(t) and y = t and let t run through I. • To parametrize a directed line segment with initial point (x 0 , y 0 ) and terminal point (x 1 , y 1 ), let x = x 0 + (x 1 −x 0 )t and y = y 0 + (y 1 −y 0 )t for 0 ≤ t ≤ 1. • To parametrize (x−h) 2 a 2 + (y−k) 2 b 2 = 1 where a, b > 0, let x = h +a cos(t) and y = k +b sin(t) for 0 ≤ t < 2π. (This will impart a counter-clockwise orientation.) The reader is encouraged to verify the above formulas by eliminating the parameter and, when indicated, checking the orientation. We put these formulas to good use in the following example. Example 11.10.3. Find a parametrization for each of the following curves and check your answers. 1. y = x 2 from x = −3 to x = 2 2. y = f −1 (x) where f(x) = x 5 + 2x + 1 3. The line segment which starts at (2, −3) and ends at (1, 5) 4. The circle x 2 + 2x +y 2 −4y = 4 5. The left half of the ellipse x 2 4 + y 2 9 = 1 Solution. 1. Since y = x 2 is written in the form y = f(x), we let x = t and y = f(t) = t 2 . Since x = t, the bounds on t match precisely the bounds on x so we get _ x = t, y = t 2 for −3 ≤ t ≤ 2. The check is almost trivial; with x = t we have y = t 2 = x 2 as t = x runs from −3 to 2. 2. We are told to parametrize y = f −1 (x) for f(x) = x 5 + 2x + 1 so it is safe to assume that f is one-to-one. (Otherwise, f −1 would not exist.) To find a formula y = f −1 (x), we follow the procedure outlined on page 384 – we start with the equation y = f(x), interchange x and y and solve for y. Doing so gives us the equation x = y 5 + 2y + 1. While we could attempt to solve this equation for y, we don’t need to. We can parametrize x = f(y) = y 5 + 2y + 1 by setting y = t so that x = t 5 + 2t + 1. We know from our work in Section 3.1 that since f(x) = x 5 + 2x + 1 is an odd-degree polynomial, the range of y = f(x) = x 5 + 2x + 1 is (−∞, ∞). Hence, in order to trace out the entire graph of x = f(y) = y 5 +2y +1, we need to let y run through all real numbers. Our final answer to this problem is _ x = t 5 + 2t + 1, y = t for −∞< t < ∞. As in the previous problem, our solution is trivial to check. 7 7 Provided you followed the inverse function theory, of course. 1052 Applications of Trigonometry 3. To parametrize line segment which starts at (2, −3) and ends at (1, 5), we make use of the formulas x = x 0 +(x 1 −x 0 )t and y = y 0 +(y 1 −y 0 )t for 0 ≤ t ≤ 1. While these equations at first glance are quite a handful, 8 they can be summarized as ‘starting point + (displacement)t’. To find the equation for x, we have that the line segment starts at x = 2 and ends at x = 1. This means the displacement in the x-direction is (1 −2) = −1. Hence, the equation for x is x = 2 + (−1)t = 2 − t. For y, we note that the line segment starts at y = −3 and ends at y = 5. Hence, the displacement in the y-direction is (5 − (−3)) = 8, so we get y = −3 + 8t. Our final answer is ¦x = 2 −t, y = −3 + 8t for 0 ≤ t ≤ 1. To check, we can solve x = 2 − t for t to get t = 2 −x. Substituting this into y = −3 + 8t gives y = −3 + 8t = −3 + 8(2 −x), or y = −8x +13. We know this is the graph of a line, so all we need to check is that it starts and stops at the correct points. When t = 0, x = 2 − t = 2, and when t = 1, x = 2 − t = 1. Plugging in x = 2 gives y = −8(2) + 13 = −3, for an initial point of (2, −3). Plugging in x = 1 gives y = −8(1) + 13 = 5 for an ending point of (1, 5), as required. 4. In order to use the formulas above to parametrize the circle x 2 +2x+y 2 −4y = 4, we first need to put it into the correct form. After completing the squares, we get (x +1) 2 +(y −2) 2 = 9, or (x+1) 2 9 + (y−2) 2 9 = 1. Once again, the formulas x = h +a cos(t) and y = k +b sin(t) can be a challenge to memorize, but they come from the Pythagorean Identity cos 2 (t) +sin 2 (t) = 1. In the equation (x+1) 2 9 + (y−2) 2 9 = 1, we identify cos(t) = x+1 3 and sin(t) = y−2 3 . Rearranging these last two equations, we get x = −1 + 3 cos(t) and y = 2 + 3 sin(t). In order to complete one revolution around the circle, we let t range through the interval [0, 2π). We get as our final answer ¦x = −1 + 3 cos(t), y = 2 + 3 sin(t) for 0 ≤ t < 2π. To check our answer, we could eliminate the parameter by solving x = −1 +3 cos(t) for cos(t) and y = 2 +3 sin(t) for sin(t), invoking a Pythagorean Identity, and then manipulating the resulting equation in x and y into the original equation x 2 +2x +y 2 −4y = 4. Instead, we opt for a more direct approach. We substitute x = −1 + 3 cos(t) and y = 2 + 3 sin(t) into the equation x 2 + 2x +y 2 −4y = 4 and show that the latter is satisfied for all t such that 0 ≤ t < 2π. x 2 + 2x +y 2 −4y = 4 (−1 + 3 cos(t)) 2 + 2(−1 + 3 cos(t)) + (2 + 3 sin(t)) 2 −4(2 + 3 sin(t)) ? = 4 1 −6 cos(t) + 9 cos 2 (t) −2 + 6 cos(t) + 4 + 12 sin(t) + 9 sin 2 (t) −8 −12 sin(t) ? = 4 9 cos 2 (t) + 9 sin 2 (t) −5 ? = 4 9 _ cos 2 (t) + sin 2 (t) _ −5 ? = 4 9 (1) −5 ? = 4 4 = 4 Now that we know the parametric equations give us points on the circle, we can go through the usual analysis as demonstrated in Example 11.10.2 to show that the entire circle is covered as t ranges through the interval [0, 2π). 8 Compare and contrast this with Exercise 65 in Section 11.8. 11.10 Parametric Equations 1053 5. In the equation x 2 4 + y 2 9 = 1, we can either use the formulas above or think back to the Pythagorean Identity to get x = 2 cos(t) and y = 3 sin(t). The normal range on the parameter in this case is 0 ≤ t < 2π, but since we are interested in only the left half of the ellipse, we restrict t to the values which correspond to Quadrant II and Quadrant III angles, namely π 2 ≤ t ≤ 3π 2 . Our final answer is ¦x = 2 cos(t), y = 3 sin(t) for π 2 ≤ t ≤ 3π 2 . Substituting x = 2 cos(t) and y = 3 sin(t) into x 2 4 + y 2 9 = 1 gives 4 cos 2 (t) 4 + 9 sin 2 (t) 9 = 1, which reduces to the Pythagorean Identity cos 2 (t) + sin 2 (t) = 1. This proves that the points generated by the parametric equations ¦x = 2 cos(t), y = 3 sin(t) lie on the ellipse x 2 4 + y 2 9 = 1. Employing the techniques demonstrated in Example 11.10.2, we find that the restriction π 2 ≤ t ≤ 3π 2 generates the left half of the ellipse, as required. We note that the formulas given on page 1051 offer only one of literally infinitely many ways to parametrize the common curves listed there. At times, the formulas offered there need to be altered to suit the situation. Two easy ways to alter parametrizations are given below. Adjusting Parametric Equations • Reversing Orientation: Replacing every occurrence of t with −t in a parametric de- scription for a curve (including any inequalities which describe the bounds on t) reverses the orientation of the curve. • Shift of Parameter: Replacing every occurrence of t with (t − c) in a parametric de- scription for a curve (including any inequalities which describe the bounds on t) shifts the start of the parameter t ahead by c units. We demonstrate these techniques in the following example. Example 11.10.4. Find a parametrization for the following curves. 1. The curve which starts at (2, 4) and follows the parabola y = x 2 to end at (−1, 1). Shift the parameter so that the path starts at t = 0. 2. The two part path which starts at (0, 0), travels along a line to (3, 4), then travels along a line to (5, 0). 3. The Unit Circle, oriented clockwise, with t = 0 corresponding to (0, −1). Solution. 1. We can parametrize y = x 2 from x = −1 to x = 2 using the formula given on Page 1051 as _ x = t, y = t 2 for −1 ≤ t ≤ 2. This parametrization, however, starts at (−1, 1) and ends at (2, 4). Hence, we need to reverse the orientation. To do so, we replace every occurrence of t with −t to get _ x = −t, y = (−t) 2 for −1 ≤ −t ≤ 2. After simplifying, we get _ x = −t, y = t 2 for −2 ≤ t ≤ 1. We would like t to begin at t = 0 instead of t = −2. The problem here is that the parametrization we have starts 2 units ‘too soon’, so we need to introduce a ‘time delay’ of 2. Replacing every occurrence of t with (t −2) gives _ x = −(t −2), y = (t −2) 2 for −2 ≤ t −2 ≤ 1. Simplifying yields _ x = 2 −t, y = t 2 −4t + 4 for 0 ≤ t ≤ 3. 1054 Applications of Trigonometry 2. When parameterizing line segments, we think: ‘starting point + (displacement)t’. For the first part of the path, we get ¦x = 3t, y = 4t for 0 ≤ t ≤ 1, and for the second part we get ¦x = 3 + 2t, y = 4 −4t for 0 ≤ t ≤ 1. Since the first parametrization leaves off at t = 1, we shift the parameter in the second part so it starts at t = 1. Our current description of the second part starts at t = 0, so we introduce a ‘time delay’ of 1 unit to the second set of parametric equations. Replacing t with (t − 1) in the second set of parametric equations gives ¦x = 3 + 2(t −1), y = 4 −4(t −1) for 0 ≤ t − 1 ≤ 1. Simplify- ing yields ¦x = 1 + 2t, y = 8 −4t for 1 ≤ t ≤ 2. Hence, we may parametrize the path as ¦x = f(t), y = g(t) for 0 ≤ t ≤ 2 where f(t) = _ 3t, for 0 ≤ t ≤ 1 1 + 2t, for 1 ≤ t ≤ 2 and g(t) = _ 4t, for 0 ≤ t ≤ 1 8 −4t, for 1 ≤ t ≤ 2 3. We know that ¦x = cos(t), y = sin(t) for 0 ≤ t < 2π gives a counter-clockwise parametrization of the Unit Circle with t = 0 corresponding to (1, 0), so the first order of business is to reverse the orientation. Replacing t with −t gives ¦x = cos(−t), y = sin(−t) for 0 ≤ −t < 2π, which simplifies 9 to ¦x = cos(t), y = −sin(t) for −2π < t ≤ 0. This parametrization gives a clockwise orientation, but t = 0 still corresponds to the point (1, 0); the point (0, −1) is reached when t = − 3π 2 . Our strategy is to first get the parametrization to ‘start’ at the point (0, −1) and then shift the parameter accordingly so the ‘start’ coincides with t = 0. We know that any interval of length 2π will parametrize the entire circle, so we keep the equations ¦x = cos(t), y = −sin(t), but start the parameter t at − 3π 2 , and find the upper bound by adding 2π so − 3π 2 ≤ t < π 2 . The reader can verify that ¦x = cos(t), y = −sin(t) for − 3π 2 ≤ t < π 2 traces out the Unit Circle clockwise starting at the point (0, −1). We now shift the parameter by introducing a ‘time delay’ of 3π 2 units by replacing every occurrence of t with _ t − 3π 2 _ . We get _ x = cos _ t − 3π 2 _ , y = −sin _ t − 3π 2 _ for − 3π 2 ≤ t − 3π 2 < π 2 . This simplifies 10 to ¦x = −sin(t), y = −cos(t) for 0 ≤ t < 2π, as required. We put our answer to Example 11.10.4 number 3 to good use to derive the equation of a cycloid. Suppose a circle of radius r rolls along the positive x-axis at a constant velocity v as pictured below. Let θ be the angle in radians which measures the amount of clockwise rotation experienced by the radius highlighted in the figure. x y P(x, y) θ r 9 courtesy of the Even/Odd Identities 10 courtesy of the Sum/Difference Formulas 11.10 Parametric Equations 1055 Our goal is to find parametric equations for the coordinates of the point P(x, y) in terms of θ. From our work in Example 11.10.4 number 3, we know that clockwise motion along the Unit Circle starting at the point (0, −1) can be modeled by the equations ¦x = −sin(θ), y = −cos(θ) for 0 ≤ θ < 2π. (We have renamed the parameter ‘θ’ to match the context of this problem.) To model this motion on a circle of radius r, all we need to do 11 is multiply both x and y by the factor r which yields ¦x = −r sin(θ), y = −r cos(θ). We now need to adjust for the fact that the circle isn’t stationary with center (0, 0), but rather, is rolling along the positive x-axis. Since the velocity v is constant, we know that at time t, the center of the circle has traveled a distance vt down the positive x-axis. Furthermore, since the radius of the circle is r and the circle isn’t moving vertically, we know that the center of the circle is always r units above the x-axis. Putting these two facts together, we have that at time t, the center of the circle is at the point (vt, r). From Section 10.1.1, we know v = rθ t , or vt = rθ. Hence, the center of the circle, in terms of the parameter θ, is (rθ, r). As a result, we need to modify the equations ¦x = −r sin(θ), y = −r cos(θ) by shifting the x-coordinate to the right rθ units (by adding rθ to the expression for x) and the y-coordinate up r units 12 (by adding r to the expression for y). We get ¦x = −r sin(θ) +rθ, y = −r cos(θ) +r , which can be written as ¦x = r(θ −sin(θ)), y = r(1 −cos(θ)). Since the motion starts at θ = 0 and proceeds indefinitely, we set θ ≥ 0. We end the section with a demonstration of the graphing calculator. Example 11.10.5. Find the parametric equations of a cycloid which results from a circle of radius 3 rolling down the positive x-axis as described above. Graph your answer using a calculator. Solution. We have r = 3 which gives the equations ¦x = 3(t −sin(t)), y = 3(1 −cos(t)) for t ≥ 0. (Here we have returned to the convention of using t as the parameter.) Sketching the cycloid by hand is a wonderful exercise in Calculus, but for the purposes of this book, we use a graphing utility. Using a calculator to graph parametric equations is very similar to graphing polar equations on a calculator. 13 Ensuring that the calculator is in ‘Parametric Mode’ and ‘radian mode’ we enter the equations and advance to the ‘Window’ screen. As always, the challenge is to determine appropriate bounds on the parameter, t, as well as for x and y. We know that one full revolution of the circle occurs over the interval 0 ≤ t < 2π, so 11 If we replace x with x r and y with y r in the equation for the Unit Circle x 2 +y 2 = 1, we obtain x r 2 + y r 2 = 1 which reduces to x 2 + y 2 = r 2 . In the language of Section 1.7, we are stretching the graph by a factor of r in both the x- and y-directions. Hence, we multiply both the x- and y-coordinates of points on the graph by r. 12 Does this seem familiar? See Example 11.1.1 in Section 11.1. 13 See page 957 in Section 11.5. 1056 Applications of Trigonometry it seems reasonable to keep these as our bounds on t. The ‘Tstep’ seems reasonably small – too large a value here can lead to incorrect graphs. 14 We know from our derivation of the equations of the cycloid that the center of the generating circle has coordinates (rθ, r), or in this case, (3t, 3). Since t ranges between 0 and 2π, we set x to range between 0 and 6π. The values of y go from the bottom of the circle to the top, so y ranges between 0 and 6. Below we graph the cycloid with these settings, and then extend t to range from 0 to 6π which forces x to range from 0 to 18π yielding three arches of the cycloid. (It is instructive to note that keeping the y settings between 0 and 6 messes up the geometry of the cycloid. The reader is invited to use the Zoom Square feature on the graphing calculator to see what window gives a true geometric perspective of the three arches.) 14 Again, see page 957 in Section 11.5. 11.10 Parametric Equations 1057 11.10.1 Exercises In Exercises 1 - 20, plot the set of parametric equations by hand. Be sure to indicate the orientation imparted on the curve by the parametrization. 1. _ x = 4t −3 y = 6t −2 for 0 ≤ t ≤ 1 2. _ x = 4t −1 y = 3 −4t for 0 ≤ t ≤ 1 3. _ x = 2t y = t 2 for −1 ≤ t ≤ 2 4. _ x = t −1 y = 3 + 2t −t 2 for 0 ≤ t ≤ 3 5. _ x = t 2 + 2t + 1 y = t + 1 for t ≤ 1 6. _ x = 1 9 _ 18 −t 2 _ y = 1 3 t for t ≥ −3 7. _ x = t y = t 3 for −∞< t < ∞ 8. _ x = t 3 y = t for −∞< t < ∞ 9. _ x = cos(t) y = sin(t) for − π 2 ≤ t ≤ π 2 10. _ x = 3 cos(t) y = 3 sin(t) for 0 ≤ t ≤ π 11. _ x = −1 + 3 cos(t) y = 4 sin(t) for 0 ≤ t ≤ 2π 12. _ x = 3 cos(t) y = 2 sin(t) + 1 for π 2 ≤ t ≤ 2π 13. _ x = 2 cos(t) y = sec(t) for 0 < t < π 2 14. _ x = 2 tan(t) y = cot(t) for 0 < t < π 2 15. _ x = sec(t) y = tan(t) for − π 2 < t < π 2 16. _ x = sec(t) y = tan(t) for π 2 < t < 3π 2 17. _ x = tan(t) y = 2 sec(t) for − π 2 < t < π 2 18. _ x = tan(t) y = 2 sec(t) for π 2 < t < 3π 2 19. _ x = cos(t) y = t for 0 ≤ t ≤ π 20. _ x = sin(t) y = t for − π 2 ≤ t ≤ π 2 In Exercises 21 - 24, plot the set of parametric equations with the help of a graphing utility. Be sure to indicate the orientation imparted on the curve by the parametrization. 21. _ x = t 3 −3t y = t 2 −4 for −2 ≤ t ≤ 2 22. _ x = 4 cos 3 (t) y = 4 sin 3 (t) for 0 ≤ t ≤ 2π 23. _ x = e t +e −t y = e t −e −t for −2 ≤ t ≤ 2 24. _ x = cos(3t) y = sin(4t) for 0 ≤ t ≤ 2π 1058 Applications of Trigonometry In Exercises 25 - 39, find a parametric description for the given oriented curve. 25. the directed line segment from (3, −5) to (−2, 2) 26. the directed line segment from (−2, −1) to (3, −4) 27. the curve y = 4 −x 2 from (−2, 0) to (2, 0). 28. the curve y = 4 −x 2 from (−2, 0) to (2, 0) (Shift the parameter so t = 0 corresponds to (−2, 0).) 29. the curve x = y 2 −9 from (−5, −2) to (0, 3). 30. the curve x = y 2 −9 from (0, 3) to (−5, −2). (Shift the parameter so t = 0 corresponds to (0, 3).) 31. the circle x 2 +y 2 = 25, oriented counter-clockwise 32. the circle (x −1) 2 +y 2 = 4, oriented counter-clockwise 33. the circle x 2 +y 2 −6y = 0, oriented counter-clockwise 34. the circle x 2 +y 2 −6y = 0, oriented clockwise (Shift the parameter so t begins at 0.) 35. the circle (x −3) 2 + (y + 1) 2 = 117, oriented counter-clockwise 36. the ellipse (x −1) 2 + 9y 2 = 9, oriented counter-clockwise 37. the ellipse 9x 2 + 4y 2 + 24y = 0, oriented counter-clockwise 38. the ellipse 9x 2 + 4y 2 + 24y = 0, oriented clockwise (Shift the parameter so t = 0 corresponds to (0, 0).) 39. the triangle with vertices (0, 0), (3, 0), (0, 4), oriented counter-clockwise (Shift the parameter so t = 0 corresponds to (0, 0).) 40. Use parametric equations and a graphing utility to graph the inverse of f(x) = x 3 + 3x −4. 41. Every polar curve r = f(θ) can be translated to a system of parametric equations with parameter θ by ¦x = r cos(θ) = f(θ) cos(θ), y = r sin(θ) = f(θ) sin(θ). Convert r = 6 cos(2θ) to a system of parametric equations. Check your answer by graphing r = 6 cos(2θ) by hand using the techniques presented in Section 11.5 and then graphing the parametric equations you found using a graphing utility. 42. Use your results from Exercises 3 and 4 in Section 11.1 to find the parametric equations which model a passenger’s position as they ride the London Eye. 11.10 Parametric Equations 1059 Suppose an object, called a projectile, is launched into the air. Ignoring everything except the force gravity, the path of the projectile is given by 15 _ _ _ x = v 0 cos(θ) t y = − 1 2 gt 2 +v 0 sin(θ) t +s 0 for 0 ≤ t ≤ T where v 0 is the initial speed of the object, θ is the angle from the horizontal at which the projectile is launched, 16 g is the acceleration due to gravity, s 0 is the initial height of the projectile above the ground and T is the time when the object returns to the ground. (See the figure below.) x y s 0 θ (x(T), 0) 43. Carl’s friend Jason competes in Highland Games Competitions across the country. In one event, the ‘hammer throw’, he throws a 56 pound weight for distance. If the weight is released 6 feet above the ground at an angle of 42 ◦ with respect to the horizontal with an initial speed of 33 feet per second, find the parametric equations for the flight of the hammer. (Here, use g = 32 ft. s 2 .) When will the hammer hit the ground? How far away will it hit the ground? Check your answer using a graphing utility. 44. Eliminate the parameter in the equations for projectile motion to show that the path of the projectile follows the curve y = − g sec 2 (θ) 2v 2 0 x 2 + tan(θ)x +s 0 Use the vertex formula (Equation 2.4) to show the maximum height of the projectile is y = v 2 0 sin 2 (θ) 2g +s 0 when x = v 2 0 sin(2θ) 2g 15 A nice mix of vectors and Calculus are needed to derive this. 16 We’ve seen this before. It’s the angle of elevation which was defined on page 753. 1060 Applications of Trigonometry 45. In another event, the ‘sheaf toss’, Jason throws a 20 pound weight for height. If the weight is released 5 feet above the ground at an angle of 85 ◦ with respect to the horizontal and the sheaf reaches a maximum height of 31.5 feet, use your results from part 44 to determine how fast the sheaf was launched into the air. (Once again, use g = 32 ft. s 2 .) 46. Suppose θ = π 2 . (The projectile was launched vertically.) Simplify the general parametric formula given for y(t) above using g = 9.8 m s 2 and compare that to the formula for s(t) given in Exercise 25 in Section 2.3. What is x(t) in this case? In Exercises 47 - 52, we explore the hyperbolic cosine function, denoted cosh(t), and the hyper- bolic sine function, denoted sinh(t), defined below: cosh(t) = e t +e −t 2 and sinh(t) = e t −e −t 2 47. Using a graphing utility as needed, verify that the domain of cosh(t) is (−∞, ∞) and the range of cosh(t) is [1, ∞). 48. Using a graphing utility as needed, verify that the domain and range of sinh(t) are both (−∞, ∞). 49. Show that ¦x(t) = cosh(t), y(t) = sinh(t) parametrize the right half of the ‘unit’ hyperbola x 2 −y 2 = 1. (Hence the use of the adjective ‘hyperbolic.’) 50. Compare the definitions of cosh(t) and sinh(t) to the formulas for cos(t) and sin(t) given in Exercise 83f in Section 11.7. 51. Four other hyperbolic functions are waiting to be defined: the hyperbolic secant sech(t), the hyperbolic cosecant csch(t), the hyperbolic tangent tanh(t) and the hyperbolic cotangent coth(t). Define these functions in terms of cosh(t) and sinh(t), then convert them to formulas involving e t and e −t . Consult a suitable reference (a Calculus book, or this entry on the hyperbolic functions) and spend some time reliving the thrills of trigonometry with these ‘hyperbolic’ functions. 52. If these functions look familiar, they should. Enjoy some nostalgia and revisit Exercise 35 in Section 6.5, Exercise 47 in Section 6.3 and the answer to Exercise 38 in Section 6.4. 11.10 Parametric Equations 1061 11.10.2 Answers 1. _ x = 4t −3 y = 6t −2 for 0 ≤ t ≤ 1 x y −3 −2 −1 1 −1 −2 1 2 3 4 2. _ x = 4t −1 y = 3 −4t for 0 ≤ t ≤ 1 x y −1 1 2 3 −1 1 2 3 3. _ x = 2t y = t 2 for −1 ≤ t ≤ 2 x y −3 −2 −1 1 2 3 4 1 2 3 4 4. _ x = t −1 y = 3 + 2t −t 2 for 0 ≤ t ≤ 3 x y −1 1 2 1 2 3 4 5. _ x = t 2 + 2t + 1 y = t + 1 for t ≤ 1 x y 1 2 3 4 5 −2 −1 1 2 6. _ x = 1 9 _ 18 −t 2 _ y = 1 3 t for t ≥ −3 x y −3 −2 −1 1 2 −1 1 2 1062 Applications of Trigonometry 7. _ x = t y = t 3 for −∞< t < ∞ x y −1 1 −4 −3 −2 −1 1 2 3 4 8. _ x = t 3 y = t for −∞< t < ∞ x y −1 1 −4 −3 −2 −1 1 2 3 4 9. _ x = cos(t) y = sin(t) for − π 2 ≤ t ≤ π 2 x y −1 1 −1 1 10. _ x = 3 cos(t) y = 3 sin(t) for 0 ≤ t ≤ π x y −3 −2 −1 1 2 3 1 2 3 11. _ x = −1 + 3 cos(t) y = 4 sin(t) for 0 ≤ t ≤ 2π x y −4 −3 −2 −1 1 2 −4 −3 −2 −1 1 2 3 4 12. _ x = 3 cos(t) y = 2 sin(t) + 1 for π 2 ≤ t ≤ 2π x y −3 −1 1 3 −1 1 2 3 11.10 Parametric Equations 1063 13. _ x = 2 cos(t) y = sec(t) for 0 < t < π 2 x y 1 2 3 4 1 2 3 4 14. _ x = 2 tan(t) y = cot(t) for 0 < t < π 2 x y 1 2 3 4 1 2 3 4 15. _ x = sec(t) y = tan(t) for − π 2 < t < π 2 x y 1 2 3 4 −4 −3 −2 −1 1 2 3 4 16. _ x = sec(t) y = tan(t) for π 2 < t < 3π 2 x y −4 −3 −2 −1 −4 −3 −2 −1 1 2 3 4 1064 Applications of Trigonometry 17. _ x = tan(t) y = 2 sec(t) for − π 2 < t < π 2 x y −2 −1 1 2 1 2 3 4 18. _ x = tan(t) y = 2 sec(t) for π 2 < t < 3π 2 x y −2 −1 1 2 −1 −2 −3 −4 19. _ x = cos(t) y = t for 0 < t < π x y π 2 π −1 1 20. _ x = sin(t) y = t for − π 2 < t < π 2 x y − π 2 π 2 −1 1 21. _ x = t 3 −3t y = t 2 −4 for −2 ≤ t ≤ 2 x y −2 −1 1 2 −4 −3 −2 −1 22. _ x = 4 cos 3 (t) y = 4 sin 3 (t) for 0 ≤ t ≤ 2π x y −4−3−2−1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 11.10 Parametric Equations 1065 23. _ x = e t +e −t y = e t −e −t for −2 ≤ t ≤ 2 x y 1 2 3 4 5 6 7 −7 −5 −3 −1 1 3 5 7 24. _ x = cos(3t) y = sin(4t) for 0 ≤ t ≤ 2π x y −1 1 −1 1 25. _ x = 3 −5t y = −5 + 7t for 0 ≤ t ≤ 1 26. _ x = 5t −2 y = −1 −3t for 0 ≤ t ≤ 1 27. _ x = t y = 4 −t 2 for −2 ≤ t ≤ 2 28. _ x = t −2 y = 4t −t 2 for 0 ≤ t ≤ 4 29. _ x = t 2 −9 y = t for −2 ≤ t ≤ 3 30. _ x = t 2 −6t y = 3 −t for 0 ≤ t ≤ 5 31. _ x = 5 cos(t) y = 5 sin(t) for 0 ≤ t < 2π 32. _ x = 1 + 2 cos(t) y = 2 sin(t) for 0 ≤ t < 2π 33. _ x = 3 cos(t) y = 3 + 3 sin(t) for 0 ≤ t < 2π 34. _ x = 3 cos(t) y = 3 −3 sin(t) for 0 ≤ t < 2π 35. _ x = 3 + √ 117 cos(t) y = −1 + √ 117 sin(t) for 0 ≤ t < 2π 36. _ x = 1 + 3 cos(t) y = sin(t) for 0 ≤ t < 2π 37. _ x = 2 cos(t) y = 3 sin(t) −3 for 0 ≤ t < 2π 38. _ _ _ x = 2 cos _ t − π 2 _ = 2 sin(t) y = −3 −3 sin _ t − π 2 _ = −3 + 3 cos(t) for 0 ≤ t < 2π 39. ¦x(t), y(t) where: x(t) = _ _ _ 3t, 0 ≤ t ≤ 1 6 −3t, 1 ≤ t ≤ 2 0, 2 ≤ t ≤ 3 y(t) = _ _ _ 0, 0 ≤ t ≤ 1 4t −4, 1 ≤ t ≤ 2 12 −4t, 2 ≤ t ≤ 3 1066 Applications of Trigonometry 40. The parametric equations for the inverse are _ x = t 3 + 3t −4 y = t for −∞< t < ∞ 41. r = 6 cos(2θ) translates to _ x = 6 cos(2θ) cos(θ) y = 6 cos(2θ) sin(θ) for 0 ≤ θ < 2π. 42. The parametric equations which describe the locations of passengers on the London Eye are _ x = 67.5 cos _ π 15 t − π 2 _ = 67.5 sin _ π 15 t _ y = 67.5 sin _ π 15 t − π 2 _ + 67.5 = 67.5 −67.5 cos _ π 15 t _ for −∞< t < ∞ 43. The parametric equations for the hammer throw are _ x = 33 cos(42 ◦ )t y = −16t 2 + 33 sin(42 ◦ )t + 6 for t ≥ 0. To find when the hammer hits the ground, we solve y(t) = 0 and get t ≈ −0.23 or 1.61. Since t ≥ 0, the hammer hits the ground after approximately t = 1.61 seconds after it was launched into the air. To find how far away the hammer hits the ground, we find x(1.61) ≈ 39.48 feet from where it was thrown into the air. 45. We solve y = v 2 0 sin 2 (θ) 2g +s 0 = v 2 0 sin 2 (85 ◦ ) 2(32) + 5 = 31.5 to get v 0 = ±41.34. The initial speed of the sheaf was approximately 41.34 feet per second. Index n th root of a complex number, 998, 999 principal, 397 n th Roots of Unity, 1004 u-substitution, 273 x-axis, 6 x-coordinate, 6 x-intercept, 25 y-axis, 6 y-coordinate, 6 y-intercept, 25 abscissa, 6 absolute value definition of, 173 inequality, 211 properties of, 173 acidity of a solution pH, 432 acute angle, 694 adjoint of a matrix, 622 alkalinity of a solution pH, 432 amplitude, 794, 879 angle acute, 694 between two vectors, 1033, 1034 central angle, 701 complementary, 696 coterminal, 698 decimal degrees, 695 definition, 693 degree, 694 DMS, 695 initial side, 698 measurement, 693 negative, 698 obtuse, 694 of declination, 761 of depression, 761 of elevation, 753 of inclination, 753 oriented, 697 positive, 698 quadrantal, 698 radian measure, 701 reference, 721 right, 694 standard position, 698 straight, 693 supplementary, 696 terminal side, 698 vertex, 693 angle side opposite pairs, 894 angular frequency, 708 annuity annuity-due, 667 ordinary definition of, 666 future value, 667 applied domain of a function, 60 arccosecant calculus friendly definition of, 831 graph of, 830 properties of, 831 1067 1068 Index trigonometry friendly definition of, 828 graph of, 827 properties of, 828 arccosine definition of, 820 graph of, 819 properties of, 820 arccotangent definition of, 824 graph of, 824 properties of, 824 arcsecant calculus friendly definition of, 831 graph of, 830 properties of, 831 trigonometry friendly definition of, 828 graph of, 827 properties of, 828 arcsine definition of, 820 graph of, 820 properties of, 820 arctangent definition of, 824 graph of, 823 properties of, 824 argument of a complex number definition of, 989 properties of, 993 of a function, 55 of a logarithm, 425 of a trigonometric function, 793 arithmetic sequence, 654 associative property for function composition, 366 matrix addition, 579 matrix multiplication, 585 scalar multiplication, 581 vector addition, 1013 scalar multiplication, 1016 asymptote horizontal formal definition of, 304 intuitive definition of, 304 location of, 308 of a hyperbola, 531 slant determination of, 312 formal definition of, 311 slant (oblique), 311 vertical formal definition of, 304 intuitive definition of, 304 location of, 306 augmented matrix, 568 average angular velocity, 707 average cost, 346 average cost function, 82 average rate of change, 160 average velocity, 706 axis of symmetry, 191 back substitution, 560 bearings, 903 binomial coefficient, 683 Binomial Theorem, 684 Bisection Method, 277 BMI, body mass index, 355 Boyle’s Law, 350 buffer solution, 478 cardioid, 949 Cartesian coordinate plane, 6 Cartesian coordinates, 6 Cauchy’s Bound, 269 center of a circle, 498 of a hyperbola, 531 of an ellipse, 516 Index 1069 central angle, 701 change of base formulas, 442 characteristic polynomial, 626 Charles’s Law, 355 circle center of, 498 definition of, 498 from slicing a cone, 495 radius of, 498 standard equation, 498 standard equation, alternate, 519 circular function, 744 cis(θ), 993 coefficient of determination, 226 cofactor, 616 Cofunction Identities, 773 common base, 420 common logarithm, 422 commutative property function composition does not have, 366 matrix addition, 579 vector addition, 1013 dot product, 1032 complementary angles, 696 Complex Factorization Theorem, 289 complex number n th root, 998, 999 n th Roots of Unity, 1004 argument definition of, 989 properties of, 993 conjugate definition of, 287 properties of, 288 definition of, 2, 286, 989 imaginary part, 989 imaginary unit, i, 286 modulus definition of, 989 properties of, 991 polar form cis-notation, 993 principal argument, 989 real part, 989 rectangular form, 989 set of, 2 complex plane, 989 component form of a vector, 1011 composite function definition of, 360 properties of, 367 compound interest, 470 conic sections definition, 495 conjugate axis of a hyperbola, 532 conjugate of a complex number definition of, 287 properties of, 288 Conjugate Pairs Theorem, 290 consistent system, 553 constant function as a horizontal line, 156 formal definition of, 101 intuitive definition of, 100 constant of proportionality, 350 constant term of a polynomial, 236 continuous, 241 continuously compounded interest, 472 contradiction, 549 coordinates Cartesian, 6 polar, 917 rectangular, 917 correlation coefficient, 226 cosecant graph of, 801 of an angle, 744, 752 properties of, 802 cosine graph of, 791 of an angle, 717, 730, 744 properties of, 791 1070 Index cost average, 82, 346 fixed, start-up, 82 variable, 159 cost function, 82 cotangent graph of, 805 of an angle, 744, 752 properties of, 806 coterminal angle, 698 Coulomb’s Law, 355 Cramer’s Rule, 619 curve orientated, 1046 cycloid, 1054 decibel, 431 decimal degrees, 695 decreasing function formal definition of, 101 intuitive definition of, 100 degree measure, 694 degree of a polynomial, 236 DeMoivre’s Theorem, 995 dependent system, 554 dependent variable, 55 depreciation, 420 Descartes’ Rule of Signs, 273 determinant of a matrix definition of, 614 properties of, 616 Difference Identity for cosine, 771, 775 for sine, 773, 775 for tangent, 775 difference quotient, 79 dimension of a matrix, 567 direct variation, 350 directrix of a conic section in polar form, 979 of a parabola, 505 discriminant of a conic, 977 of a quadratic equation, 195 trichotomy, 195 distance definition, 10 distance formula, 11 distributive property matrix matrix multiplication, 585 scalar multiplication, 581 vector dot product, 1032 scalar multiplication, 1016 DMS, 695 domain applied, 60 definition of, 45 implied, 58 dot product commutative property of, 1032 definition of, 1032 distributive property of, 1032 geometric interpretation, 1033 properties of, 1032 relation to orthogonality, 1035 relation to vector magnitude, 1032 work, 1040 Double Angle Identities, 776 earthquake Richter Scale, 431 eccentricity, 522, 979 eigenvalue, 626 eigenvector, 626 ellipse center, 516 definition of, 516 eccentricity, 522 foci, 516 from slicing a cone, 496 guide rectangle, 519 major axis, 516 minor axis, 516 Index 1071 reflective property, 523 standard equation, 519 vertices, 516 ellipsis (. . . ), 31, 651 empty set, 2 end behavior of f(x) = ax n , n even, 240 of f(x) = ax n , n odd, 240 of a function graph, 239 polynomial, 243 entry in a matrix, 567 equation contradiction, 549 graph of, 23 identity, 549 linear of n variables, 554 linear of two variables, 549 even function, 95 Even/Odd Identities, 770 exponential function algebraic properties of, 437 change of base formula, 442 common base, 420 definition of, 418 graphical properties of, 419 inverse properties of, 437 natural base, 420 one-to-one properties of, 437 solving equations with, 448 extended interval notation, 756 Factor Theorem, 258 factorial, 654, 681 fixed cost, 82 focal diameter of a parabola, 507 focal length of a parabola, 506 focus of a conic section in polar form, 979 focus (foci) of a hyperbola, 531 of a parabola, 505 of an ellipse, 516 free variable, 552 frequency angular, 708, 879 of a sinusoid, 795 ordinary, 708, 879 function (absolute) maximum, 101 (absolute, global) minimum, 101 absolute value, 173 algebraic, 399 argument, 55 arithmetic, 76 as a process, 55, 378 average cost, 82 circular, 744 composite definition of, 360 properties of, 367 constant, 100, 156 continuous, 241 cost, 82 decreasing, 100 definition as a relation, 43 dependent variable of, 55 difference, 76 difference quotient, 79 domain, 45 even, 95 exponential, 418 Fundamental Graphing Principle, 93 identity, 168 increasing, 100 independent variable of, 55 inverse definition of, 379 properties of, 379 solving for, 384 uniqueness of, 380 linear, 156 local (relative) maximum, 101 local (relative) minimum, 101 logarithmic, 422 1072 Index notation, 55 odd, 95 one-to-one, 381 periodic, 790 piecewise-defined, 62 polynomial, 235 price-demand, 82 product, 76 profit, 82 quadratic, 188 quotient, 76 range, 45 rational, 301 revenue, 82 smooth, 241 sum, 76 transformation of graphs, 120, 135 zero, 95 fundamental cycle of y = cos(x), 791 Fundamental Graphing Principle for equations, 23 for functions, 93 for polar equations, 936 Fundamental Theorem of Algebra, 289 Gauss-Jordan Elimination, 571 Gaussian Elimination, 557 geometric sequence, 654 geometric series, 669 graph hole in, 305 horizontal scaling, 132 horizontal shift, 123 of a function, 93 of a relation, 20 of an equation, 23 rational function, 321 reflection about an axis, 126 transformations, 135 vertical scaling, 130 vertical shift, 121 greatest integer function, 67 growth model limited, 475 logistic, 475 uninhibited, 472 guide rectangle for a hyperbola, 532 for an ellipse, 519 Half-Angle Formulas, 779 harmonic motion, 883 Henderson-Hasselbalch Equation, 446 Heron’s Formula, 912 hole in a graph, 305 location of, 306 Hooke’s Law, 350 horizontal asymptote formal definition of, 304 intuitive definition of, 304 location of, 308 horizontal line, 23 Horizontal Line Test (HLT), 381 hyperbola asymptotes, 531 branch, 531 center, 531 conjugate axis, 532 definition of, 531 foci, 531 from slicing a cone, 496 guide rectangle, 532 standard equation horizontal, 534 vertical, 534 transverse axis, 531 vertices, 531 hyperbolic cosine, 1060 hyperbolic sine, 1060 hyperboloid, 542 identity function, 367 matrix, additive, 579 Index 1073 matrix, multiplicative, 585 statement which is always true, 549 imaginary axis, 989 imaginary part of a complex number, 989 imaginary unit, i, 286 implied domain of a function, 58 inconsistent system, 553 increasing function formal definition of, 101 intuitive definition of, 100 independent system, 554 independent variable, 55 index of a root, 397 induction base step, 673 induction hypothesis, 673 inductive step, 673 inequality absolute value, 211 graphical interpretation, 209 non-linear, 643 quadratic, 215 sign diagram, 214 inflection point, 477 information entropy, 477 initial side of an angle, 698 instantaneous rate of change, 161, 472, 707 integer definition of, 2 greatest integer function, 67 set of, 2 intercept definition of, 25 location of, 25 interest compound, 470 compounded continuously, 472 simple, 469 Intermediate Value Theorem polynomial zero version, 241 interrobang, 321 intersection of two sets, 4 interval definition of, 3 notation for, 3 notation, extended, 756 inverse matrix, additive, 579, 581 matrix, multiplicative, 602 of a function definition of, 379 properties of, 379 solving for, 384 uniqueness of, 380 inverse variation, 350 invertibility function, 382 invertible function, 379 matrix, 602 irrational number definition of, 2 set of, 2 irreducible quadratic, 290 joint variation, 350 Kepler’s Third Law of Planetary Motion, 355 Kirchhoff’s Voltage Law, 605 latus rectum of a parabola, 507 Law of Cosines, 908 Law of Sines, 895 leading coefficient of a polynomial, 236 leading term of a polynomial, 236 Learning Curve Equation, 315 least squares regression line, 225 lemniscate, 948 lima¸con, 948 line horizontal, 23 least squares regression, 225 linear function, 156 of best fit, 225 parallel, 166 1074 Index perpendicular, 167 point-slope form, 155 slope of, 151 slope-intercept form, 155 vertical, 23 linear equation n variables, 554 two variables, 549 linear function, 156 local maximum formal definition of, 102 intuitive definition of, 101 local minimum formal definition of, 102 intuitive definition of, 101 logarithm algebraic properties of, 438 change of base formula, 442 common, 422 general, “base b”, 422 graphical properties of, 423 inverse properties of, 437 natural, 422 one-to-one properties of, 437 solving equations with, 459 logarithmic scales, 431 logistic growth, 475 LORAN, 538 lower triangular matrix, 593 main diagonal, 585 major axis of an ellipse, 516 Markov Chain, 592 mathematical model, 60 matrix addition associative property, 579 commutative property, 579 definition of, 578 properties of, 579 additive identity, 579 additive inverse, 579 adjoint, 622 augmented, 568 characteristic polynomial, 626 cofactor, 616 definition, 567 determinant definition of, 614 properties of, 616 dimension, 567 entry, 567 equality, 578 invertible, 602 leading entry, 569 lower triangular, 593 main diagonal, 585 matrix multiplication associative property of, 585 definition of, 584 distributive property, 585 identity for, 585 properties of, 585 minor, 616 multiplicative inverse, 602 product of row and column, 584 reduced row echelon form, 570 rotation, 984 row echelon form, 569 row operations, 568 scalar multiplication associative property of, 581 definition of, 580 distributive properties, 581 identity for, 581 properties of, 581 zero product property, 581 size, 567 square matrix, 586 sum, 578 upper triangular, 593 maximum formal definition of, 102 intuitive definition of, 101 measure of an angle, 693 Index 1075 midpoint definition of, 12 midpoint formula, 13 minimum formal definition of, 102 intuitive definition of, 101 minor, 616 minor axis of an ellipse, 516 model mathematical, 60 modulus of a complex number definition of, 989 properties of, 991 multiplicity effect on the graph of a polynomial, 245, 249 of a zero, 244 natural base, 420 natural logarithm, 422 natural number definition of, 2 set of, 2 negative angle, 698 Newton’s Law of Cooling, 421, 474 Newton’s Law of Universal Gravitation, 351 oblique asymptote, 311 obtuse angle, 694 odd function, 95 Ohm’s Law, 350, 605 one-to-one function, 381 ordered pair, 6 ordinary frequency, 708 ordinate, 6 orientation, 1046 oriented angle, 697 oriented arc, 704 origin, 7 orthogonal projection, 1036 orthogonal vectors, 1035 overdetermined system, 554 parabola axis of symmetry, 191 definition of, 505 directrix, 505 focal diameter, 507 focal length, 506 focus, 505 from slicing a cone, 496 graph of a quadratic function, 188 latus rectum, 507 reflective property, 510 standard equation horizontal, 508 vertical, 506 vertex, 188, 505 vertex formulas, 194 paraboloid, 510 parallel vectors, 1028 parameter, 1046 parametric equations, 1046 parametric solution, 552 parametrization, 1046 partial fractions, 628 Pascal’s Triangle, 688 password strength, 477 period circular motion, 708 of a function, 790 of a sinusoid, 879 periodic function, 790 pH, 432 phase, 795, 879 phase shift, 795, 879 pi, π, 700 piecewise-defined function, 62 point of diminishing returns, 477 point-slope form of a line, 155 polar coordinates conversion into rectangular, 922 definition of, 917 equivalent representations of, 921 polar axis, 917 pole, 917 1076 Index polar form of a complex number, 993 polar rose, 948 polynomial division dividend, 258 divisor, 258 factor, 258 quotient, 258 remainder, 258 synthetic division, 260 polynomial function completely factored over the complex numbers, 290 over the real numbers, 290 constant term, 236 definition of, 235 degree, 236 end behavior, 239 leading coefficient, 236 leading term, 236 variations in sign, 273 zero lower bound, 274 multiplicity, 244 upper bound, 274 positive angle, 698 Power Reduction Formulas, 778 power rule for absolute value, 173 for complex numbers, 995 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 991 price-demand function, 82 principal, 469 principal n th root, 397 principal argument of a complex number, 989 principal unit vectors, ˆı, ˆ , 1022 Principle of Mathematical Induction, 673 product rule for absolute value, 173 for complex numbers, 995 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 991 Product to Sum Formulas, 780 profit function, 82 projection x−axis, 45 y−axis, 46 orthogonal, 1036 Pythagorean Conjugates, 751 Pythagorean Identities, 749 quadrantal angle, 698 quadrants, 8 quadratic formula, 194 quadratic function definition of, 188 general form, 190 inequality, 215 irreducible quadratic, 290 standard form, 190 quadratic regression, 228 Quotient Identities, 745 quotient rule for absolute value, 173 for complex numbers, 995 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 991 radian measure, 701 radical properties of, 398 radicand, 397 radioactive decay, 473 radius of a circle, 498 range definition of, 45 rate of change average, 160 Index 1077 instantaneous, 161, 472 slope of a line, 154 rational exponent, 398 rational functions, 301 rational number definition of, 2 set of, 2 Rational Zeros Theorem, 269 ray definition of, 693 initial point, 693 real axis, 989 Real Factorization Theorem, 291 real number definition of, 2 set of, 2 real part of a complex number, 989 Reciprocal Identities, 745 rectangular coordinates also known as Cartesian coordinates, 917 conversion into polar, 922 rectangular form of a complex number, 989 recursion equation, 654 reduced row echelon form, 570 reference angle, 721 Reference Angle Theorem for cosine and sine, 722 for the circular functions, 747 reflection of a function graph, 126 of a point, 10 regression coefficient of determination, 226 correlation coefficient, 226 least squares line, 225 quadratic, 228 total squared error, 225 relation algebraic description, 23 definition, 20 Fundamental Graphing Principle, 23 Remainder Theorem, 258 revenue function, 82 Richter Scale, 431 right angle, 694 root index, 397 radicand, 397 Roots of Unity, 1004 rotation matrix, 984 rotation of axes, 972 row echelon form, 569 row operations for a matrix, 568 scalar multiplication matrix associative property of, 581 definition of, 580 distributive properties of, 581 properties of, 581 vector associative property of, 1016 definition of, 1015 distributive properties of, 1016 properties of, 1016 scalar projection, 1037 secant graph of, 800 of an angle, 744, 752 properties of, 802 secant line, 160 sequence n th term, 652 alternating, 652 arithmetic common difference, 654 definition of, 654 formula for n th term, 656 sum of first n terms, 666 definition of, 652 geometric common ratio, 654 definition of, 654 formula for n th term, 656 sum of first n terms, 666 1078 Index recursive, 654 series, 668 set definition of, 1 empty, 2 intersection, 4 roster method, 1 set-builder notation, 1 sets of numbers, 2 union, 4 verbal description, 1 set-builder notation, 1 Side-Angle-Side triangle, 908 Side-Side-Side triangle, 908 sign diagram algebraic function, 399 for quadratic inequality, 214 polynomial function, 242 rational function, 321 simple interest, 469 sine graph of, 792 of an angle, 717, 730, 744 properties of, 791 sinusoid amplitude, 794, 879 baseline, 879 frequency angular, 879 ordinary, 879 graph of, 795, 880 period, 879 phase, 879 phase shift, 795, 879 properties of, 879 vertical shift, 879 slant asymptote, 311 slant asymptote determination of, 312 formal definition of, 311 slope definition, 151 of a line, 151 rate of change, 154 slope-intercept form of a line, 155 smooth, 241 sound intensity level decibel, 431 square matrix, 586 standard position of a vector, 1017 standard position of an angle, 698 start-up cost, 82 steady state, 592 stochastic process, 592 straight angle, 693 Sum Identity for cosine, 771, 775 for sine, 773, 775 for tangent, 775 Sum to Product Formulas, 781 summation notation definition of, 661 index of summation, 661 lower limit of summation, 661 properties of, 664 upper limit of summation, 661 supplementary angles, 696 symmetry about the x-axis, 9 about the y-axis, 9 about the origin, 9 testing a function graph for, 95 testing an equation for, 26 synthetic division tableau, 260 system of equations back-substitution, 560 coefficient matrix, 590 consistent, 553 constant matrix, 590 definition, 549 dependent, 554 free variable, 552 Gauss-Jordan Elimination, 571 Gaussian Elimination, 557 Index 1079 inconsistent, 553 independent, 554 leading variable, 556 linear n variables, 554 two variables, 550 linear in form, 646 non-linear, 637 overdetermined, 554 parametric solution, 552 triangular form, 556 underdetermined, 554 unknowns matrix, 590 tangent graph of, 804 of an angle, 744, 752 properties of, 806 terminal side of an angle, 698 Thurstone, Louis Leon, 315 total squared error, 225 transformation non-rigid, 129 rigid, 129 transformations of function graphs, 120, 135 transverse axis of a hyperbola, 531 Triangle Inequality, 183 triangular form, 556 underdetermined system, 554 uninhibited growth, 472 union of two sets, 4 Unit Circle definition of, 501 important points, 724 unit vector, 1021 Upper and Lower Bounds Theorem, 274 upper triangular matrix, 593 variable dependent, 55 independent, 55 variable cost, 159 variation constant of proportionality, 350 direct, 350 inverse, 350 joint, 350 variations in sign, 273 vector x-component, 1010 y-component, 1010 addition associative property, 1013 commutative property, 1013 definition of, 1012 properties of, 1013 additive identity, 1013 additive inverse, 1013, 1016 angle between two, 1033, 1034 component form, 1010 Decomposition Theorem Generalized, 1038 Principal, 1022 definition of, 1010 direction definition of, 1018 properties of, 1018 dot product commutative property of, 1032 definition of, 1032 distributive property of, 1032 geometric interpretation, 1033 properties of, 1032 relation to magnitude, 1032 relation to orthogonality, 1035 work, 1040 head, 1010 initial point, 1010 magnitude definition of, 1018 properties of, 1018 relation to dot product, 1032 normalization, 1022 orthogonal projection, 1036 1080 Index orthogonal vectors, 1035 parallel, 1028 principal unit vectors, ˆı, ˆ , 1022 resultant, 1011 scalar multiplication associative property of, 1016 definition of, 1015 distributive properties, 1016 identity for, 1016 properties of, 1016 zero product property, 1016 scalar product definition of, 1032 properties of, 1032 scalar projection, 1037 standard position, 1017 tail, 1010 terminal point, 1010 triangle inequality, 1042 unit vector, 1021 velocity average angular, 707 instantaneous, 707 instantaneous angular, 707 vertex of a hyperbola, 531 of a parabola, 188, 505 of an angle, 693 of an ellipse, 516 vertical asymptote formal definition of, 304 intuitive definition of, 304 location of, 306 vertical line, 23 Vertical Line Test (VLT), 43 whole number definition of, 2 set of, 2 work, 1039 wrapping function, 704 zero multiplicity of, 244 of a function, 95 upper and lower bounds, 274 ii Acknowledgements While the cover of this textbook lists only two names, the book as it stands today would simply not exist if not for the tireless work and dedication of several people. First and foremost, we wish to thank our families for their patience and support during the creative process. We would also like to thank our students - the sole inspiration for the work. Among our colleagues, we wish to thank Rich Basich, Bill Previts, and Irina Lomonosov, who not only were early adopters of the textbook, but also contributed materials to the project. Special thanks go to Katie Cimperman, Terry Dykstra, Frank LeMay, and Rich Hagen who provided valuable feedback from the classroom. Thanks also to David Stumpf, Ivana Gorgievska, Jorge Gerszonowicz, Kathryn Arocho, Heather Bubnick, and Florin Muscutariu for their unwaivering support (and sometimes defense!) of the project. From outside the classroom, we wish to thank Don Anthan and Ken White, who designed the electric circuit applications used in the text, as well as Drs. Wendy Marley and Marcia Ballinger for the Lorain CCC enrollment data used in the text. The authors are also indebted to the good folks at our schools’ bookstores, Gwen Sevtis (Lakeland CC) and Chris Callahan (Lorain CCC), for working with us to get printed copies to the students as inexpensively as possible. We would also like to thank Lakeland folks Jeri Dickinson, Mary Ann Blakeley, Jessica Novak, and Corrie Bergeron for their enthusiasm and promotion of the project. The administration at both schools have also been very supportive of the project, so from Lakeland, we wish to thank Dr. Morris W. Beverage, Jr., President, Dr. Fred Law, Provost, Deans Don Anthan and Dr. Steve Oluic, and the Board of Trustees. From Lorain County Community College, we which to thank Dr. Roy A. Church, Dr. Karen Wells, and the Board of Trustees. From the Ohio Board of Regents, we wish to thank former Chancellor Eric Fingerhut, Darlene McCoy, Associate Vice Chancellor of Affordability and Efficiency, and Kelly Bernard. From OhioLINK, we wish to thank Steve Acker, John Magill, and Stacy Brannan. We also wish to thank the good folks at WebAssign, most notably Chris Hall, COO, and Joel Hollenbeck (former VP of Sales.) Last, but certainly not least, we wish to thank all the folks who have contacted us over the interwebs, most notably Dimitri Moonen and Joel Wordsworth, who gave us great feedback, and Antonio Olivares who helped debug the source code. Table of Contents 10 Foundations of Trigonometry 10.1 Angles and their Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.1 Applications of Radian Measure: Circular Motion . . . . . . . . . 10.1.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.3 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 The Unit Circle: Cosine and Sine . . . . . . . . . . . . . . . . . . . . . . . 10.2.1 Beyond the Unit Circle . . . . . . . . . . . . . . . . . . . . . . . . 10.2.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.3 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 The Six Circular Functions and Fundamental Identities . . . . . . . . . . . 10.3.1 Beyond the Unit Circle . . . . . . . . . . . . . . . . . . . . . . . . 10.3.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3.3 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5 Graphs of the Trigonometric Functions . . . . . . . . . . . . . . . . . . . . 10.5.1 Graphs of the Cosine and Sine Functions . . . . . . . . . . . . . . 10.5.2 Graphs of the Secant and Cosecant Functions . . . . . . . . . . . 10.5.3 Graphs of the Tangent and Cotangent Functions . . . . . . . . . . 10.5.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.5 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6 The Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . 10.6.1 Inverses of Secant and Cosecant: Trigonometry Friendly Approach 10.6.2 Inverses of Secant and Cosecant: Calculus Friendly Approach . . . 10.6.3 Calculators and the Inverse Circular Functions. . . . . . . . . . . . 10.6.4 Solving Equations Using the Inverse Trigonometric Functions. . . 10.6.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6.6 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.7 Trigonometric Equations and Inequalities . . . . . . . . . . . . . . . . . . 10.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.7.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Applications of Trigonometry 11.1 Applications of Sinusoids . . 11.1.1 Harmonic Motion . 11.1.2 Exercises . . . . . . 11.1.3 Answers . . . . . . . 11.2 The Law of Sines . . . . . . 11.2.1 Exercises . . . . . . 11.2.2 Answers . . . . . . . 11.3 The Law of Cosines . . . . . vii 693 693 706 709 712 717 730 736 740 744 752 759 766 770 782 787 790 790 800 804 809 811 819 827 830 833 838 841 849 857 871 874 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 879 . 879 . 883 . 889 . 892 . 894 . 902 . 906 . 908 viii 11.3.1 Exercises . . . . . . . . . 11.3.2 Answers . . . . . . . . . . 11.4 Polar Coordinates . . . . . . . . . 11.4.1 Exercises . . . . . . . . . 11.4.2 Answers . . . . . . . . . . 11.5 Graphs of Polar Equations . . . . 11.5.1 Exercises . . . . . . . . . 11.5.2 Answers . . . . . . . . . . 11.6 Hooked on Conics Again . . . . . 11.6.1 Rotation of Axes . . . . . 11.6.2 The Polar Form of Conics 11.6.3 Exercises . . . . . . . . . 11.6.4 Answers . . . . . . . . . . 11.7 Polar Form of Complex Numbers 11.7.1 Exercises . . . . . . . . . 11.7.2 Answers . . . . . . . . . . 11.8 Vectors . . . . . . . . . . . . . . . 11.8.1 Exercises . . . . . . . . . 11.8.2 Answers . . . . . . . . . . 11.9 The Dot Product and Projection 11.9.1 Exercises . . . . . . . . . 11.9.2 Answers . . . . . . . . . . 11.10 Parametric Equations . . . . . . 11.10.1 Exercises . . . . . . . . . 11.10.2 Answers . . . . . . . . . . Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Table of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 914 916 917 928 930 936 956 961 971 971 979 984 985 989 1002 1005 1010 1025 1029 1032 1041 1043 1046 1057 1061 1067 Preface Thank you for your interest in our book, but more importantly, thank you for taking the time to read the Preface. I always read the Prefaces of the textbooks which I use in my classes because I believe it is in the Preface where I begin to understand the authors - who they are, what their motivation for writing the book was, and what they hope the reader will get out of reading the text. Pedagogical issues such as content organization and how professors and students should best use a book can usually be gleaned out of its Table of Contents, but the reasons behind the choices authors make should be shared in the Preface. Also, I feel that the Preface of a textbook should demonstrate the authors’ love of their discipline and passion for teaching, so that I come away believing that they really want to help students and not just make money. Thus, I thank my fellow Preface-readers again for giving me the opportunity to share with you the need and vision which guided the creation of this book and passion which both Carl and I hold for Mathematics and the teaching of it. Carl and I are natives of Northeast Ohio. We met in graduate school at Kent State University in 1997. I finished my Ph.D in Pure Mathematics in August 1998 and started teaching at Lorain County Community College in Elyria, Ohio just two days after graduation. Carl earned his Ph.D in Pure Mathematics in August 2000 and started teaching at Lakeland Community College in Kirtland, Ohio that same month. Our schools are fairly similar in size and mission and each serves a similar population of students. The students range in age from about 16 (Ohio has a Post-Secondary Enrollment Option program which allows high school students to take college courses for free while still in high school.) to over 65. Many of the “non-traditional” students are returning to school in order to change careers. A majority of the students at both schools receive some sort of financial aid, be it scholarships from the schools’ foundations, state-funded grants or federal financial aid like student loans, and many of them have lives busied by family and job demands. Some will be taking their Associate degrees and entering (or re-entering) the workforce while others will be continuing on to a four-year college or university. Despite their many differences, our students share one common attribute: they do not want to spend $200 on a College Algebra book. The challenge of reducing the cost of textbooks is one that many states, including Ohio, are taking quite seriously. Indeed, state-level leaders have started to work with faculty from several of the colleges and universities in Ohio and with the major publishers as well. That process will take considerable time so Carl and I came up with a plan of our own. We decided that the best way to help our students right now was to write our own College Algebra book and give it away electronically for free. We were granted sabbaticals from our respective institutions for the Spring x Preface semester of 2009 and actually began writing the textbook on December 16, 2008. Using an opensource text editor called TexNicCenter and an open-source distribution of LaTeX called MikTex 2.7, Carl and I wrote and edited all of the text, exercises and answers and created all of the graphs (using Metapost within LaTeX) for Version 0.9 in about eight months. (We choose to create a text in only black and white to keep printing costs to a minimum for those students who prefer a printed edition. This somewhat Spartan page layout stands in sharp relief to the explosion of colors found in most other College Algebra texts, but neither Carl nor I believe the four-color print adds anything of value.) I used the book in three sections of College Algebra at Lorain County Community College in the Fall of 2009 and Carl’s colleague, Dr. Bill Previts, taught a section of College Algebra at Lakeland with the book that semester as well. Students had the option of downloading the book as a .pdf file from our website www.stitz-zeager.com or buying a low-cost printed version from our colleges’ respective bookstores. (By giving this book away for free electronically, we end the cycle of new editions appearing every 18 months to curtail the used book market.) During Thanksgiving break in November 2009, many additional exercises written by Dr. Previts were added and the typographical errors found by our students and others were √ corrected. On December 10, 2009, Version 2 was released. The book remains free for download at our website and by using Lulu.com as an on-demand printing service, our bookstores are now able to provide a printed edition for just under $19. Neither Carl nor I have, or will ever, receive any royalties from the printed editions. As a contribution back to the open-source community, all of the LaTeX files used to compile the book are available for free under a Creative Commons License on our website as well. That way, anyone who would like to rearrange or edit the content for their classes can do so as long as it remains free. The only disadvantage to not working for a publisher is that we don’t have a paid editorial staff. What we have instead, beyond ourselves, is friends, colleagues and unknown people in the opensource community who alert us to errors they find as they read the textbook. What we gain in not having to report to a publisher so dramatically outweighs the lack of the paid staff that we have turned down every offer to publish our book. (As of the writing of this Preface, we’ve had three offers.) By maintaining this book by ourselves, Carl and I retain all creative control and keep the book our own. We control the organization, depth and rigor of the content which means we can resist the pressure to diminish the rigor and homogenize the content so as to appeal to a mass market. A casual glance through the Table of Contents of most of the major publishers’ College Algebra books reveals nearly isomorphic content in both order and depth. Our Table of Contents shows a different approach, one that might be labeled “Functions First.” To truly use The Rule of Four, that is, in order to discuss each new concept algebraically, graphically, numerically and verbally, it seems completely obvious to us that one would need to introduce functions first. (Take a moment and compare our ordering to the classic “equations first, then the Cartesian Plane and THEN functions” approach seen in most of the major players.) We then introduce a class of functions and discuss the equations, inequalities (with a heavy emphasis on sign diagrams) and applications which involve functions in that class. The material is presented at a level that definitely prepares a student for Calculus while giving them relevant Mathematics which can be used in other classes as well. Graphing calculators are used sparingly and only as a tool to enhance the Mathematics, not to replace it. The answers to nearly all of the computational homework exercises are given in the xi text and we have gone to great lengths to write some very thought provoking discussion questions whose answers are not given. One will notice that our exercise sets are much shorter than the traditional sets of nearly 100 “drill and kill” questions which build skill devoid of understanding. Our experience has been that students can do about 15-20 homework exercises a night so we very carefully chose smaller sets of questions which cover all of the necessary skills and get the students thinking more deeply about the Mathematics involved. Critics of the Open Educational Resource movement might quip that “open-source is where bad content goes to die,” to which I say this: take a serious look at what we offer our students. Look through a few sections to see if what we’ve written is bad content in your opinion. I see this opensource book not as something which is “free and worth every penny”, but rather, as a high quality alternative to the business as usual of the textbook industry and I hope that you agree. If you have any comments, questions or concerns please feel free to contact me at jeff@stitz-zeager.com or Carl at [email protected]. Jeff Zeager Lorain County Community College January 25, 2010 xii Preface . When two rays share a common initial point they form an angle and the common initial point is called the vertex of the angle.1 Angles and their Measure This section begins our study of Trigonometry and to get started. in some sense. we recall some basic definitions from Geometry. The point from which the ray originates is called the initial point of the ray. in the second. In the first case. extreme cases. However.Chapter 10 Foundations of Trigonometry 10. the two figures below also depict angles . P A ray with initial point P . An angle with vertex Q. . Two examples of what are commonly thought of as angles are Q P An angle with vertex P . the two rays are directly opposite each other forming what is known as a straight angle. Q P A straight angle. as pictured below. A ray is usually described as a ‘half-line’ and can be thought of as a line segment in which one of the two endpoints is pushed off infinitely distant from the other.albeit these are. The measure of an angle is a number which indicates the amount of rotation that separates the rays of the angle. There is one immediate problem with this. as pictured below. the rays are identical so the ‘angle’ is indistinguishable from the ray itself. as is commonplace in Geometry. we have β α One commonly used system to measure angles is degree measure.1 Clearly these two angles have different measures because one appears to represent a larger rotation than the other. so we must label them differently. One complete revolution as shown below is 360◦ . and parts of a revolution are measured proportionately. Recall that if an angle measures strictly between 0◦ and 90◦ it is called an acute angle and if it measures strictly between 90◦ and 180◦ it is called an obtuse angle. The choice of ‘360’ is most often attributed to the Babylonians.2 Thus half of a revolution (a straight angle) measures 1 1 ◦ ◦ ◦ ◦ 2 (360 ) = 180 . Quantities measured in degrees are denoted by the familiar ‘◦ ’ symbol. we can know the measure of any angle as long as we 1 2 The phrase ‘at least’ will be justified in short order. γ (gamma) and θ (theta) to label angles. In this book. theoretically. for instance. a quarter of a revolution (a right angle) measures 4 (360 ) = 90 and so on. So.694 Foundations of Trigonometry Which amount of rotation are we attempting to quantify? What we have just discovered is that we have at least two angles described by this diagram. One revolution ↔ 360◦ 180◦ 90◦ Note that in the above figure. β (beta). we use lower case Greek letters such as α (alpha). we have used the small square ‘ ’ to denote a right angle. It is important to note that. . to convert 117◦ 15 45 to decimal degrees.3 For instance. Awesome math pun aside. Converting the partial amount of degrees to minutes. is decimal degrees. Even though it may be hard to draw. To convert a measure of 42. or equivalently. Then we find 3 4 = 1◦ 4 and This is how a protractor is graded.10. an angle with a measure of 30. the measure of an angle which 2 2 represents a rotation of 3 of a revolution would measure 3 (360◦ ) = 240◦ . one degree is divided equally into sixty minutes. the measure of an angle 1 1 which constitutes only 12 of a revolution measures 12 (360◦ ) = 30◦ and an angle which indicates no rotation at all is measured as 0◦ . For example.125◦ . This √ 360 ◦ can be taken to the limit using Calculus so that measures like 2 make sense. 30. from which it follows that 1◦ = 3600 . and most familiar. and in turn. we write 1◦ = 60 and 1 = 60 . we start by noting that 42.4 The second way to divide degrees is the Degree .125◦ = = = = = 60 1 = 30 .1. Putting it all together yields 42◦ + 0. this is the same idea behind defining irrational exponents in Section 6. 5 Does this kind of system seem familiar? .5 42.125◦ to the DMS system.125◦ = 42◦ + 0. each minute is divided equally into sixty seconds.5◦ would 61 represent a rotation halfway between 30◦ and 31◦ .Minute .125◦ 42◦ + 7.1 Angles and their Measure 695 know the proportion it represents of entire revolution. Converting the 1◦ partial amount of minutes to seconds gives 0. it is nonetheless not difficult to imagine an angle with measure smaller than 1◦ . we first compute 15 1◦ 1 ◦ 45 3600 = 80 .5 42◦ + 7 + 30 42◦ 7 30 1◦ 60 On the other hand. The first.5 42◦ + 7 + 0. we find 0. 240◦ 30◦ 0◦ Using our definition of degree measure.5 .5 In symbols.125◦ 60 = 7. we have that 1◦ represents the measure of an angle which 1 constitutes 360 of a revolution. There are two ways to subdivide degrees.5 = 7 + 0.5 = 720 of a full rotation.Second (DMS) system. In this system. 2625◦ Recall that two acute angles are called complementary angles if their measures add to 90◦ . β θ α γ Supplementary Angles Complementary Angles In practice. are called supplementary angles if their measures add to 180◦ .371◦ 111◦ + 22.26 . 1.371◦ .6 Rounding to seconds. Round your answer to the nearest thousandth of a degree. Sketch α and β.26 = 22 + 0. 1◦ 1 111. the distinction between the angle itself and its measure is blurred so that the sentence ‘α is an angle measuring 42◦ ’ is often abbreviated as ‘α = 42◦ . 2. we convert 0. . To convert α to the DMS system.26 60 = 15.6 .1. Solution.6 111◦ 22 15.1.26 111◦ + 22 + 15.371◦ and β = 37◦ 28 17 .371◦ = 111◦ + 0. we start with 111. 1.26 111◦ + 22 + 0.371◦ = = = = = 111◦ + 0. 5. 3. 4. Hence.26 .371◦ 60 = 22. Find a complementary angle for β. Convert β to decimal degrees.696 Foundations of Trigonometry 117◦ 15 45 = 117◦ + 15 + 45 = 117◦ + = 9381 ◦ 80 1◦ 4 + 1 ◦ 80 = 117. Convert α to the DMS system. In the diagram below.’ It is now time for an example. Example 10. we obtain α ≈ 111◦ 22 16 . Next we convert 0. Let α = 111. Two angles. Writing 22. Round your answer to the nearest second. the angles α and β are supplementary angles while the pair γ and θ are complementary angles. either a pair of right angles or one acute angle and one obtuse angle. Find a supplementary angle for α. we are ‘borrowing’ 1◦ = 60 from the degree place. If we divide this range in half. In essence. we seek an angle γ so that β + γ = 90◦ . To sketch α. 7 Like ‘latus rectum. then 0◦ < β < 45◦ . While we could reach for the calculator to obtain an approximate answer. To find a complementary angle for β. in an oriented If this process seems hauntingly familiar.8 This yields. we first note that 90◦ < α < 180◦ . 22. To convert β to decimal degrees. we have discussed only angles which measure between 0◦ and 360◦ . 5. we get 90◦ < α < 135◦ .6 Proceeding similarly for β. Compare this method to the Bisection Method introduced in Section 3. we choose instead to do a bit of sexagesimal7 arithmetic. inclusive. 6 . We get θ = 180◦ − α = 180◦ − 111. Back then. This gives us a pretty good estimate for α. We get γ = 90◦ − β = 90◦ − 37◦ 28 17 . 33. but also to extend their applicability to other real-world phenomena. we have 90◦ < α < 112. we convert 28 it all together.371◦ = 68. we have 37◦ 28 17 1◦ 60 697 = 7 ◦ 15 and 17 1◦ 3600 = 17 ◦ 3600 . As its name suggests. Angle α Angle β 4. Up to this point. To find a supplementary angle for α.471◦ + 17 ◦ 3600 3. as shown below. we seek an angle θ so that α + θ = 180◦ . 8 This is the exact same kind of ‘borrowing’ you used to do in Elementary School when trying to find 300 − 125. and lastly.10. Ultimately. it is base sixty. and once more. we want to use the arsenal of Algebra which we have stockpiled in Chapters 1 through 9 to not only solve geometric problems involving angles. we introduce the concept of an oriented angle.3.1 Angles and their Measure 2. and then borrowing 1 = 60 from the minutes place. here. it should.75◦ < β < 45◦ .’ this is also a real math term.629◦ . We first rewrite 90◦ = 90◦ 0 0 = 89◦ 60 0 = 89◦ 59 60 . you were working in a base ten system. To that end. Putting = 37◦ + 28 + 17 = 37◦ + = ≈ 7 ◦ 15 134897 ◦ 3600 37. we find 0◦ < β < 90◦ .5◦ < β < 45◦ .5◦ . A first step in this direction is to extend our notion of ‘angle’ from merely measuring an extent of rotation to quantities which can be associated with real numbers. γ = 90◦ − 37◦ 28 17 = 89◦ 59 60 − 37◦ 28 17 = 52◦ 31 43 . ±1. If the terminal side of an angle lies on one of the coordinate axes. ±2. Angles in standard position are classified according to where their terminal side lies. Two angles in standard position are called coterminal if they share the same terminal side. β = α − 360◦ . an angle in standard position whose terminal side lies in Quadrant I is called a ‘Quadrant I angle’. . 11 It is worth noting that all of the pathologies of Analytic Trigonometry result from this innocuous fact. For instance.. When the rotation is counter-clockwise9 from initial side to terminal side. we also extend our allowable rotations to include angles which encompass more than one revolution. α = 120◦ and β = −240◦ are two coterminal Quadrant II angles drawn in standard position. we shall often overlay an angle diagram on the coordinate plane. Note that α = β + 360◦ . the direction of the rotation is important. we say that the angle is negative. . For example. −45◦ At this point.12 ‘widdershins’ Note that by being in standard position they automatically share the same initial side which is the positive x-axis. to sketch an angle with measure 450◦ we start with an initial side.10 In the figure below.11 More precisely. 450◦ To further connect angles with the Algebra which has come before.698 Foundations of Trigonometry angle. 10 9 . rotate counter-clockwise one complete revolution (to take care of the ‘first’ 360◦ ) then continue with an additional 90◦ counter-clockwise rotation. Initial Side Si m in al de m er T T er al in d Si e Initial Side A positive angle. 45◦ A negative angle. we say that the angle is positive. We imagine the angle being swept out starting from an initial side and ending at a terminal side. as seen below. 12 Recall that this means k = 0. when the rotation is clockwise. An angle is said to be in standard position if its vertex is the origin and its initial side coincides with the positive x-axis. . as shown below. or equivalently. We leave it as an exercise to the reader to verify that coterminal angles always differ by a multiple of 360◦ . then β = α + 360◦ · k where k is an integer. it is called a quadrantal angle. if α and β are coterminal angles. Substituting k = −1 gives θ = 60◦ −360◦ = −300◦ . 1. we draw an angle with its initial side on the positive x-axis and rotate 60◦ counter-clockwise 360◦ = 1 of a revolution. Graph each of the (oriented) angles below in standard position and classify them according to where their terminal side lies. −585◦ and 495◦ are all coterminal with −225◦ . When k = 1. we proceed as before and compute θ = −225◦ + 360◦ · k for integer values of k. We see that β is a Quadrant II angle. 5 2. α = 120◦ and β = −240◦ . we start at the positive x-axis and rotate clockwise 225◦ = 8 of 360 a revolution. we get θ = 60◦ +360◦ = 420◦ . We see that α is a Quadrant I angle. y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 α = 60◦ 3 4 x 4 3 2 1 −4 −3 −2 −1 −1 β = −225◦ −2 −3 −4 1 2 3 4 x y ◦ 2. Find three coterminal angles. in standard position. Example 10. α = 60◦ Solution. To find angles 6 which are coterminal. We find 135◦ . Since β = −225◦ is negative. To find coterminal angles. for some integer k. we look for angles θ of the form θ = α + 360◦ · k. To graph α = 60◦ . φ = −750◦ α = 60◦ in standard position. 1.1 Angles and their Measure y 4 3 2 1 −4 −3 −2 −1 −1 β = −240◦ −2 −3 −4 1 2 3 4 x α = 120◦ 699 Two coterminal angles. . at least one of which is positive and one of which is negative. γ = 540◦ 4.2. β = −225◦ 3. we get θ = 60◦ + 720◦ = 780◦ . if we let k = 2. Finally.10.1. β = −225◦ in standard position. Since the diameter of a circle is twice its radius. we rotate counter-clockwise from the positive x-axis. any given angle has infinitely many coterminal angles. The Greek letter φ is pronounced ‘fee’ or ‘fie’ and since φ is negative. Since γ = 540◦ is positive. with just 30◦ or 12 of a revolution to go. or 1 of a revolution remaining. it doesn’t matter which circle is selected. given a circle of circumference C and diameter d.that is. Working through the arithmetic. As the reader is probably aware. We are now just one step away from completely marrying angles with the real numbers and the rest of Algebra. where k is an integer. the number π is a mathematical constant . π= C d While Definition 10. we begin our rotation 1 clockwise from the positive x-axis. with 180◦ . Note that since there are infinitely many integers.1. In symbols.1. Definition 10.2 to see this.700 Foundations of Trigonometry 3. namely: 2π = r . 4. The real number π is defined to be the ratio of a circle’s circumference to its diameter. −30◦ and 330◦ . we recall this definition from Geometry. it is far from obvious and leads to a counterintuitive scenario which is explored in the Exercises. All angles coterminal with γ are of the form θ = 540◦ + 360◦ · k. To that end. we compute θ = −750◦ + 360◦ · k for a few integers k and obtain −390◦ . and the reader is encouraged to plot the few sets of coterminal angles found in Example 10. y 4 γ = 540◦ 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x y φ = −750◦ γ = 540◦ in standard position. we find three such angles: 180◦ . Two full revolutions account for 720◦ . We find that φ is a Quadrant IV angle. φ = −750◦ in standard position. One full revolution accounts for 360◦ . To find coterminal angles. the authors would be remiss if we didn’t mention that buried in this definition is actually a theorem. −180◦ and 900◦ .1 to get a formula more useful for our purposes. Since the terminal 2 side of γ lies on the negative x-axis. γ is a quadrantal angle. While this is indeed true.1 is quite possibly the ‘standard’ definition of π. the ratio of its circumference to its diameter will have the same value as any other circle. we can quickly rearrange the C equation in Definition 10. hence s = r. so instead of comparing the entire circumference C to the radius. When the radian measure is 2. From this we can find the radian measure of other central angles using proportions. when the radian measure is 3. it stands to reason that the ratio should also be a r constant among all circles.10. Thus the radian measure of an angle θ tells us how many ‘radius lengths’ we need to sweep out along the circle to subtend the angle θ. in this case the constant is 2π. Using proportionality arguments. Let θ be the central angle subtended by this arc. s = 3r. r . we note that an angle with radian measure 1 means the corresponding arc length s equals the radius of the circle r. one revolution has radian measure 2πr = 2π. that is. and so forth. an angle whose vertex is the center of the circle and whose determining rays pass through the endpoints of s the arc. r r r r α r r r r β r α has radian measure 1 β has radian measure 4 Since one revolution sweeps out the entire circumference 2πr.1 Angles and their Measure 701 This tells us that for any circle. Suppose now we take a portion of the circle. and it is this ratio which defines the radian measure of an angle. the ratio of its circumference to its radius is also always constant. we compare some arc measuring s units in length to the radius. r To get a better feel for radian measure. as depicted below. we have s = 2r. s θ r r The radian measure of θ is s . α = π 6 2. we say one revolution measures ‘2π radians. for some integer k. Example 10. The angle α = π is positive. φ = − 5π 2 Solution. so we draw an angle with its initial side on the positive x-axis and 6 1 rotate counter-clockwise (π/6) = 12 of a revolution. by definition. 6 6 6 6 6 6 2. 1. We obtain 2π . Graph each of the (oriented) angles below in standard position and classify them according to where their terminal side lies. Substituting 6 6 6 6 k = −1 gives θ = π − 12π = − 11π and when we let k = 2. We will justify this shortly. two angles 2 α and β are coterminal if and only if β = α + 2πk for some integer k. a 2 quarter revolution has radian measure 1 (2π) = π . half of a revolution has radian measure 1 (2π) = π. but we use the word ‘radians’ to denote these dimensionless units as needed.702 Foundations of Trigonometry just like we did with degrees. we leave it to the reader to show that when using radian measure. endows the subtended arcs with an orientation as well. γ = 9π 4 4. We find β to be a Quadrant II angle. we get θ = π + 24π = 25π . thus when k = 1. 3 3 3 3 − 10π and 8π as coterminal angles. We address this in short order. Find three coterminal angles.3. As with degree measure. Note that. and compute θ = − 4π + 6π · k for integer values of k. we do not use any symbols to denote radian measure. To make the arithmetic a bit easier. we mean θ is an angle which measures π radians. 3 3 13 14 The authors are well aware that we are now identifying radians with real numbers.’ half of a revolution measures ‘π radians. just as we did with degrees beforehand. the distinction between the angle itself and its measure is often blurred in practice. we note that 2π = 12π . at least one of which is positive and one of which is negative. For this reason.14 Much like before. 1. To find coterminal angles. we proceed as before using 2π = 6π .1. so when we write ‘θ = π ’. two positive angles α and β are supplementary if α + β = π and complementary if α + β = π . we start at the positive x-axis and rotate clockwise (4π/3) = 2 of 3 2π 3 a revolution. Finally. This.’ and so forth. Coterminal 2π angles θ are of the form θ = α + 2π · k. we get θ = π + 12π = 13π . Thus α is a Quadrant I angle. so that a positive measure indicates counter-clockwise rotation and a negative measure indicates clockwise rotation.13 We 2 2 extend radian measure to oriented angles. in turn. the radian 4 2 measure of an angle is a length divided by another length so that these measurements are actually dimensionless and are considered ‘pure’ numbers. For instance. β = − 4π 3 3. and so forth. For instance. Since β = − 4π is negative. . however.1. Since the 2 terminal side of φ lies on the negative y-axis. 2 2 2 2 2 y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 9π 4 y 4 3 2 1 x −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x φ = − 5π 2 γ= γ= 9π 4 in standard position. it is best that you learn to ‘think in radians’ as well as you can ‘think in degrees’. φ = − 5π in standard position. 2 2 1 after one full revolution clockwise. be .2. 3 3. φ is a quadrantal angle. To graph φ = − 5π . One full 4 revolution accounts for 2π = 8π of the radian measure with π or 1 of a revolution remaining.1. The authors would. While converting back and forth from degrees and radians is certainly a good skill to have. 4 4 8 We have γ as a Quadrant I angle.1 Angles and their Measure y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 α= 3 π 6 703 y 4 3 2 1 x −4 −3 −2 −1 −1 β = − 4π 3 −2 −3 −4 1 2 3 4 x 4 α= π 6 in standard position. we find: π . All angles coterminal with γ are of the form θ = 9π + 8π · k. Since γ = 9π is positive. β = − 4π in standard position. we begin our rotation clockwise from the positive x-axis.3 by first converting them to degree measure and following the procedure set forth in Example 10. To find coterminal angles. we have π or 4 of a revolution remaining. 4 4 4 4. Working through the arithmetic.10. As 2π = 4π . we compute θ = − 5π + 4π · k for a few integers k and obtain − π . 4 4 where k is an integer. − 7π and 17π . we rotate counter-clockwise from the positive x-axis. 2 It is worth mentioning that we could have plotted the angles in Example 10. 3π and 7π . t] around the Unit Circle in a counter-clockwise fashion. Equation 10. given a real number t > 0. or its reduced equivalent. 15 Note that the negative sign indicates clockwise rotation in both systems. the s s radian measure of θ is = = s so that. we multiply by the ratio π 180 .3 and Equation 10. once again blurring the distinction between an angle r 1 and its measure.Radian Conversion: • To convert degree measure to radian measure. multiply by • To convert radian measure to degree measure. 0] clockwise around the Unit Circle.704 Foundations of Trigonometry derelict in our duties if we ignored the basic conversion between these systems altogether.1. . If t = 0. If t < 0. multiply by π radians 180◦ 180◦ π radians In light of Example 10. − 5π radians is equal to − 5π radians π radians = −150◦ . π radians . For example. Viewing the vertical line x = 1 as another real number line demarcated like the y-axis. and the fact that the Unit Circle has radius 1. we can radians use the proportion 2π 360◦ . as the conversion factor between 180◦ the two systems. In this way. 0) on the x-axis which corresponds to an angle with radian measure 0. In order to identify real numbers with oriented angles. x2 +y 2 = 1. Since we have defined clockwise rotation as having negative radian measure. we identify each real number t with the corresponding angle with radian measure t. By definition. the angle determined by this arc has radian measure equal to t. the reader may well wonder what the allure of radian measure is. to convert 60◦ to radians we find 60◦ π radians = π radians.1. or 180◦ 3 ◦ π simply 3 . and so it is carried along accordingly. admittedly.15 Of particular interest is the 6 6 ◦ fact that an angle which measures 1 in radian measure is equal to 180 ≈ 57. The resulting arc has a length of t units and therefore the corresponding angle has radian measure equal to t.2958◦ . To convert from radian measure back to degrees. For radian 180◦ example. The numbers involved are. as drawn below. 0). we are at the point (1.1. we ‘wrap’ the (vertical) interval [0. Consider the Unit Circle. Degree . Since one revolution counter-clockwise measures 360◦ and the same angle measures 2π radians. π We summarize these conversions below. we have θ = s. we wrap the interval [t. much more complicated than degree measure. The answer lies in how easily angles in radian measure can be identified with real numbers. we make good use of this fact by essentially ‘wrapping’ the real number line around the Unit Circle and associating to each real number t an oriented arc on the Unit Circle with initial point (1. the angle θ in standard position and the corresponding arc measuring s units in length. 0) and proceeds clockwise for one full revolution. Example 10.1. Identifying t < 0 with an angle. To more accurately place the endpoint. The arc associated with t = 3π is the arc on the Unit Circle which subtends the angle 3π in 4 4 radian measure. we find that rotating 2 radians clockwise from 2 the point (1. t = −2 is negative. y 1 y 1 1 x 1 x t= 3π 4 t = −2π 3.4. Sketch the oriented arc on the Unit Circle corresponding to each of the following real numbers. . t = 117 Solution. we proceed as we did in Example 10. 1.1 Angles and their Measure y 1 y 1 y 1 705 s θ 1 x t t 1 x 1 x t t On the Unit Circle. Since 3π is 3 of a revolution. successively halving the angle measure until we find 5π ≈ 1.14 and π ≈ 1. Since π ≈ 3. 0) and proceed clockwise around the unit circle. 0) lands us in Quadrant III. θ = s. Like t = −2π. Identifying t > 0 with an angle. t = −2 4.1. 0) 4 8 proceeds counter-clockwise up to midway through Quadrant II. t = −2π 3. and t = −2π is negative. t = 3π 4 2.57. we graph the arc which begins at (1.10. 2.96 8 which tells us our arc extends just a bit beyond the quarter mark into Quadrant III. 1. we have an arc which begins at the point (1.1. so we begin our arc at (1. Since one revolution is 2π radians. 62. the arc corresponding to t = 117 begins at (1. Q s θ r P Here s represents a displacement so that s > 0 means the object is traveling in a counter-clockwise direction and s < 0 indicates movement in a clockwise direction.1 Applications of Radian Measure: Circular Motion Now that we have paired angles with real numbers via radian measure.3 in Section 2.16 As a result. namely θ = . or just shy of 5 of a revolution to spare. We approximate 117 as 18. we wrap around the Unit Circle several times before finally reaching our endpoint. Suppose an object is moving as pictured below along a circular path of radius r from the point P to the point Q in an amount of time t. . Since 117 is positive.1 for a review of this concept. y 1 y 1 1 x 1 x t = −2 t = 117 10. denoted v and read as ‘v-bar’.706 Foundations of Trigonometry 4. we 16 See Definition 2. As 117 is much greater than 2π. is defined as the average rate of change of the position of the object with respect to time. still holds since a negative value r of s incurred from a clockwise displacement matches the negative we assign to θ for a clockwise rotation. Note that with this convention s the formula we used to define radian measure. In other 8 words. a whole world of applications awaits us. the terminal side of the angle which measures 117 radians in standard position is just short of being midway through Quadrant III.62 which tells us we complete 2π 18 revolutions counter-clockwise with 0. In Physics. Our first excursion into this realm comes by way of circular motion. 0) and proceeds counterclockwise. the average velocity of the object.1. Putting together the ideas of the previous paragraph. so ω = 2π radians = 12 hours . are the same as their instantaneous counterparts. and time the units of ω are radians . . The contribution of direction in the quantity v is either to make it positive (in the case of counter-clockwise motion) or negative (in the case of clockwise motion). we get the following. then the velocity of the object on that line segment would be the same as the velocity on the circle. 19 We will discuss how we arrived at this approximation in Example 10. the units of r are length only. Equation 10. Solution. and one revolution is 2π radians. The quantity v has units of length and conveys two ideas: the direction time time t in which the object is moving and how fast the position of the object is changing. the quantity v is often called the linear velocity of the object in order to distinguish it from the angular velocity.it is the speed of the object. See the discussion on Page 161 for more details on the idea of an ‘instantaneous’ rate of change. For this reason. Lakeland Community College is at 41.6.628◦ north latitude. the (linear) velocity of the object is given by v = rω. We need to talk about units here. any point on the Earth can be thought of as an object traveling on a circle which completes one revolution in (approximately) 24 hours. Find the linear velocity. v is simply called the ‘velocity’ of the object and is the instantaneous rate of change of the position of the object with respect to time. so that the quantity |v| quantifies how fast the object s is moving . whereas time time the right hand side has units length · radians = length·radians .1 Angles and their Measure 707 s have v = displacement = . of Lakeland Community College as the world turns.2. In this case. Velocity for Circular Motion: For an object moving on a circular path of radius r with constant angular velocity ω. 24 hours 17 18 You guessed it. Example 10. To use the formula v = rω. The path traced out by the point during this 24 hour period is the Latitude of that point. respectively. . ω is called the ‘angular velocity’ and is the instantaneous rate of change of the angle with respect to time. Measuring θ in radians we have θ = thus s = rθ and r v= s rθ θ = =r· t t t θ The quantity is called the average angular velocity of the object. using Calculus . we first need to compute the angular velocity ω. v and ω.10. If ω is constant throughout the duration of the motion. The quantity ω is the average rate of change of the angle θ with respect to time and thus has units radians . It is denoted by ω and is t read ‘omega-bar’. then it can be time shown17 that the average velocities involved. Thus the left hand side of the equation v = rω has units length . namely v and ω.2. We are time time long overdue for an example. The supposed contradiction in units is time time resolved by remembering that radians are a dimensionless quantity and angles in radian measure are identified with real numbers so that the units length·radians reduce to the units length . Assuming that the surface of the Earth is a sphere. in miles per hour. The units of v are length .5.18 Similarly. . The earth π makes one revolution in 24 hours. If the path of the object were ‘uncurled’ from a circle to form a line segment.1. ω. and it can be shown19 that the radius of the earth at this Latitude is approximately 2960 miles. We will have more to say about frequencies and periods in Section 11. the period of the motion is 24 hours.1. The ordinary frequency is a measure of how often an object makes a complete cycle of the motion. and is the ‘magnification factor’ which relates ω and v.708 Foundations of Trigonometry where.1. points which are farther from the center of rotation need to travel faster to maintain the same angular frequency since they have farther to travel to make one revolution in one period’s time.1. the linear velocity is v = 2960 miles · π miles ≈ 775 12 hours hour It is worth noting that the quantity 1 revolution in Example 10. In the scenario of Example 10. (For simplicity’s sake.5 is called the ordinary frequency 24 hours of the motion and is usually denoted by the variable f . or one day. it is called the angular frequency of the motion.5. While we have exhaustively discussed velocities associated with circular motion. what is the position of the object at time t? The answer to this question is the very heart of Trigonometry and is answered in the next section. 1 the quantity T = is called the period of the motion and is the amount of time it takes for the f object to complete one cycle of the motion. we are using the fact that radians are real numbers and are dimensionless. r. The fact that ω = 2πf suggests that ω is also a frequency. once again. On a related note.) Hence. we have yet to discuss a more natural question: if an object is moving on a circular path of radius r with a fixed angular velocity (frequency) ω. The distance of the object to the center of rotation is the radius of the circle. . That is. if ω is fixed. The concepts of frequency and period help frame the equation v = rω in a new light. we are also assuming that we are viewing the rotation of the earth as counterclockwise so ω > 0. Indeed. 7π 2 5π 3 21. 37. −270◦ 17. 63. 2π 3 39. −225◦ In Exercises 37 . 11π 3 12. − 15π 4 In Exercises 29 . 3π 4 π 2 10. 44. one of which is positive and the other negative.10. 330◦ 13. − 13π 6 25. − 42. convert the angles into decimal degrees. 240◦ 34. 0◦ 33.75◦ 2. −32◦ 10 12 7. −270◦ 36.1 Angles and their Measure 709 10. Round each of your answers to three decimal places. 24. 45◦ 32.4. 502◦ 35 8. 29. 5π 4 π 4 18. − 7π 6 π 4 23.2 Exercises In Exercises 1 . −315◦ 30. convert the angle from radian measure into degree measure. − 27. 9. −135◦ 14. − 19. 20. 3π 28. −2π 26. 125◦ 50 6. 5.8. 120◦ 15. 405◦ 16. − . 5π 6 π 3 11. π 41. giving the exact value in terms of π. convert the angle from degree measure into radian measure. 7π 6 π 6 40.28.325◦ 3.999◦ In Exercises 5 . − 22. 1.1. 150◦ 31. −317.36. graph the oriented angle in standard position.06◦ 4. 135◦ 35. Classify each angle according to where its terminal side lies and then give two coterminal angles. π 3 38. 237◦ 58 43 In Exercises 9 . Round each of your answers to the nearest second.44. convert the angles into the DMS system. 179. 200. 11π 6 π 2 5π 3 43. Prove that the area of the shaded sector is A = 2 r2 θ.5 inches and spins at a rate of 7200 RPM (revolutions per minute). t = −π 47. t = 5π 6 46. 51. in miles per hour. 53. was the rock traveling when it came out of the tread? (The tire has a diameter of 23 inches. how fast are they traveling in miles per hour? 56. (Remember this from Exercise 17 in Section 7. How fast is the edge of the yo-yo spinning in miles per hour? Round your answer to two decimal places. 52.) r θ r s 20 Source: Cedar Point’s webpage.2?) It completes two revolutions in 2 minutes and 7 seconds. 54. the rock came loose and hit the inside of the wheel well of the car. A rock got stuck in the tread of my tire and when I was driving 70 miles per hour. . t = 6 48.20 Assuming the riders are at the edge of the circle. If the yo-yo string is 28 inches long and the yo-yo takes 3 seconds to complete one revolution of the circle. Find the linear speed of a point on the edge of the disk in miles per hour.25 inches in diameter spins at a rate of 4500 revolutions per minute. How many revolutions per minute would the yo-yo in exercise 50 have to complete if the edge of the yo-yo is to be spinning at a rate of 42 miles per hour? Round your answer to two decimal places. compute the speed of the yo-yo in miles per hour. 45. Consider the circle of radius r pictured below with central angle θ.710 Foundations of Trigonometry In Exercises 45 . In the yo-yo trick ‘Around the World. (Hint: Use the proportion A area of the circle = s circumference of the circle . sketch the oriented arc on the Unit Circle which corresponds to the given real number.) 55.49. t = 12 50. t = −2 49.’ the performer throws the yo-yo so it sweeps out a vertical circle whose radius is the yo-yo string. measured in radians. A yo-yo which is 2. A computer hard drive contains a circular disk with diameter 2. Round your answer to two decimal places. and 1 subtended arc of length s. The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height is 136 feet. How fast. 3 62. θ = 1◦ . (You do NOT need to know the radius of the Earth to show this. θ = 240◦ . r = 12 6 58. . r = 9. Imagine a rope tied around the Earth at the equator. r = 1 61.10. θ = 330◦ . r = 100 4 59.) 64. θ = π. With the help of your classmates. r = 5 63. 57. use the result of Exercise 56 to compute the areas of the circular sectors with the given central angles and radii. Show that you need to add only 2π feet of length to the rope in order to lift it one foot above the ground around the entire equator. look for a proof that π is indeed a constant.62.1 Angles and their Measure 711 In Exercises 57 . θ = π . r = 117 60. θ = 5π . −317◦ 3 36 7. 330◦ is a Quadrant IV angle coterminal with 690◦ and −30◦ y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 10. 237.979◦ 1. 200◦ 19 30 6. −270◦ lies on the positive y-axis coterminal with 90◦ and −630◦ y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 14.1.712 Foundations of Trigonometry 10. 120◦ is a Quadrant II angle coterminal with 480◦ and −240◦ y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 12. 405◦ is a Quadrant I angle coterminal with 45◦ and −315◦ y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 13. −135◦ is a Quadrant III angle coterminal with 225◦ and −495◦ y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 11.833◦ 9.17◦ 3. 125. 63◦ 45 5. 502.583◦ 4.3 Answers 2. −32. 5π is a Quadrant II angle 6 17π 7π coterminal with and − 6 6 y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x . 179◦ 59 56 8. − 11π is a Quadrant I angle 3 5π π coterminal with and − 3 3 y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 713 16. 5π is a Quadrant III angle 4 3π 13π and − coterminal with 4 4 y 17. 7π lies on the negative y-axis 2 3π π coterminal with and − 2 2 y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 20. − π is a Quadrant IV angle 3 7π 5π and − coterminal with 3 3 y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 19. π is a Quadrant I angle 4 9π 7π coterminal with and − 4 4 y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x .1 Angles and their Measure 15. 3π is a Quadrant II angle 4 5π 11π and − coterminal with 4 4 y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 18.10. 714 21. 7π is a Quadrant III angle 6 5π 19π and − coterminal with 6 6 y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 23. − π lies on the negative y-axis 2 5π 3π and − coterminal with 2 2 y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x Foundations of Trigonometry 22. − π is a Quadrant IV angle 4 7π 9π coterminal with and − 4 4 y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x . − 5π is a Quadrant I angle 3 11π π coterminal with and − 3 3 y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 24. 3π lies on the negative x-axis coterminal with π and −π y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 25. −2π lies on the positive x-axis coterminal with 2π and −4π y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 26. 10. 4π 3 5π 34. t = −2 y 1 1 x 1 x . 6 3π 4 π 35. 0 33. 330◦ 44. 180◦ 41. 4 31. t = −π 3π 2 5π 36. 15π is a Quadrant IV angle 4 π 7π and − coterminal with 4 4 y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 715 28. −30◦ 46. − 7π 4 30. 300◦ y 1 1 x 1 x 47. − 13π is a Quadrant IV angle 6 π 11π and − coterminal with 6 6 y 29. − 40. 210◦ 43. 90◦ 37. t = 6 y 1 48. t = 5π 6 y 1 38. 60◦ 45. 39. −120◦ 42.1 Angles and their Measure 27. − 4 32. 55 miles per hour 55. 70 miles per hour 57.52 revolutions per minute 53.2825π ≈ 249.07 square units 61. About 30. π square units 2 62.025π ≈ 119.32 miles per hour 58. About 53.33 miles per hour 54. 38.12 miles per hour 52. 50π square units 3 51.716 49. 12π square units 59. t = 12 (between 1 and 2 revolutions) y 1 Foundations of Trigonometry 1 x 50. About 3. 6250π square units 60. About 6274. 79.46 square units . About 4. 1. θ = π 6 5. The x-coordinate of P is called the cosine of θ.1.2. there is only one associated value of cos(θ) and only one associated value of sin(θ). The angle θ = −π represents one half of a clockwise revolution so its terminal side lies on the negative x-axis. written cos(θ). θ = 60◦ . Find the cosine and sine of the following angles.2 The Unit Circle: Cosine and Sine In Section 10. 2 2 2.1 The reader is encouraged to verify that these rules used to match an angle with its cosine and sine do.4.1. we plot the angle θ = 270◦ in standard position and find the point on the terminal side of θ which lies on the Unit Circle. 1 2. The etymology of the name ‘sine’ is quite colorful.10. written sin(θ). the point we seek is (0. That is. To that end. while the y-coordinate of P is called the sine of θ. −1) so that cos 3π = 0 and sin 3π = −1. for each angle θ. we are assigning a position on the Unit Circle to the angle θ. By associating the point P with the angle θ. the ‘co’ in ‘cosine’ is explained in Section 10.1. The point on the Unit Circle that lies on the negative x-axis is (−1. Since 270◦ represents 3 of a 4 counter-clockwise revolution. consider an angle θ in standard position and let P denote the point where the terminal side of θ intersects the Unit Circle. Hence. sin(θ)) θ 1 x θ 1 x Example 10. satisfy the definition of a function. 1. we introduced circular motion and derived a formula which describes the linear velocity of an object moving on a circular path at a constant angular velocity. and the interested reader is invited to research it. the terminal side of θ lies along the negative y-axis.2 The Unit Circle: Cosine and Sine 717 10. To find cos (270◦ ) and sin (270◦ ). θ = −π 3. θ = 270◦ Solution. in fact. y y 1 1 P (cos(θ). One of the goals of this section is describe the position of such an object. θ = 45◦ 4. 0) which means cos(−π) = −1 and sin(−π) = 0. we obtain a 45◦ − 45◦ − 90◦ right triangle whose legs have lengths x and y units. When we sketch θ = 45◦ in standard position.2 we get y = x. If we drop a perpendicular line segment from P to the x-axis. we see that its terminal does not lie along any of the coordinate axes which makes our job of finding the cosine and sine values a bit more difficult.718 y Foundations of Trigonometry y 1 θ = 270◦ 1 P (−1. y 1 P (x. x > 0. By definition. x = cos (45◦ ) and y = sin (45◦ ). y) lies in the first quadrant. y) denote the point on the terminal side of θ which lies on the Unit Circle. From Geometry. y) lies on the Unit Circle. y) 45◦ y θ = 45◦ x 2 Can you show this? . 0) 1 x 1 θ = −π x P (0. Let P (x. or x = ± Since P (x. −1) Finding cos (270◦ ) and sin (270◦ ) Finding cos (−π) and sin (−π) 3. so x = cos (45◦ ) = y= sin (45◦ ) √ √ 2 2 1 2 √ = ± 2 2 . y) θ = 45◦ x 1 P (x. Since P (x. and with y = x we have = 2 2 . we have x2 + y 2 = 1. Substituting y = x into this equation yields 2x2 = 1. y) P (x. we substitute y = 1 into x2 + y 2 = 1 to get x2 = 3 . so we proceed 6 using a triangle approach. y) 60◦ x 1 y θ= π 6 = 30◦ x 5. 2 6 2 Since P (x. we get a 30◦ − 60◦ − 90◦√right triangle and. we drop a perpendicular line segment from P to the x-axis to form a 30◦ − 60◦ − 90◦ right triangle. y) denote the point on the terminal side of θ which lies on the Unit Circle. y) 30◦ θ= 60◦ x 1 y θ = 60◦ x 3 Again. After a bit of Geometry3 we find y = 1 so sin π = 1 .10. y) lies on the Unit Circle. Here. As before. x > 0 so x = cos y 1 π 6 √ = 3 2 . Plotting θ = 60◦ in standard position. 2 y 1 P (x.2 The Unit Circle: Cosine and Sine 719 4. P (x. y) θ= π 6 P (x. the terminal side of θ = π does not lie on any of the coordinate axes. find x = cos (60◦ ) = 1 and y = sin (60◦ ) = 23 . Once again. can you show this? . after the usual computations. or 2 4 √ x=± 3 2 . Letting P (x. we find it is not a quadrantal angle and set about using a triangle approach. find cos(θ). we made good use of the fact that the point P (x. Hence. sin(θ)) lies on the Unit Circle. See Section 10. cos2 (θ) + sin2 (θ) = 1. find sin(θ). we know where the terminal side of θ lies when in standard position. An unfortunate4 convention. See Sections 1. If θ is a Quadrant II angle with sin(θ) = 5 . Hence both its sine and cosine 3. The moniker ‘Pythagorean’ brings to mind the Pythagorean Theorem. Rewriting the identity using this convention results in the following theorem. This means 2 This is unfortunate from a ‘function notation’ perspective. regardless of the angle θ. for instance.2. If π < θ < 3π 2 √ with cos(θ) = − 5 5 . √ 2 1 gives sin(θ) = ± √5 = ± 2 5 5 .1. Another tool which helps immensely in determining cosines and sines of angles is the symmetry inherent in the Unit Circle. y) denote the point on the terminal side of θ which lies on the Unit Circle. when plotted in standard position. cos(θ) = − 4 . its terminal side. lies in Quadrant II. find the indicated value. which is without a doubt one of the most important results in Trigonometry. Since the x-coordinates are negative in Quadrant II. it was quite easy to find the cosine and sine of the quadrantal angles. the equation in Theorem 10. we wish to know the cosine and sine of θ = 5π . 1. Solution. the task was much more involved. we know θ is a Quadrant 2 √ conclude sin(θ) = − 2 5 5 . is to write (cos(θ))2 as cos2 (θ) and (sin(θ))2 as sin2 (θ). Suppose. 5 2. Substituting cos(θ) = − are given that π < θ are negative and we 5 2 2 5 into cos (θ) + sin (θ) = < 3π .1. but for non-quadrantal angles.2 for details.5 The word ‘Identity’ reminds us that. we get (cos(θ))2 + (sin(θ))2 = 1. from which both the Distance Formula and the equation for a circle are ultimately derived. we find cos(θ) = ± 5 . cos2 (θ) + sin2 (θ) = 1.1.2.1 is always true. Since we √ III angle. then we can remove the ambiguity of the (±) and completely determine the missing value as the next example illustrates. 6 We plot θ in standard position below and. Using the given information about θ. as usual. which the authors are compelled to perpetuate. If. cos(θ) is too. find cos(θ).6. we find cos(θ) = 0. y) = (cos(θ). Solving. In Example 10. In these latter cases. we determined that cos 4 5 π 6 √ = 3 2 and sin π 6 = 1 . x2 + y 2 = 1. Theorem 10.1 and 7.2. The Pythagorean Identity: For any angle θ. Note that the terminal side of θ lies π radians short of one 6 half revolution. When we substitute sin(θ) = 3 into The Pythagorean Identity. If sin(θ) = 1.2. Example 10. 2. Since θ is a Quadrant II angle.720 Foundations of Trigonometry In Example 10.1 can be used to determine the other. let P (x. up to a (±) sign. 3. 3 1. If one of cos(θ) or sin(θ) is known. When we substitute sin(θ) = 1 into cos2 (θ) + sin2 (θ) = 1. If we substitute x = cos(θ) and y = sin(θ) into x2 + y 2 = 1. we 5 4 9 obtain cos2 (θ) + 25 = 1. . in addition. Theorem 10. 10. the reference angle for θ (usually denoted α) is the acute angle made between the terminal side of θ and the x-axis. 1 . y y 1 1 P =Q α 1 x P α Q α 1 x Reference angle α for a Quadrant I angle Reference angle α for a Quadrant II angle .2 2 x 1 1 x In the above scenario. α as seen below. there is a point Q symmetric with P which determines θ’s reference angle. If we let P denote the point (cos(θ). the angle π is called the reference angle for the angle 5π . In general. y) π 6 θ= 5π 6 √ P − 3 1 . If θ is a Quadrant I or IV angle. y-axis and origin. 6 6 2 y 1 y 1 P (x. α is the angle between the terminal side of θ and the positive x-axis.2 The Unit Circle: Cosine and Sine √ 721 that the point on the terminal side of the angle π . then P lies on the Unit Circle. sin(θ)). it is clear that the point P (x. y) we seek can be obtained by reflecting that √ point about the y-axis. cos 5π = − 23 and sin 5π = 1 . Hence. if θ is a Quadrant II or III angle. when plotted in standard position. Since the Unit Circle possesses symmetry with respect to the x-axis. α is the angle between the terminal side of θ and the negative x-axis. for 6 6 a non-quadrantal angle θ. regardless of where the terminal side of θ lies. 6 2 From the figure below.2 2 π 6 θ= 5π 6 π 6 √ 3 1 . is 23 . θ = − 5π 4 4. Reference Angle Theorem. Cosine and Sine Values of Common Angles θ(degrees) θ(radians) 0◦ 30◦ 45◦ 60◦ 90◦ 0 π 6 π 4 π 3 π 2 cos(θ) sin(θ) √ 1 0 1 2 √ 2 2 √ 3 2 3 2 √ 2 2 1 2 0 1 Example 10. θ = 11π 6 3.2. We begin by plotting θ = 225◦ in standard position and find its terminal side overshoots the negative x-axis to land in Quadrant III. we summarize the values which we consider essential and must be memorized. Suppose α is the reference angle for θ. both cos(θ) < 0 and 2. it pays to know the cosine and sine values for certain common angles. 1. 1. In light of Theorem 10. Find the cosine and sine of the following angles.3. Theorem 10. where the choice of the (±) depends on the quadrant in which the terminal side of θ lies.2. Since θ is a Quadrant III angle.2. θ = 225◦ Solution. we obtain θ’s reference angle α by subtracting: α = θ − 180◦ = 225◦ − 180◦ = 45◦ . Then cos(θ) = ± cos(α) and sin(θ) = ± sin(α). θ = 7π 3 . In the table below. Hence.722 y Foundations of Trigonometry y 1 1 Q α α P 1 x Q α α P 1 x Reference angle α for a Quadrant III angle Reference angle α for a Quadrant IV angle We have just outlined the proof of the following theorem. 6 6 6 6 2 y 1 y 1 θ = 225◦ 45◦ 1 x θ= 11π 6 π 6 1 x Finding cos (225◦ ) and sin (225◦ ) Finding cos 11π 6 and sin 11π 6 3. Since the angle θ = 7π measures more than 2π = 6π . lies in Quadrant IV. The terminal side of θ = 11π . To plot θ = − 5π .10. we rotate clockwise an angle of 5π from the positive x-axis. lies in Quadrant II making an angle of α = 5π − π = π radians with 4 4 respect to the negative x-axis. The terminal 4 4 side of θ. 4 4 4 4 4. the Reference Angle Theorem √ √ gives: cos − 5π = − cos π = − 22 and sin − 5π = sin π = 22 . y 1 π 4 θ= 7π 3 π 3 1 θ = − 5π 4 x 1 x Finding cos − 5π and sin − 5π 4 4 Finding cos 7π 3 and sin 7π 3 . we find the terminal side of θ by rotating 3 3 one full revolution followed by an additional α = 7π − 2π = π radians.2 The Unit Circle: Cosine and Sine sin(θ) < 0. Since θ and α are 3 3 coterminal. The Reference Angle Theorem yields: cos (225◦ ) = − cos (45◦ ) = − sin (225◦ ) = − sin (45◦ ) √ √ 2 2 723 and =− 2 2 . just shy 6 of the positive x-axis. when plotted in standard position. cos(θ) > 0 and sin(θ) < 0. therefore. 2. 6 6 Since θ is a Quadrant IV angle. To find θ’s reference angle α. so the Reference Angle Theorem √ gives: cos 11π = cos π = 23 and sin 11π = − sin π = − 1 . cos 7π 3 = cos y 1 π 3 = 1 2 and sin 7π 3 = sin π 3 √ = 3 2 . we subtract: α = 2π − θ = 2π − 11π = π . Since θ is a Quadrant II angle. 2π (1. − √ 3 2 3π 2 (0.6 The Reference Angle Theorem 3 in conjunction with the table of cosine and sine values on Page 722 can be used to generate the following figure. − 22 2 4π 3 1 −2. 2 π 3 √ √ 2 . 1) 1 −2. √ √ 3 2 − √ √ 2 . Reduced fraction multiples of π with a denominator of 6 have π as a reference angle. those with a denominator of 4 have π as their reference angle. 22 2 π 4 π 6 √ 3 1 2 . −2 √ − √ 2 .2 3π 4 5π 6 2π 3 π 2 √ 3 1 2. the reference angle for a non-quadrantal angle is easy to spot. we have something convenient about using radian measure in contrast to the abstract theoretical nonsense about using them as a ‘natural’ way to match oriented angles with real numbers! 6 . 0) 0.724 Foundations of Trigonometry The reader may have noticed that when expressed in radian measure. − 22 2 √ 1 . − 23 2 Important Points on the Unit Circle For once.2 π (−1. and 6 4 those with a denominator of 3 have π as their reference angle. 22 2 − 3 1 2 . 0) x √ − 3 1 2 . y (0. −2 7π 6 5π 4 7π 4 11π 6 √ 3 1 2 . −1) 5π 3 √ √ 2 . which the authors feel should be committed to memory. Hence. 12 . . You should do this. Since the terminal side of θ falls in Quadrant III.2. (a) To find the cosine and sine of θ = π + α. 5 12 hence. sin(α) is positive. Find sin(α) and use this to plot α in standard position.2.2 The Unit Circle: Cosine and Sine The next example summarizes all of the important ideas discussed thus far in the section. We can imagine the sum of the angles π+α as a sequence of two rotations: a rotation of π radians followed by a rotation of α radians. (b) θ = 2π − α (c) θ = 3π − α (d) θ = π 2 5 13 .10. we begin our rotation on the positive 13 5 x-axis to the ray which contains the point (cos(α). Suppose α is an acute angle with cos(α) = 1. θ may be plotted by reversing the order of rotations given here. Since α is an acute (and therefore Quadrant I) angle. Proceeding as in Example 10. sin(α)) = 13 . cos(θ) = ± cos(α) = ± 13 and sin(θ) = ± sin(α) = ± 13 . 13 y 1 5 12 13 . so by 5 12 The Reference Angle Theorem. we first plot θ in standard position. sin(α) = 12 . 725 +α 5 1. To plot α in standard position. 7 Since π + α = α + π. 13 α 1 x Sketching α 2.2. Example 10. both cos(θ) and sin(θ) are negative. 2.4. we substitute cos(α) = 13 into cos2 (α) + sin2 (α) = 1 and 12 find sin(α) = ± 13 . Find the sine and cosine of the following angles: (a) θ = π + α Solution.7 We see that α is the reference angle for θ. cos(θ) = − 13 and sin(θ) = − 13 . Using the Reference Angle Theorem. we get 5 cos(θ) = − 13 and sin(θ) = 12 .726 y Foundations of Trigonometry y 1 1 θ π x θ α 1 α 1 x Visualizing θ = π + α θ has reference angle α (b) Rewriting θ = 2π − α as θ = 2π + (−α). The angle 3π represents one and a half revolutions counter-clockwise. or ‘backing up.’ of α radians. 13 . we rewrite θ = 3π − α as θ = 3π + (−α). We see that α is θ’s reference angle. we can plot θ by visualizing one complete revolution counter-clockwise followed by a clockwise revolution. so that when we ‘back up’ α radians. and since θ is a Quadrant IV angle. the 5 Reference Angle Theorem gives: cos(θ) = 13 and sin(θ) = − 12 . we end up in Quadrant II. 13 y y 1 1 θ θ 2π 1 x 1 x −α α Visualizing θ = 2π − α θ has reference angle α (c) Taking a cue from the previous problem. The reference 2 2 angle here is not α. so The Reference Angle Theorem is not immediately applicable. we find y = sin(θ) = 13 . we use the fact that equal angles subtend equal chords to show that the dotted lines in 12 5 the figure below are equal.10. Hence. Similarly. y) .) Let Q(x. (It’s important that you see why this is the case. x = cos(θ) = − 13 . we first rotate π radians and follow up with α radians. y) be the point on the terminal side of θ which lies on the Unit Circle so that x = cos(θ) and y = sin(θ). y y 1 1 π 2 θ α α Q (x. 13 α 1 x Visualizing θ = π 2 +α Using symmetry to determine Q(x. Take a moment to think about this before reading on. Once we graph α in standard position. y) 1 x P 5 12 13 .2 The Unit Circle: Cosine and Sine y y 727 1 1 θ −α 3π 1 x α x 1 Visualizing 3π − α θ has reference angle α (d) To plot θ = π + α. 5. radian measure will be the only appropriate angle measure so it is worth the time to become “fluent in radians” now. we will default to using radian measure in our answers to each of these problems. we find the solution to cos(θ) = 2 here is 5π .7. If cos(θ) = 1 . we add 2πk for integers k = 0.9 Proceeding similarly for the Quadrant 3 3 1 IV case. we find θ = π + 2πk for integers k. . Find all of the angles which satisfy the given equation. If sin(θ) = − 1 . 1. This choice will be justified later in the text when we study what is known as Analytic Trigonometry. . so our answer in this Quadrant is 3 5π θ = 3 + 2πk for integers k. 1. Enjoy these relatively straightforward exercises while they last! 9 Recall in Section 10. This means θ is a Quadrant I or IV angle with reference angle π . . 2. 3 y 1 y 1 π 3 1 2 1 2 x 1 π 3 1 x One solution in Quadrant I is θ = π . In those sections to come. then when θ is plotted in standard position. we determine θ is a Quadrant III or Quadrant IV angle π with reference angle 6 .8 Example 10. when plotted in standard position. . Hence to describe all angles coterminal with a given angle. From this. two angles in radian measure are coterminal if and only if they differ by an integer multiple of 2π. and since all other Quadrant I solutions must be 3 coterminal with π . ±1. sin(θ) = − 1 2 3.728 Foundations of Trigonometry Our next example asks us to solve some very basic trigonometric equations.2. ±2. Solution. 8 We will more formally study of trigonometric equations in Section 10. then the terminal side of θ. its terminal side intersects the 2 1 Unit Circle at y = − 2 . Since there is no context in the problem to indicate whether to use degrees or radians. cos(θ) = 0. cos(θ) = 1 2 2.1. intersects the 2 1 Unit Circle at x = 2 . . technically speaking. The angles with cos(θ) = 0 are quadrantal angles whose terminal sides. In Quadrant IV. another Quadrant IV solution to sin(θ) = − 1 is θ = − π . solutions to trigonometric equations consist of infinitely many answers. the reader is encouraged to follow our mantra from Chapter 9 . To get a feel for these answers.5 is that. one solution is 7π . in general. While on the surface. ‘When in doubt.2 The Unit Circle: Cosine and Sine y 1 y 1 729 π 6 x 1 π 6 1 −2 x 1 −1 2 In Quadrant III. when plotted in standard position. While this solution is correct. k. (Can you see why this works from the diagram?) 2 One of the key items to take from Example 10. this family may look 6 . one solution is 11π so all the solutions here 6 6 are of the form θ = 11π + 2πk for integers k.10.that is. so we capture all Quadrant III solutions by adding integer 6 multiples of 2π: θ = 7π + 2πk. 6 3. π isn’t a reference angle we can nonetheless use it to find our 2 answers. If we follow the procedure set forth in the previous examples. the family of Quadrant IV answers to number 2 above could just 2 6 have easily been written θ = − π + 2πk for integers k. it can be shortened to 2 θ = π + πk for integers k.2. we find θ = π + 2πk 2 and θ = 3π + 2πk for integers. For example. lie along the y-axis. y 1 y 1 π 2 π 2 π x 1 π 2 1 x While. Hence. write it out!’ This is especially important when checking answers to the exercises. Let Q(x.1 Beyond the Unit Circle We began the section with a quest to describe the position of a particle experiencing circular motion. Consider for the moment the acute angle θ drawn below in standard position. and we leave it to the reader to verify these formulas hold for the quadrantal angles. by definition. Now consider dropping perpendiculars from P and Q to create two right triangles. If Q(x. 11π 6 Foundations of Trigonometry + 2πk for integers k. y-axis and origin.10 thus it follows x r that x = 1 = r. which lies on the circle x2 + y 2 = r2 then x = r cos(θ) and y = r sin(θ). These triangles are similar. ∆OP A and ∆OQB. y) be the point on the terminal side of θ which lies on the circle x2 + y 2 = r2 . Moreover. y ) be the point on the terminal side of θ which lies on the Unit Circle. and let P (x . we find y = ry . These results are summarized in the following theorem. 0) B(x. Theorem 10.2. so x = rx and. similarly. By reflecting these points through the x-axis.3. we can also express cos(θ) and sin(θ) in terms of the coordinates of Q. we broaden our scope to include circles of radius r centered at the origin. y ) θ 1 1 r x P (x . cos(θ) = x = r x x2 + y2 and sin(θ) = y = r y x2 + y2 10 Do you remember why? . In defining the cosine and sine functions. we get the coordinates of Q to be x = r cos(θ) and y = r sin(θ). y) Q(x. plotted in standard position. 0) x Not only can we describe the coordinates of Q in terms of cos(θ) and sin(θ) but since the radius of the circle is r = x2 + y 2 . r sin(θ)) 1 P (x .730 different than the stated solution of θ = they represent the same list of angles. y ) θ O A(x . we leave it to the reader to show 10. we obtain the result for all non-quadrantal angles θ. y r y Q (x. we assigned to each angle a position on the Unit Circle. y) = (r cos(θ). x = cos(θ) and y = sin(θ). Since. In this subsection. y) is the point on the terminal side of an angle θ. Justify this approximation if the radius of the Earth at the Equator is approximately 3960 miles. y) 41.628◦ is approximately 2960 miles. y) indicated in the figure below. Viewing the Equator as the x-axis. 1.3 reduces to 1.628◦ north latitude to be 2960 miles. 731 x2 + y 2 = 1. r (4)2 + (−2)2 = √ √ 20 = 2 5 so 2.2 The Unit Circle: Cosine and Sine Note that in the case of the Unit Circle we have r = our definitions of cos(θ) and sin(θ). we find 3960 cos (41. so Theorem 10. In Example 10.6. Suppose that the terminal side of an angle θ.2.3. Using Theorem 10. the radius of the Earth at North Latitude 41. . −2). −2) The terminal side of θ contains Q(4. Find sin(θ) and cos(θ). Hence. we get x = 3960 cos (41.1.628◦ N Using Theorem 10. we approximated the radius of the earth at 41. y y 3960 4 2 Q (x. the value we seek is the x-coordinate of the point Q(x.628◦ ). −2) A point on the Earth at 41. 2.1. Using a calculator in ‘degree’ mode. we find r = x √ 2 5 x 4 √ that cos(θ) = r = 2√5 = 5 and y = y = 2−25 = − 55 .5 in Section 10. when plotted in standard position. contains the point Q(4.10.628◦ ) ≈ 2960. Solution.3 with √ = 4 and y = −2. Example 10. Assuming the Earth is a sphere.628◦ −4 −2 −2 −4 2 4 x 3960 x Q(4. a cross-section through the poles produces a circle of radius 3960 miles. the angle θ is in standard position. at time t. Indeed.732 Foundations of Trigonometry Theorem 10. the location of the object Q(x. Equation 10.2. we made good use of some properties of right triangles to find the exact values of the cosine and sine of many of the angles in Example 10. r sin(ωt)). many applications in trigonometry involve finding the measures of the angles in. Hence. From Example 10.1.3. and the remaining side of length c (the side opposite the 11 You may have been exposed to this in High School. the object has swept out an angle measuring θ radians. Find the equations of motion of Lakeland Community College as the earth rotates. the side with length b is called the side of the triangle opposite θ.11 As we shall see in the sections to come. 0) when t = 0.5. y r Q (x.3 is also the key to developing what is usually called ‘right triangle’ trigonometry. the object is at the point (r cos(ωt). the equations π π of motion are x = r cos(ωt) = 2960 cos 12 t and y = r sin(ωt) = 2960 sin 12 t . In addition to circular motion.1. we take r = 2960 miles and and ω = 12 hours . 0).5.7. If we assume that the object is at the point (r.3. Consider the generic right triangle below with corresponding acute angle θ. By definition. y) = (r cos(ωt). right triangles.1.2. r sin(ωt)) 1 θ = ωt 1 r x Equations for Circular Motion Example 10. We have just argued the following. Theorem 10. Suppose an object is traveling in a circular path of radius r centered at the origin with constant angular velocity ω. where x and y are measured in miles and t is measured in hours. Suppose we are in the situation of Example 10. Here. Hence. so the following development shouldn’t be that much of a surprise. ω > 0 indicates a counter-clockwise direction and ω < 0 indicates a clockwise direction. According t to Theorem 10. The side with length a is called the side of the triangle adjacent to θ. and lengths of the sides of. . Suppose that at time t.3 gives us what we need to describe the position of an object traveling in a circular path of radius r with constant angular velocity ω. ω = θ which we rewrite as θ = ωt. y) on the circle is found using the equations x = r cos(θ) = r cos(ωt) and y = r sin(θ) = r sin(ωt). then the x and y coordinates of the object are functions of t and are given by x = r cos(ωt) and y = r sin(ωt). π Solution. If t = 0 corresponds to the point (r. and the length of the hypotenuse b . Theorem 10. we have two ways to proceed to find the length of the side opposite the 30◦ angle. we know that the missing angle has measure 180◦ − 30◦ − 90◦ = 60◦ . We now imagine drawing this triangle in Quadrant I so that the angle θ is in standard position with the adjacent side to θ lying along the positive x-axis. c Example 10. so that the point P (a. Suppose θ is an side adjacent to θ is a. If the length of the of the side opposite θ is b. Find the measure of the missing angle and the lengths of the missing sides of: 30◦ 7 Solution. so we have determined the cosine c and sine of θ in terms of the lengths of the sides of the right triangle. then cos(θ) = and sin(θ) = c acute angle residing in a right triangle.4.4.10. or c = cos(30◦ ) . c Since cos (30◦ ) = 23 . y c P (a. we have cos (30◦ ) = 7 . a2 + b2 = c2 . Let c denote the 7 length of the hypotenuse of the triangle. after the usual fraction gymnastics. Since the sum of angles of a triangle is 180◦ .3 tells us that cos(θ) = a and sin(θ) = c .2 The Unit Circle: Cosine and Sine 733 right angle) is called the hypotenuse.2. which we’ll denote √ b. We now proceed to find the lengths of the remaining two sides of the triangle. c = 143 3 . Thus we have the following theorem.8. b) c b θ a θ x c According to the Pythagorean Theorem. the length a is c. so we √ √ . At this point. By Theorem 10. we have. b) lies on a circle of b radius c. Theorem 10. We know the length of the adjacent side is 7 and the length of the hypotenuse is 143 3 . The first and easiest task is to find the measure of the missing angle. sin(t)) θ=t t θ=t 1 x 1 x In the same way we studied polynomial. y 1 y 1 P (cos(t). Since cos(t) and sin(t) represent x. we could use Theorem 10. For each real number t. the range of f (t) = cos(t) and of g(t) = sin(t) is the interval [−1. we find The triangle with all of its data is recorded below. ∞). Choosing the latter. To summarize: .4. and logarithmic functions. inclusive. If we trace the identification of real numbers t with angles θ in radian measure to its roots on page 704. In other words. In practice this means expressions like cos(π) and sin(2) can be found by regarding the inputs as angles in radian measure or real numbers. · 1 2 = √ 7 3 3 . c= √ 14 3 3 60◦ b= √ 7 3 3 30◦ 7 We close this section by noting that we can easily extend the functions cosine and sine to real numbers by identifying a real number t with the angle θ = t radians. 0) and endpoint P (cos(t). rational. of points on the Unit Circle. the domain of f (t) = cos(t) and of g(t) = sin(t) is (−∞. the choice is the reader’s. we associate an oriented arc t units in length with initial point (1. 1].and y-coordinates.734 Foundations of Trigonometry √ 2 14 3 3 could use the Pythagorean Theorem to find the missing side and solve (7)2 + b2 = Alternatively. sin (30◦ ) = b c. Whether we think of identifying the real number t with the angle θ = t radians. Using this identification. namely that b= c sin (30◦ ) = √ 14 3 3 for b. respectively. exponential. or think of wrapping an oriented arc around the Unit Circle to find coordinates on the Unit Circle. sin(t)). we define cos(t) = cos(θ) and sin(t) = sin(θ). The first order of business is to find the domains and ranges of these functions. we can spell out this correspondence more precisely. they both take on all of the values between −1 an 1. we will study the trigonometric functions f (t) = cos(t) and g(t) = sin(t). it should be clear that both the cosine and sine functions are defined for all real numbers t. In other words. k. Until then. Indeed.10.2 The Unit Circle: Cosine and Sine Theorem 10. Our solution is only cosmetically different in that the variable used is t rather than θ: t = 7π + 2πk or t = 11π + 2πk for integers. Well. we would be technically using ‘reference numbers’ or ‘reference arcs’ instead of ‘reference angles’ – but the idea is the same.5 2 number 2. 1] • The function g(t) = sin(t) – has domain (−∞. Domain and Range of the Cosine and Sine Functions: • The function f (t) = cos(t) – has domain (−∞. 1] 735 1 Suppose. We will study the cosine and sine functions in greater 6 6 detail in Section 10.5. 12 . As we have already mentioned.2. ∞) – has range [−1. we are asked to solve an equation such as sin(t) = − 2 . ∞) – has range [−1.5. to be pedantic. keep in mind that any properties of cosine and sine developed in the following sections which regard them as functions of angles in radian measure apply equally well if the inputs are regarded as real numbers. the distinction between t as a real number and as an angle θ = t radians is often blurred. as in the Exercises. we solve sin(t) = − 1 in the exact same manner12 as we did in Example 10. 20. θ = 8.42 and π < θ < 30. θ = − 17. θ = 7. θ = − 10π 3 20.30. θ = 3π 2 13π 2 15. θ = 3. what is sin(θ)? 10 2 29. θ = 14. what is cos(θ)? 5 2 √ 10 5π 28. If sin(θ) = and < θ < π. θ = 2π 3 5π 4 7π 4 3π 4 2. use the results developed throughout the section to find the requested value. If cos(θ) = − 25. If cos(θ) = 23. θ = π 11.98 and 3π . 21. what is sin(θ)? 11 24. what is sin(θ)? 9 5 with θ in Quadrant II. θ = 9. what is sin(θ)? 53 √ 2 5 π 27. θ = 117π In Exercises 21 . If cos(θ) = and 2π < θ < . what is cos(θ)? 25 4 with θ in Quadrant I. θ = 13. If sin(θ) = − 22. If sin(θ) = − 26. what is cos(θ)? 2 π < θ < π. what is cos(θ)? 3 28 with θ in Quadrant IV.2 Exercises π 4 3π 4 4π 3 23π 6 π 6 π 3 π 2 7π 6 5π 3 43π 6 In Exercises 1 . θ = 0 5. If cos(θ) = −0. If sin(θ) = 7 with θ in Quadrant IV. θ = 6. θ = − 19. what is cos(θ)? 13 2 with θ in Quadrant III. θ = − 18. θ = 16. 1. what is sin(θ)? 2 . θ = 4. If sin(θ) = −0. θ = 12. θ = 10.2.736 Foundations of Trigonometry 10. find the exact value of the cosine and sine of the given angle. If cos(θ) = 2 with θ in Quadrant III. 10. cos(−2.001 In Exercises 40 . c 30◦ 1 θ b a 45◦ c θ 3 . Find θ. 56. cos(e) In Exercises 55 .2 The Unit Circle: Cosine and Sine In Exercises 31 . (See Example 10. sin(78. (See the comments following Theorem 10. Find θ. cos(t) = 1 47. b.) √ 40.8) 55. cos(θ) = − √ 3 2 33. Make sure your calculator is in the proper angle measurement mode! 49.01) 53. sin(t) = − 44. sin(t) = 1 48.54. sin (π ◦ ) 51. find the measurement of the missing angle and the lengths of the missing sides. a. sin(θ) = 1 2 √ √ 32. cos(t) = 3 45. cos(t) = 0 43.48.95◦ ) 52.2.5. sin(t) = − 1 2 41. find all of the angles which satisfy the given equation. sin(θ) = 3 2 √ 3 38.994) 54. cos(t) = 1 2 2 2 42. sin(t) = −2 √ 46. sin(392. cos(θ) = 2 36. cos(t) = − 2 2 In Exercises 49 . use your calculator to approximate the given value to three decimal places.39. 31. solve the equation for t. sin(θ) = 0 737 2 34.58. sin(θ) = −1 35. cos(θ) = −1. cos(θ) = −1 39. cos(207◦ ) 50. and c. cos(θ) = 2 37. and c. how long is the side opposite θ? 63. how long is the hypotenuse? 60.) 69.25 inches and it spins at 4500 revolutions per minute. 65. 59. Recall: The radius of the circle is 28 inches and it completes one revolution in 3 seconds. how long is the side adjacent to θ? 64.1. the motion is counter-clockwise and that t = 0 corresponds to a position along the positive x-axis. Assume that the center of the motion is the origin. find the equations of motion for the given scenario.738 57.1. assume that θ is an acute angle in a right triangle and use Theorem 10. 4) 67. a. . 70. −9) 68.5 inches and it spins at 7200 revolutions per minute. Recall: The diameter of the yo-yo is 2. and c.5◦ and the side opposite θ has length 306. R(5. If θ = 5◦ and the hypotenuse has length 10. If θ = 12◦ and the side adjacent to θ has length 4.3 and Example 10. 24) 66. Find β. how long is the side adjacent to θ? 61. a. and b. a 48◦ a 8 33◦ c 6 β In Exercises 59 .5. If θ = 59◦ and the side opposite θ has length 117. P (−7. If θ = 37. A point on the edge of the spinning yo-yo in Exercise 50 from Section 10. A point on the edge of the hard drive in Exercise 53 from Section 10. how long is the hypotenuse? 62.72.1. b α Foundations of Trigonometry 58. −11) In Exercises 69 .1.68. Q(3. T (−2. Recall: The diameter of the hard disk is 2.4 to find the requested side.123◦ and the hypotenuse has length 5280. If θ = 78. how long is the side adjacent to θ? In Exercises 65 . let θ be the angle in standard position whose terminal side contains the given point then compute cos(θ) and sin(θ).42. The yo-yo in exercise 52 from Section 10. (See Equation 10. If θ = 5◦ and the hypotenuse has length 10.64. Find α. 71. Take the square root of each of these numbers. A passenger on the Big Wheel in Exercise 55 from Section 10. 4. we can adjust the equations of motion by introducing a ‘time delay. 7 seconds. show.) 74. The resulting numbers should look hauntingly familiar. with the help of your classmates. 0). then divide each by 2. the object was at the point (r. Recall: The diameter is 128 feet and completes 2 revolutions in 2 minutes. the equations of motion are x = r cos(ω(t − t0 )) and y = r sin(ω(t − t0 )). (See the values in the table on 722.’ If t0 > 0 is the first time the object passes through the point (r.3.10. If this is not the case. In the scenario of Equation 10.2 The Unit Circle: Cosine and Sine 72. The fact that co-functions of complementary angles are equal in this case is not an accident and a more general result will be given in Section 10. we assumed that at t = 0. Let α and β be the two acute angles of a right triangle. 739 73. 75. 2. Consider the numbers: 0. 1. 3. .4.) Show that sin(α) = cos(β) and sin(β) = cos(α). (Thus α and β are complementary angles. 0).1. sin 2 10π 3 √ 1 5π 3 = .2. sin − 2 4 1 = − . If cos(θ) = − with θ in Quadrant III. cos 2π 3 1 π = . If cos(θ) = with θ in Quadrant I. cos(0) = 1. sin √ π 2 √ √ 2 = 2 1. then cos(θ) = − . cos 5π 4 3π 2 7π 4 2 =− . sin 2 15. cos √ 6. sin − √ 13π 2 = −1 3π 17. sin 2 1 = − . then cos(θ) = − . then sin(θ) = − .3 Answers π 2. sin 2 √ 3 =− . sin 2 =− √ =− 3 2 = −1 13. If cos(θ) = 28 45 with θ in Quadrant IV. sin(117π) = 0 7 24 with θ in Quadrant IV. sin =− 6 2 6 2 √ 3 43π 43π 1 − =− . cos 12. cos 8. cos 10π 3 3π 2 =− . cos − 4 19. If sin(θ) = 26. sin √ 3π 2 7π 4 √ 2 =− . sin − =− 6 2 6 2 20. then sin(θ) = . sin 2 √ = 2π 3 3 2 =1 3π 4 7π 6 4π 3 √ = 2 2 1 2 3 = 2 7. cos 11. 11 11 √ 5 2 25. cos − 13π 2 = 0. 25 25 √ 4 65 22. sin(π) = 0 9. 13 13 √ 2 117 24. cos π 2 3π 4 7π 6 4π 3 5π 3 2 π = . cos √ =− √ =− 3 2 2 2 18. 53 53 . cos 16. sin 2 4 = 0. cos 5π 4 2 =− 2 √ 10. cos(117π) = −1.740 Foundations of Trigonometry 10. sin(0) = 0 π 3. then cos(θ) = . sin − = 6 2 6 2 √ π π 1 3 − = . sin 2 = 0. 9 9 21. cos 3 5. If sin(θ) = − with θ in Quadrant III. sin 2 3 1 = − . cos 2 = . cos 4 4. If sin(θ) = − 5 12 with θ in Quadrant II. cos(π) = −1. sin =− 2 3 2 √ 23π 3 23π 1 = . cos √ =− 2 2 14. 3 3 23. then sin(θ) = − . cos(t) = 1 when t = 2πk for any integer k. If cos(θ) = −0. If cos(θ) = and 2π < θ < .001 never happens 40. sin(t) = − 44.2 The Unit Circle: Cosine and Sine √ √ 2 5 π 5 27. If sin(θ) = −0. sin(θ) = 0 when θ = πk for any integer k. then sin(θ) = 0. 2 6 6 1 π 5π when t = + 2πk or t = + 2πk for any integer k.8236 ≈ −0.1990. 2 4 4 42.10. If sin(θ) = and < θ < π. 2 6 6 √ 3 5π 7π 32.42 and π < θ < . 2 3 3 45. then cos(θ) = − 0. cos(θ) = −1 when θ = (2k + 1)π for any integer k. cos(t) = 3 never happens. then cos(θ) = − . √ 2 π 7π 34. sin(θ) = when θ = + 2πk or θ = + 2πk for any integer k. sin(θ) = 1 π 5π when θ = + 2πk or θ = + 2πk for any integer k. 10 2 10 √ 3π 29. cos(θ) = when θ = + 2πk or θ = + 2πk for any integer k. 2 6 6 741 33. 2 6 6 π + πk for any integer k.98 and < θ < π. 5 2 5 √ √ 10 5π 3 10 28. cos(θ) = − when θ = + 2πk or θ = + 2πk for any integer k. 37. 43. cos(t) = 0 when t = √ 41. sin(t) = − 2 5π 7π when t = + 2πk or t = + 2πk for any integer k.0396 ≈ 0.9075. 2 3 π 11π when θ = + 2πk or θ = + 2πk for any integer k. 2 √ π 30. cos(θ) = −1. 2 31. then sin(θ) = . sin(t) = −2 never happens 46. 2 4 4 √ 3 π 2π 35. 2 3 3 36. 2 39. . sin(θ) = −1 when θ = √ 38. cos(t) = 1 7π 11π when t = + 2πk or t = + 2πk for any integer k. cos(θ) = 3π + 2πk for any integer k. 68. sin(59◦ ) 62. The hypotenuse has length 60. y = 1. Here x minute and y are measured in inches and t is measured in minutes.994) ≈ −0.891 √ √ 3 2 3 ◦.125 cos(9000π t). . sin(392. x = 1.981 52. sin (π ◦ ) ≈ 0. b = 55.425 53. 306 64.123◦ ) ≈ 1086. 61. sin(t) = 1 when t = √ 48. cos(t) = − π + 2πk for any integer k. c = 3 2 57. 65. so the side adjacent to θ has length sin(37. 63.709.089.055 51.99. The hypotenuse has length c = ≈ 502.402 sin(48◦ ) 4 ≈ 4. cos(θ) = − . The side opposite θ has length 10 sin(5◦ ) ≈ 0. cos(12◦ ) 59.742 47. The side adjacent to θ has length 5280 cos(78. sin(θ) = − 25 25 69. a = c2 − 62 ≈ 5.5◦ ) √ c2 − 3062 ≈ 398. sin(78.962. The hypotenuse has length 117. sin(θ) = 5 5 √ √ 5 106 9 106 67.074. α = 57◦ . ω = 9000π radians .872. a = 8 cos(33◦ ) ≈ 6. θ = 45◦ . cos(θ) = . a = 3.125 sin(9000π t). c = √ 6 ≈ 8. cos(θ) = − 7 24 . cos(e) ≈ −0.c= 3 3 √ 56. sin(θ) = − 106 106 √ √ 2 5 11 5 68. β = 42◦ .797. The side adjacent to θ has length 10 cos(5◦ ) ≈ 9.291 54.660. 2 4 4 50.42 ≈ 136.95◦ ) ≈ 0. b = 8 sin(33◦ ) ≈ 4. r = 1. cos(θ) = .01) ≈ −0. cos(207◦ ) ≈ −0. cos(−2. 2 Foundations of Trigonometry 3π 5π 2 when t = + 2πk or t = + 2πk for any integer k.125 inches. θ = 60 . sin(θ) = 25 25 3 4 66.912 49.357 58. x = 64 cos second in feet and t is measured in seconds 4π 127 t .2 The Unit Circle: Cosine and Sine 70. ω = 14400π radians .25 cos(14400π t). x = 1. 4π 72. y = 28 sin 2π 3 t . Here x minute and y are measured in inches and t is measured in minutes. r = 64 feet. Here x and y are measured 71.10. x = 28 cos 3 second in inches and t is measured in seconds.25 inches. y = 1. Here x and y are measured . ω = 2π radians . ω = 127 radians . r = 1. r = 28 inches. 2π 3 743 t . y = 64 sin 4π 127 t .25 sin(14400π t). denoted sec(θ). As such. you will find that we do indeed use the phrase ‘trigonometric function’ interchangeably with the term ‘circular function’. y) denote. y Q(1. y ) = (1. • The sine of θ. denoted tan(θ). provided x = 0. denoted csc(θ). • The cosine of θ. In later sections. Consider the acute angle θ below in standard position. • The secant of θ.4 we also showed cosine and sine to be functions of an angle residing in a right triangle so we could just as easily call them trigonometric functions. as usual. is defined by tan(θ) = y . tan(θ)) 1 P (x. 0) x In Theorem 10. 1 . y • The cosecant of θ. the names ‘tangent’ and ‘secant’ can be explained using the diagram below. denoted cos(θ). 0) B(1. is defined by sec(θ) = 1 .2. the point on the terminal side of θ which lies on the Unit Circle and let Q(1. x 1 . we defined cos(θ) and sin(θ) for angles θ using the coordinate values of points on the Unit Circle. provided y = 0. these functions earn the moniker circular functions. is defined by sin(θ) = y. is defined by cos(θ) = x. y While we left the history of the name ‘sine’ as an interesting research project in Section 10. is defined by cot(θ) = . x x • The cotangent of θ. Let P (x.2. y) θ O A(x. provided x = 0.2. Definition 10.744 Foundations of Trigonometry 10. is defined by csc(θ) = • The tangent of θ.1 It turns out that cosine and sine are just two of the six commonly used circular functions which we define below. y ) denote the point on the terminal side of θ which lies on the vertical line x = 1. denoted cot(θ). The Circular Functions: Suppose θ is an angle plotted in standard position and P (x. provided y = 0.3 The Six Circular Functions and Fundamental Identities In section 10. denoted sin(θ). y) is the point on the terminal side of θ which lies on the Unit Circle. Theorem 10. With the point P lying on the Unit Circle.2. . cot(3) 4.3 The Six Circular Functions and Fundamental Identities 745 The word ‘tangent’ comes from the Latin meaning ‘to touch.2. 0). if sin(θ) = 0. csc(θ) is undefined. 1. sec(θ) is undefined. Compare this with the definition given in Section 2. the circle at only one point. sin(θ) sin(θ) .1. cos (θ). if cos(θ) = 0. cos(θ) 1 . cot(θ) is undefined. csc 7π 4 3. Dropping perpendiculars from P and Q creates a pair of similar triangles y 1 ∆OP A and ∆OQB.3. 6. Not only do these observations help explain the names of these functions. where tan(θ) = 3 and π < θ < 2 3π 2 . so a secant line is any line that ‘cuts through’ a circle at two points. tan(θ) is the y-coordinate of the point on the terminal side of θ which lies on the line x = 1 which is tangent to the Unit Circle. it is customary to rephrase the remaining four circular functions in terms of cosine and sine. provided sin(θ) = 0. tan(θ) is undefined.2. namely (1. if sin(θ) = 0.’ and for this reason. or h = x = sec(θ). only cosine and sine are defined for all angles. sin(θ) It is high time for an example. where this last equality comes from y applying Definition 10. they serve as the basis for a fundamental inequality needed for Calculus which we’ll explore in the Exercises.6. If we let h denote the length of the hypotenuse of ∆OQB. Find the indicated value.1. Thus y = x which gives y = x = tan(θ). provided cos(θ) = 0. sin (θ). Since cos(θ) = x and sin(θ) = y in Definition 10. if it exists. sec (60◦ ) 2. Of the six circular functions. the length of the hypotenuse of ∆OP A is 1. 2 √ 5. where csc(θ) = − 5 and θ is a Quadrant IV angle. Hence for an acute angle θ. The following theorem is a result of simply replacing x with cos(θ) and y with sin(θ) in Definition 10. we have from similar 1 1 triangles that h = x . provided cos(θ) = 0. sec(θ) is the length of the line 1 segment which lies on the secant line determined by the terminal side of θ and ‘cuts off’ the tangent line x = 1. if cos(θ) = 0. We have just shown that for acute angles θ.2 The line containing the terminal side of θ is a secant line since it intersects the Unit Circle in Quadrants I and III. Reciprocal and Quotient Identities: • sec(θ) = • csc(θ) = • tan(θ) = • cot(θ) = 1 . cos(θ) cos(θ) . or ‘touches’. provided sin(θ) = 0.10. Example 10. where θ is any angle coterminal with 3π . Now the word ‘secant’ means ‘to cut’. tan (θ). the line x = 1 is called a tangent line to the Unit Circle since it intersects. √ θ is a Quadrant IV angle. cos2 (θ) + sin2 (θ) = 1. Substituting. Foundations of Trigonometry Hence. cos(θ) > 0. 1 (1/2) = 2. we usually convert back to cosines and sines. it is not always convenient to do so. then cos(θ) = cos 3π = 0 and sin(θ) = sin 2 2 sin(θ) to compute tan(θ) = cos(θ) results in −1 . Substituting this into the Pythagorean Identity. sec (60◦ ) = √ 2 = − √2 = − 2. However.6 allow us to always reduce problems involving secant. θ is a Quadrant III angle. we get cos2 (θ) + − = 1. we once again employ the Pythagorean Identity. √ 3π 2 . so cos(θ) = 6.6.2. If θ is coterminal with 3π . Since 5 √ 2 5 5 . Attempting √ √ 1 = − 5 so sin(θ) = − √5 = − 55 . 1. √ 5 5 2 we can use the Pythagorean Identity. so our final answer is sin(θ) = While the Reciprocal and Quotient Identities presented in Theorem 10.746 Solution. =− 2 2 . we resort to the calculator for a decimal approximation. Since π < θ < √ 10 − 3 10 . Solving. As we saw in Section 10. tangent and cotangent to problems involving cosine and sine. we get sin2 (θ) = 9 10 10 so sin(θ) = ± 3 10 .3 It is worth taking the time to memorize the tangent and cotangent values of the common angles summarized below. csc 7π 4 = 1 sin( 7π 4 ) = 1 √ − 2/2 3.this does NOT mean we can take sin(θ) = 3 and sin(θ) cos(θ) = 1. 3 . Ensuring that the calculator is in radian mode. sec (60◦ ) = 2. As we shall see shortly. we usually stick with what we’re dealt.2. Instead. This means sin(θ) < 0. sin(3) 4. we find 3 sin2 (θ) + 1 3 sin(θ) 2 = 1. when solving equations involving secant and cosecant. we find cos(θ) = 1 sin(θ). then sin(θ) cos(θ) = 3. According to Theorem 10. Be careful . from cos(θ) = 3 we get: sin(θ) = 3 cos(θ). 0 5. cosecant. we find cot(3) = cos(3) ≈ −7. We are given that csc(θ) = 1 sin(θ) 3π 2 = −1. or cos(θ) = ± 2 5 5 . If tan(θ) = 3. which gives cos2 (θ) = 4 . cos2 (θ) + sin2 (θ) = 1. Since sin 7π 4 √ 1 cos(60◦ ) . when solving for tangent or cotangent. Since θ = 3 radians is not one of the ‘common angles’ from Section 10. Solving sin(θ) = 3 cos(θ) for cos(θ).015. so tan(θ) is undefined. To relate cos(θ) and sin(θ). to find cos(θ) by knowing sin(θ). we get the solutions: θ = π + 2πk for integers k. sec(θ) = ± sec(α).5. of the corresponding circular functions of its reference angle. number 1. and for 3 Quadrant III. tangent is positive when x and y have the same sign (i. 3 2. Generalized Reference Angle Theorem. tan(θ) = √ 3 3. we see tan π = 3.7. when they are both positive or both negative. sec(θ) = 2 Solution. Example 10. Theorem 10. √ 2. In Quadrant I. More specifically. . we know 3 √ the solutions to tan(θ) = 3 must.. up to a sign.) This happens in Quadrants I and III.3.2 for another example of this kind of simplification of the solution. sin(θ) = ± sin(α). This is the exact 2 same equation we solved in Example 10.3 The Six Circular Functions and Fundamental Identities Tangent and Cotangent Values of Common Angles θ(degrees) θ(radians) 0◦ 30◦ 45◦ 60◦ 90◦ 0 π 6 π 4 π 3 π 2 747 tan(θ) √ cot(θ) undefined √ 3 1 √ 3 3 0 3 3 1 √ 3 undefined 0 Coupling Theorem 10. we get the following.10. they can be combined into one list as θ = π + πk for integers k.2. While these descriptions of the solutions 3 are correct. The values of the circular functions of an angle.2.2. We put Theorem 10. so we know the answer is: θ = π + 2πk 3 or θ = 5π + 2πk for integers k. 1 1. have a reference angle of π . we convert to cosines and get cos(θ) = 2 or cos(θ) = 1 . if they exist. The choice of the (±) depends on the quadrant in which the terminal side of θ lies. are the same.2. csc(θ) = ± csc(α).7 to good use in the following example. tan(θ) = ± tan(α) and cot(θ) = ± cot(α). 1. therefore. Find all angles which satisfy the given equation.7.6 with the Reference Angle Theorem.5 number 3 in Section 10. y) on the Unit Circle with x = 0. if α is the reference angle for θ. we get θ = 4π + 2πk for integers k.e. Theorem 10. To solve sec(θ) = 2. Our next task is 3 to determine in which quadrants the solutions to this equation lie.4 4 See Example 10. From the table of common values. then: cos(θ) = ± cos(α). cot(θ) = −1. Since tangent is defined y as the ratio x of points (x. According to Theorem 10. The latter form 3 of the solution is best understood looking at the geometry of the situation in the diagram below. 6 once again.one such way to capture all the solutions is: θ = 3π + πk for integers k. along with some of their other common forms. Using properties of exponents along with the Reciprocal 2 (θ) and Quotient Identities. Our Quadrant II solution is θ = 3π + 2πk.748 y 1 Foundations of Trigonometry y 1 π π 3 x 1 π 3 1 π 3 x 3. we may start with cos2 (θ) + sin2 (θ) = 1 and divide both sides sin2 by cos2 (θ) to obtain 1 + cos2(θ) = cos1(θ) . which means the 4 solutions to cot(θ) = −1 have a reference angle of π . are summarized in the following theorem. To find the quadrants in which our 4 solutions lie. we note that cot(θ) = x for a point (x..6 coupled with the Pythagorean Identity found in Theorem 10. These three Pythagorean Identities are worth memorizing and they. and obtain cot2 (θ) + 1 = csc2 (θ).) Hence. If y cot(θ) is negative. one positive and one negative.1 to derive new Pythagorean-like identities for the remaining four circular functions. we get θ = 7π +2πk for integers k. we see that π has a cotangent of 1. From the table of common values. 4 and for Quadrant IV. 4 y 1 y 1 π 4 x 1 π 4 π 4 x 1 π We have already seen the importance of identities in trigonometry. our solutions lie in Quadrants II and IV. y) on the Unit Circle where y = 0. . we can divide both sides of the identity cos2 (θ) + sin2 (θ) = 1 by sin2 (θ). this reduces to 1 + tan2 (θ) = sec2 (θ). If sin(θ) = 0.e. Assuming cos(θ) = 0. apply Theorem 10. Our next task is to use use the Reciprocal and Quotient Identities found in Theorem 10. Can these lists be combined? Indeed 4 they can . then x and y must have different signs (i. 6 sec(θ) tan(θ) = 3 3 − 1 − sin(θ) 1 + sin(θ) Solution. In Calculus. Common Alternate Forms: • 1 − sin2 (θ) = cos2 (θ) • 1 − cos2 (θ) = sin2 (θ) 2. In verifying identities. The Pythagorean Identities: 1. Common Alternate Forms: • csc2 (θ) − cot2 (θ) = 1 • csc2 (θ) − 1 = cot2 (θ) 749 Trigonometric identities play an important role in not just Trigonometry. but in Calculus as well.3.8. 1 = sin(θ) csc(θ) 2. Common Alternate Forms: • sec2 (θ) − tan2 (θ) = 1 • sec2 (θ) − 1 = tan2 (θ) 3. 1 + tan2 (θ) = sec2 (θ). Example 10. provided cos(θ) = 0. cos2 (θ) + sin2 (θ) = 1. .6 and 10. they are needed to simplify otherwise complicated expressions. 1. tan(θ) = sin(θ) sec(θ) 4. we get: = sin(θ). We’ll use them in this book to find the values of the circular functions of an angle and solve equations and inequalities. we typically start with the more complicated side of the equation and use known identities to transform it into the other side of the equation. 6. (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = 1 5. Verify the following identities. To verify 1 csc(θ) = sin(θ). In the next example.10. Using csc(θ) = 1 = csc(θ) 1 1 sin(θ) 1 sin(θ) .3.3 The Six Circular Functions and Fundamental Identities Theorem 10. which is what we were trying to prove. we start with the left side. 1. 1 + cot2 (θ) = csc2 (θ). we make good use of the Theorems 10. sec(θ) 1 = 1 − tan(θ) cos(θ) − sin(θ) sin(θ) 1 + cos(θ) = 1 − cos(θ) sin(θ) 3.8. Assume that all quantities are defined. provided sin(θ) = 0. 750 Foundations of Trigonometry 1 cos(θ) 2. we use sec(θ) = sin(θ) sec(θ) = sin(θ) 1 sin(θ) = = tan(θ). 4.5 Substituting sec(θ) = cos(θ) and tan(θ) = cos(θ) . sec2 (θ) − tan2 (θ) = 1. cos(θ) cos(θ) and find: where the last equality is courtesy of Theorem 10. The right hand side of the equation seems to hold more promise. Expanding the left hand side of the equation gives: (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = sec2 (θ) − tan2 (θ).6. Starting with the right hand side of tan(θ) = sin(θ) sec(θ). 5. While both sides of our last identity contain fractions. earn more partial credit if this were an exam question! . cos(θ) − sin(θ) = = (cos(θ)) = which is exactly what we had set out to show. we get: sec(θ) 1 − tan(θ) 1 1 cos(θ) cos(θ) cos(θ) = · sin(θ) sin(θ) cos(θ) 1− 1− cos(θ) cos(θ) 1 (cos(θ)) 1 cos(θ) = sin(θ) sin(θ) (1)(cos(θ)) − (cos(θ)) 1− cos(θ) cos(θ) 1 . 3. Putting it all together. (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = sec2 (θ) − tan2 (θ) = 1. We get common denominators and add: 3 3 − 1 − sin(θ) 1 + sin(θ) = = = = 5 3(1 + sin(θ)) 3(1 − sin(θ)) − (1 − sin(θ))(1 + sin(θ)) (1 + sin(θ))(1 − sin(θ)) 3 + 3 sin(θ) 3 − 3 sin(θ) − 1 − sin2 (θ) 1 − sin2 (θ) (3 + 3 sin(θ)) − (3 − 3 sin(θ)) 1 − sin2 (θ) 6 sin(θ) 1 − sin2 (θ) Or.8. According to Theorem 10. the left side affords us more opportusin(θ) 1 nities to use our identities. to put to another way. This suggests the strategy of starting with the left hand side and multiplying the numerator and denominator by the quantity 1 + cos(θ): sin(θ) 1 − cos(θ) = = = sin(θ) (1 + cos(θ)) sin(θ)(1 + cos(θ)) · = (1 − cos(θ)) (1 + cos(θ)) (1 − cos(θ))(1 + cos(θ)) sin(θ)(1 + cos(θ)) sin(θ)(1 + cos(θ)) = 2 (θ) 1 − cos sin2 (θ) 1 + cos(θ) sin(θ)(1 $$$ + cos(θ)) = $$ sin(θ) sin(θ) sin(θ) $ In Example 10.3.3 number 6 above. the quantities (1 − cos(θ)) and (1 + cos(θ)) are called ‘Pythagorean Conjugates. we need to get cosines in our denominator. We wish to transform this expression into 6 sec(θ) tan(θ). Pythagorean Conjugates • 1 − cos(θ) and 1 + cos(θ): (1 − cos(θ))(1 + cos(θ)) = 1 − cos2 (θ) = sin2 (θ) • 1 − sin(θ) and 1 + sin(θ): (1 − sin(θ))(1 + sin(θ)) = 1 − sin2 (θ) = cos2 (θ) • sec(θ) − 1 and sec(θ) + 1: (sec(θ) − 1)(sec(θ) + 1) = sec2 (θ) − 1 = tan2 (θ) • sec(θ)−tan(θ) and sec(θ)+tan(θ): (sec(θ)−tan(θ))(sec(θ)+tan(θ)) = sec2 (θ)−tan2 (θ) = 1 • csc(θ) − 1 and csc(θ) + 1: (csc(θ) − 1)(csc(θ) + 1) = csc2 (θ) − 1 = cot2 (θ) • csc(θ) − cot(θ) and csc(θ) + cot(θ): (csc(θ) − cot(θ))(csc(θ) + cot(θ)) = csc2 (θ) − cot2 (θ) = 1 . (Can you recall instances from Algebra where we did such things?) For this reason.3 The Six Circular Functions and Fundamental Identities 751 At this point. Using a reciprocal and quotient identity. In other words. Theorem 10. it is worth pausing to remind ourselves of our goal. It is debatable which side of the identity is more complicated. while the numerator of the right hand side is 1 + cos(θ).8.10. we see that multiplying 1 − cos(θ) by 1 + cos(θ) produces a difference of squares that can be simplified to one term using Theorem 10. we find sin(θ) 1 6 sec(θ) tan(θ) = 6 cos(θ) cos(θ) . This√ exactly the is √ same kind of phenomenon that occurs when we multiply expressions such as 1 − 2 by 1 + 2 or 3 − 4i by 3 + 4i.’ Below is a list of other common Pythagorean Conjugates.8 tells us 1 − sin2 (θ) = cos2 (θ) so we get: 3 3 − 1 − sin(θ) 1 + sin(θ) = 6 sin(θ) 6 sin(θ) = 2 cos2 (θ) 1 − sin (θ) 1 cos(θ) sin(θ) cos(θ) = 6 sec(θ) tan(θ) = 6 6. One thing which stands out is that the denominator on the left hand side is 1 − cos(θ). 10.8. a summary of some strategies which may be helpful (depending on the situation) is provided below and ample practice is provided for you in the Exercises. y . Nevertheless. Simplify the resulting complex fractions. Theorem 10. provided x = 0. x x2 + y 2 . and simplify sums or differences of squares to one term. provided y = 0. • Add rational expressions with unlike denominators by obtaining common denominators. • Use the Reciprocal and Quotient Identities in Theorem 10.3. secants and tangents.752 Foundations of Trigonometry Verifying trigonometric identities requires a healthy mix of tenacity and inspiration.1 Beyond the Unit Circle In Section 10. x x • cot(θ) = . You will need to spend many hours struggling with them just to become proficient in the basics.2. y y . • Use the Pythagorean Identities in Theorem 10. we generalize the remaining circular functions in kind. Strategies for Verifying Identities • Try working on the more complicated side of the identity. provided x = 0. • If you find yourself stuck working with one side of the identity. cosecants and cotangents.3 in conjunction with Theorem 10.8. Using Theorem 10. Then: • sec(θ) = • csc(θ) = • tan(θ) = r = x r = y x2 + y 2 . provided y = 0. x2 + y 2 = r2 . • Multiply numerator and denominator by Pythagorean Conjugates in order to take advantage of the Pythagorean Identities in Theorem 10. there is no short-cut here – there is no complete algorithm for verifying identities. Like many things in life.6 to write functions on one side of the identity in terms of the functions on the other side of the identity.8 to ‘exchange’ sines and cosines.9. Suppose Q(x. try starting with the other side of the identity and see if you can find a way to bridge the two parts of your work. y) is the point on the terminal side of an angle θ (plotted in standard position) which lies on the circle of radius r. we generalized the cosine and sine functions from coordinates on the Unit Circle to coordinates on circles of radius r. If the length of the side adjacent to θ is a. the ground). csc(θ) = − 5 . 753 1.9 tells us cos(θ) = 3 . −4).9. Find the values of the six circular functions of θ. Viewing −4 = −1 . and whose terminal side is the line-of-sight to an object above the base-line. sec(θ) = 5 .’ The angle of inclination (or angle of elevation) of an object refers to the angle whose initial side is some kind of base-line (say. In order to use Theorem 10.9 once √ 17 17 17 more. when θ is plotted in standard position. This is represented schematically below. 4 . Solution. we could choose x = 8 y and y = −2. √ 1. sin(θ) = − 4 . we find cos(θ) = √4 = 4 17 . Suppose the terminal side of θ. Suppose θ is an acute angle residing in a right triangle. sin(θ) = − √1 = − 17 . when plotted in standard position.10. c b θ a Theorem 10. The fact that all such points lie on the terminal side of θ is a consequence of the fact that the terminal side of θ is the portion of the line with slope − 1 which extends from the origin into Quadrant IV. 5 5 3 4 3 4 2.10. tan(θ) = − 4 and cot(θ) = − 3 . r = x2 + y 2 = (3)2 + (−4)2 = 25 = 5. we may choose6 x = 4 √ y and y = −1 so that r = x2 +√ 2 = (4)2 + (−1)2 = √17. Applying√Theorem 10. and since θ is a y 4 Quadrant IV angle. then c c a b tan(θ) = sec(θ) = csc(θ) = cot(θ) = a a b b The following example uses Theorem 10. 6 We may choose any values x and y so long as x > 0. We have that cot(θ) = −4 = x . the length of the side opposite θ is b.9 to the case of acute angles θ which reside in a right triangle.10 as well as the concept of an ‘angle of inclination. sec(θ) = 4 . Theorem 10. Since x = 3 and y = −4. 2. Suppose θ is a Quadrant IV angle with cot(θ) = −4. we also know x > 0 and y < 0. 4 We may also specialize Theorem 10. y) which lies on the terminal side of θ.3 The Six Circular Functions and Fundamental Identities Example 10. Find the values of the five remaining circular functions of θ. csc(θ) = − 17 17 17 and tan(θ) = − 1 . For example.3.4. contains the point Q(3. y < 0 and x = −4. we need to find a point Q(x. as visualized below. and the length of the hypotenuse is c. how tall is the tree to the nearest foot? Solution. 60◦ 30 ft. In order to determine the height of a California Redwood tree. Armington. are made. we find ourselves with two unknowns: the height h of the tree and the distance x from the base of the tree to the first observation point. From this we get √ ◦ ) = 30 3 ≈ 51.3. then Theorem 10.10 gives tan (60◦ ) = 30 .96. If the angles of inclination were 45◦ and 30◦ . 2. h = 30 tan (60 h ft.754 Foundations of Trigonometry object θ ‘base line’ The angle of inclination from the base line to the object is θ Example 10. 1. 1. respectively. We can represent the problem situation using a right triangle as shown below. If we let h h denote the height of the tower. Sketching the problem situation below. Finding the height of the Clocktower 2. Find the height of the Clocktower to the nearest foot. 7 Named in honor of Raymond Q. . the Clocktower is approximately 52 feet tall. two sightings from the ground. The angle of inclination from a point on the ground 30 feet away to the top of Lakeland’s Armington Clocktower7 is 60◦ . Hence. one 200 feet directly behind the other. Lakeland’s Clocktower has been a part of campus since 1972.5. 2 2 2 Graphing this set on the number line we get . Running through a few values of k. First.10. we could go through the formality of the wrapping function on page 704 and define these functions as the appropriate ratios of x and y coordinates of points on the Unit Circle. the tree is approximately 273 feet tall. Since x tan (45◦ ) = 1. tan(t) and cot(t) for real numbers t. . the first equation gives h = 1.5 number 3. there are three equivalent ways to define the functions sec(t). ± 3π .2. csc(t). for integers k}. The result is a linear equation for h. we know cos(t) = 0 whenever t = π + πk for integers k. As we did in Section 10.3 The Six Circular Functions and Fundamental Identities 755 h ft. we could simply define them using the Reciprocal and Quotient Identities as combinations of the functions f (t) = cos(t) and g(t) = sin(t). we find the domain to be {t : t = ± π . Presently. We know F is undefined whenever cos(t) = 0.2. we may consider all six circular functions as functions of real numbers. .20 h = 3− 3 Hence. Hence. . we get a pair of equations: tan (45◦ ) = h and tan (30◦ ) = x+200 . we could define them by associating the real number t with the angle θ = t radians so that the value of the trigonometric function of t coincides with that of θ. 45◦ x ft. or x = h. so we proceed to expand the right hand side and gather all the terms involving h to one side. Substituting this into the second √ x √ h equation gives h+200 = tan (30◦ ) = 33 . We now set about determining the domains and ranges of the remaining four circular functions. second. ± 5π . 30◦ 200 ft. To get a better 2 understanding what set of real numbers we’re dealing with. in set builder notation is {t : t = π + πk. Clearing fractions. Finding the height of a California Redwood h Using Theorem 10. it pays to write out and graph this set. lastly. Consider the function 1 F (t) = sec(t) defined as F (t) = sec(t) = cos(t) . At this stage.10. √ 3h = (h + 200) 3 √ √ 3h = h 3 + 200 3 √ √ 3h − h 3 = 200 3 √ √ (3 − 3)h = 200 3 √ 200 3 √ ≈ 273.1. we adopt the last approach. we get 3h = (h + 200) 3.}. our 2 domain for F (t) = sec(t). From Example 10. on the other hand. as cos(t) → 0+ . the kth point excluded from the domain. ±1. but rather. ∪ − 5π 3π . 2 2 ∪ 3π 5π . . to say the least! In order to write this in a more compact way. but rather.. . This is cumbersome. we get . this conveys the idea that k ranges through all of the integers. using the notation 1 1 introduced in Section 4.2. . the index k in the union ∞ (2k + 1)π (2k + 3)π . we have that as cos(t) → 0+ . Now that we have painstakingly determined the domain of F (t) = sec(t). it is time to 1 discuss the range.. which we’ll call xk . we split our discussion into two cases: when 0 < cos(t) ≤ 1 and when −1 ≤ cos(t) < 0. Once again. 1]. (We are using sequence notation from Chapter 9. The way we denote taking the union of infinitely many intervals like this is to use what we call in this text extended interval notation. that is.2. Since .. If. sec(t) → ∞.− 2 2 ∪ − 3π π . if −1 ≤ cos(t) < 0. so that as cos(t) → 0− . and since F (t) = sec(t) is undefined when cos(t) = 0. ranges through all of the natural numbers. 2 2 The reader should compare this notation with summation notation introduced in Section 9. as cos(t) → 0− . 1 1 sec(t) = cos(t) ≈ very small (−) ≈ very big (−). Moreover. If 0 < cos(t) ≤ 1. 2 2 In order to capture all of the intervals in the domain.− 2 2 π π ∪ − . k = 0. xk + 1 ) = (2k+1)π . can be found by the formula xk = π +πk. The range of f (t) = cos(t) is [−1. (2k+3)π . then we can 1 divide the inequality cos(t) ≤ 1 by cos(t) to obtain sec(t) = cos(t) ≥ 1.756 Foundations of Trigonometry − 5π 2 − 3π 2 −π 0 2 π 2 3π 2 5π 2 Using interval notation to describe this set. sec(t) = cos(t) ≈ very small (+) ≈ very big (+).2. In this case.) 2 Getting a common denominator and factoring out the π in the numerator. then dividing 1 by cos(t) causes a reversal of the inequality so that sec(t) = sec(t) ≤ −1. 2 2 ∪ .. we get sec(t) → −∞. we note that from the set-builder description of the domain. k must run through all of the integers. . 2 2 k=−∞ can never actually be ∞ or −∞. in particular the notation used to describe geometric series in Theorem 9. In other words. The 2 domain consists of the intervals determined by successive points xk : (xk . In the same way the index k in the series ∞ ark−1 k=1 can never equal the upper limit ∞. ±2. we appeal to the definition F (t) = sec(t) = cos(t) . ∪ 2 2 π 3π . we get xk = (2k+1)π . The domain of F (t) = sec(t) can now be written as ∞ k=−∞ (2k + 1)π (2k + 3)π . ∞) • The function J(t) = tan(t) = – has domain {t : t = – has range (−∞. the range of F (t) = sec(t) can be written as {u : u ≤ −1 or u ≥ 1}. for integers k} = (2k + 1)π (2k + 3)π . (See the Exercises. ∞). −1] ∪ [1. 2 2 – has range {u : |u| ≥ 1} = (−∞. The reader is encouraged to do so.9 Similar arguments can be used to determine the domains and ranges of the remaining three circular functions: csc(t). for integers k} = (2k + 1)π (2k + 3)π . ∞) • The function K(t) = cot(t) = cos(t) sin(t) π 2 sin(t) cos(t) ∞ k=−∞ + πk. ∞) • The function G(t) = csc(t) = 1 sin(t) ∞ – has domain {t : t = πk. we gather these facts into the theorem below. ∞) – has range [−1. −1] ∪ [1.) For now. or.10. 1] • The function F (t) = sec(t) = – has domain {t : t = π 2 • The function g(t) = sin(t) – has domain (−∞.8 as {u : |u| ≥ 1}. Domains and Ranges of the Circular Functions • The function f (t) = cos(t) – has domain (−∞.4. Using set-builder notation. (k + 1)π) – has range (−∞. ∞) – has range [−1. for integers k} = k=−∞ (kπ. 9 8 . Notice we have used the variable ‘u’ as the ‘dummy variable’ to describe the range elements. −1] ∪ [1. 1].3 The Six Circular Functions and Fundamental Identities 757 f (t) = cos(t) admits all of the values in [−1.11. could use t again) we choose u to help solidify the idea that these real numbers are the outputs from the inputs. 1] 1 cos(t) ∞ k=−∞ + πk. which we have been calling t. Theorem 10. tan(t) and cot(t). (k + 1)π) – has range {u : |u| ≥ 1} = (−∞. the function F (t) = sec(t) admits all of the values in (−∞.4 from Section 2. as such. ∞) Using Theorem 2. for integers k} = k=−∞ (kπ. 2 2 ∞ – has domain {t : t = πk. While there is no mathematical reason to do this (we are describing a set of real numbers. more succinctly. and. 3. number 3 to solve cot(θ) = −1 for angles θ in radian measure – we just need to remember to write our answers using the variable t as opposed to θ. For 1 example. you will need to 2 review the notions of reference angles and coterminal angles so that you can see why csc(t) = −42 has an infinite set of solutions in Quadrant III and another infinite set of solutions in Quadrant IV. In particular. to solve the equation cot(t) = −1 for real numbers t. not just f (t) = cos(t) and g(t) = sin(t). Next. sec(t) = 2 has no solution because 1 is not in the range of secant.2. . the discussion on page 735 in Section 10. First. we can use the same thought process we used in Example 10. Finally. it is critical that you know the domains and ranges of the six circular functions so that you know which equations have no solutions.1 concerning solving equations applies to all six circular functions.758 Foundations of Trigonometry We close this section with a few notes about solving equations which involve the circular functions.2. π . csc π 2 7π 4 12. cot 4π 3 13π 2 5.902) 36. cos(θ) = 1 with θ in Quadrant I. tan(θ) = √ 10 with π < θ < In Exercises 35 .20.01) 40. 2 24.672◦ ) 37. cot(3◦ ) 38. csc(78. tan(39. 21.3. cot(392. csc 5π 6 π 3 4. tan(−2. csc(5. sec(θ) = 7 with θ in Quadrant IV 26. tan − 6.994) 41. sin(θ) = 23. csc(θ) = − with θ in Quadrant III 91 27. tan (117π) 13. 1. sec (−7π) In Exercises 21 .42. use your calculator to approximate the given value to three decimal places. cot 7π 6 3π 4 18.2 Exercises In Exercises 1 . cot 8. csc(θ) = 5 with < θ < π. csc(θ) = 3 with θ in Quadrant II 5 22. cot (−5π) 16. use the given the information to find the exact values of the remaining circular functions of θ. sec π 4 7. 3 π 32. sec(θ) = 2 5 with < θ < 2π. 30. cot(θ) = √ 5 with θ in Quadrant III. cot(θ) = 2 with 0 < θ < 33. tan π 4 11π 6 2. tan 17. tan(θ) = 12 with θ in Quadrant III 5 25 with θ in Quadrant I 24 √ 10 91 25. tan(θ) = −2 with θ in Quadrant IV. 2 31. sec − 14. 29. csc − 19.34. sec(θ) = −4 with θ in Quadrant II. cot 20. Make sure your calculator is in the proper angle measurement mode! 35. sec − 10.10. sec π 6 3π 2 5π 3 3. cot(θ) = −23 with θ in Quadrant II 28.95◦ ) 39. find the exact value or state that it is undefined. csc − 9. sec(0.3 The Six Circular Functions and Fundamental Identities 759 10. 2 3π .45) . sec(207◦ ) 42. tan 31π 2 2π 3 11. 2 √ 3π 34. csc (3π) 15. find all of the angles which satisfy the equation. Find θ. csc(θ) = −2 45. sec(θ) = −1 3 3 47. 66. a. b β a 47◦ c c θ 6 2.65. b. cot(θ) = 50. cot(θ) = −1 In Exercises 58 . tan(θ) = 0 51. sec(t) = 62. cot(θ) = 0 54. sec(θ) = 1 52. Give exact values. tan(t) = 60. and c. a. csc(θ) = −1 49. 43. b. 69. sec(θ) = 2 48. csc(θ) = 2 53. and c. solve the equation for t.5 50◦ . sec(t) = − 3 3 √ √ √ 2 3 3 63. b 60◦ α a 12 c 61. csc(t) = 0 √ 2 3 65.760 Foundations of Trigonometry In Exercises 43 . tan(θ) = √ √ 3 44. Find θ. and c. Find α. and c. cot(t) = − 3 3 3 In Exercises 66 . c θ 9 34◦ 67. cot(t) = 1 59.69. tan(t) = − 64. use Theorem 10. tan(θ) = − 3 57. √ √ 3 2 3 58. tan(θ) = −1 √ 55.57.10 to find the requested quantities. csc(t) = 3 68. Find β. sec(θ) = 0 56. csc(θ) = − 1 2 46. 4◦ .the angle of depression (also known as the angle of declination). .05◦ and the hypotenuse has length 3. If θ = 30◦ and the side opposite θ has length 4. If θ = 2.125◦ . If θ = 38. If θ = 87◦ and the side adjacent to θ has length 2. A tree standing vertically on level ground casts a 120 foot long shadow. The broadcast tower for radio station WSAZ (Home of “Algebra in the Morning with Carl and Jeff”) has two enormous flashing red lights on it: one at the very top and one a few feet below the top. how long is the side adjacent to θ? 75. how long is the side opposite θ? 76. how long is the side adjacent to θ? 71. horizontal θ observer object The angle of depression from the horizontal to the object is θ (a) Show that if the horizontal is above and parallel to level ground then the angle of depression (from observer to object) and the angle of inclination (from object to observer) will be congruent because they are alternate interior angles.970◦ and to the second light is 7. With the help of your classmates. This is represented schematically below. If θ = 42◦ and the side adjacent to θ has length 31. The angle of elevation from the end of the shadow to the top of the tree is 21.98.3 The Six Circular Functions and Fundamental Identities 761 In Exercises 70 . If θ = 15◦ and the hypotenuse has length 10. 78. From a point 5000 feet away from the base of the tower on level ground the angle of elevation to the top light is 7. On page 753 we defined the angle of inclination (also known as the angle of elevation) and in this exercise we introduce a related angle .75.10. how long is the hypoteneuse? 74. Find the height of the tree to the nearest foot. use Theorem 10. 70. how long is the side opposite θ? 72. 77. Assume that θ is an angle in a right triangle. The angle of depression of an object refers to the angle whose initial side is a horizontal line above the object and whose terminal side is the line-of-sight to the object below the horizontal.2◦ and the side opposite θ has lengh 14. how long is the side opposite θ? 73.10 to answer the question. Find the distance between the lights to the nearest foot. research the term umbra versa and see what it has to do with the shadow in this problem. A guy wire 1000 feet long is attached to the top of a tower. tan(θ) cos(θ) = sin(θ) 85. 91. tan(θ) cot(θ) = 1 87. How far had the boat traveled between the sightings? 81. From the observation deck of the lighthouse at Sasquatch Point 50 feet above the surface of Lake Ippizuti. Assume that all quantities are defined. sin(θ) csc(θ) = 1 86. 89. how long will it take for the Sasquatch to reach the firetower from his location at the second sighting? Round your answer to the nearest minute. taken just 10 seconds later. a lifeguard spots a boat out on the lake sailing directly toward the lighthouse. 80. What is the height of the tree? Round your answer to the nearest foot. verify the identity.9◦ . At the first sighthing. If the Sasquatch keeps up this pace. 90.5◦ . the angle of depression from the tower to the Sasquatch is 6◦ . cos(θ) = csc(θ) cot(θ) sin2 (θ) 1 − cos(θ) = csc(θ) − cot(θ) sin(θ) sin(θ) = csc(θ) 1 − cos2 (θ) 83. How far did the Saquatch travel in those 10 seconds? Round your answer to the nearest foot. the angle of elevation to the top of the tree is 50◦ and the angle of depression to the base of the tree is 10◦ . 79. csc(θ) cos(θ) = cot(θ) 88. How tall is the tower? How far away from the base of the tower does the wire hit the ground? In Exercises 82 . The angle of depression to the fire is 2. How far away from the base of the tower is the fire? (c) The ranger in part 78b sees a Sasquatch running directly from the fire towards the firetower. sin(θ) = sec(θ) tan(θ) cos2 (θ) 1 + sin(θ) = sec(θ) + tan(θ) cos(θ) cos(θ) = sec(θ) 1 − sin2 (θ) sec(θ) = cos(θ) 1 + tan2 (θ) . When I stand 30 feet away from a tree at home. The first sighting had an angle of depression of 8. 92.5◦ . How fast is it running in miles per hour? Round your answer to the nearest mile per hour. cos(θ) sec(θ) = 1 84. gives the the angle of depression as 6. The second sighting. The ranger takes two sightings.2◦ and the second sighting had an angle of depression of 25. 93.128. When pulled taut it makes a 43◦ angle with the ground.762 Foundations of Trigonometry (b) From a firetower 200 feet above level ground in the Sasquatch National Forest. a ranger spots a fire off in the distance. 82. 124. 97. 1 = sec(θ) + tan(θ) sec(θ) − tan(θ) 1 = csc(θ) − cot(θ) csc(θ) + cot(θ) 1 = sec2 (θ) − sec(θ) tan(θ) 1 + sin(θ) 1 = csc2 (θ) − csc(θ) cot(θ) 1 + cos(θ) sin(θ) 1 + cos(θ) 127. sin5 (θ) = 1 − cos2 (θ) 2 sin(θ) sec2 (θ) 102. 123. tan(θ) + cot(θ) = sec(θ) csc(θ) 110.3 The Six Circular Functions and Fundamental Identities csc(θ) = sin(θ) 1 + cot2 (θ) cot(θ) = tan(θ) csc2 (θ) − 1 tan(θ) = cot(θ) sec2 (θ) − 1 763 94. sec4 (θ) − sec2 (θ) = tan2 (θ) + tan4 (θ) 105. 106. sin(θ)(tan(θ) + cot(θ)) = sec(θ) 114. 126. 9 − cos2 (θ) − sin2 (θ) = 8 100. 125. 4 cos2 (θ) + 4 sin2 (θ) = 4 99. cos(θ) − sec(θ) = − tan(θ) sin(θ) 112. cos(θ)(tan(θ) + cot(θ)) = csc(θ) 113. 96. 120. 1 = sec(θ) − tan(θ) sec(θ) + tan(θ) 1 = csc(θ) + cot(θ) csc(θ) − cot(θ) 1 = sec2 (θ) + sec(θ) tan(θ) 1 − sin(θ) 1 = csc2 (θ) + csc(θ) cot(θ) 1 − cos(θ) cos(θ) 1 − sin(θ) = 1 + sin(θ) cos(θ) 1 − sin(θ) = (sec(θ) − tan(θ))2 1 + sin(θ) 119.10. 109. 117. 1 + sec(θ) cos(θ) + 1 = cos(θ) − 1 1 − sec(θ) 1 − cot(θ) tan(θ) − 1 = 1 + cot(θ) tan(θ) + 1 103. 95. csc(θ) − cot(θ) = . sec10 (θ) = 1 + tan2 (θ) 4 98. 116. tan3 (θ) = tan(θ) sec2 (θ) − tan(θ) 101. 128. 115. 122. csc(θ) − sin(θ) = cot(θ) cos(θ) 111. 1 1 + = 2 csc2 (θ) 1 − cos(θ) 1 + cos(θ) 1 1 + = 2 sec(θ) tan(θ) csc(θ) + 1 csc(θ) − 1 sin(θ) cos(θ) + = sin(θ) + cos(θ) 1 − tan(θ) 1 − cot(θ) 1 1 + = 2 csc(θ) cot(θ) sec(θ) + 1 sec(θ) − 1 1 1 − = 2 cot(θ) csc(θ) − cot(θ) csc(θ) + cot(θ) 118. 107. 121. cos2 (θ) tan3 (θ) = tan(θ) − sin(θ) cos(θ) 104. sin(θ) + 1 1 + csc(θ) = sin(θ) − 1 1 − csc(θ) 1 − tan(θ) cos(θ) − sin(θ) = 1 + tan(θ) cos(θ) + sin(θ) 108. let α and β be the two acute angles of a right triangle.4. 2 sin(θ) π < 1 for 0 < θ < . Use the diagram from θ 2 the beginning of the section. 2 1 θ. 134. . 2 (d) Comparing areas. As we did in Exercise 74 in Section 10. sin(θ) π 135. y Q 1 P θ O B(1.) Show that sec(α) = csc(β) and tan(α) = cot(β).2 for a review of the properties of absolute value and logarithms before proceeding. ln | sec(θ)| = − ln | cos(θ)| 130. 129.2 and 6. verify the identity.2. Combine this θ 2 with the previous part to complete the proof.11. − ln | csc(θ)| = ln | sin(θ)| 132. The fact that co-functions of complementary angles are equal in this case is not an accident and a more general result will be given in Section 10. − ln | sec(θ) − tan(θ)| = ln | sec(θ) + tan(θ)| 133.132. partially reproduced below. θ 2 sin(θ) π (f) Use the inequality θ < tan(θ) to show that cos(θ) < for 0 < θ < . to answer the following. cosecant and cotangent functions as presented in Theorem 10.764 Foundations of Trigonometry In Exercises 129 . Verify the domains and ranges of the tangent. We wish to establish the inequality cos(θ) < < 1 for 0 < θ < . 2 (b) Show that the circular sector OP B with central angle θ has area (c) Show that triangle OQB has area 1 tan(θ). show that sin(θ) < θ < tan(θ) for 0 < θ < (e) Use the inequality sin(θ) < θ to show that π . (Thus α and β are complementary angles. 0) x (a) Show that triangle OP B has area 1 sin(θ). You may need to consult Sections 2. − ln | csc(θ) + cot(θ)| = ln | csc(θ) − cot(θ)| 131. 1.10. Explain why the fact that tan(θ) = 3 = 1 does not mean sin(θ) = 3 and cos(θ) = 1? (See the solution to number 6 in Example 10. θ 2 765 136.3.) .3 The Six Circular Functions and Fundamental Identities π sin(θ) < 1 also holds for − < θ < 0. Show that cos(θ) < 3 137. cot(θ) = −2 6 5 √ √ √ √ √ √ 110 11 110 10 − 11 . tan(θ) = 10. sin(θ) = 26. cos(θ) = 3 . csc(θ) = 3 4 2 . cos(θ) = − 23530 . sin(θ) = 28. csc(θ) = 4 15 . cot(θ) = − 19 . csc(θ) = 530. cos(θ) = − 1 . cot π 4 2π 3 3π 4 = √ 2 31π 2 7π 6 π 2 √ =− 3 = −1 18. sec(θ) = − 5 . sec 17. tan(θ) = − 23 . tan − 6 8. csc(θ) = 24 . csc 5π 3 11. tan (117π) = 0 12. sec(θ) = − 10 . tan(θ) = − 15. tan(θ) = 12 5 . cot(θ) = 5 √ √ √ √ 2 2 1 . csc(θ) = − 6. sin(θ) = 34. csc(θ) = − 10 . csc(θ) = 5. cos(θ) = 25 . tan(θ) = − 4 . cos(θ) = − 13 . cot 4π 3 π 3 11π 5. sec − 7. sec 6 √ 2 3 = 3 = =0 3. cos(θ) = − 10 . sin(θ) = 32. sin(θ) = 31. cos(θ) = 10 . sec(θ) = 7 . tan(θ) = 7 . cot 13π 2 6. cos(θ) = 2 5 5 . sec(θ) = − 5 . tan(θ) = 55 . sec(θ) = − 23 . tan(θ) = 2 2. cot(θ) = 42 3 √ √ √ √ 5 . sin(θ) = 25. tan(θ) = 1 . sec(θ) = − 11. sec(θ) = −4. cot(θ) = − 12 √ √ √ √ 91 91 91 3 − 10 . cot(θ) = 12 5 5 12 23. sin(θ) = 7 24 25 25 7 24 25 . sec (−7π) = −1 3 21. csc(θ) = − 12 . cos(θ) = 55 . cot(θ) = 3 91 3 √ √ √ √ 530 530 530 1 .3 Answers =1 3 = 3 √ 2 3 =− 3 =2 is undefined = =1 √ 3 √ π 2. cos(θ) = − 11 . tan(θ) = −2. sec(θ) = 3. cot (−5π) is undefined 15. cot(θ) = 10 √ √ √ √ √ √ 95 5 95 19 − 10 . csc(θ) = − 109191 .766 Foundations of Trigonometry 10. csc (3π) is undefined 14. sin(θ) = 30. csc − 9. sin(θ) = − 13 . sec(θ) = 5. sec − 13. tan(θ) = − 19. tan 4 4. cot(θ) = − 15 4 4 √ √ √ √ √ √ 30 30 − 66 . tan(θ) = − 12 . sin(θ) = 3 . cot(θ) = −23 530 √ √ √ √ − 2 5 5 . cos(θ) = − 5 . tan 16. cot 19. csc(θ) = 5 . sec(θ) = − 5126 . sin(θ) = 29. cot(θ) = − 4 5 5 3 4 3 12 5 22. sec(θ) = 25 . cot(θ) = 24 √ √ √ √ −4 3 3 7 3 1 7 . cot(θ) = 2 5 2 √ √ √ √ 2 6 6 1 . sec(θ) = − 13 . cos(θ) = − 6 . tan(θ) = −4 3. csc(θ) = − 25 . csc(θ) = − 2 19 . sec(θ) = 7. sin(θ) = 33. sec(θ) = 2 5. tan 20. csc − 7π 4 = √ 2 10. csc(θ) = − 13 . sin(θ) = 24. csc √ 3 3 5π 6 3π 2 =2 is undefined π 1. cos(θ) = 7 . csc(θ) = 5. sin(θ) = 27.3. cos(θ) = − 4 . cot(θ) = − 1 2 √ √ √ √ 15 15 15 . tan(θ) = 3 . sec(θ) = 0 never happens 53.3 The Six Circular Functions and Fundamental Identities 35.902) ≈ −2.688 36.081 42. tan(θ) = − 3 when θ = + πk for any integer k 3 56.129 38.01) ≈ 2.95◦ ) ≈ 1. csc(θ) = −1 when θ = √ 46. tan(39. tan(θ) = 0 when θ = πk for any integer k 48. csc(θ) = 2 when θ = 5π π + 2πk or θ = + 2πk for any integer k. cot(θ) = 3 π when θ = + πk for any integer k 3 3 47. cot(3◦ ) ≈ 19. tan(θ) = −1 when θ = 52. csc(θ) = −2 when θ = 57. cot(392.292 39. csc(θ) = − 1 never happens 2 54.122 40.672◦ ) ≈ 0.10. sec(0.829 767 41.019 37. cot(θ) = 0 when θ = + πk for any integer k 2 3π + πk for any integer k 4 51. csc(5. 2 45. tan(θ) = 3 when θ = + πk for any integer k 3 44. sec(θ) = 1 when θ = 2πk for any integer k 49. sec(θ) = 2 when θ = π 5π + 2πk or θ = + 2πk for any integer k 3 3 3π + 2πk for any integer k. cot(t) = 1 when t = 7π 11π + 2πk or θ = + 2πk for any integer k 6 6 3π + πk for any integer k 4 √ 3 π 59.111 √ π 43. sec(207◦ ) ≈ −1.994) ≈ 3. cot(θ) = −1 when θ = 58.45) ≈ 1. tan(t) = when t = + πk for any integer k 3 6 π + πk for any integer k 4 . csc(78. tan(−2. 6 6 π 50. sec(θ) = −1 when θ = π + 2πk = (2k + 1)π for any integer k √ 2π 55. c = 2. tan(t) = − when t = + πk for any integer k 3 6 √ 2 3 π 11π 64.595. (b) The fire is about 4581 feet from the base of the tower. (c) The Sasquatch ran 200 cot(6◦ ) − 200 cot(6. a = 3 3. The side opposite θ has length 31 tan(42◦ ) ≈ 27. 77. The side opposite θ has length 10 sin(15◦ ) ≈ 2.912 76. The side adjacent to θ has length 4 3 ≈ 6. c = 6 csc(47◦ ) = ≈ 8. sec(t) = when t = + 2πk or t = + 2πk for any integer k 3 6 6 √ 2 3 π 2π 65. At the scene of the second sighting. the Sasquatch was ≈ 1755 feet from the tower.977 75.588 72. The lights are about 75 feet apart. . csc(t) = when t = + 2πk or t = + 2πk for any integer k 3 3 3 √ √ √ 66. The side opposite θ is 2 tan(87◦ ) ≈ 38. c = 108 = 6 3 67.475 cos(34◦ ) 6 6 ≈ 5. it will reach the tower in about 2 minutes.5 sec(50◦ ) = √ 70. a = 6 cot(47◦ ) = 12 ≈ 14.2◦ ) 74. α = 56◦ .204 ◦) tan(47 sin(47◦ ) 2.5 ≈ 3.889 cos(50◦ ) 69. sec(t) = − when t = + 2πk or t = + 2πk for any integer k 3 6 6 61. This translates to ≈ 10 miles per hour.639 sin(38. b = 2.5 tan(50◦ ) ≈ 2.928 71.5◦ ) ≈ 147 feet in those 10 seconds. c = 12 sec(34◦ ) = 68.979.05◦ ) ≈ 3.98 cos(2. θ = 43◦ . if it keeps up this pace.2◦ ) = 14 ≈ 22. The hypoteneuse has length 14 csc(38.768 Foundations of Trigonometry √ 2 3 5π 7π 60. β = 40◦ . b = 12 tan(34◦ ) = 8. The side adjacent to θ has length 3. The tree is about 47 feet tall.094. cot(t) = − 3 when t = + πk for any integer k 6 √ 3 5π 63.162 73. which means. 78. θ = 30◦ . csc(t) = 0 never happens √ 5π 62. The guy wire hits the ground about 731 feet away from the base of the tower. 769 81.3 The Six Circular Functions and Fundamental Identities 79. . The tree is about 41 feet tall.10. The boat has traveled about 244 feet. 80. The tower is about 682 feet tall. Since θ is coterminal with θ0 .770 Foundations of Trigonometry 10. it suffices to show cos(−θ) = cos(θ) and sin(−θ) = − sin(θ). Let P and Q denote the points on the terminal sides of θ0 and −θ0 . which lie on the Unit Circle. Since θ0 and −θ0 sweep out congruent central sectors of the Unit Circle. it 1 As mentioned at the end of Section 10. sin(θ0 )) 1 x θ0 θ Q(cos(−θ0 ).3.6.6. Not surprisingly.1 Theorem 10. Therefore.) Since θ and θ0 are coterminal. there is some integer k so that θ = θ0 + 2π · k. Hence. sin(−θ0 )). • cos(−θ) = cos(θ) • sec(−θ) = sec(θ) • sin(−θ) = − sin(θ) • csc(−θ) = − csc(θ) • tan(−θ) = − tan(θ) • cot(−θ) = − cot(θ) In light of the Quotient and Reciprocal Identities. Since k is an integer. they were also useful in simplifying expressions involving the circular functions. the Even / Odd properties of the circular functions are so named because they identify cosine and secant as even functions. sin(θ0 )) and the coordinates of Q are (cos(−θ0 ). (See Section 1. which means −θ is coterminal with −θ0 . The remaining four circular functions can be expressed in terms of cos(θ) and sin(θ) so the proofs of their Even / Odd Identities are left as exercises. Even / Odd Identities: For all applicable angles θ. properties of the circular functions when thought of as functions of angles in radian measure hold equally well if we view these functions as functions of real numbers.12.2. sin(−θ0 )) 1 x −θ0 We now consider the angles −θ and −θ0 . we saw the utility of the Pythagorean Identities in Theorem 10. the coordinates of P are (cos(θ0 ).8 along with the Quotient and Reciprocal Identities in Theorem 10. Let θ0 be the angle coterminal with θ with 0 ≤ θ0 < 2π. y 1 y 1 θ0 P (cos(θ0 ). −θ = −θ0 − 2π · k = −θ0 + 2π · (−k). Our first set of identities is the ‘Even / Odd’ identities. Theorem 10. By definition. Not only did these identities help us compute the values of the circular functions for angles. Consider an angle θ plotted in standard position. we introduce several collections of identities which have uses in this course and beyond. cos(−θ) = cos(−θ0 ) and sin(−θ) = sin(−θ0 ). (We can construct the angle θ0 by rotating counter-clockwise from the positive x-axis to the terminal side of θ as pictured below. cos(θ) = cos(θ0 ) and sin(θ) = sin(θ0 ). In this section. so is (−k).) .4 Trigonometric Identities In Section 10.6. respectively. while the remaining four circular functions are odd. Thus. respectively. coterminal with α and β. However. The Even / Odd Identities are readily demonstrated using any of the ‘common angles’ noted in Section 10. the triangles P OQ and AOB are congruent. respectively. sin(α0 − β0 )) α0 − β 0 B(1. sin(α0 )) α0 − β0 α0 β0 O 1 x O Q(cos(β0 ).10. Consider the case below where α0 ≥ β0 . which makes a triangle. α0 − β0 could be 0 or it could be π. we expand the left hand side of this equation as (cos(α0 ) − cos(β0 ))2 + (sin(α0 ) − sin(β0 ))2 = cos2 (α0 ) − 2 cos(α0 ) cos(β0 ) + cos2 (β0 ) + sin2 (α0 ) − 2 sin(α0 ) sin(β0 ) + sin2 (β0 ) = cos2 (α0 ) + sin2 (α0 ) + cos2 (β0 ) + sin2 (β0 ) −2 cos(α0 ) cos(β0 ) − 2 sin(α0 ) sin(β0 ) 2 In the picture we’ve drawn. yields (cos(α0 ) − cos(β0 ))2 + (sin(α0 ) − sin(β0 ))2 = (cos(α0 − β0 ) − 1)2 + (sin(α0 − β0 ) − 0)2 Squaring both sides. It could also be larger than π. . which is even better. we can reduce the proof for general angles α and β to angles α0 and β0 . Sum and Difference Identities for Cosine: For all angles α and β. Since the cosines and sines of θ0 and −θ0 are the same as those for θ and −θ. it follows that α − β is coterminal with α0 − β0 . cos(−θ0 ) = cos(θ0 ) and sin(−θ0 ) = − sin(θ0 ). 0) x Since the angles P OQ and AOB are congruent. In fact. but in simplifying expressions involving the circular functions. just not the one we’ve drawn.1. however. As in the proof of the Even / Odd Identities. our next batch of identities makes heavy use of the Even / Odd Identities.4 Trigonometric Identities 771 follows that the points P and Q are symmetric about the x-axis. sin(β0 )) A(cos(α0 − β0 ). You should think about those three cases. as required.2. Theorem 10. • cos(α + β) = cos(α) cos(β) − sin(α) sin(β) • cos(α − β) = cos(α) cos(β) + sin(α) sin(β) We first prove the result for differences. Their true utility. lies not in computation. the distance between P and Q is equal to the distance between A and B. Since α and α0 are coterminal. each of which measure between 0 and 2π radians. as are β and β0 . neither of which makes a triangle.2 The distance formula.13. we get cos(−θ) = cos(θ) and sin(−θ) = − sin(θ). Equation 1. y y 1 P (cos(α0 ). Example 10. cos2 (α0 ) + sin2 (α0 ) = 1 and cos2 (β0 ) + sin2 (β0 ) = 1.13 to find cos (15◦ ). In order to use Theorem 10. One way to do so is to write 15◦ = 45◦ − 30◦ . we can apply the above argument to the angle β0 − α0 to obtain the identity cos(β0 − α0 ) = cos(β0 ) cos(α0 ) + sin(β0 ) sin(α0 ). so that (cos(α0 − β0 ) − 1)2 + (sin(α0 − β0 ) − 0)2 = 2 − 2 cos(α0 − β0 ) Putting it all together. we get cos(β0 − α0 ) = cos(−(α0 − β0 )) = cos(α0 − β0 ). we need to write 15◦ as a sum or difference of angles whose cosines and sines we know. Verify the identity: cos Solution. we use the difference formula along with the Even/Odd Identities cos(α + β) = cos(α − (−β)) = cos(α) cos(−β) + sin(α) sin(−β) = cos(α) cos(β) − sin(α) sin(β) We put these newfound identities to good use in the following example. we have cos(α − β) = cos(α) cos(β) + sin(α) sin(β). Find the exact value of cos (15◦ ). 1. Applying the Even Identity of cosine. so (cos(α0 ) − cos(β0 ))2 + (sin(α0 ) − sin(β0 ))2 = 2 − 2 cos(α0 ) cos(β0 ) − 2 sin(α0 ) sin(β0 ) Turning our attention to the right hand side of our equation. we get 2 − 2 cos(α0 ) cos(β0 ) − 2 sin(α0 ) sin(β0 ) = 2 − 2 cos(α0 − β0 ).1. we find (cos(α0 − β0 ) − 1)2 + (sin(α0 − β0 ) − 0)2 = cos2 (α0 − β0 ) − 2 cos(α0 − β0 ) + 1 + sin2 (α0 − β0 ) = 1 + cos2 (α0 − β0 ) + sin2 (α0 − β0 ) − 2 cos(α0 − β0 ) Once again. For the case where α0 ≤ β0 . To get the sum identity for cosine. we simplify cos2 (α0 − β0 ) + sin2 (α0 − β0 ) = 1. which simplifies to: cos(α0 − β0 ) = cos(α0 ) cos(β0 ) + sin(α0 ) sin(β0 ). . and we get the identity in this case. Since α and α0 .772 Foundations of Trigonometry From the Pythagorean Identities. 1. 2.4. β and β0 and α − β and α0 − β0 are all coterminal pairs of angles. too. cos (15◦ ) = cos (45◦ − 30◦ ) = cos (45◦ ) cos (30◦ ) + sin (45◦ ) sin (30◦ ) √ √ √ 2 3 2 1 = + 2 2 2 2 √ √ 6+ 2 = 4 π 2 − θ = sin(θ). 2 2 2 which says.14. we find cos π −θ 2 π π cos (θ) + sin sin (θ) 2 2 = (0) (cos(θ)) + (1) (sin(θ)) = cos = sin(θ) 773 The identity verified in Example 10. in words. • cos π − θ = sin(θ) 2 π • sin − θ = cos(θ) 2 π − θ = csc(θ) 2 π • csc − θ = sec(θ) 2 • sec • tan π − θ = cot(θ) 2 π • cot − θ = tan(θ) 2 With the Cofunction Identities in place. cos π − θ = sin(θ). In a straightforward application of Theorem 10. Now that these identities have been established for cosine and sine.2. To derive the sum formula for sine. we leave the details to the reader. namely.4 Trigonometric Identities 2.4.1. These identities were first hinted at in Exercise 74 in Section 10. we convert to cosines using a cofunction identity. sin Theorem 10.13. is the first of the celebrated 2 ‘cofunction’ identities.10. the remaining circular functions follow suit. Again. Cofunction Identities: For all applicable angles θ. we get: 2 π π π − θ = cos − − θ = cos(θ).15. then expand using the difference formula for cosine sin(α + β) = cos π − (α + β) 2 π = cos −α −β 2 π π − α cos(β) + sin − α sin(β) = cos 2 2 = sin(α) cos(β) + cos(α) sin(β) We can derive the difference formula for sine by rewriting sin(α − β) as sin(α + (−β)) and using the sum formula and the Even / Odd Identities. Theorem 10. The remaining proofs are left as exercises. • sin(α + β) = sin(α) cos(β) + cos(α) sin(β) • sin(α − β) = sin(α) cos(β) − cos(α) sin(β) . we are now in the position to derive the sum and difference formulas for sine. that the ‘co’sine of an angle is the sine of its ‘co’mplement. Sum and Difference Identities for Sine: For all angles α and β. From sin(θ) = cos π − θ . As in Example 10.15. we have cos2 (α) + 13 = 1. Since β is a √ 1 1 Quadrant III angle. Since 5 5 2 sin(α) = 13 .4. 13 cos(α) = − 12 . To find cos(α). we get 1 + 22 = sec2 (β) so that sec(β) = ± √5. We now need to determine sin(β). We now set about finding cos(β) and sin(β). Foundations of Trigonometry 2.1. we need to write the angle 19π as a sum or difference of common angles. and β is a Quadrant III angle with tan(β) = 2. We could use The Pythagorean Identity cos2 (β) + sin2 (β) = 1. or cos(α) = ± 12 . If α is a Quadrant II angle with sin(α) = find sin(α − β).4. we choose sec(β) = − 5 so cos(β) = sec(β) = −√5 = − 55 . we have sin(β) = tan(β) cos(β) √ so we get sin(β) = (2) − 5 5 = − 2 5 5 . We now have all the pieces needed to find sin(α − β): √ sin(α − β) = sin(α) cos(β) − cos(α) sin(β) √ √ 5 5 12 2 5 = − − − − 13 5 13 5 √ 29 5 = − 65 . 1.15. 12 The denominator of 12 suggests a combination of angles with denominators 3 and 4. 13 but the Pythagorean Identity 1 + tan2 (β) = sec2 (β) is a quick way to get sec(β). Solution. we use the Pythagorean Identity cos2 (α) + sin2 (α) = 1. but we sin(β) opt instead to use a quotient identity. With tan(β) = 2.2. Since α is a Quadrant II angle. Find the exact value of sin 19π 12 5 13 . Applying Theorem 10. 3. We have several ways to proceed. and hence.774 Example 10. we need to find cos(α) and both cos(β) and sin(β). we get 12 3 4 sin 19π 12 4π π + 3 4 4π π 4π π = sin cos + cos sin 3 4 3 4 √ √ √ 3 2 1 2 = − + − 2 2 2 2 √ √ − 6− 2 = 4 = sin 2. Derive a formula for tan(α + β) in terms of tan(α) and tan(β). √ cos(β). 1. In order to find sin(α − β) using Theorem 10. From tan(β) = cos(β) . One such combination is 19π = 4π + π . 10.4 Trigonometric Identities 3. We can start expanding tan(α + β) using a quotient identity and our sum formulas tan(α + β) = = sin(α + β) cos(α + β) sin(α) cos(β) + cos(α) sin(β) cos(α) cos(β) − sin(α) sin(β) 775 sin(α) sin(β) Since tan(α) = cos(α) and tan(β) = cos(β) , it looks as though if we divide both numerator and denominator by cos(α) cos(β) we will have what we want tan(α + β) = 1 sin(α) cos(β) + cos(α) sin(β) cos(α) cos(β) · 1 cos(α) cos(β) − sin(α) sin(β) cos(α) cos(β) sin(α) cos(β) cos(α) sin(β) + cos(α) cos(β) cos(α) cos(β) cos(α) cos(β) sin(α) sin(β) − cos(α) cos(β) cos(α) cos(β) $ $ sin(α)$$$ cos(β) $$$ sin(β) cos(α) $ $ + cos(α) cos(β) cos(α)$$$ $$$ cos(β) $ $ $ $ cos(α)cos(β) sin(α) sin(β) $$ $$ $cos(β) − cos(α) cos(β) $ $$ $$$ cos(α) $ = = = tan(α) + tan(β) 1 − tan(α) tan(β) Naturally, this formula is limited to those cases where all of the tangents are defined. The formula developed in Exercise 10.4.2 for tan(α +β) can be used to find a formula for tan(α −β) by rewriting the difference as a sum, tan(α+(−β)), and the reader is encouraged to fill in the details. Below we summarize all of the sum and difference formulas for cosine, sine and tangent. Theorem 10.16. Sum and Difference Identities: For all applicable angles α and β, • cos(α ± β) = cos(α) cos(β) sin(α) sin(β) • sin(α ± β) = sin(α) cos(β) ± cos(α) sin(β) • tan(α ± β) = tan(α) ± tan(β) 1 tan(α) tan(β) In the statement of Theorem 10.16, we have combined the cases for the sum ‘+’ and difference ‘−’ of angles into one formula. The convention here is that if you want the formula for the sum ‘+’ of 776 Foundations of Trigonometry two angles, you use the top sign in the formula; for the difference, ‘−’, use the bottom sign. For example, tan(α) − tan(β) tan(α − β) = 1 + tan(α) tan(β) If we specialize the sum formulas in Theorem 10.16 to the case when α = β, we obtain the following ‘Double Angle’ Identities. Theorem 10.17. Double Angle Identities: For all applicable angles θ,  2 2  cos (θ) − sin (θ)  2 cos2 (θ) − 1 • cos(2θ) =   1 − 2 sin2 (θ) • sin(2θ) = 2 sin(θ) cos(θ) • tan(2θ) = 2 tan(θ) 1 − tan2 (θ) The three different forms for cos(2θ) can be explained by our ability to ‘exchange’ squares of cosine and sine via the Pythagorean Identity cos2 (θ) + sin2 (θ) = 1 and we leave the details to the reader. It is interesting to note that to determine the value of cos(2θ), only one piece of information is required: either cos(θ) or sin(θ). To determine sin(2θ), however, it appears that we must know both sin(θ) and cos(θ). In the next example, we show how we can find sin(2θ) knowing just one piece of information, namely tan(θ). Example 10.4.3. 1. Suppose P (−3, 4) lies on the terminal side of θ when θ is plotted in standard position. Find cos(2θ) and sin(2θ) and determine the quadrant in which the terminal side of the angle 2θ lies when it is plotted in standard position. 2. If sin(θ) = x for − π ≤ θ ≤ π , find an expression for sin(2θ) in terms of x. 2 2 3. Verify the identity: sin(2θ) = 2 tan(θ) . 1 + tan2 (θ) 4. Express cos(3θ) as a polynomial in terms of cos(θ). Solution. 1. Using Theorem 10.3 from Section 10.2 with x = −3 and y = 4, we find r = x2 + y 2 = 5. 3 Hence, cos(θ) = − 5 and sin(θ) = 4 . Applying Theorem 10.17, we get cos(2θ) = cos2 (θ) − 5 2 2 7 4 3 sin2 (θ) = − 3 − 4 = − 25 , and sin(2θ) = 2 sin(θ) cos(θ) = 2 5 − 5 = − 24 . Since both 5 5 25 cosine and sine of 2θ are negative, the terminal side of 2θ, when plotted in standard position, lies in Quadrant III. 10.4 Trigonometric Identities 777 2. If your first reaction to ‘sin(θ) = x’ is ‘No it’s not, cos(θ) = x!’ then you have indeed learned something, and we take comfort in that. However, context is everything. Here, ‘x’ is just a variable - it does not necessarily represent the x-coordinate of the point on The Unit Circle which lies on the terminal side of θ, assuming θ is drawn in standard position. Here, x represents the quantity sin(θ), and what we wish to know is how to express sin(2θ) in terms of x. We will see more of this kind of thing in Section 10.6, and, as usual, this is something we need for Calculus. Since sin(2θ) = 2 sin(θ) cos(θ), we need to write cos(θ) in terms of x to finish the problem. We substitute x = sin(θ) into the Pythagorean Identity, cos2 (θ) + sin2 (θ) = 1, √ to get cos2 (θ) + x2 = 1, or cos(θ) = ± 1 − x2 . Since − π ≤ θ ≤ π , cos(θ) ≥ 0, and thus 2 2 √ √ cos(θ) = 1 − x2 . Our final answer is sin(2θ) = 2 sin(θ) cos(θ) = 2x 1 − x2 . 3. We start with the right hand side of the identity and note that 1 + tan2 (θ) = sec2 (θ). From this point, we use the Reciprocal and Quotient Identities to rewrite tan(θ) and sec(θ) in terms of cos(θ) and sin(θ): 2 tan(θ) 1 + tan2 (θ) 2 tan(θ) = sec2 (θ) 2 sin(θ) cos(θ) 1 cos2 (θ) sin(θ) cos(θ) = =2 cos2 (θ) = 2 sin(θ) cos(θ) $$$ cos(θ) = 2 sin(θ) cos(θ) = sin(2θ) cos(θ) $$$ 4. In Theorem 10.17, one of the formulas for cos(2θ), namely cos(2θ) = 2 cos2 (θ) − 1, expresses cos(2θ) as a polynomial in terms of cos(θ). We are now asked to find such an identity for cos(3θ). Using the sum formula for cosine, we begin with cos(3θ) = cos(2θ + θ) = cos(2θ) cos(θ) − sin(2θ) sin(θ) Our ultimate goal is to express the right hand side in terms of cos(θ) only. We substitute cos(2θ) = 2 cos2 (θ) − 1 and sin(2θ) = 2 sin(θ) cos(θ) which yields cos(3θ) = cos(2θ) cos(θ) − sin(2θ) sin(θ) = 2 cos2 (θ) − 1 cos(θ) − (2 sin(θ) cos(θ)) sin(θ) = 2 cos3 (θ) − cos(θ) − 2 sin2 (θ) cos(θ) Finally, we exchange sin2 (θ) for 1 − cos2 (θ) courtesy of the Pythagorean Identity, and get cos(3θ) = = = = and we are done. 2 cos3 (θ) − cos(θ) − 2 sin2 (θ) cos(θ) 2 cos3 (θ) − cos(θ) − 2 1 − cos2 (θ) cos(θ) 2 cos3 (θ) − cos(θ) − 2 cos(θ) + 2 cos3 (θ) 4 cos3 (θ) − 3 cos(θ) 778 Foundations of Trigonometry In the last problem in Example 10.4.3, we saw how we could rewrite cos(3θ) as sums of powers of cos(θ). In Calculus, we have occasion to do the reverse; that is, reduce the power of cosine and sine. Solving the identity cos(2θ) = 2 cos2 (θ) − 1 for cos2 (θ) and the identity cos(2θ) = 1 − 2 sin2 (θ) for sin2 (θ) results in the aptly-named ‘Power Reduction’ formulas below. Theorem 10.18. Power Reduction Formulas: For all angles θ, • cos2 (θ) = • sin2 (θ) = 1 + cos(2θ) 2 1 − cos(2θ) 2 Example 10.4.4. Rewrite sin2 (θ) cos2 (θ) as a sum and difference of cosines to the first power. Solution. We begin with a straightforward application of Theorem 10.18 sin2 (θ) cos2 (θ) = = = 1 − cos(2θ) 2 1 1 − cos2 (2θ) 4 1 1 − cos2 (2θ) 4 4 1 + cos(2θ) 2 Next, we apply the power reduction formula to cos2 (2θ) to finish the reduction sin2 (θ) cos2 (θ) = = = = 1 4 1 4 1 4 1 8 1 cos2 (2θ) 4 1 1 + cos(2(2θ)) − 4 2 1 1 − − cos(4θ) 8 8 1 − cos(4θ) 8 − Another application of the Power Reduction Formulas is the Half Angle Formulas. To start, we θ apply the Power Reduction Formula to cos2 2 cos2 θ 2 = 1 + cos 2 2 θ 2 = 1 + cos(θ) . 2 θ We can obtain a formula for cos 2 by extracting square roots. In a similar fashion, we may obtain a half angle formula for sine, and by using a quotient formula, obtain a half angle formula for tangent. We summarize these formulas below. 10.4 Trigonometric Identities Theorem 10.19. Half Angle Formulas: For all applicable angles θ, • cos • sin • tan θ 2 θ 2 θ 2 =± 1 + cos(θ) 2 1 − cos(θ) 2 1 − cos(θ) 1 + cos(θ) θ lies. 2 779 =± =± where the choice of ± depends on the quadrant in which the terminal side of Example 10.4.5. 1. Use a half angle formula to find the exact value of cos (15◦ ). 2. Suppose −π ≤ θ ≤ 0 with cos(θ) = − 3 . Find sin 5 θ 2 . 3. Use the identity given in number 3 of Example 10.4.3 to derive the identity tan Solution. 1. To use the half angle formula, we note that 15◦ = its cosine is positive. Thus we have cos (15◦ ) = + = 30◦ 2 θ 2 = sin(θ) 1 + cos(θ) and since 15◦ is a Quadrant I angle, √ 1 + cos (30◦ ) = 2 √ 1+ 2 3 2 · 2 = 2 1 + 23 2 √ √ 2+ 3 2+ 3 = 4 2 Back in Example 10.4.1, we found√ √ ◦ ) by using the difference formula for cosine. In that cos (15 6+ 2 ◦) = case, we determined cos (15 . The reader is encouraged to prove that these two 4 expressions are equal. 2. If −π ≤ θ ≤ 0, then − π ≤ 2 sin θ 2 ≤ 0, which means sin = − = − θ 2 < 0. Theorem 10.19 gives 3 1 − −5 2 √ 8 2 5 =− 10 5 θ 2 1 − cos (θ) =− 2 1+ 2 3 5 · 5 =− 5 780 Foundations of Trigonometry 3. Instead of our usual approach to verifying identities, namely starting with one side of the equation and trying to transform it into the other, we will start with the identity we proved in number 3 of Example 10.4.3 and manipulate it into the identity we are asked to prove. The 2 tan(θ) identity we are asked to start with is sin(2θ) = 1+tan2 (θ) . If we are to use this to derive an identity for tan θ 2 , it seems reasonable to proceed by replacing each occurrence of θ with sin 2 θ 2 θ 2 = 2 tan 1 + tan2 2 tan 1 + tan2 θ 2 θ 2 θ 2 θ 2 sin(θ) = We now have the sin(θ) we need, but we somehow need to get a factor of 1 + cos(θ) involved. θ θ To get cosines involved, recall that 1 + tan2 2 = sec2 2 . We continue to manipulate our given identity by converting secants to cosines and using a power reduction formula sin(θ) = sin(θ) = 2 tan 1 + tan2 2 tan sec2 θ 2 θ 2 θ 2 θ 2 θ 2 θ 2 θ 2 sin(θ) = 2 tan sin(θ) = 2 tan sin(θ) = tan tan θ 2 = cos2 θ 2 θ 2 1 + cos 2 2 (1 + cos(θ)) sin(θ) 1 + cos(θ) Our next batch of identities, the Product to Sum Formulas,3 are easily verified by expanding each of the right hand sides in accordance with Theorem 10.16 and as you should expect by now we leave the details as exercises. They are of particular use in Calculus, and we list them here for reference. Theorem 10.20. Product to Sum Formulas: For all angles α and β, • cos(α) cos(β) = • sin(α) sin(β) = • sin(α) cos(β) = 3 1 2 1 2 1 2 [cos(α − β) + cos(α + β)] [cos(α − β) − cos(α + β)] [sin(α − β) + sin(α + β)] These are also known as the Prosthaphaeresis Formulas and have a rich history. The authors recommend that you conduct some research on them as your schedule allows. 10.4 Trigonometric Identities 781 Related to the Product to Sum Formulas are the Sum to Product Formulas, which we will have need of in Section 10.7. These are easily verified using the Product to Sum Formulas, and as such, their proofs are left as exercises. Theorem 10.21. Sum to Product Formulas: For all angles α and β, • cos(α) + cos(β) = 2 cos • cos(α) − cos(β) = −2 sin • sin(α) ± sin(β) = 2 sin Example 10.4.6. 1. Write cos(2θ) cos(6θ) as a sum. 2. Write sin(θ) − sin(3θ) as a product. Solution. 1. Identifying α = 2θ and β = 6θ, we find cos(2θ) cos(6θ) = = = 1 2 1 2 1 2 α+β 2 α+β 2 α±β 2 cos sin cos α−β 2 α−β 2 α 2 β [cos(2θ − 6θ) + cos(2θ + 6θ)] cos(−4θ) + 1 cos(8θ) 2 cos(4θ) + 1 cos(8θ), 2 where the last equality is courtesy of the even identity for cosine, cos(−4θ) = cos(4θ). 2. Identifying α = θ and β = 3θ yields sin(θ) − sin(3θ) = 2 sin θ + 3θ θ − 3θ cos 2 2 = 2 sin (−θ) cos (2θ) = −2 sin (θ) cos (2θ) , where the last equality is courtesy of the odd identity for sine, sin(−θ) = − sin(θ). The reader is reminded that all of the identities presented in this section which regard the circular functions as functions of angles (in radian measure) apply equally well to the circular (trigonometric) functions regarded as functions of real numbers. In Exercises 38 - 43 in Section 10.5, we see how some of these identities manifest themselves geometrically as we study the graphs of the these functions. In the upcoming Exercises, however, you need to do all of your work analytically without graphs. If α is a Quadrant IV angle with cos(α) = . sin(105◦ ) 12.1 Exercises In Exercises 1 . cot(255◦ ) 14. where < β < π. where 0 < α < . sec(165◦ ) 11.6. where 0 < α < (a) cos(α + β) (d) cos(α − β) (b) sin(α + β) (e) sin(α − β) (c) tan(α + β) (f) tan(α − β) π . and cos(β) = where < β < 2π. sec − √ 5 10 π 22. 7. and β is a Quadrant II angle with tan(β) = −7. sin(3π − 2θ) = − sin(2θ − 3π) 3. tan(−t2 + 1) = − tan(t2 − 1) 5. You may have need of the Quotient. tan 18. sec(−6t) = sec(6t) 2. and sin(β) = . find 5 2 13 2 (a) sin(α + β) (b) cos(α − β) (c) tan(α − β) .4. sin 13π 12 π 12 π 12 21. csc(−θ − 5) = − csc(θ + 5) 6. If sin(α) = . use the Sum and Difference Identities to find the exact value. sin 17. 1. Reciprocal or Even / Odd Identities as well. cos 19. use the Even / Odd Identities to verify the identity. csc(195◦ ) 13. cos 16. cot 13π 12 7π 12 11π 12 8. cot(9 − 7θ) = − cot(7θ − 9) In Exercises 7 . find 5 10 2 (a) cos(α + β) (d) cos(α − β) 23. tan 20. cos − π π − 5t = cos 5t + 4 4 4. tan(375◦ ) 15. Assume all quantities are defined. find 2 (b) sin(α + β) (e) sin(α − β) (c) tan(α + β) (f) tan(α − β) π 12 3π 3 24.782 Foundations of Trigonometry 10.21. csc 11π 12 17π 12 5π 12 √ 9. If csc(α) = 3. cos(75◦ ) 10. where < α < π. sin(157. cos(θ − π) = − cos(θ) 28.48. cos(α + β) − cos(α − β) = −2 sin(α) sin(β) 34. You may have need of the Quotient. sin(α + β) + sin(α − β) = 2 sin(α) cos(β) 31. tan 7π 12 π 8 7π 8 (compare with Exercise 16) . cos(75◦ ) (compare with Exercise 7) 41. tan θ + π 2 = − cot(θ) 27. sin(105◦ ) (compare with Exercise 9) 42.5◦ ) 44. sin π 12 5π 8 (compare with Exercise 18) 40. sin(α + β) − sin(α − β) = 2 cos(α) sin(β) 32.5◦ ) 43. Reciprocal or Even / Odd Identities as well. cos 48. 36. cos(h) − 1 h − sin(t) 38. 26.10. cos(67. and tan(β) = . verify the identity. find 3 2 7 2 (a) csc(α − β) (b) sec(α + β) (c) cot(α + β) 783 In Exercises 26 . sin(π − θ) = sin(θ) 29. 39. cos(α + β) + cos(α − β) = 2 cos(α) cos(β) 33. cos 46. + sin(t) sin(α + β) 1 + cot(α) tan(β) = sin(α − β) 1 − cot(α) tan(β) tan(α + β) sin(α) cos(α) + sin(β) cos(β) = tan(α − β) sin(α) cos(α) − sin(β) cos(β) 30.38. If sec(α) = − .4 Trigonometric Identities 5 π 24 3π 25. use the Half Angle Formulas to find the exact value.5◦ ) 45. tan(h) h sec2 (t) 1 − tan(t) tan(h) In Exercises 39 . where π < β < . sin 47. tan(112. cos(α + β) 1 − tan(α) tan(β) = cos(α − β) 1 + tan(α) tan(β) sin(t + h) − sin(t) = cos(t) h cos(t + h) − cos(t) = cos(t) h tan(t + h) − tan(t) = h sin(h) h cos(h) − 1 h sin(h) h 37. 35. sin(θ) = − In Exercises 59 . 53. 32 sin4 (θ) cos2 (θ) = 2 − cos(2θ) − 2 cos(4θ) + cos(6θ) 69. 8 cos4 (θ) = cos(4θ) + 4 cos(2θ) + 3 66. 59. use the given information about θ to find the exact values of • sin(2θ) • sin θ 2 • cos(2θ) • cos θ 2 50. 73. tan(θ) = −2 where < θ < π 2 54. sin(3θ) = 3 sin(θ) − 4 sin3 (θ) 64. cos(4θ) = 8 cos4 (θ) − 8 cos2 (θ) + 1 70. csc(2θ) = cot(θ) + tan(θ) 2 63.) 71. 7 3π where < θ < 2π 25 2 12 3π tan(θ) = where π < θ < 5 2 3 π cos(θ) = where 0 < θ < 5 2 12 3π cos(θ) = where < θ < 2π 13 2 √ 3π sec(θ) = 5 where < θ < 2π 2 28 π where 0 < θ < 53 2 π 52. sec(2θ) = 72. Assume all quantities are defined. cos(θ) sin(θ) + cos(θ) + sin(θ) cos(θ) − sin(θ) 1 1 2 cos(θ) + = cos(θ) − sin(θ) cos(θ) + sin(θ) cos(2θ) 1 1 2 sin(θ) − = cos(θ) − sin(θ) cos(θ) + sin(θ) cos(2θ) . (cos(θ) + sin(θ))2 = 1 + sin(2θ) 61. 8 sin4 (θ) = cos(4θ) − 4 cos(2θ) + 3 65. csc(θ) = 4 where < θ < π 2 4 3π where π < θ < 5 2 5 π 56. cos(8θ) = 128 cos8 (θ) − 256 cos6 (θ) + 160 cos4 (θ) − 32 cos2 (θ) + 1 (HINT: Use the result to 69. (cos(θ) − sin(θ))2 = 1 − sin(2θ) 62. sin(4θ) = 4 sin(θ) cos3 (θ) − 4 sin3 (θ) cos(θ) 67. 55. tan(2θ) = 1 1 − 1 − tan(θ) 1 + tan(θ) 60. cos(θ) = • tan(2θ) • tan θ 2 49.784 Foundations of Trigonometry In Exercises 49 . sin(θ) = − 51.58. 32 sin2 (θ) cos4 (θ) = 2 + cos(2θ) − 2 cos(4θ) − cos(6θ) 68. verify the identity. 57.73. sin(θ) = where < θ < π 13 2 π 58. Verify the following formulas (a) cos(θ) = √ (c) sin(2θ) = 1 x2 + 1 2x +1 (b) sin(θ) = √ (d) cos(2θ) = x x2 + 1 x2 1 − x2 x2 + 1 89. You may need to use an Even/Odd Identity. or IV angle. find an expression for cos(2θ) in terms of x. cos(5θ) − cos(6θ) 85. 7 2 2 x π 92. Suppose θ is a Quadrant I angle with sin(θ) = x. cos(θ) − sin(θ) 86. if any. 2 2 2 x π π 91. sin(2θ) sin(7θ) 78. if any. Show that cos2 (θ) − sin2 (θ) = 2 cos2 (θ) − 1 = 1 − 2 sin2 (θ) for all θ. Discuss with your classmates how each of the formulas. III. or IV angle. in Exercise 88 change if we change assume θ is a Quadrant II. sin(3θ) sin(2θ) 76. cos(3θ) + cos(5θ) 83.10. cos(2θ) cos(6θ) 75. 1 94. sin(2θ) − sin(7θ) 84. Verify the following formulas (a) cos(θ) = √ 1 − x2 √ (b) sin(2θ) = 2x 1 − x2 (c) cos(2θ) = 1 − 2x2 87. You may need to use an Even/Odd or Cofunction Identity. Let θ be a Quadrant III angle with cos(θ) = − . III. If sec(θ) = for 0 < θ < . sin(9θ) − sin(−θ) 81.85. 88.4 Trigonometric Identities 785 In Exercises 74 . Suppose θ is a Quadrant I angle with tan(θ) = x. in Exercise 86 change if we change assume θ is a Quadrant II. sin(θ) + cos(θ) 82. 2 . sin(9θ) cos(θ) 79. write the given product as a sum. cos(3θ) cos(5θ) 77. Discuss with your classmates how each of the formulas. x π π for − < θ < .79. If sin(θ) = 93. find an expression for sin(2θ) in terms of x. cos(θ) sin(3θ) In Exercises 80 . Show that this is not enough information to 5 7π 3π θ determine the sign of sin by first assuming 3π < θ < and then assuming π < θ < 2 2 2 θ and computing sin in both cases. 74. If tan(θ) = for − < θ < . find an expression for ln | sec(θ) + tan(θ)| in terms of x. write the given sum as a product. 4 2 90. 80. we had you verify an identity which expresses cos(4θ) as a polynomial in terms of cos(θ). secant. Verify the Product to Sum Identities.786 2+ 2 √ 3 Foundations of Trigonometry √ = 6+ 4 √ 2 95. 100. Can you do the same for sin(5θ)? What about for sin(4θ)? If not. Verify the Even / Odd Identities for tangent. 101. what goes wrong? 98. Without using your calculator. Verify the Difference Identities for sine and tangent. In part 4 of Example 10.3. we wrote cos(3θ) as a polynomial in terms of cos(θ). secant. In Exercise 69. cosecant and cotangent. 99. In Exercise 65.4. show that 96. 102. . cosecant and cotangent. Verify the Cofunction Identities for tangent. Verify the Sum to Product Identities. Can you find a polynomial in terms of cos(θ) for cos(5θ)? cos(6θ)? Can you find a pattern so that cos(nθ) could be written as a polynomial in cosine for any natural number n? 97. we has you verify an identity which expresses sin(3θ) as a polynomial in terms of sin(θ). sec − (c) tan(α + β) = −7 √ (e) sin(α − β) = 2 2 √ 7 2 (b) sin(α + β) = 10 √ 2 (d) cos(α − β) = − 2 (f) tan(α − β) = −1 √ 28 − 2 (b) sin(α + β) = 30 √ −4 + 7 2 (d) cos(α − β) = 30 √ √ 28 + 2 63 + 100 2 √ =− (f) tan(α − β) = 41 4−7 2 33 65 (c) tan(α − β) = 56 33 √ 4+7 2 23. √ 6− 2 = 4 √ √ 6+ 2 ◦) = sin(105 4 √ √ 3−1 cot(255◦ ) = √ =2− 3 3+1 √ √ 13π 6+ 2 =− cos 12 4 √ √ 13π 3− 3 √ =2− 3 tan = 12 3+ 3 cos(75◦ ) 17π 12 11π 12 =2+ √ 3 √ 3) 12. (a) cos(α + β) = − 30 √ √ −28 + 2 63 − 100 2 √ = (c) tan(α + β) = 41 4+7 2 √ 28 + 2 (e) sin(α − β) = − 30 24.10. csc(195◦ ) = √ √ √ 4 √ = 2− 6 2+ 6 13. cot 18. 15. Answers √ 8. csc √ √ π = 6− 2 12 √ 2 22.4. √ √ 4 √ = −( 2 + 6) 2− 6 √ √ ◦ ) = 3 − √3 = 2 − 3 tan(375 3+ 3 √ √ 11π 6− 2 = sin 12 4 √ √ 7π 2− 6 cos = 12 4 √ √ 6− 2 π = sin 12 4 5π 12 = √ 6− √ 2 = −(2 + 20.4 Trigonometric Identities 787 10. 17. 16. 9. tan 19. 14. 11. (a) sin(α + β) = 16 65 (b) cos(α − β) = .2 7. (a) cos(α + β) = − 10 21. sec(165◦ ) = − √ 10. 788 25.5◦ ) 2− 2 √ sin(157. sin 5π 8 = π 46. sin = 12 47. tan • cos(2θ) = • cos θ 2 = 2+ 2 √ 7π 8 √ √ 2− 2 √ =1− 2 2+ 2 • tan(2θ) = − • tan θ 2 49. 120 169 √ 3 13 θ • sin 2 = 13 √ 15 • sin(2θ) = − 8 • sin(2θ) = • sin θ 2 119 169 √ 2 13 =− 13 7 8 √ 8 − 2 15 4 • cos(2θ) = • cos 120 119 3 θ • tan 2 = − 2 √ 15 • tan(2θ) = − 7 • tan(2θ) = − • tan tan θ 2 θ 2 = √ 8 + 2 15 4 θ 2 = √ 8 + 2 15 √ = 8 − 2 15 √ = 4 + 15 24 7 53. 41. cos =− 2+ 2 =− 2− 2 √ 2 √ π 45. 336 • sin(2θ) = − 625 √ 2 θ • sin 2 = 10 • sin(2θ) = • sin θ 2 527 625 √ 7 2 =− 10 336 527 1 =− 7 2520 1241 50. • sin(2θ) = • sin θ 2 24 25 √ 5 = 5 7 • cos(2θ) = − 25 √ 2 5 θ • cos 2 = 5 • tan(2θ) = − • tan θ 2 = 1 2 .5◦ ) =− 2− 2 √ √ 2+ 2 √ = −1 − 2 2− 2 √ 3 2 44. sin(105◦ ) = = cos(67. cos 8 48. 2520 2809 √ 5 106 = 106 1241 • cos(2θ) = − 2809 √ 9 106 θ • cos 2 = 106 • cos(2θ) = − • cos θ 2 • tan(2θ) = − • tan θ 2 = 5 9 51. 52. (a) csc(α − β) = − 2− 2 √ 5 4 (b) sec(α + β) = 125 117 Foundations of Trigonometry (c) cot(α + β) = 2+ 2 √ 3 2 3 117 44 39. cos(75◦ ) = = 3 2 40. tan(112.5◦ ) 7π 12 2− 2 √ 43. 42. 10. • sin(2θ) = − • sin θ 2 4 5 √ 50 + 10 5 10 • cos(2θ) = − • cos θ 2 3 5 √ 50 − 10 5 10 = = √ 5− 5 √ • tan =− 5+ 5 √ 5−5 5 θ tan 2 = 10 4 • tan(2θ) = 3 √ 5+ 5 θ √ • tan 2 = 5− 5 √ 5+5 5 θ tan 2 = 10 76. 2 sin 2 cos θ − 14x + 49 x2 √ π 85. • • 56.4 Trigonometric Identities 54. sin(8θ) + sin(10θ) 2 sin(2θ) + sin(4θ) 2 11 θ sin 2 1 θ 2 74. • sin(2θ) = • 55. 2 cos(4θ) cos(θ) 83. √ 82. − 2 sin θ − 4 √ 92. • • 57. • 24 25 √ 2 5 θ sin 2 = 5 120 sin(2θ) = − 169 √ 26 θ sin 2 = 26 120 sin(2θ) = − 169 √ 5 26 θ sin 2 = 26 4 sin(2θ) = − 5 θ 2 789 • cos(2θ) = − • • • • • • 7 25 √ 5 θ cos 2 = − 5 119 cos(2θ) = 169 √ 5 26 θ cos 2 = − 26 119 cos(2θ) = 169 √ 26 θ cos 2 = 26 3 cos(2θ) = − 5 θ 2 • tan(2θ) = − • tan θ 2 24 7 = −2 120 119 1 θ • tan 2 = − 5 120 • tan(2θ) = − 119 • tan(2θ) = − • tan θ 2 =5 4 3 • tan(2θ) = √ θ 2 • sin = √ 50 − 10 5 10 • cos =− 50 + 10 5 10 58. 79. 91. −2 cos 84. 1 − x2 2 81. 78. 2 cos(4θ) sin(5θ) 90. ln |x + x2 + 16| − ln(4) . cos(5θ) − cos(9θ) 2 cos(θ) − cos(5θ) 2 9 θ sin 2 π 4 5 θ 2 80. 77. cos(2θ) + cos(8θ) 2 cos(4θ) + cos(8θ) 2 75. To determine the period of f . Recall from Section 3. Then. 2π]. cos(0 + p) = cos(0) so that cos(p) = 1.2 Having period 2π essentially means that we can completely understand everything about the functions f (t) = cos(t) and g(t) = sin(t) by studying one interval of length 2π. 2 1 .790 Foundations of Trigonometry 10. while g(t) = sin(t) is an odd function. The Even / Odd Identities in Theorem 10.1 that geometrically this means the graphs of the cosine and sine functions have no jumps. ∞).5 in Section 10.12 tell us cos(−t) = cos(t) for all real numbers t and sin(−t) = − sin(t) for all real numbers t.3 One last property of the functions f (t) = cos(t) and g(t) = sin(t) is worth pointing out: both of these functions are continuous and smooth.5 Graphs of the Trigonometric Functions In this section. cos(α) = cos(β) and sin(α) = sin(β). Said differently.1. the smallest positive real number p such that cos(t + p) = cos(t) for all real numbers t. asymptotes. say [0.3.6 for a review of these concepts. The smallest positive number p for which f (t + p) = f (t) for all real numbers t in the domain of f . Suppose p > 0 and cos(t + p) = cos(t) for all real numbers t. we can show g(t) = sin(t) is also periodic with 2π as its period.) Returning to the circular functions. (We’ll leave that odd gem as an exercise for you. holes in the graph. As usual. since the smallest positive multiple of 2π is 2π itself. This means f (t) = cos(t) is an even function. However. See section 1.1 Graphs of the Cosine and Sine Functions From Theorem 10. We know that cos(t + 2π) = cos(t) for all real numbers t but the question remains if any smaller real number will do the trick. we begin our study with the functions f (t) = cos(t) and g(t) = sin(t).3. 4 In some advanced texts. We have already seen a family of periodic functions in Section 2.14 to show that g(t) = sin(t) is periodic with period 2π since g(t) = sin(t) = cos π − t = f π − t . we know that the domain of f (t) = cos(t) and of g(t) = sin(t) is all real numbers. 1]. cos(t+2πk) = cos(t) and sin(t+2πk) = sin(t) for all real numbers t and any integer k. we return to our discussion of the circular (trigonometric) functions as functions of real numbers and pick up where we left off in Sections 10.1 and 10. in particular. we need to find the smallest real number p so that f (t + p) = f (t) for all real numbers t or. if it exists. Similarly. Definition 10. Alternatively. gaps.2.2.1: the constant functions. π). From this we know p is a multiple of 2π and. (−∞. we should study the interval [0. t = 2π gives us an extra ‘check’ when we go to graph these functions. is called the period of f .1 Another important property of these functions is that for coterminal angles α and β.5. f (t) = cos(t) is periodic. and the range of both functions is [−1. we have the result. 10.1. said differently. since cos(t + 2πk) = cos(t) for any integer k. the interval of choice is [−π. we can use the Cofunction Identities in Theorem 10. 2 2 3 Technically. This last property is given a special name. As we will see shortly. despite being periodic a constant function has no period.3. we see that by Definition 10.4 since whatever happens at t = 2π is the same as what happens at t = 0. Periodic Functions: A function f is said to be periodic if there is a real number c so that f (t + c) = f (t) for all real numbers t in the domain of f . 2π). we imagine ‘copying and pasting’ this graph end to end infinitely in both directions (left and right) on the x-axis. First. ∞) – has range [−1. the graphs of both f (t) = cos(t) and g(t) = sin(t) meander nicely and don’t cause any trouble. To graph y = cos(x).6 and used x as the independent variable and y as the dependent variable.0 √ 3π . 1] – is continuous and smooth – is odd – has period 2π In the chart above. The 5 The use of x and y in this context is not to be confused with the x. Theorem 10. Properties of the Cosine and Sine Functions • The function f (x) = cos(x) – has domain (−∞. A few things about the graph above are worth mentioning. we make a table as we did in Section 1.6 using some of the ‘common values’ of x in the interval [0.5 Graphs of the Trigonometric Functions 791 corners or cusps.and y-coordinates of points on the Unit Circle which define cosine and sine.5 This allows us to turn our attention to graphing the cosine and sine functions in the Cartesian Plane. which we call the ‘fundamental cycle’ of y = cos(x). 2π]. −1) −1 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x 2π (2π. As we shall see.22. − 22 4 3π 2 . We summarize these facts in the following theorem. this graph represents only part of the graph of y = cos(x). x 0 π 4 π 2 3π 4 cos(x) 1 √ 2 2 (x. Secondly. cos(x)) (0. 1] – is continuous and smooth – is even – has period 2π • The function g(x) = sin(x) – has domain (−∞. the vertical scale here has been greatly exaggerated for clarity and aesthetics.10. − 22 4 1 y √ 0 − 2 2 π 5π 4 3π 2 7π 4 −1 √ − 2 2 √ 0 1 2 2 √ 5π . This generates a portion of the cosine graph. 22 4 π 2. Below is an accurate-to-scale graph of y = cos(x) showing several cycles with the ‘fundamental cycle’ plotted thicker than the others. where’s the triangle?” . ∞) – has range [−1. To get the entire graph. “Hey. but one could then ask. 1) The ‘fundamental cycle’ of y = cos(x). Using the term ‘trigonometric function’ as opposed to ‘circular function’ can help with that. 22 4 (π. we followed the convention established in Section 1.0 √ 7π . 1) √ π . 7. we see from this formula that the graph of y = sin(x) is the result of shifting the graph of y = cos(x) to the right π units. we provide an accurately scaled graph of y = sin(x) below with the fundamental cycle highlighted. x 0 π 4 π 2 3π 4 sin(x) 0 √ 2 2 (x. 22 4 1 y √ 1 0 2 2 π 5π 4 3π 2 7π 4 √ − 2 2 −1 √ − 2 2 √ 2 5π 4 . − 22 4 (π. −1 √ 7π . 0) √ 2 π 4. We can plot the fundamental cycle of the graph of y = sin(x) similarly. we can use Theorem 1. 2 π 2. with similar results. we have π π π − x = cos − x − = cos x − 2 2 2 Recalling Section 1. To do so. we need to keep track of .7 to graph more complicated curves. A visual inspection confirms this. y x An accurately scaled graph of y = cos(x).7 in Section 1.− 2 3π 2 . It is no accident that the graphs of y = cos(x) and y = sin(x) are so similar. many of the applications involving the cosine and sine functions feature modeling wavelike phenomena. Using a cofunction identity along with the even property of cosine. sin(x)) (0. y x An accurately scaled graph of y = sin(x).792 Foundations of Trigonometry graph of y = cos(x) is usually described as ‘wavelike’ – indeed. 0) −1 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x 2π 0 (2π. 2 sin(x) = cos Now that we know the basic shapes of the graphs of y = cos(x) and y = sin(x). As with the graph of y = cos(x). 0) The ‘fundamental cycle’ of y = sin(x).1 √ 3π . g(x) = 1 2 sin(π − 2x) + 3 2 π. which in this case is 2x − π. One cycle is graphed on [1. πx−π . the argument of a trigonometric function is the expression ‘inside’ the function.10. f (x) = 3 cos Solution. −2) (4. 5] so the period is the length of that interval which is 4. We choose to track the values x = 0.5. Loosely stated. we set the argument of the sine. We shall have occasion. 1. Proceeding as above. 3π 2 . 4) 4 3 2 1 1 −1 −2 2 3 4 5 x One cycle of y = f (x). πx−π 2 +1 2. equal to each of the values: 0. 4) (2. Graph one cycle of the following functions. equal to each of our quarter marks and solve for x. a 0 π 2 πx−π =a 2 πx−π =0 2 πx−π =π 2 2 πx−π =π 2 πx−π = 3π 2 2 πx−π = 2π 2 π 2. Before we begin our next example. y x 1 2 3 4 5 f (x) (x. 2.4. f (x)) 4 1 −2 1 4 (1. . We set the argument of the cosine. For the function f (x) = 1 − 5 cos(2x − π).5 Graphs of the Trigonometric Functions 793 the movement of some key points on the original graphs. however. 1) (5. π − 2x. These ‘quarter marks’ correspond to quadrantal angles. 1) (3. π. 2 3π 2 and 2π. the argument of f is x. 1. and as such. 2 solve for x. mark the location of the zeros and the local extrema of these functions over exactly one period. State the period of each. We summarize the results below.1. 2π and x 1 2 3 4 5 π 3π 2 2π Next. to refer to the argument of the cosine. π . we substitute each of these x values into f (x) = 3 cos πx−π + 1 to determine the 2 corresponding y-values and connect the dots in a pleasing wavelike fashion. Example 10. we need to review the concept of the ‘argument’ of a function as first introduced in Section 1. the reverse is true: f (x) = cos(x) can be written as a transformed version of g(x) = sin(x).794 Foundations of Trigonometry a 0 π 2 π − 2x = a π − 2x = 0 π − 2x = π − 2x = π 2 x π 2 π 4 π 3π 2 π − 2x = π 3π 2 0 −π 4 2π π − 2x = 2π − π 2 We now find the corresponding y-values on the graph by substituting each of these x-values into g(x) = 1 sin(π − 2x) + 3 . the sinusoids are always written in terms of sine functions. 6 . The amplitude of the standard cosine and sine functions is 1. as indicated in the figure below.1 are examples of sinusoids.7. 2 The functions in Example 10. We have already discussed period. phase shift and vertical shift. 2 2 1 − π 2 − π 4 π 4 π 2 x One cycle of y = g(x). One cycle was graphed on the interval − π . a sinusoid is the result of taking the basic graph of f (x) = cos(x) or g(x) = sin(x) and performing any of the transformations6 mentioned in Section 1. amplitude. Roughly speaking. we connect the dots in a wavelike fashion. we will keep our options open. how long it takes for the sinusoid to complete one cycle. Once again. so of course. The authors have seen some instances where sinusoids are always converted to cosine functions while in other disciplines. The amplitude of the sinusoid is a measure of how ‘tall’ the wave is. 2 2 x π 2 π 4 g(x) (x. We will discuss the applications of sinusoids in greater detail in Chapter 11. Until then. 1 4 π 3 −2. but vertical scalings can alter this.5. 2 π 4. g(x)) 3 2 y 2 3 2 0 −π 4 −π 2 1 3 2 π 3 2. We have already seen how the Even/Odd and Cofunction Identities can be used to rewrite g(x) = sin(x) as a transformed version of f (x) = cos(x). that is. 3 2 −π. The standard period of both f (x) = cos(x) and g(x) = sin(x) is 2π.2 0. Sinusoids can be characterized by four properties: period. but horizontal scalings will change the period of the resulting sinusoid. π so the period is 2 2 π 2 − − π = π. 23 is called the phase of the sinusoid. but we state the more general case below. a phase shift of π to the left takes g(x) = sin(x) to 2 2 f (x) = cos(x).23 using the functions f and g featured in Example 10.7.23. The parameter ω.7 in Section 1.7 We now test out Theorem 10. The vertical shift of a sinusoid is exactly the same as the vertical shifts in Section 1. The proof of Theorem 10. Try using the formulas in Theorem 10.5 Graphs of the Trigonometric Functions 795 amplitude baseline period The phase shift of the sinusoid is the horizontal shift experienced by the fundamental cycle. We can always ensure ω > 0 using the Even/Odd Identities.23 is a direct application of Theorem 1. First.10.1. we write f (x) in the form prescribed in Theorem 10.23.5.7 and is left to the reader. In most contexts. . Theorem 10. We have seen that a phase (horizontal) shift of π to the right takes f (x) = cos(x) to g(x) = sin(x) since 2 cos x − π = sin(x). As the reader can verify. which is stipulated to be positive. is called the (angular) frequency of the sinusoid and is the number of cycles the sinusoid completes over a 2π interval. f (x) = 3 cos 7 πx − π 2 + 1 = 3 cos π π x+ − 2 2 + 1.7 in Section 1. shows how to find these four fundamental quantities from the formula of the given sinusoid.7.23 applied to C(x) = cos(−x + π) to see why we need ω > 0. we focus our attention on the horizontal shift − ω induced by φ. which is reminiscent of Theorem 1. the functions C(x) = A cos(ωx + φ) + B • have period 2π ω and S(x) = A sin(ωx + φ) + B • have phase shift − φ ω • have amplitude |A| • have vertical shift B We note that in some scientific and engineering circles. For ω > 0. the quantity φ mentioned in Theorem 10. the vertical shift of a sinusoid is assumed to be 0. Since our interest in this book is primarily with graphing φ sinusoids. The following theorem. 23. we get one complete cycle of the graph. 1 2 2 −1 −1 1 2 3 4 5 x −2 3 2.5. the amplitude is |A| = |3| = 3. ω = π . all of the data match our graph of y = g(x) with the exception of the phase shift. Find a cosine function whose graph matches the graph of y = f (x).23. shift it 1 unit to the right. 1 2 2 7. Note that whether we graph y = g(x) using the ‘quarter marks’ approach or using the Theorem 10.1 is only one portion of the graph of y = g(x). we first need to use the odd property of the sine function to write it in the form required by Theorem 10. however. y 3 −1.1.23. Indeed. Instead of the graph starting at x = π . another complete cycle begins at x = π . 1 unit up. According to Theorem 10. if we start with the basic shape of the cosine graph. Below is the graph of one complete cycle of a sinusoid y = f (x). that the 2 graph presented in Example 10. we had to rewrite the formula for g(x) using the odd property of the sine function. The period is then 2π = π. stretch the amplitude to 3 and shrink the period to 4. the phase shift is − ω = − −π/2 = 1 (indicating ω π/2 a shift to the right 1 unit) and the vertical shift is B = 1 (indicating a shift up 1 unit. 1. we will have reconstructed one period of the graph of y = f (x). ω = 2. the period of f is 2 2 φ 2π 2π = π/2 = 4. Note that. φ = −π and B = 3 . it ends there. In other words.5. and this is the cycle Theorem 10. φ = − π and B = 1. we can use five other pieces of information: the phase shift.) All of these match with our graph of y = f (x).23 g(x) = 1 3 1 3 1 3 1 3 sin(π − 2x) + = sin(−(2x − π)) + = − sin(2x − π) + = − sin(2x + (−π)) + 2 2 2 2 2 2 2 2 We find A = − 1 . the phase shift is − −π = π (indicating a shift right π units) and the vertical shift is up 2 2 2 3 2 . The reason for the 2 discrepancy is that. in this case. 5 2 5.5. amplitude. in order to apply Theorem 10. Example 10. .23 is detecting. 5 2 2 1 1. vertical shift.796 Foundations of Trigonometry so that A = 3. Turning our attention now to the function g in Example 10. Moreover.2. − 2 One cycle of y = f (x). period and basic shape of the cosine curve. instead of tracking the five ‘quarter marks’ through the transformations to plot y = f (x). Remember. the amplitude is 2 2 2 1 1 − 2 = 2 . which means we have completely determined the sinusoid. 5]. Note that each of the answers given in Example 10. we see that the phase shift is −1. 2 6 3 6 2 Alternatively. if we use the sum identity for cosine. 1 .10. To that end. we can expand the formula to yield C(x) = A cos(ωx + φ) + B = A cos(ωx) cos(φ) − A sin(ωx) sin(φ) + B. we get − ω = 7 . and so on. so that ω = π . Next. Taking the phase shift to be 7 . 3 6 2 y 3 5. or ω 3 π 3 5 φ = ω = 3 . we 2 2 2 1 average the endpoints of the range to find B = 1 5 + − 3 = 2 (1) = 1 . or φ = − 2 ω = − 7 π = − 7π .23. Find a sine function whose graph matches the graph of y = f (x). 797 1. we imagine extending the graph of the given sinusoid as in the figure below so that we can identify a cycle beginning φ 7 at 7 . the phase shift is 1 so φ = − π for an answer of S(x) = −2 sin π x − π + 1 . 1 the amplitude A = 1 5 − − 3 = 2 (4) = 2. 1 2 2 19 . note that the range of the sinusoid is − 2 . amplitude and vertical shift are the same as before with ω = π . According to Theorem 10. 1 2 2 13 .5. 1 . − 3 2 Extending the graph of y = f (x). To find the amplitude. 2 taking A = −2. to determine the vertical shift. We fit the data to a function of the form C(x) = A cos(ωx + φ) + B. φ 6 = 2π . 5 2 2 −1 −1 1 2 3 4 5 6 7 8 9 10 x −2 8. Solution. Finally. In this case. . 2 2 2 2 2 3 6 Hence. The period. Our final answer is 2 2 2 2 π π 1 C(x) = 2 cos 3 x + 3 + 2 . so we have − ω = −1. Each of these formulas determine the same sinusoid curve 2 and their formulas are all equivalent using identities. A = 2 and 3 1 B = 2 . 2 . 1 1 For example.2 is one choice out of many possible answers.5 Graphs of the Trigonometric Functions 2. when fitting a sine function to the data. 2. our answer is S(x) = 2 sin π x − 7π + 1 . Speaking of identities. The trickier part is finding the phase shift. its period is 5 − (−1) = 6. Since one cycle is graphed over the interval [−1. 5 2 2 1 7. we could have chosen to start at 2 . we could have extended the graph of y = f (x) to the left and considered a sine 5 function starting at − 2 . Most of the work to fit the data to a function of the form S(x) = A sin(ωx + φ) + B is done. As a result. using the sum identity for sine. we have 2 cos(φ) = 1 and 2 sin(φ) = 3 or. We get started by equating the coefficients of the trigonometric functions on either side of the equation. one way to write f (x) as a sinusoid is f (x) = 2 cos 2x + π .5. Choosing A = 2. On the left hand side. we get cos(2x) − √ 3 sin(2x) = A cos(ωx) cos(φ) − A sin(ωx) sin(φ) + B It should be clear that we can take ω = 2 and B = 0 to get cos(2x) − √ 3 sin(2x) = A cos(2x) cos(φ) − A sin(2x) sin(φ) To determine A and φ. in the form S(x) = A sin(ωx + φ) + B for ω > 0 Check your answers analytically using identities and graphically using a calculator. Similarly. Since this equation is to hold for all real numbers. Making these observations allows us to recognize (and graph) functions as sinusoids which. we find by A equating the coefficients of sin(2x) that A sin(φ) = 3. we get Foundations of Trigonometry S(x) = A sin(ωx + φ) + B = A sin(ωx) cos(φ) + A cos(ωx) sin(φ) + B. a bit more work is involved.798 Similarly.6. so multiplying this by A√ gives √ 2 cos2 (φ)+A2 sin2 (φ) = A2 . don’t appear to fit the forms of either C(x) or S(x). the coefficient of cos(2x) is 1. Equating f (x) = cos(2x) − 3 sin(2x) with the expanded form of C(x) = A cos(ωx + φ) + B. Find a formula for f (x): 1. . Hence. we get√ A 4 or A = ±2. √ Example 10. Since A cos(φ) = 1 and A sin(φ) = 2 = 12 +( 3)2 = A 3. The key to this problem is to use the expanded forms of the sinusoid formulas and match up √ corresponding coefficients. at first glance. We know cos2 (φ) + sin2 (φ) = 1. we must have8 that √ cos(φ) = 1. One such angle φ which satisfies this criteria is φ = π . cos(φ) = 2 and sin(φ) = 23 . Solution. Consider the function f (x) = cos(2x) − 3 sin(2x). What we have here is a system of nonlinear equations! We can temporarily eliminate the dependence on φ by using the 2 Pythagorean Identity. it is A cos(φ).3. We can easily 3 3 check our answer using the sum formula for cosine f (x) = 2 cos 2x + = 2 cos(2x) = cos(2x) − 8 π 3 π 3 = 2 cos(2x) cos √ 1 2 − sin(2x) sin √ 3 2 π 3 − sin(2x) 3 sin(2x) This should remind you of equation coefficients of like powers of x in Section 8. after some √ 1 rearrangement. in the form C(x) = A cos(ωx + φ) + B for ω > 0 2. while on the right hand side. 1. and 2 cos(φ) = − 3.10 It is also worth mentioning that. The general equations to fit a function of the form f (x) = a cos(ωx) + b sin(ωx) + B into one of the forms in Theorem 10. verifying our analytic work. while f (x) = cos(2x) − 3 sin(2x) is a sinusoid. g(x) = cos(2x) − 3 sin(3x) is not.3 to fit a function into one of the forms in Theorem √ 10. the arguments of the cosine and sine√ function much match. we have f (x) = 2 sin 2x + 5π . exhibit sinusoid-like characteristics! Check it out! . It is important to note that in order for the technique presented in Example 10.5. we have f (x) = 2 sin 2x + 5π 6 √ 5π 6 3 2 = 2 sin(2x) cos + cos(2x) sin 1 2 5π 6 + cos(2x) = 2 sin(2x) − √ = cos(2x) − 3 sin(2x) Graphing the three formulas for f (x) result in the identical curve.5 Graphs of the Trigonometric Functions 2. we equate f (x) = cos(2x) − S(x) = A sin(ωx + φ) + B to get cos(2x) − √ √ 799 3 sin(2x) with the expanded form of 3 sin(2x) = A sin(ωx) cos(φ) + A cos(ωx) sin(φ) + B Once again.3.3 using A = −2 to see what these differences are. and then for a challenging exercise. 9 10 Be careful here! This graph does.10. Proceeding as before.23 are explored in Exercise 35. 2 One such angle which meets these criteria is φ = 5π .5. or sin(φ) = 1 . however. Hence. we may take ω = 2 and B = 0 so that cos(2x) − √ 3 sin(2x) = A sin(2x) cos(φ) + A cos(2x) sin(φ) √ We equate9 the coefficients of cos(2x) on either side and get A sin(φ) = 1 and A cos(φ) = − 3. 6 6 Checking our work analytically. and again we choose A = 2. had we chosen A = −2 instead of A = 2 as we worked through Example 10. That is.5. Using A2 cos2 (φ) + A2 sin2 (φ) = A2 as before. use identities to show that the formulas are all equivalent. which means cos(φ) = − 23 . our final answers would have looked different. The reader is encouraged to rework Example 10. we get A = ±2. √ √ This means 2 sin(φ) = 1.23. 2 x = π and x = 3π .3. we argued the range of F (x) = sec(x) is (−∞. 1) The ‘fundamental cycle’ of y = sec(x). (See Section 10.3.− (π. sec(x)) (0.3.2.800 Foundations of Trigonometry 10. We know from Section 10. it follows that sec(x) is also. sec(x) → −∞. Since cos(x) is periodic with period 2π. sec(x) → −∞. and as 2 2 + x → 3π .− 3 2 1 0 undefined √ − 22 − 2 √ √ 2 −1 −2 π 5π 4 3π 2 7π 4 −1 √ − 2 2 −1 √ − 2 5π 4 . −1) √ 2 √ 2 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x √ 0 undefined √ 2 2 2 1 1 7π 4 . 12 In Section 10.’12 y x 0 π 4 π 2 3π 4 cos(x) 1 √ 2 2 sec(x) 1 √ 2 (x. sec(x) → ∞. we find that as x → π + . 2 3π 4 .11 Below 2 2 we graph a fundamental cycle of y = sec(x) along with a more complete graph obtained by the usual ‘copying and pasting. and sure enough. sec(x) inherits its period from cos(x). Using the notation introduced in Section 4. so sec(x) → ∞.5. we run into trouble at 2 x = π and x = 3π since cos(x) = 0 at these values. −1] ∪ [1. This means we have a pair of vertical asymptotes to the graph of y = sec(x). 1) √ π 4. Hence. sec(α) = sec(β) if and only if cos(α) = cos(β).2 Graphs of the Secant and Cosecant Functions 1 We now turn our attention to graphing y = sec(x). 11 Provided sec(α) and sec(β) are defined. y x The graph of y = sec(x).) Similarly.1. 2 2 we have that as x → π − . .1 that the domain of F (x) = sec(x) excludes all odd multiples of π .1 for a more detailed 2 − analysis. we can use our table of values for the graph of y = cos(x) and take reciprocals. as x → 3π . Since sec(x) = cos(x) . cos(x) → 0+ . ∞). −3 2π (2π. We can now see this graphically. −1 √ 7π .− 2 3π 2 . Proceeding with the usual analysis.5 Graphs of the Trigonometric Functions 801 As one would expect.3. Since y = sin(x) and y = cos(x) are merely phase shifts of each other. so too are y = csc(x) and y = sec(x). csc(x)) π 4. .10. x = π and x = 2π. both F (x) = sec(x) and G(x) = csc(x) are continuous and smooth. Here. 2 π 2. our domain and range work in Section 10.1 is verified geometrically in the graph of y = G(x) = csc(x). 2 3 2 1 √ 1 2 2 √ 1 2 π 5π 4 3π 2 7π 4 0 undefined √ √ − 22 − 2 −1 √ π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x − 2 2 −1 √ − 2 5π 4 . y x The graph of y = csc(x). we encounter issues at x = 0. they are continuous and smooth on their domains. In other words. to graph y = csc(x) we begin with y = sin(x) and take reciprocals of the corresponding y-values.− 2 4 √ −1 −2 −3 2π 0 undefined The ‘fundamental cycle’ of y = csc(x). on the intervals between the vertical asymptotes. Once again. the secant and cosecant functions are continuous and smooth on their domains since the cosine and sine functions are continuous and smooth everywhere.13 The following theorem summarizes the properties of the secant and cosecant functions. Note that. y x 0 π 4 π 2 3π 4 sin(x) csc(x) 0 undefined √ √ 2 2 2 √ (x. Note that 13 Just like the rational functions in Chapter 4 are continuous and smooth on their domains because polynomials are continuous and smooth everywhere.1 √ 3π 4 . we graph the fundamental cycle of y = csc(x) below along with the dotted graph of y = sin(x) for reference. 4. equal to the ‘quarter marks’ 0. π .802 Foundations of Trigonometry all of these properties are direct results of them being reciprocals of the cosine and sine functions.5. Theorem 10. ∞) – is continuous and smooth on its domain – is even – has period 2π • The function G(x) = csc(x) ∞ – has domain {x : x = πk. −1] ∪ [1. π. To graph y = 1 − 2 sec(2x). we discuss graphing more general secant and cosecant curves. 3π and 2π and solve for x.5. (k + 1)π) – has range {y : |y| ≥ 1} = (−∞. First. k is an integer} = k=−∞ (kπ. State the period of each. we set the argument of secant. k is an integer = k=−∞ (2k + 1)π (2k + 3)π . 2x. Graph one cycle of the following functions. ∞) – is continuous and smooth on its domain – is odd – has period 2π In the next example. g(x) = csc(π − πx) − 5 3 2x = a 2x = 0 2x = 2x = π 2 x 0 π 4 π 2 3π 4 π 3π 2 2x = π 3π 2 2π 2x = 2π π . Example 10.1. we follow the same procedure as in Example 10. f (x) = 1 − 2 sec(2x) Solution. 2 2 – has range {y : |y| ≥ 1} = (−∞. respectively. Properties of the Secant and Cosecant Functions • The function F (x) = sec(x) ∞ – has domain x : x = π 2 + πk.24. 1. 2 2 a 0 π 2 2. −1] ∪ [1. 1. we generate the graph below and find the period to be 1 − (−1) = 2. In addition to these points and asymptotes. otherwise. −2 2 −2 One cycle of y = csc(π−πx)−5 . If f (x) exists. g(x)) 1 4 2. we have graphed the associated cosine curve – in this case y = 1 − 2 cos(2x) – dotted in the picture below. 3 y x 1 1 2 g(x) undefined 4 −3 (x. we have found a vertical asymptote. π]. The associated sine curve. Since one cycle is graphed over the interval [0. Proceeding as before.3 3 2 1 −1 π 4 π 2 3π 4 π x π One cycle of y = 1 − 2 sec(2x). −1) undefined 3 undefined −1 (π. a 0 π 2 csc(π−πx)−5 3 equal to the π − πx = a π − πx = 0 π − πx = π − πx = π 2 x 1 1 2 π 3π 2 π − πx = π 3π 2 0 −1 2 −1 2π π − πx = 2π Substituting these x-values into g(x). y x 0 π 4 π 2 3π 4 f (x) (x. −3 −1 − 1 2 −1 1 2 1 x 0 −1 2 −1 undefined −2 undefined − 1 . −1) π 2. 2.5 Graphs of the Trigonometric Functions 803 Next. is dotted in as a reference.10. we set the argument of cosecant in g(x) = quarter marks and solve for x. we have a point on the graph. 3 . we substitute these x values into f (x). y = sin(π−πx)−5 . the period is π − 0 = π. f (x)) −1 (0. 804 Foundations of Trigonometry Before moving on. and as x → 3π . As x → π − . 0) 5π 4 . tan(x) → ∞. Since these quantities match those of the corresponding cosine and sine curves.1. tan(x) → −∞.3 Graphs of the Tangent and Cotangent Functions Finally. 0) The graph of y = tan(x) over [0. so that tan(x) = cos(x) → ∞ producing a vertical asymptote at x = π . Using a 2 similar analysis. . sin(x) → 1− 2 2 2 sin(x) and cos(x) → 0+ . we get that as x → π + .3. in accordance with our findings in Section 10. we note that it is possible to speak of the period. phase shift and vertical shift of secant and cosecant graphs and use even/odd identities to put them in a form similar to the sinusoid forms mentioned in Theorem 10.1 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x −1 2π (2π.1 1 π 5π 4 3π 2 7π 4 (π. tan(x)) 0 (0. since the ranges of secant and cosecant are unbounded. 0) 1 undefined −1 0 1 undefined −1 0 7π 4 . we do not spell this out explicitly. y x The graph of y = tan(x). we see that J(x) = tan(x) is undefined at x = π and x = 3π . 2 2 2 tan(x) → −∞. we turn our attention to the graphs of the tangent and cotangent functions. as x → 3π . When constructing a table of values for the tangent function.23. −1 3π 4 . 2π]. 10. Finally. −1 π 4.5. there is no amplitude associated with these curves. Plotting this information and performing the usual ‘copy and paste’ produces: y − + x 0 π 4 π 2 3π 4 tan(x) (x. and we leave it to the reader to prove this. 2 . − π . we 2 2 2 4 4 2 see confirmation of our domain and range work in Section 10. we see the domain and range of K(x) = cot(x) as read from the graph matches with what we found analytically in Section 10. −1 1 π 5π 4 3π 2 7π 4 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x 2π From these data. 2π]. we have tan(p) = tan(0 + p) = tan(0) = 0. π π 3π 4 . To prove that this is the case.1. for another approach. 4 and π. 0.0 3π 4 .1 3π 2 . cot(x)) undefined 1 0 −1 undefined 1 0 −1 undefined The graph of y = cot(x) over [0. From the graph. and use as our ‘quarter marks’ x = − π .10. A more complete graph of y = cot(x) is below. it clearly appears as if the period of cot(x) is π. 5π 4 .0 7π 4 . 2π] results in the graph below. We have: tan(x) + tan(π) tan(x) + 0 = = tan(x).5 Graphs of the Trigonometric Functions 805 From the graph. mimicking the proof that the period of tan(x) is an option.1 π 2. we appeal to the sum formula for tangents. It should be no surprise that K(x) = cot(x) behaves similarly to J(x) = tan(x). −1 −1 π 4.3. Once again.3.1. which means p is a multiple of π. π . Plotting cot(x) over the interval [0. y x 0 π 4 π 2 3π 4 cot(x) (x. For x = 0. consider transforming tan(x) to cot(x) using identities. Certainly. suppose p is a positive real number so that tan(x + p) = tan(x) for all real numbers x. it appears as if the tangent function is periodic with period π. We take as our fundamental cycle for y = tan(x) the interval − π . so we have established the result. To show that it is exactly π.14 We take as one fundamental cycle the interval (0. π) with quarter marks: x = 0. 14 . 1 − tan(x) tan(π) 1 − (tan(x))(0) tan(x + π) = which tells us the period of tan(x) is at most π. The smallest positive multiple of π is π itself. along with the fundamental cycle highlighted as usual. π and π . k is an integer = k=−∞ (2k + 1)π (2k + 3)π . (k + 1)π) – has range (−∞.24. 2 2 – is continuous and smooth on its domain – is odd – has period π • The function K(x) = cot(x) ∞ – has domain {x : x = πk.806 y Foundations of Trigonometry x The graph of y = cot(x). k is an integer} = k=−∞ (kπ. The properties of the tangent and cotangent functions are summarized below. Properties of the Tangent and Cotangent Functions • The function J(x) = tan(x) ∞ – has domain x : x = – has range (−∞. ∞) π 2 + πk. ∞) – is continuous and smooth on its domain – is odd – has period π . As with Theorem 10.25. Theorem 10. each of the results below can be traced back to properties of the cosine and sine functions and the definition of the tangent and cotangent functions as quotients thereof. 10.5. and solving for x. 2.5 Graphs of the Trigonometric Functions Example 10. x 2 . π . Graph one cycle of the following functions. 4 2 4 To graph g(x) = 2 cot π x + π + 1.0 −π −π 2 2 1 π 2 π x −1 −2 π One cycle of y = 1 − tan We see that the period is π − (−π) = 2π. x 2 807 . equal to each of the ‘quarter marks’ − π . − π . . 1. g(x) = 2 cot π 2x + π + 1. The ‘quarter marks’ for the fundamental cycle of the cotangent curve are 0. Find the period. we find points on the graph and the vertical asymptotes. 2 2 (0. π 2 2 2 4 4 and π . π . We proceed as we have in all of the previous graphing examples by setting the argument of tangent in f (x) = 1 − tan x . y x −π −π 2 0 π 2 f (x) (x. 1) π 2. 2 a −π 2 −π 4 0 π 4 π 2 x 2 x 2 x 2 =a −π 4 x −π −π 2 0 π 2 = −π 2 = x 2 =0 x π 2 = 4 x π 2 = 2 π Substituting these x-values into f (x). namely x . 3π and π. 0. we begin by setting π x + π equal to each quarter mark 2 2 and solving for x.5. 1. f (x)) undefined 2 1 0 undefined −π. 2. f (x) = 1 − tan Solution. it is possible to extend the notion of period. As with the secant and cosecant functions. however. We find the period to be 0 − (−2) = 2. y x −2 −3 2 −1 −1 2 0 g(x) undefined 3 1 −1 undefined (x. g(x)) 3 −2. The ambitious reader is invited to formulate such a theorem. 1) − 1 . −1 2 One cycle of y = 2 cot π x 2 + π + 1.808 Foundations of Trigonometry a 0 π 4 π 2 3π 4 π π 2x + π = a π 2x + π = 0 π π 2x + π = 4 π π 2x + π = 2 π 3π 2x + π = 4 π 2x + π = π x −2 3 −2 −1 1 −2 0 We now use these x-values to generate our graph. . Since the number of classical applications involving sinusoids far outnumber those involving tangent and cotangent functions. we omit this.23. 3 3 2 1 −2 −1 −1 x (−1. phase shift and vertical shift to the tangent and cotangent functions as we did for the cosine and sine functions in Theorem 10. 10.5 Graphs of the Trigonometric Functions 809 10.5.4 Exercises In Exercises 1 - 12, graph one cycle of the given function. State the period, amplitude, phase shift and vertical shift of the function. 1. y = 3 sin(x) 4. y = cos x − 1 7. y = − cos 3 10. y = π 2 1 π x+ 2 3 2. y = sin(3x) 5. y = − sin x + π 3 3. y = −2 cos(x) 6. y = sin(2x − π) 9. y = sin −x − π −2 4 8. y = cos(3x − 2π) + 4 3 π 1 11. y = − cos 2x + − 2 3 2 2 π cos − 4x + 1 3 2 12. y = 4 sin(−2πx + π) In Exercises 13 - 24, graph one cycle of the given function. State the period of the function. 13. y = tan x − 16. y = sec x − π 3 π 2 14. y = 2 tan 1 x −3 4 π 3 15. y = 1 tan(−2x − π) + 1 3 1 π x+ 2 3 π −2 4 +1 17. y = − csc x + 1 18. y = − sec 3 19. y = csc(2x − π) 22. y = cot x + π 6 20. y = sec(3x − 2π) + 4 23. y = −11 cot 1 x 5 21. y = csc −x − 24. y = 1 3π cot 2x + 3 2 In Exercises 25 - 34, use Example 10.5.3 as a guide to show that the function is a sinusoid by rewriting it in the forms C(x) = A cos(ωx + φ) + B and S(x) = A sin(ωx + φ) + B for ω > 0 and 0 ≤ φ < 2π. 25. f (x) = √ 2 sin(x) + √ 2 cos(x) + 1 √ 26. f (x) = 3 3 sin(3x) − 3 cos(3x) √ 1 3 28. f (x) = − sin(2x) − cos(2x) 2 2 √ 3 3 3 30. f (x) = cos(2x) − sin(2x) + 6 2 2 √ 32. f (x) = −6 3 cos(3x) − 6 sin(3x) − 3 27. f (x) = − sin(x) + cos(x) − 2 √ 29. f (x) = 2 3 cos(x) − 2 sin(x) √ 1 3 31. f (x) = − cos(5x) − sin(5x) 2 2 810 √ √ 5 2 5 2 33. f (x) = sin(x) − cos(x) 2 2 Foundations of Trigonometry √ x x − 3 3 cos 6 6 34. f (x) = 3 sin 35. In Exercises 25 - 34, you should have noticed a relationship between the phases φ for the S(x) and C(x). Show that if f (x) = A sin(ωx + α) + B, then f (x) = A cos(ωx + β) + B where π β =α− . 2 36. Let φ be an angle measured in radians and let P (a, b) be a point on the terminal side of φ when it is drawn in standard position. Use Theorem 10.3 and the sum identity for sine in Theorem 10.15 to show that f (x) = a sin(ωx) + b cos(ωx) + B (with ω > 0) can be rewritten √ as f (x) = a2 + b2 sin(ωx + φ) + B. 37. With the help of your classmates, express the domains of the functions in Examples 10.5.4 and 10.5.5 using extended interval notation. (We will revisit this in Section 10.7.) In Exercises 38 - 43, verify the identity by graphing the right and left hand sides on a calculator. 38. sin2 (x) + cos2 (x) = 1 41. tan(x + π) = tan(x) 39. sec2 (x) − tan2 (x) = 1 42. sin(2x) = 2 sin(x) cos(x) 40. cos(x) = sin π −x 2 x sin(x) 43. tan = 2 1 + cos(x) In Exercises 44 - 50, graph the function with the help of your calculator and discuss the given questions with your classmates. 44. f (x) = cos(3x) + sin(x). Is this function periodic? If so, what is the period? 45. f (x) = sin(x) x . What appears to be the horizontal asymptote of the graph? 46. f (x) = x sin(x). Graph y = ±x on the same set of axes and describe the behavior of f . 47. f (x) = sin 1 x . What’s happening as x → 0? 48. f (x) = x − tan(x). Graph y = x on the same set of axes and describe the behavior of f . 49. f (x) = e−0.1x (cos(2x) + sin(2x)). Graph y = ±e−0.1x on the same set of axes and describe the behavior of f . 50. f (x) = e−0.1x (cos(2x) + 2 sin(x)). Graph y = ±e−0.1x on the same set of axes and describe the behavior of f . 51. Show that a constant function f is periodic by showing that f (x + 117) = f (x) for all real numbers x. Then show that f has no period by showing that you cannot find a smallest number p such that f (x + p) = f (x) for all real numbers x. Said another way, show that f (x + p) = f (x) for all real numbers x for ALL values of p > 0, so no smallest value exists to satisfy the definition of ‘period’. 10.5 Graphs of the Trigonometric Functions 811 10.5.5 Answers y 3 1. y = 3 sin(x) Period: 2π Amplitude: 3 Phase Shift: 0 Vertical Shift: 0 π 2 π 3π 2 2π x −3 2. y = sin(3x) 2π Period: 3 Amplitude: 1 Phase Shift: 0 Vertical Shift: 0 y 1 π 6 π 3 π 2 2π 3 x −1 3. y = −2 cos(x) Period: 2π Amplitude: 2 Phase Shift: 0 Vertical Shift: 0 y 2 π 2 π 3π 2 2π x −2 π 4. y = cos x − 2 Period: 2π Amplitude: 1 π Phase Shift: 2 Vertical Shift: 0 y 1 π 2 π 3π 2 2π 5π x 2 −1 812 π 5. y = − sin x + 3 Period: 2π Amplitude: 1 π Phase Shift: − 3 Vertical Shift: 0 y 1 Foundations of Trigonometry −π 3 −1 π 6 2π 3 7π 6 5π 3 x 6. y = sin(2x − π) Period: π Amplitude: 1 π Phase Shift: 2 Vertical Shift: 0 y 1 π 2 3π 4 π 5π 4 3π 2 x −1 1 π 1 x+ 7. y = − cos 3 2 3 Period: 4π 1 Amplitude: 3 2π Phase Shift: − 3 Vertical Shift: 0 y 1 3 − 2π 3 −1 3 y 5 4 3 π 3 4π 3 7π 3 10π 3 x 8. y = cos(3x − 2π) + 4 2π Period: 3 Amplitude: 1 2π Phase Shift: 3 Vertical Shift: 4 2π 3 5π 6 π 7π 6 4π 3 x 10.5 Graphs of the Trigonometric Functions π 9. y = sin −x − −2 4 Period: 2π Amplitude: 1 π Phase Shift: − (You need to use 4 π y = − sin x + − 2 to find this.)15 4 Vertical Shift: −2 2 π cos − 4x + 1 3 2 π Period: 2 2 Amplitude: 3 π Phase Shift: (You need to use 8 2 π y = cos 4x − + 1 to find this.)16 3 2 Vertical Shift: 1 y π − 9π − 7π − 5π − 3π − 4 4 4 4 4 −1 π 4 3π 4 5π 4 7π 4 813 x −2 −3 10. y = y 5 3 1 1 3 π π − 3π − 4 − 8 8 π 8 π 4 3π 8 π 2 5π 8 x π 1 3 − 11. y = − cos 2x + 2 3 2 Period: π 3 Amplitude: 2 π Phase Shift: − 6 1 Vertical Shift: − 2 y 1 −π 6 π 12 π 3 7π 12 5π 6 −1 2 x −2 12. y = 4 sin(−2πx + π) Period: 1 Amplitude: 4 1 Phase Shift: (You need to use 2 y = −4 sin(2πx − π) to find this.)17 Vertical Shift: 0 y 4 1 −2 −1 4 1 4 1 2 3 4 1 5 4 3 2 x −4 15 16 Two cycles of the graph are shown to illustrate the discrepancy discussed on page 796. Again, we graph two cycles to illustrate the discrepancy discussed on page 796. 17 This will be the last time we graph two cycles to illustrate the discrepancy discussed on page 796. 814 π 13. y = tan x − 3 Period: π Foundations of Trigonometry y 1 −π 6 −1 π 12 π 3 7π 12 5π 6 x 14. y = 2 tan Period: 4π 1 x −3 4 y −2π −π −1 −3 −5 π 2π x 1 15. y = tan(−2x − π) + 1 3 is equivalent to 1 y = − tan(2x + π) + 1 3 via the Even / Odd identity for tangent. π Period: 2 π − 3π − 5π − 2 − 3π 4 8 8 y 4 3 2 3 1 −π 4 x 10.5 Graphs of the Trigonometric Functions 16. y = sec x − π 2 Start with y = cos x − Period: 2π y π 2 815 1 π 2 π −1 3π 2 2π 5π x 2 17. y = − csc x + π 3 π Start with y = − sin x + 3 Period: 2π 1 −π 3 y −1 π 6 2π 3 7π 6 5π 3 x 1 18. y = − sec 3 1 π x+ 2 3 1 Start with y = − cos 3 Period: 4π y 1 π x+ 2 3 1 3 − 2π 3 −1 3 π 3 4π 3 7π 3 10π x 3 y = csc −x − π −2 4 π Start with y = sin −x − −2 4 Period: 2π y −π 4 −1 −2 −3 π 4 3π 4 5π 4 7π 4 x . y = sec(3x − 2π) + 4 Start with y = cos(3x − 2π) + 4 2π Period: 3 5 4 3 y 2π 3 5π 6 π 7π 6 4π 3 x 21. y = csc(2x − π) Start with y = sin(2x − π) Period: π 1 y Foundations of Trigonometry −1 π 2 3π 4 π 5π 4 3π 2 x 20.816 19. y = 1 3π cot 2x + 3 2 π Period: 2 y +1 4 3 1 2 3 π π − 3π − 5π − 2 − 3π − 4 4 8 8 x .5 Graphs of the Trigonometric Functions π 22.10. y = −11 cot Period: 5π 1 x 5 y 11 5π 4 5π 2 15π 4 5π −11 x 24. y = cot x + 6 Period: π y 817 1 −π 6 −1 π 12 π 3 7π 12 5π 6 x 23. f (x) = 2 3 cos(x) − 2 sin(x) = 4 sin x + 3 6 √ 3 3 3 π 5π sin(2x) + 6 = 3 sin 2x + + 6 = 3 cos 2x + +6 30. f (x) = − cos(5x) − sin(5x) = sin 5x + = cos 5x + 2 2 6 3 √ 4π 5π 32.818 √ √ Foundations of Trigonometry π 7π + 1 = 2 cos x + 4 4 = 6 cos 3x + √ 4π 3 π −2 4 25. f (x) = − sin(2x) − cos(2x) = sin 2x + 2 2 3 = cos 2x + √ π 2π = 4 cos x + 29. f (x) = − sin(x) + cos(x) − 2 = √ 2 sin x + 3π 4 −2= 2 cos x + 5π 6 √ 1 3 4π 28. f (x) = sin(x) − cos(x) = 5 sin x + = 5 cos x + 2 2 4 4 34. f (x) = cos(2x) − 2 2 6 3 √ 1 3 7π 2π 31. f (x) = 3 sin √ x x − 3 3 cos = 6 sin 6 6 x 5π + 6 3 = 6 cos x 7π + 6 6 −3 . f (x) = 3 3 sin(3x) − 3 cos(3x) = 6 sin 3x + 6 27. f (x) = −6 3 cos(3x) − 6 sin(3x) − 3 = 12 sin 3x + − 3 = 12 cos 3x + 3 6 √ √ 5 2 5 2 7π 5π 33. f (x) = 2 sin(x) + 2 cos(x) + 1 = 2 sin x + +1 √ 11π 26. Recall from Section 5. Choosing the interval [0.6 The Inverse Trigonometric Functions As the title indicates. The function f −1 . The function f (t) = cos(t) takes a real number input t. We first consider f (x) = cos(x). Digging deeper. in accordance with Theorem 5. reverses the process of the original function. y x Restricting the domain of f (x) = cos(x) to [0. hence the ‘arc’ in arccosine. 0 ≤ x ≤ π −− − − − −→ −−−−−− switch x and y coordinates f −1 (x) = arccos(x).2 that the inverse of a function f is typically denoted f −1 . where we obtain the latter from the former by reflecting it across the line y = x. would take x-coordinates on the Unit Circle and return oriented arcs. then. 0).6 The Inverse Trigonometric Functions 819 10. we may view the inputs to f (t) = cos(t) as oriented arcs and the outputs as x-coordinates on the Unit Circle. owing to their periodic nature.3 in Section 5. none of the six circular functions is one-to-one.2 to obtain a one-to-one function. by definition. It 1 is far too easy to confuse cos−1 (x) with cos(x) = sec(x) so we will not use this notation in our text. To remedy this. For this reason. read ‘arc-cosine of x’. (cos(x))3 as cos3 (x) and so on. in this section we concern ourselves with finding inverses of the (circular) trigonometric functions. Our immediate problem is that. y π 1 π 2 y π 2 π x −1 reflect across y = x −1 1 x f (x) = cos(x). and returns the value cos(θ). recall that an inverse function. π]. we use the notation f −1 (x) = arccos(x). be sure to check the context! See page 704 if you need a review of how we associate real numbers with angles in radian measure. Hence. . some textbooks use the notation f −1 (x) = cos−1 (x) for the inverse of f (x) = cos(x). 1] as well as the properties of being smooth and continuous.2. associates it with the angle θ = t radians.1 Instead. 1 2 But be aware that many books do! As always. To understand the ‘arc’ in ‘arccosine’. The obvious pitfall here is our convention of writing (cos(x))2 as cos2 (x). we restrict the domains of the circular functions in the same way we restricted the domain of the quadratic function in Example 5. Below are the graphs of f (x) = cos(x) and f −1 (x) = arccos(x).2 we have that cos(θ) = cos(t) is the x-coordinate of the terminal point on the Unit Circle of an oriented arc of length |t| whose initial point is (1.3.10. π] allows us to keep the range as [−1. 820 Foundations of Trigonometry We restrict g(x) = sin(x) in a similar manner. π 2 2 – arcsin(x) = t if and only if − π ≤ t ≤ 2 π 2 and sin(t) = x π 2 – sin(arcsin(x)) = x provided −1 ≤ x ≤ 1 – arcsin(sin(x)) = x provided − π ≤ x ≤ 2 – additionally. which is read ‘arc-sine of x’. π . Properties of the Arccosine and Arcsine Functions • Properties of F (x) = arccos(x) – Domain: [−1.26. y π 2 y 1 −π 2 −1 π 2 x −1 1 x g(x) = sin(x). 2 −− − − − −→ −−−−−− switch x and y coordinates reflect across y = x −π 2 g −1 (x) = arcsin(x). We list some important facts about the arccosine and arcsine functions in the following theorem. although the interval of choice is − π . 1] – Range: [0. − π ≤ x ≤ 2 π . Theorem 10. 2 2 It should be no surprise that we call g −1 (x) = arcsin(x). π] – arccos(x) = t if and only if 0 ≤ t ≤ π and cos(t) = x – cos(arccos(x)) = x provided −1 ≤ x ≤ 1 – arccos(cos(x)) = x provided 0 ≤ x ≤ π • Properties of G(x) = arcsin(x) – Domain: [−1. arcsine is odd . 2 2 y x Restricting the domain of f (x) = sin(x) to − π . 1] – Range: − π . π . arccos 23 is the real number t with 0 ≤ t ≤ π 6 √ and cos(t) = 3 2 . It’s about time for an example. π] with cos(t) = − Our answer is arccos − 2 2 = 3π 4 . so that arccos cos 6 π 6 = π. an angle measuring 2 1 t radians) which lies between 0 and π with cos(t) = 2 . Working from the inside out. we seek the number t in the interval − π . (a) To find arccos 1 . Example 10. 1.1. 1.6. 6 6 6 However. Now. Hence. Find the exact values of the following. π with sin(t) = − 2 . The number we seek is t = π . 6 2 6 (e) Since 0 ≤ π ≤ π. we choose to work the example through using the definition of arccosine. The 2 2 2 answer is t = − π so that arcsin − 1 = − π . we could simply invoke Theorem 10. (c) The number t = arccos − √ 2 2 lies in the interval [0. 1 (d) To find arcsin − 1 . √ √ arccos cos π = arccos 23 . We know t = π meets these 3 π 1 criteria. 6 . (a) tan (arccos (x)) Solution. so arccos 2 = 3 . Rewrite the following as algebraic expressions of x and state the domain on which the equivalence is valid. (a) arccos 1 2 √ 2 2 π 6 √ (b) arcsin 2 2 (c) arccos − (e) arccos cos 1 (d) arcsin − 2 (f) arccos cos 11π 6 (g) cos arccos − 3 5 3 (h) sin arccos − 5 2.10.26 is a direct consequence of the facts that f (x) = cos(x) for 0 ≤ x ≤ π and F (x) = arccos(x) are inverses of each other as are g(x) = sin(x) for − π ≤ x ≤ π and 2 2 G(x) = arcsin(x). arcsin 4 √ = π.6 The Inverse Trigonometric Functions 821 Everything in Theorem 10. We find t = π . √ (b) cos (2 arcsin(x)) (b) The value of arcsin 2 2 is a real number t between − π and 2 √ 2 2 π 2 √ with sin(t) = √ 2 2 .26 to get arccos cos π = π . 4 2 2 . in order to make sure we understand why this is the case. we need to find the real number t (or. equivalently. 2 x (b) We proceed as in the previous problem by writing t = arcsin(x) so that t lies in the interval − π . Using the Pythagorean Identity cos2 (t) + sin2 (t) = 1. we know we must restrict −1 ≤ x ≤ 1. we know arccos 3 2 = arccos 11π 6 3 2 .26 does not apply. arccos cos 6 = π. 5 5 2. to really understand why this cancellation occurs. Theorem 10. then. tan (arccos (x)) = 1−x is valid for x in [−1. we need to discard 2 x = cos π = 0. (h) As in the previous example. . Additionally. as 5 5 3 before. sin(t) ≥ 0. t corresponds to an angle in Quadrant I or Quadrant II. Since t = arccos(x). cos(t) = − 5 . Thus. We are forced to √ 11π 6 √ work through from the inside out starting with arccos cos the previous problem. sin arccos − 3 = 4 . but since we are after an expression for tan(t). Hence. we let t = arccos(x). Then. we let t = arccos − 5 . 0) ∪ (0. it is easiest to use the last form: cos (2 arcsin(x)) = cos(2t) = 1 − 2 sin2 (t) = 1 − 2x2 Alternatively. we consider our substitution t = arccos(x). so we choose sin(t) = 1 − x2 . 2 2 2 geometrically. we 4 choose sin(t) = 5 . Since − 3 is 5 5 between −1 and 1. To help us see the forest for the trees. Since t corresponds to a Quadrants II angle. sec(t) = is invited to work through this approach to see what. cos arccos − 5 = cos(t) = − 5 . One approach3 sin(t) to finding tan(t) is to use the quotient identity tan(t) = cos(t) . Since we know x = sin(t). 6 (g) One way to simplify cos arccos − 3 is to use Theorem 10. difficulties arise. we can narrow this down a bit and conclude that π < t < π. since we had to discard t = π . either 0 ≤ t < π or π < t ≤ π so that. We have three choices for rewriting cos(2t): cos2 (t) − sin2 (t). Since cos(2t) is defined everywhere. Since the domain of arccos(x) is [−1. if any. we get no additional restrictions on t as we did in the previous problem.822 (f) Since 11π 6 Foundations of Trigonometry does not fall between 0 and π. Since x = cos(t). we could use the identity: 1 + tan2 (t) = sec2 (t). so our goal is to find a way to express tan (arccos (x)) = tan(t) in terms of x. From = π . 2 cos2 (t) − 1 and 1 − 2 sin2 (t).26 directly. we get 5 4 3 2 − 5 + sin2 (t) = 1 or sin(t) = ± 5 . π with sin(t) = x. (a) We begin this problem in the same manner we began the previous two problems. We aim to express cos (2 arcsin(x)) = cos(2t) in terms 2 2 of x. we know cos(t) = x where 0 ≤ t ≤ π. √ 1 − x2 sin(t) = tan(t) = cos(t) x To determine the values of x for which this equivalence is valid. 1]. x The reader . 3 3 3 by definition. from which we √ get sin(t) = ± 1 − x2√ Since t corresponds to angles in Quadrants I and II. 2 so that t corresponds to an angle in Quadrant II. we have that cos arccos − 3 = − 3 and we are done. we let t = arccos − 3 so that cos(t) = − 3 for some t where 5 5 0 ≤ t ≤ π. Since cos(t) < 0. Hence. In terms of t. Hence. Hence. However. and we are finished in (nearly) the same amount of time. we know we need to throw out t = π from consideration. Substituting cos(t) = x into the Pythagorean Identity cos2 (t) + sin2 (t) = 1 gives x2 + sin2 (t) = 1. 3 1 cos(t) √ 2 = 1 . we need to find sin arccos − 3 = sin(t). Hence. 1]. which are named arctangent and arccotangent. Most of the common errors encountered in dealing with the inverse circular functions come from the need to restrict the domains of the original functions so that they are one-to-one. Among other things. respectively. Next. This is the exact same phenomenon discussed in Section 6 6 6 5. π) to obtain g −1 (x) = arccot(x). even though the expression we arrived at in part 2b above. 2 −− − − − −→ −−−−−− switch x and y coordinates reflect across y = x −π 2 f −1 (x) = arctan(x). For this reason. Since arcsin(x) is defined only for −1 ≤ x ≤ 1. We show these graphs on the next page and list some of the basic properties of the arctangent and arccotangent functions.1 are in order. . the vertical asymptotes x = 0 and x = π of the graph of g(x) = cot(x) become the horizontal asymptotes y = 0 and y = π of the graph of g −1 (x) = arccot(x).2 when we saw (−2)2 = 2 as opposed to −2. note that the 2 2 vertical asymptotes x = − π and x = π of the graph of f (x) = tan(x) become the horizontal 2 2 asymptotes y = − π and y = π of the graph of f −1 (x) = arctan(x). The next pair of functions we wish to discuss are the inverses of tangent and cotangent. 1]. A few remarks about Example 10. − π < x < 2 π . the equivalence cos (2 arcsin(x)) = 1 − 2x2 is valid for only −1 ≤ x ≤ 1. we once again appeal to our substitution t = arcsin(x). it pays to be careful when we determine the intervals where such equivalences are valid. First. the equivalence cos (2 arcsin(x)) = 1−2x2 is valid only on [−1. This is akin to the fact that while the √ 2 expression x is defined for all real numbers. One instance of this phenomenon is the fact that arccos cos 11π = π as opposed to 11π . Once again.6 The Inverse Trigonometric Functions 823 To find the restrictions on x. 2 2 y 1 −π −π 2 4 −1 −1 π 4 π 2 y π 2 x π 4 1 −π 4 x f (x) = tan(x). is defined for all real numbers. Additionally. namely 1 − 2x2 .10.6. π to obtain f −1 (x) = arctan(x). the equivalence ( x) = x is valid only for x ≥ 0. we restrict g(x) = cot(x) to its fundamental cycle on (0. we restrict f (x) = tan(x) to its fundamental cycle on − π . Theorem 10. arctan(x) → − π + . Properties of the Arctangent and Arccotangent Functions • Properties of F (x) = arctan(x) – Domain: (−∞. as x → ∞.824 Foundations of Trigonometry y y π 3π 4 π 2 π 4 1 π 4 π 2 3π 4 π x −1 −1 1 x −− − − − −→ −−−−−− g(x) = cot(x). arctangent is odd • Properties of G(x) = arccot(x) – Domain: (−∞. as x → ∞.27. 0 < x < π. π 2 2 – as x → −∞. π) – as x → −∞. switch x and y coordinates reflect across y = x g −1 (x) = arccot(x). arccot(x) → π − . arccot(x) → 0+ – arccot(x) = t if and only if 0 < t < π and cot(t) = x – arccot(x) = arctan 1 x for x > 0 – cot (arccot(x)) = x for all real numbers x – arccot(cot(x)) = x provided 0 < x < π . ∞) – Range: − π . arctan(x) → 2 – arctan(x) = t if and only if − π < t < 2 – arctan(x) = arccot 1 x π 2 π− 2 and tan(t) = x for x > 0 π 2 – tan (arctan(x)) = x for all real numbers x – arctan(tan(x)) = x provided − π < x < 2 – additionally. ∞) – Range: (0. Were the identities we used there valid for all t under consideration? A pedantic point. √ (a) arctan( 3) (c) cot(arccot(−5)) √ (b) arccot(− 3) (d) sin arctan − 3 4 825 2. in fact. Substituting. is valid for all the values of t under consideration. where k is 4 2 an integer. − 2 < t < 0. Returning 2 4 4 4 4 2 2 tan(t) to arctan(2t). (a) tan(2 arctan(x)) Solution. then − π < t < π and tan(t) = x. 3 3 √ √ (b) The real √ number t = arccot(− 3) lies in the interval (0. we choose csc(t) = − 3 . 3 3 (d) We start simplifying sin arctan − 4 by letting t = arctan − 4 . so arctan( 3) = π . π) and cot(t) = −5. Before we get started using identities. However. we note the double angle identity tan(2t) = 1−tan2 (t) . We find 2 2 √ t = π . but what else do you expect from this book? . we note that tan(2t) is undefined when 2t = π + πk for integers k.4 From tan(t) = − 3 .1. we know. Hence. we have that t belongs to the interval (0. One way to proceed is to use The Pythagorean Identity. Letting t = arccot(−5). We look for a way to express 2 2 tan(2 arctan(x)) = tan(2t) in terms of x. π are t = ± π . √ √ 1. π . sin arctan − 3 = − 3 . to be sure. Since 3 3 5 − π < t < 0. 1. π ∪ π . (a) If we let t = arctan(x).6 The Inverse Trigonometric Functions Example 10. so sin(t) = − 3 . − π ∪ − π . The only members of this family which lie in − π . working it through provides us with yet another opportunity to understand why this is the case. Find the exact values of the following. We get arccot(− 3) = 5π . Dividing both sides of this 2 equation by 2 tells us we need to exclude values of t where t = π + π k.27 directly and obtain cot(arccot(−5)) = −5. we 4 2 5 get cot(t) = − 4 . 6 (c) We can apply Theorem 10.2. since this relates the reciprocals of tan(t) and sin(t) and is valid for all t under consideration. Hence. we get 1 + − 4 = csc2 (t) so that csc(t) = ± 3 .6. Then tan(t) = − 3 for 4 π π π some − 2 < t < 2 . Check our work back in Example 10.10. 2 5 4 5 2. (a) We know arctan( 3) is the real number t between − π and π with tan(t) = 3. Rewrite the following as algebraic expressions of x and state the domain on which the equivalence is valid. cot(arccot(−5)) = cot(t) = −5. π) with cot(t) = − 3. 1+cot2 (t) = csc2 (t).6. which 2 2 4 means the values of t under consideration are − π . Since tan(t) < 0. hence we get tan(2 arctan(x)) = tan(2t) = 2 tan(t) 2x 2 (t) = 1 − x2 1 − tan (b) cos(arccot(2x)) 4 It’s always a good idea to make sure the identities used in these situations are valid for all values t under consideration. which were first discussed in Subsection 10. are given below with the fundamental cycles highlighted. but complicates the Calculus. cos(arccot(2x)) = cos(t). One approach simplifies the Trigonometry associated with the inverse functions. The graph of y = csc(x). t = ± π . the only exclusions come from the values of t we discarded earlier. we have cos (arccot(2x)) = √4x2 +1 for all real numbers x. namely [1. ∞). and another piece to cover the bottom. −1] ∪ [1. −1) ∪ (−1. cos(arccot(2x)) = cos(t) = cot(t) sin(t) = √ 2x 4x2 + 1 Since arccot(2x) is defined for all real numbers x and we encountered no additional 2x restrictions on t. there are no additional restrictions on t. the other makes the Calculus easier. In terms of t. The 1 identity 1 + cot2 (t) = csc2 (t) holds for all t in (0. Hence. We present both points of view. The last two functions to invert are secant and cosecant. (b) To get started. The identity cot(t) = cos(t) is valid for t in (0. −1]. ∞) and restricts the domain of the function so that it is one-to-one. ∞). The same is true for cosecant. Since t is √ 1 between 0 and π. so csc(t) = 4x2 + 1 which gives sin(t) = √4x2 +1 .2. It is clear from the graph of secant that we cannot find one single continuous piece of its graph which covers its entire range of (−∞. the equivalence tan(2 arctan(x)) = 1−x2 holds for all x in 4 (−∞. A portion of each of their graphs. π). sin(t) so our strategy is to obtain sin(t) in terms of x. Since the domain of arctan(x) is all real numbers. Since x = tan(t).826 Foundations of Trigonometry To find where this equivalence is valid we check back with our substitution t = arctan(x). √ Substituting cot(t) = 2x. we let t = arccot(2x) so that cot(t) = 2x where 0 < t < π. π) and relates cot(t) and csc(t) = sin(t) . Thus in order to define the arcsecant and arccosecant functions. y y x x The graph of y = sec(x). . 1) ∪ (1. namely (−∞. There are two generally accepted ways make these choices which restrict the domains of these functions so that they are one-to-one. or csc(t) = ± 4x2 + 1.5. this means we exclude 4 2x x = tan ± π = ±1. Hence. we get 1 + (2x)2 = csc2 (t). Since cos(t) is always defined. then write cos(t) = cot(t) sin(t). so we can begin using identities to relate cot(t) to cos(t). but the Trigonometry less so. csc(t) > 0. and our goal is to express the latter in terms of x. we must settle for a piecewise approach wherein we choose one piece to cover the top of the range. This is often done in Calculus textbooks. For f (x) = sec(x). .6. π . we restrict the secant and cosecant functions to coincide with the restrictions on cosine and sine. the domain is (−∞.1 Inverses of Secant and Cosecant: Trigonometry Friendly Approach In this subsection. −1] ∪ [1. we can rewrite this as {x : |x| ≥ 1}. 0 ∪ 0. 2 π 2 −− − − − −→ −−−−−− switch x and y coordinates reflect across y = x −π 2 g −1 (x) = arccsc(x) Note that for both arcsecant and arccosecant. 2 2 y 1 π 2 y π 2 −π 2 −1 x −1 1 x g(x) = csc(x) on − π . π 2 2 y y 1 π π 2 π x π 2 −1 f (x) = sec(x) on 0.π 2 −− − − − −→ −−−−−− switch x and y coordinates reflect across y = x −1 1 x f −1 (x) = arcsec(x) and we restrict g(x) = csc(x) to − π . we restrict the domain to 0. Using these definitions. π 2 ∪ π . so we include it here for completeness. respectively.10. Taking a page from Section 2.2. ∞). π ∪ π . we get the following properties of the arcsecant and arccosecant functions.6 The Inverse Trigonometric Functions 827 10. 0 ∪ 0. as x → ∞. −1] ∪ [1. π ∪ 2 π 2. arcsec(x) → – arcsec(x) = arccos 1 x as x → ∞. assuming the “Trigonometry Friendly” ranges are used.π π+ 2 . arccsc(x) → 0− . – as x → −∞. Rewrite the following as algebraic expressions of x and state the domain on which the equivalence is valid. 0 ∪ 0. arccosecant is odd a . 1. . ∞) – Range: − π .28. arcsec(x) → π 2 π− 2 – arcsec(x) = t if and only if 0 ≤ t < – sec (arcsec(x)) = x provided |x| ≥ 1 or π 2 < t ≤ π and sec(t) = x provided |x| ≥ 1 π 2 – arcsec(sec(x)) = x provided 0 ≤ x < • Properties of G(x) = arccsc(x) or π 2 <x≤π – Domain: {x : |x| ≥ 1} = (−∞. Properties of the Arcsecant and Arccosecant Functionsa • Properties of F (x) = arcsec(x) – Domain: {x : |x| ≥ 1} = (−∞. arccsc(x) → 0+ – arccsc(x) = t if and only if − π ≤ t < 0 or 0 < t ≤ 2 – arccsc(x) = arcsin 1 x π 2 and csc(t) = x provided |x| ≥ 1 π 2 – csc (arccsc(x)) = x provided |x| ≥ 1 – arccsc(csc(x)) = x provided − π ≤ x < 0 or 0 < x ≤ 2 – additionally. Example 10. . (a) tan(arcsec(x)) (b) cos(arccsc(4x)) . (a) arcsec(2) (b) arccsc(−2) (c) arcsec sec 5π 4 (d) cot (arccsc (−3)) 2. ∞) – Range: 0.6.828 Foundations of Trigonometry Theorem 10. −1] ∪ [1. Find the exact values of the following. π 2 2 – as x → −∞.3. 6 (c) Since 5π doesn’t fall between 0 and π or π and π. 1. 4 4 (d) One way to begin to simplify cot (arccsc (−3)) is to let t = arccsc(−3). As a result. 2 2 Since cos(t) is defined for all t. −1] ∪ [1. on the the other hand. Since tan(t) is defined for all t values 2 2 under consideration. (a) We begin simplifying tan(arcsec(x)) by letting t = arcsec(x). Substituting sin(t) = 4x gives cos2 (t) + 4x = 1. x ≤ −1. we have 1 + cot2 (t) = (−3)2 . Since we encountered no further 2 2 restrictions on t. if π < t ≤ π − x 2 √ Now we need to determine what these conditions on t mean for x. if 0 ≤ t < π 2 √ 2 − 1. π ∪ π . we get a piecewise defined function for tan(t) tan(t) = x2 − 1. or cot(t) = ± 8 = ±2 2. Solving. 0 . Then.28. Then csc(t) = 4x for t in − π .28 comes to our aid giving arccsc(−2) = arcsin − 2 = − π .28. sec(t) = x for t in 0. π . 3 1 (b) Once again.10. Then. We are after 2 cot (arccsc (−3)) = cot(t). we have arcsec(2) = arccos 1 2 829 = π. 1 From csc(t) = 4x. If t belongs to 0. ∞). we have that t lies in the interval − π . so we choose cos(t) = 16−x . tan(arcsec(x)) = x2 − 1. we use the identity 1 + tan2 (t) = sec2 (t).6 The Inverse Trigonometric Functions Solution. the equivalence below holds for all x in (−∞. π . we cannot use the inverse property 4 2 2 stated in Theorem 10. Since − π ≤ t < 0. when 0 ≤ t < π . 0 ∪ 0. This is valid for all values of t under consideration. so we get cot (arccsc (−3)) = −2 2. if. we do not encounter any additional restrictions on t. if x ≥ 1 √ − x2 − 1. π then 2 2 tan(t) ≤ 0. x ≥ 1. and we now set about finding an expression for cos(arccsc(4x)) = cos(t). we start by letting t = arccsc(4x). so to find cos(t). we can make use if the identity 1 1 2 cos2 (t) + sin2 (t) = 1. we have no additional restrictions on t. Hence. 1 Substituting. Theorem 10. and when π < t ≤ π. we get 1 + tan2 (t) = x2 . To relate sec(t) to tan(t). (a) Using Theorem 10. Since x = sec(t). since this is negative. begin by working ‘inside out’ which √ √ yields arcsec sec 5π = arcsec(− 2) = arccos − 22 = 3π . since x could be negative. we get cos(t) = ± √ 16x2 − 1 16x2 − 1 =± 16x2 4|x| √ 2 Since t belongs to − π . csc(t) = −3 and. 2. and we seek a formula for tan(t).) To find the values for . 0 ∪ 0. We can. we get sin(t) = 4x . 2 √ cot(t) < 0. t belongs to π . nevertheless. π then tan(t) ≥ 0. 2 2 4|x| (The absolute values here are necessary. we know cos(t) ≥ 0. √ and when we substitute sec(t) = x. so we use the Pythagorean √ Identity √ + cot2 (t) = csc2 (t). tan(t) = ± x2 − 1. if x ≤ −1 √ (b) To simplify cos(arccsc(4x)). π . we restrict f (x) = sec(x) to 0. . ∞ .830 Foundations of Trigonometry which this equivalence is valid. 2 2 y y 3π 2 1 π π 2 π 3π 2 x π 2 −1 g(x) = csc(x) on 0. π ∪ π. 3π 2 −− − − − −→ −−−−−− switch x and y coordinates reflect across y = x −1 1 x f −1 (x) = arcsec(x) and we restrict g(x) = csc(x) to 0.4. 3π 2 −− − − − −→ −−−−−− switch x and y coordinates reflect across y = x −1 g −1 (x) 1 = arccsc(x) x Using these definitions. Using Theorem 2.2 Inverses of Secant and Cosecant: Calculus Friendly Approach In this subsection. 3π . we rewrite this inequality and solve 1 to get x ≤ − 1 or x ≥ 4 . π ∪ π. 3π 2 2 y y 3π 2 1 π π 2 π 3π 2 x π 2 −1 f (x) = sec(x) on 0. t = arccsc(4x). we get the following result. the domain of arccsc(4x) requires |4x| ≥ 1. π 2 ∪ π. we look back at our original substution.6. − 1 ∪ 4 1 4. 10. π 2 ∪ π. Since we had no additional restrictions on t. Since the domain of arccsc(x) requires its argument x to satisfy |x| ≥ 1. the equivalence 4 √ cos(arccsc(4x)) = 16x2 −1 4|x| holds for all x in −∞. 28. arcsec(x) → – arcsec(x) = arccos 1 x 3π − 2 . 831 as x → ∞. (a) arcsec(2) (b) arccsc(−2) (c) arcsec sec 5π 4 (d) cot (arccsc (−3)) 2. . .6 The Inverse Trigonometric Functions Theorem 10.6. ∞) – Range: 0.28. Rewrite the following as algebraic expressions of x and state the domain on which the equivalence is valid. π ∪ π. ∞) – Range: 0. 1. 3π 2 2 – as x → −∞. π ∪ π. 3π 2 2 – as x → −∞.4. Example 10. −1] ∪ [1. as x → ∞. arcsec(x) → π 2 π− 2 – arcsec(x) = t if and only if 0 ≤ t < for x ≥ 1 – sec (arcsec(x)) = x provided |x| ≥ 1 or π ≤ t < 3π 2 and sec(t) = x onlyb π 2 – arcsec(sec(x)) = x provided 0 ≤ x < • Properties of G(x) = arccsc(x) or π ≤ x < 3π 2 – Domain: {x : |x| ≥ 1} = (−∞. arccsc(x) → π + . (a) tan(arcsec(x)) (b) cos(arccsc(4x)) . −1] ∪ [1.6. arccsc(x) → 0+ – arccsc(x) = t if and only if 0 < t ≤ – arccsc(x) = arcsin 1 x π 2 or π < t ≤ 3π 2 and csc(t) = x for x ≥ 1 onlyc π 2 – csc (arccsc(x)) = x provided |x| ≥ 1 – arccsc(csc(x)) = x provided 0 < x ≤ a b or π < x ≤ 3π 2 . Properties of the Arcsecant and Arccosecant Functionsa • Properties of F (x) = arcsec(x) – Domain: {x : |x| ≥ 1} = (−∞. The interested reader is invited to compare and contrast the solution to each. Our next example is a duplicate of Example 10.29.10. Compare this with the similar result in Theorem 10. assuming the “Calculus Friendly” ranges are used. Find the exact values of the following. c Compare this with the similar result in Theorem 10.3. so arccsc(−2) = 7π . the equivalence tan(arcsec(x)) = x2 − 1 holds for all x in the domain of t = arcsec(x). sec(t) = x for t in 0. tan(t) = ± x2 − 1. and we choose cos(t) = 2 to π. t belongs 4|x| √ 2 −1 − 16x This leads us to a 4|x| If t lies in 0. To relate sec(t) to tan(t). Then csc(t) = 4x for t in 0. Foundations of Trigonometry 1.29 directly to simplify 4 2 arcsec sec 5π = 5π . 3π . ∞). tan(t) ≥ 0. we can make use if the identity 1 2 1 cos2 (t) + sin2 (t) = 1. 3π . Since tan(t) is defined for all t values 2 2 under consideration. We know csc(t) = −3. 2 2. We encourage the reader to work this through using the defini4 4 tion as we have done in the previous examples to see how it goes. we know cot(t) > 0.832 Solution. 1 From csc(t) = 4x. √ and when we substitute sec(t) = x. 2 2 Since cos(t) is defined for all t.29 to arccsc(−2) and convert this into an arcsine problem. Then. we find 1 + cot2 (t) = (−3)2 so that cot(t) = ± 8 = ±2 2. so. Substituting sin(t) = 4x gives cos2 (t) + 4x = 1. (a) Since 2 ≥ 1. The real number t = arccsc(−2) lies in 0. we choose cos(t) = 2 (momentarily) piecewise defined function for cos(t)     √ 16x2 − 1 . The t 2 2 we’re after is t = 7π . (a) We begin simplifying tan(arcsec(x)) by letting t = arcsec(x). we get cos(t) = ± √ 16x2 − 1 16x2 − 1 =± 16x2 4|x| √ 16x2 −1 . and we now set about finding an expression for cos(arccsc(4x)) = cos(t). so to find cos(t). π ∪ π. π ∪ π. π . we appeal to the definition.29 to get arcsec(2) = arccos 1 2 = π. t lies in π. 6 6 (c) Since 5π lies between π and 3π . we use the identity 1 + tan2 (t) = sec2 (t). Solving. we have no additional restrictions on t. if 0 ≤ t ≤ π 2 4|x| √ cos(t) = 2−1   − 16x  . Using the identity 2 √ √ 1 + cot2 (t) = csc2 (t). −1] ∪ [1. and since this is negative. −2 is not greater to or equal to 1. we may apply Theorem 10. 3π . then cos(t) ≥ 0. Instead. √ Since t lies in 0. Hence. so we choose tan(t) = x2 − 1. 3π . (d) To simplify cot (arccsc (−3)) we let t = arccsc (−3) so that cot (arccsc (−3)) = cot(t). This is valid for all values of t under consideration. and we seek a formula for tan(t). 3 (b) Unfortunately. Our answer is cot (arccsc (−3)) = 2 2. namely (−∞. 3π and satisfies csc(t) = −2. so we cannot apply Theorem 10. we get 1 + tan2 (t) = x2 . Otherwise. we start by letting t = arccsc(4x). π ∪ π. Since √ t is in the interval π. π ∪ π. we may invoke Theorem 10. we do not encounter any additional restrictions on t. we get sin(t) = 4x . (b) To simplify cos(arccsc(4x)). if π < t ≤ 3π 2 4|x| . Since we found 2 2 √ no additional restrictions on t. 3π in which case cos(t) ≤ 0. 3π . arcsecant or arccosecant. On most calculators. 1. (a) arccot(2) (b) arcsec(5) (c) arccot(−2) (d) arccsc − 3 2 2. for π < t ≤ get 3π 2 . however. we will have need to approximate the values of the inverse circular functions. If we are asked to approximate these values. 1.4636. − 4 ∪ 1 . 16x2 − 1 = 4|x| √ 16x2 − 1 4x we get 4x ≤ −1. 2 2 (b) Since 5 ≥ 1. namely −∞. we find arccot(2) = arctan 1 ≈ 0. Since 4x = csc(t). 4x ≥ 1. Check your answers using a calculator. In ‘radian’ mode.6. (a) f (x) = Solution. we can use the property from either Theorem 10. 5 x π − arccos 2 5 (b) f (x) = 3 arctan (4x). it is a simple matter to punch up the appropriate decimal on the calculator. (c) f (x) = arccot x +π 2 . If we are asked for an arccotangent.6.6 The Inverse Trigonometric Functions 833 We now see what these restrictions mean in terms of x. Example 10. In this case. In this case. as our next example illustrates. or x ≤ − 1 . In the sections to come. or x ≥ 1 . we get that for 0 ≤ t ≤ π . ∞ 4 10.10. Use a calculator to approximate the following values to four decimal places.27 to rewrite arccot(2) as arccot(2) = arctan 1 .29 to write arcsec(5) = arccos 1 ≈ 1. cos−1 and tan−1 . cos(arccsc(4x)) = 16x −1 . respectively. Find the domain and range of the following functions.3694.28 or Theorem 10. we can use the property listed in Theorem 10. arccosine and arctangent functions are available and they are usually labeled as sin−1 . so we also 4 √ √ √ 16x2 − 1 16x2 − 1 16x2 − 1 cos(t) = − =− = 4|x| 4(−x) 4x √ 2 Hence. |x| = −x.5. we often need to employ some ingenuity. we can simplify |x| = x so 2 4 √ cos(t) = Similarly. (a) Since 2 > 0.3 Calculators and the Inverse Circular Functions. in all cases. and this equivalence is valid for all x in 4x 1 the domain of t = arccsc(4x). only the arcsine. y 1 θ = arccot(−2) radians α 1 x Another way to attack the problem is to use arctan − 1 . we once again visualize the angle θ = arccot(−2) radians 1 and note that it is a Quadrant II angle with tan(θ) = − 2 . This means it is exactly π units away from β. Moreover. as pictured below. 2 we get arccot(−2) ≈ 2. the real number 2 1 t = arctan − 2 satisfies tan(t) = − 1 with − π < t < π . we know 2 2 2 more specifically that − π < t < 0.27 to help us find arccot(−2). we can use Theorem 10. and we get θ = π + β = π + arctan − 1 ≈ 2. Then t is a real number such that 0 < t < π and cot(t) = −2.27 to get 1 α = arccot(2) = arctan 2 radians. Geometrically. Since θ = π − α = π − arctan 1 ≈ 2. 2 as before. and the Reference Angle 2 Theorem. we cannot directly apply Theorem 10. This allows us to proceed using a ‘reference angle’ approach.6779 radians.6779. α is an acute angle so 0 < α < π . Let t = arccot(−2). so t corresponds to an angle β in Quadrant IV. . we know π < t < π. This means α = arccot(2) radians. To 2 find the value of arccot(−2). By definition. Hence.6779. Since tan(t) < 0. since cot(t) < 0. Consider α.6779 radians. Theorem 10. the reference angle for θ. Since the argument of arccotangent is now a positive 2. arccot(−2) ≈ 2. this 2 means t corresponds to a Quadrant II angle θ = t radians.2. By definition. tells us that cot(α) = 2.834 Foundations of Trigonometry (c) Since the argument −2 is negative. 10. so t corresponds to a Quadrant III angle. Since csc(t) < 0. θ.8713 3 3 radians. arccsc − 3 ≈ 3. π ∪ π. 0 ∪ 0. If. Since the argument of arccosecant is now positive. we can use Theorem 10. we may use Theorem 10. we let α be 2 the reference angle for θ. 3π . which means α = arccsc 3 2 2 2 radians. we have 2 that π < θ ≤ 3π . π . Then 0 < α < π and csc(α) = 3 . Since θ = π + α = π + arcsin 2 ≈ 3.7297.8713. the range of arccosecant 2 3 is taken to be 0. 2 y 1 θ = arccsc − 3 radians 2 α 1 x . As above.28 to 2 2 get arccsc − 3 = arcsin − 2 ≈ −0. on the other hand.6 The Inverse Trigonometric Functions y 835 1 θ = arccot(−2) radians β 1 x π (d) If the range of arccosecant is taken to be − π . then we proceed as in the previous problem by letting 2 2 3 t = arccsc − 2 . Then t is a real number with csc(t) = − 3 .29 3 to get α = arccsc 2 = arcsin 2 radians. 836 Foundations of Trigonometry 2. (a) Since the domain of F (x) = arccos(x) is −1 ≤ x ≤ 1, we can find the domain of f (x) = π − arccos x by setting the argument of the arccosine, in this case x , between 2 5 5 −1 and 1. Solving −1 ≤ x ≤ 1 gives −5 ≤ x ≤ 5, so the domain is [−5, 5]. To determine 5 the range of f , we take a cue from Section 1.7. Three ‘key’ points on the graph of F (x) = arccos(x) are (−1, π), 0, π and (1, 0) . Following the procedure outlined in 2 Theorem 1.7, we track these points to −5, − π , (0, 0) and 5, π . Plotting these values 2 2 tells us that the range5 of f is − π , π . Our graph confirms our results. 2 2 (b) To find the domain and range of f (x) = 3 arctan (4x), we note that since the domain of F (x) = arctan(x) is all real numbers, the only restrictions, if any, on the domain of f (x) = 3 arctan (4x) come from the argument of the arctangent, in this case, 4x. Since 4x is defined for all real numbers, we have established that the domain of f is all real numbers. To determine the range of f , we can, once again, appeal to Theorem 1.7. Choosing our ‘key’ point to be (0, 0) and tracking the horizontal asymptotes y = − π 2 and y = π , we find that the graph of y = f (x) = 3 arctan (4x) differs from the graph of 2 y = F (x) = arctan(x) by a horizontal compression by a factor of 4 and a vertical stretch by a factor of 3. It is the latter which affects the range, producing a range of − 3π , 3π . 2 2 We confirm our findings on the calculator below. x π − arccos y = f (x) = 3 arctan (4x) 2 5 (c) To find the domain of g(x) = arccot x + π, we proceed as above. Since the domain of 2 G(x) = arccot(x) is (−∞, ∞), and x is defined for all x, we get that the domain of g is 2 (−∞, ∞) as well. As for the range, we note that the range of G(x) = arccot(x), like that of F (x) = arctan(x), is limited by a pair of horizontal asymptotes, in this case y = 0 and y = π. Following Theorem 1.7, we graph y = g(x) = arccot x + π starting with 2 y = G(x) = arccot(x) and first performing a horizontal expansion by a factor of 2 and following that with a vertical shift upwards by π. This latter transformation is the one which affects the range, making it now (π, 2π). To check this graphically, we encounter a bit of a problem, since on many calculators, there is no shortcut button corresponding to the arccotangent function. Taking a cue from number 1c, we attempt to rewrite g(x) = arccot x +π in terms of the arctangent function. Using Theorem 10.27, we have 2 2 that arccot x = arctan x when x > 0, or, in this case, when x > 0. Hence, for x > 0, 2 2 2 we have g(x) = arctan x + π. When x < 0, we can use the same argument in number 2 2 1c that gave us arccot(−2) = π + arctan − 1 to give us arccot x = π + arctan x . 2 2 y = f (x) = 5 It also confirms our domain! 10.6 The Inverse Trigonometric Functions 837 2 2 Hence, for x < 0, g(x) = π + arctan x + π = arctan x + 2π. What about x = 0? We know g(0) = arccot(0) + π = π, and neither of the formulas for g involving arctangent will produce this result.6 Hence, in order to graph y = g(x) on our calculators, we need to write it as a piecewise defined function: x g(x) = arccot +π =  2  We show the input and the result below.   arctan  arctan 2 x + 2π, if x < 0 π, if x = 0 2 x + π, if x > 0 y = g(x) in terms of arctangent y = g(x) = arccot x 2 +π The inverse trigonometric functions are typically found in applications whenever the measure of an angle is required. One such scenario is presented in the following example. Example 10.6.6. 7 The roof on the house below has a ‘6/12 pitch’. This means that when viewed from the side, the roof line has a rise of 6 feet over a run of 12 feet. Find the angle of inclination from the bottom of the roof to the top of the roof. Express your answer in decimal degrees, rounded to the nearest hundredth of a degree. Front View Side View Solution. If we divide the side view of the house down the middle, we find that the roof line forms the hypotenuse of a right triangle with legs of length 6 feet and 12 feet. Using Theorem 10.10, we 6 7 Without Calculus, of course . . . The authors would like to thank Dan Stitz for this problem and associated graphics. 838 Foundations of Trigonometry 6 find the angle of inclination, labeled θ below, satisfies tan(θ) = 12 = 1 . Since θ is an acute angle, 2 1 we can use the arctangent function and we find θ = arctan 2 radians ≈ 26.56◦ . 6 feet θ 12 feet 10.6.4 Solving Equations Using the Inverse Trigonometric Functions. In Sections 10.2 and 10.3, we learned how to solve equations like sin(θ) = 1 for angles θ and 2 tan(t) = −1 for real numbers t. In each case, we ultimately appealed to the Unit Circle and relied on the fact that the answers corresponded to a set of ‘common angles’ listed on page 724. If, on the other hand, we had been asked to find all angles with sin(θ) = 1 or solve tan(t) = −2 for 3 real numbers t, we would have been hard-pressed to do so. With the introduction of the inverse trigonometric functions, however, we are now in a position to solve these equations. A good parallel to keep in mind is how the square root function can be used to solve certain quadratic equations. 1 The equation x2 = 4 is a lot like sin(θ) = 2 in that it has friendly, ‘common value’ answers x = ±2. The equation x2 = 7, on the other hand, is a lot like sin(θ) = 1 . We know8 there are answers, but 3 we can’t express them using ‘friendly’ numbers.9 To solve x2 = 7, we make use of the square root √ function and write x = ± 7. We can certainly approximate these answers using a calculator, but √ as far as exact answers go, we leave them as x = ± 7. In the same way, we will use the arcsine function to solve sin(θ) = 1 , as seen in the following example. 3 Example 10.6.7. Solve the following equations. 1. Find all angles θ for which sin(θ) = 1 . 3 2. Find all real numbers t for which tan(t) = −2 3. Solve sec(x) = − 5 for x. 3 Solution. 1 1. If sin(θ) = 3 , then the terminal side of θ, when plotted in standard position, intersects the Unit Circle at y = 1 . Geometrically, we see that this happens at two places: in Quadrant I 3 and Quadrant II. If we let α denote the acute solution to the equation, then all the solutions 8 9 How do we know this again? √ This is all, of course, a matter of opinion. For the record, the authors find ± 7 just as ‘nice’ as ±2. 10.6 The Inverse Trigonometric Functions 839 to this equation in Quadrant I are coterminal with α, and α serves as the reference angle for all of the solutions to this equation in Quadrant II. y y 1 1 1 3 α = arcsin 1 x 1 3 radians α 1 3 1 x Since 1 isn’t the sine of any of the ‘common angles’ discussed earlier, we use the arcsine 3 1 functions to express our answers. The real number t = arcsin 3 is defined so it satisfies π 1 1 0 < t < 2 with sin(t) = 3 . Hence, α = arcsin 3 radians. Since the solutions in Quadrant I 1 are all coterminal with α, we get part of our solution to be θ = α + 2πk = arcsin 3 + 2πk for integers k. Turning our attention to Quadrant II, we get one solution to be π − α. Hence, the Quadrant II solutions are θ = π − α + 2πk = π − arcsin 1 + 2πk, for integers k. 3 2. We may visualize the solutions to tan(t) = −2 as angles θ with tan(θ) = −2. Since tangent is negative only in Quadrants II and IV, we focus our efforts there. y y 1 1 1 x β = arctan(−2) radians 1 x π β Since −2 isn’t the tangent of any of the ‘common angles’, we need to use the arctangent function to express our answers. The real number t = arctan(−2) satisfies tan(t) = −2 and − π < t < 0. If we let β = arctan(−2) radians, we see that all of the Quadrant IV solutions 2 840 Foundations of Trigonometry to tan(θ) = −2 are coterminal with β. Moreover, the solutions from Quadrant II differ by exactly π units from the solutions in Quadrant IV, so all the solutions to tan(θ) = −2 are of the form θ = β + πk = arctan(−2) + πk for some integer k. Switching back to the variable t, we record our final answer to tan(t) = −2 as t = arctan(−2) + πk for integers k. 5 3. The last equation we are asked to solve, sec(x) = − 3 , poses two immediate problems. First, we are not told whether or not x represents an angle or a real number. We assume the latter, but note that we will use angles and the Unit Circle to solve the equation regardless. Second, as we have mentioned, there is no universally accepted range of the arcsecant function. For that reason, we adopt the advice given in Section 10.3 and convert this to the cosine problem 3 cos(x) = − 5 . Adopting an angle approach, we consider the equation cos(θ) = − 3 and note 5 3 that our solutions lie in Quadrants II and III. Since − 5 isn’t the cosine of any of the ‘common angles’, we’ll need to express our solutions in terms of the arccosine function. The real number 3 t = arccos − 3 is defined so that π < t < π with cos(t) = − 5 . If we let β = arccos − 3 5 2 5 radians, we see that β is a Quadrant II angle. To obtain a Quadrant III angle solution, 3 we may simply use −β = − arccos − 5 . Since all angle solutions are coterminal with β 3 or −β, we get our solutions to cos(θ) = − 3 to be θ = β + 2πk = arccos − 5 + 2πk or 5 3 θ = −β + 2πk = − arccos − 5 + 2πk for integers k. Switching back to the variable x, we 5 3 record our final answer to sec(x) = − 3 as x = arccos − 5 + 2πk or x = − arccos − 3 + 2πk 5 for integers k. y y 1 1 β = arccos − 3 radians 5 β = arccos − 3 radians 5 1 x 1 x 3 −β = − arccos − 5 radians The reader is encouraged to check the answers found in Example 10.6.7 - both analytically and with the calculator (see Section 10.6.3). With practice, the inverse trigonometric functions will become as familiar to you as the square root function. Speaking of practice . . . 10.6 The Inverse Trigonometric Functions 841 10.6.5 Exercises √ √ 3. arcsin − √ 7. arcsin In Exercises 1 - 40, find the exact value. 1. arcsin (−1) 3 2. arcsin − 2 6. arcsin 1 2 2 2 4. arcsin − 1 2 5. arcsin (0) 2 2 √ 8. arcsin √ 3 2 9. arcsin (1) 1 13. arccos − 2 √ 17. arccos 3 2 √ 10. arccos (−1) 3 11. arccos − 2 15. arccos 1 2 √ 2 12. arccos − 2 √ 16. arccos 2 2 14. arccos (0) 18. arccos (1) √ 19. arctan − 3 √ 3 3 20. arctan (−1) 3 21. arctan − 3 25. arctan √ 3 22. arctan (0) √ 26. arccot − 3 √ 3 3 23. arctan 24. arctan (1) √ 3 3 27. arccot (−1) 28. arccot − √ √ 29. arccot (0) 33. arcsec (2) 37. arcsec √ 2 3 3 30. arccot 34. arccsc (2) 38. arccsc 31. arccot (1) 35. arcsec √ 2 32. arccot 36. arccsc 3 2 √ 2 3 3 39. arcsec (1) 40. arccsc (1) In Exercises 41 - 48, assume that the range of arcsecant is 0, π ∪ π, 3π and that the range of 2 2 arccosecant is 0, π ∪ π, 3π when finding the exact value. 2 2 41. arcsec (−2) √ 42. arcsec − 2 √ 46. arccsc − 2 √ 2 3 43. arcsec − 3 √ 2 3 47. arccsc − 3 44. arcsec (−1) 45. arccsc (−2) 48. arccsc (−1) 842 Foundations of Trigonometry π 2,π In Exercises 49 - 56, assume that the range of arcsecant is 0, π ∪ 2 arccosecant is − π , 0 ∪ 0, π when finding the exact value. 2 2 49. arcsec (−2) √ 50. arcsec − 2 √ 54. arccsc − 2 √ 2 3 51. arcsec − 3 √ 2 3 55. arccsc − 3 and that the range of 52. arcsec (−1) 53. arccsc (−2) 56. arccsc (−1) In Exercises 57 - 86, find the exact value or state that it is undefined. 57. sin arcsin 1 2 √ 58. sin arcsin − 5 4 5 13 2 2 59. sin arcsin 3 5 √ 62. cos arccos 2 2 60. sin (arcsin (−0.42)) 1 2 61. sin arcsin 63. cos arccos − 66. cos (arccos (π)) 69. tan arctan 64. cos arccos 65. cos (arccos (−0.998)) 68. tan arctan √ 3 67. tan (arctan (−1)) 70. tan (arctan (0.965)) √ 73. cot arccot − 3 76. cot arccot 79. sec arcsec 17π 4 1 2 √ 2 5 12 71. tan (arctan (3π)) 74. cot arccot − 77. sec (arcsec (2)) 80. sec (arcsec (0.75)) √ 2 3 83. csc arccsc − 3 86. csc arccsc π 4 7 24 72. cot (arccot (1)) 75. cot (arccot (−0.001)) 78. sec (arcsec (−1)) 81. sec (arcsec (117π)) √ 84. csc arccsc 2 2 82. csc arccsc 85. csc (arccsc (1.0001)) In Exercises 87 - 106, find the exact value or state that it is undefined. 87. arcsin sin π 6 88. arcsin sin − π 3 89. arcsin sin 3π 4 arcsec sec 114. arcsec sec − 125. 2 2 119. arccsc csc 122. arccsc csc 128.6 The Inverse Trigonometric Functions 11π 6 2π 3 5π 4 4π 3 3π 2 π 3 π 2 π 4 π 4 π 6 π 4 843 90. arcsec sec 124. arccot cot 104. arccsc csc 5π 6 π 6 π 2 110.10. arccsc csc π 2. arcsec sec − 113. arcsec sec 4π 3 5π 3 2π 3 11π 12 109. arctan tan − 101. 2 2 107. assume that the range of arcsecant is 0. π ∪ π. arcsec sec 111. arccot cot π 3 π 2 2π 3 103. arccsc csc 116. arcsec sec π 4 π 2 120. arccsc csc 129. arcsec sec 126. arctan tan 92. arccsc csc − 130. arcsin sin 94. arcsec sec 4π 3 5π 3 2π 3 11π 12 and that the range of 5π 6 π 6 π 2 121. 0 ∪ 0. 3π and that the range of 2 2 arccosecant is 0. 3π when finding the exact value.118. arcsec sec π 4 π 2 108. arccsc csc 117. arccsc csc 9π 8 . π when finding the exact value. arccos cos 91. arctan tan 100. arcsec sec 123. arcsec sec 112. arccot cot − 106. π ∪ π. arccos cos 95. arccos cos 96. arccsc csc − 118. assume that the range of arcsecant is 0.130. arctan (tan (π)) 102. arctan tan 99. π ∪ 2 arccosecant is − π . arccot (cot (π)) 2π 3 In Exercises 107 . arccsc csc 5π 4 11π 6 127. arccos cos 97. arccsc csc 5π 4 11π 6 115. arccot cot 105. arccos cos − 98.π 9π 8 In Exercises 119 . arcsin sin 93. sec (arctan (10)) 3 5 151. find the exact value or state that it is undefined. tan arccos − 1 2 136.154. find the exact value or state that it is undefined. sin (2 arctan (2)) 25 7 161. csc arcsin 154. cos (arcsec(3) + arctan(2)) 4 5 157. sin (arccsc (−3)) 138. sin (arctan (−2)) 5 13 134. tan arctan(3) + arccos − 13 5 3 5 158.844 Foundations of Trigonometry In Exercises 131 . 131. tan arcsin − 5 143. cos arctan 5 135. cot arccos 3 2 √ 5 147. cos (arccot (3)) 141. sec arccos 146. cot (arctan (0. sec arccot − 153. csc (arccot (9)) 150. cos 2arccot − 5 164. tan (arccot (12)) √ 144. cos 2 arcsin 162. cos 2arcsec arctan(2) 2 √ 163. sin 2 arcsin − 159.164. sec arcsin − 13 152. cot arccsc 3 2 √ 10 10 12 149. tan arcsec 5 3 √ 7 √ 2 5 140. cos (arcsec (5)) 142. 155. csc arctan − 2 3 In Exercises 155 . sin arccot 137. sin . cot arcsin 12 13 √ 145. sin arccos − √ 1 2 132.25)) 148. sin arcsin 5 13 + π 4 3 5 156. sin arccos 3 5 133. cos arcsin − 139. sin 2arccsc 160. sin(arccos(2x)) 176. cos (arctan (x)) 169. csc (arccos (x)) 172. cos(x) = 0. csc(x) = − 192. sin(2 arcsin(7x)) 179. cos(x) = − 2 9 190. tan (arcsin (x)) 170. cot(x) = −12 √ 198. solve the equation using the techniques discussed in Example 10. tan (2 arcsin(x)) 185. cos(x) = − 16 197. If sec(θ) = for 0 < θ < . sin(x) = 0.7 then approximate the solutions which lie in the interval [0. sin arccos x 5 166. 165. sec (arctan (x)) 171. cos (2 arctan (x)) 175. sin (2 arctan (x)) 173. sin(x) = 7 11 189.207.117 194. sec(x) = 2 3 199. cos arcsin x 2 167. find an expression for θ + sin(2θ) in terms of x. sin (arcsin(x) + arccos(x)) 183. sin(x) = −0. sin 1 arctan(x) 2 π π x for − < θ < . sin (arccos (x)) 168. find an expression for θ − sin(2θ) in terms of x. cos(2 arcsin(4x)) 181. sin 2 arcsin 180. cos (arcsin(x) + arctan(x)) 184. 2 2 2 x π π 1 1 for − < θ < . sin(x) = 8 193.10. find an expression for 4 tan(θ) − 4θ in terms of x. tan(x) = 0.008 195. 2π) to four decimal places. tan(x) = − 10 201.569 359 360 3 196. 4 2 In Exercises 188 . sin (2 arccos (x)) 174.184. If tan(θ) = 178. cos(x) = 202.03 . If sin(θ) = 186. 188. 7 2 2 2 2 x π 187. tan(x) = 117 90 17 7 200. cos (arctan (3x)) √ x 3 3 177.6. sin(x) = 0.6 The Inverse Trigonometric Functions 845 In Exercises 165 . rewrite the quantity as algebraic expressions of x and state the domain on which the equivalence is valid.3502 191. sec(arctan(2x)) tan(arctan(2x)) 182. 846 203. f (x) = 3 cos(2x) + 4 sin(2x) 219.5637 In Exercises 208 . What is the angle of elevation from the boat to the parasailor? Express your answer using degree measure rounded to one decimal place. 5. Approximate the value of φ (which is in radians.5 for reference. find the two acute angles in the right triangle whose sides have the given lengths. At Cliffs of Insanity Point. A parasailor is being pulled by a boat on Lake Ippizuti. A tag-and-release program to study the Sasquatch population of the eponymous Sasquatch National Park is begun.721 206. cos(x) = 0. 4 and 5 209. a fire is seen at a location known to be 10 miles away from the base of the sheer canyon wall.6109 Foundations of Trigonometry 205. In Exercises 216 . will the rod make with the wall? 214. f (x) = 5 sin(3x) + 12 cos(3x) 218. f (x) = cos(x) − 3 sin(x) 217. From that point. 336. If the range of the rifle with a tranquilizer dart is 300 feet. 213. What angle. Round your answer to the nearest hundreth of a degree. What angle does the wire make with the ground? Express your answer using degree measure rounded to one decimal place. sin(x) = −0. tan(x) = −0.9824 207. An 18-inch rod will be attached to the wall and the underside of the shelf at its edge away from the wall. 215. cos(x) = −0.210. 212. When pulled taut it touches level ground 360 feet from the base of the tower. From a 200 foot tall tower. a ranger spots a Sasquatch lumbering through the wilderness directly towards the tower. forming a right triangle under the shelf to support it. 216. The cable is 300 feet long and the parasailor is 100 feet above the surface of the water. 208. of course) to four decimal places. The Great Sasquatch Canyon is 7117 feet deep. Let θ denote the angle of depression from the top of the tower to a point on the ground.221. Express your answers using degree measure rounded to two decimal places. 527 and 625 211. Shelving is being built at the Utility Muffin Research Library which is to be 14 inches deep. A guy wire 1000 feet long is attached to the top of a tower. find the smallest value of θ for which the corresponding point on the ground is in range of the rifle. to the nearest degree. 12 and 13 210. 3. f (x) = 7 sin(10x) − 24 cos(10x) . rewrite the given function as a sinusoid of the form S(x) = A sin(ωx + φ) using Exercises 35 and 36 in Section 10. cot(x) = 1 117 204. What angle of depression is made by the line of sight from the canyon edge to the fire? Express your answer using degree measure rounded to one decimal place. f (x) = 2 sin(x) − cos(x) 847 In Exercises 222 . π π for |x| ≥ 1 as long as we use − . f (x) = arccos 1 −4 223. 235. f (x) = arccos 3x − 1 2 224. 0 ∪ 0. π as the range 2 2 234. f (x) = − cos(x) − 2 2 sin(x) 221.10. Discuss with your classmates why arcsin 238. 236. Use the following picture and the series of exercises on the next page to show that arctan(1) + arctan(2) + arctan(3) = π y D(2. for |x| ≥ 1 as long as we use 0. f (x) = arctan(4x) √ 229. 0) C(2. f (x) = arccot( 2x − 1) 232. Show that arccsc(x) = arcsin of f (x) = arccsc(x). Show that arcsec(x) = arccos of f (x) = arcsec(x). f (x) = arccsc e2x π π ∪ . as the range 2 2 π for −1 ≤ x ≤ 1. 3) A(0. Write your answers in interval notation. f (x) = arctan(ln(2x − 1)) 231. 1) α β γ x O(0. 2 1 2 = 30◦ . f (x) = arcsin(5x) 225. f (x) = arcsec(12x) 233. f (x) = arccsc(x + 5) 230. find the domain of the given function. 222.6 The Inverse Trigonometric Functions √ 220. f (x) = arcsec 1 x 1 x x3 8 x2 228. Show that arcsin(x) + arccos(x) = 237. f (x) = arccot 2x −9 x2 226. 0) .233. f (x) = arcsin 2x2 227. 0) B(1. (b) Use (c) Use (d) Use AOB to show that α = arctan(1) BAD to show that β = arctan(2) BCD to show that γ = arctan(3) (e) Use the fact that O. Thus arctan(1) + arctan(2) + arctan(3) = π. . Use the distance formula to show that BAD is also a right triangle (with ∠BAD being the right angle) by showing that the sides of the triangle satisfy the Pythagorean Theorem.848 Foundations of Trigonometry (a) Clearly AOB and BCD are right triangles because the line through O and A and the line through C and D are perpendicular to the x-axis. B and C all lie on the x-axis to conclude that α + β + γ = π. arcsec (2) = 36. arccot 3 = 41.6 The Inverse Trigonometric Functions 849 10. arccsc (2) = 6 √ 2 3 π 37.6. arcsin − 1 2 2 2 = π 2 = π 3 3π 4 =− π 6 π 4 π 1. arctan − 3 3 π 4 3π 4 = π 3 π 4 π 3 π 6 √ π 19. arcsin = 9. arctan (0) = 0 25. arcsec (1) = 0 √ 5π 42. arctan − 3 = − 3 22.10. arcsin 7. arccos − 1 2 10. arctan (1) = √ 5π 26. arccot − 3 = 6 π 29. arccot (−1) = √ 30. arccsc = 3 3 32. arcsin (−1) = − 2 4. arccsc (1) = π 2 33. arccos − 2 14. arcsec 2 = 4 √ 2 3 π 38. arcsec − 2 = 4 . arccos π = 4 17. arcsec (−2) = 4π 3 √ 27.6 Answers 3 2. arctan 3 3 = π 6 =− 24. arcsin (0) = 0 √ 8. arccot − 3 31. arccot (0) = 2 π 6 √ π 35. arccsc √ 2 = 39. arccos − = 15. arctan √ 3 = √ π 3 2π = 3 π 20. arcsec = 3 6 40. arccos (1) = 0 √ 21. arcsin − 2 π 6 5. arcsin − 1 2 √ 2 2 =− π = 4 6. arcsin 3 2 √ π 3 5π = 6 √ π =− 3 √ 3. arccos π 6 16. arccos = 18. arccos (−1) = π 1 2 √ 2 2 2π 3 3 11. arccos (0) = √ 3 2 π 2 2 2 = 13. arctan (−1) = − 4 √ 23. arccot (1) = π 4 π 34. arccot 3 3 3 28. arcsin (1) = √ 12. tan (arctan (3π)) = 3π √ 73.42 √ 2 2 √ = 2 2 61. arccsc (−1) = √ 3π 50. arcsec − 3 √ π 54.998 67. arccsc − 2 = 4 2π 49. cos (arccos (−0. sin arcsin 60. tan arctan √ 3 = √ 3 70. arcsec (−2) = 3 52. cot (arccot (1)) = 1 74. cos arccos 63. arccsc (−1) = − π 6 π 2 √ 2 3 51. arcsec (−1) = π √ 2 3 47. arcsec − 2 = 4 53. sin (arcsin (−0. arccsc − 3 57. arccsc − 2 = − 4 1 = 2 = 3 5 2 58. cos arccos 5 13 = 5 13 65.42)) = −0. cot arccot − 3 √ =− 3 75. tan arctan 5 12 = 5 12 66.965 72. cot (arccot (−0. sin arcsin − 2 √ √ =− 2 2 59.965)) = 0. sin arcsin is undefined. tan (arctan (0. 1 2 62.998)) = −0. cot arccot − 76. arccsc (−2) = − 56. arcsec − 3 7π 6 Foundations of Trigonometry 7π 6 3π 2 = 5π 6 = 44. cos arccos − =− 64. cot arccot 7 24 =− = 7 24 71.001 77. arccsc − 3 4π 3 45. 68.850 √ 2 3 43. tan (arctan (−1)) = −1 69. sin arcsin 1 2 3 5 5 4 1 2 =− π 3 = 48. arccsc (−2) = √ 5π 46. sec (arcsec (2)) = 2 17π 4 17π 4 78. cos (arccos (π)) is undefined. arcsec (−1) = π √ 2 3 55.001)) = −0. sec (arcsec (−1)) = −1 . arctan tan 3π 4 π 4 π 3 π 3 98. sec arcsec 81. arccos cos π 6 3π 4 4π 3 2π 3 π 6 = π 6 = π 4 π 3 88. arcsec sec − 112.10. arccos cos − 97. arcsec sec 104. arcsin sin 92.0001 87. arccot cot 107. arccos cos 94. arcsin sin 91.0001)) = 1. csc (arccsc (1. arcsin sin 93. arccot (cot (π)) is undefined 106. csc arccsc 86. π 6 . arccot cot 103. arccsc csc π 6 is undefined. arcsec sec 111. arcsec sec 2π 3 4π 3 π 2 = = = 2π 3 4π 3 3π 2 π 4 5π 6 5π 3 110. arcsec sec 109. csc arccsc 84. arctan tan − 100. arcsin sin − 90. =− π 3 π 6 85. csc arccsc π 4 π 3 √ 2 2 2 = √ 2 √ 2 3 83. = π 2 √ 2 3 =− 3 80. 82. arctan tan π 2 π 3 =− 99.75)) is undefined. sec arcsec is undefined. arccot cot − 105. arccos cos 11π 6 π 4 3π 2 5π 4 π 4 = =− π 4 = = π 2 =− = = = 2π 3 π 6 95. arccot cot 108. is undefined. sec (arcsec (0. arccos cos 96.6 The Inverse Trigonometric Functions 1 2 π 2 851 79. arcsin sin 89. arctan (tan (π)) = 0 101. csc arccsc − 3 √ is undefined. arctan tan 2π 3 π 4 =− = = = π 4 = = 7π 6 π 3 3π 4 π 2 π 3 is undefined = π 3 102. 852 5π 4 π 2 5π 4 3π 2 13π 12 Foundations of Trigonometry 2π 3 11π 6 9π 8 4π 3 π 2 = π 3 7π 6 113. arccsc csc 126. arcsec sec 123. arccsc csc 116. arccsc csc 130. arcsec sec − 124. cos (arccot (3)) = 10 √ 2 5 140. arccsc csc − 129. arcsec sec 121. arcsec sec π 2 =− =− 4 5 √ 6 = 6 = = 11π 12 √ 1 3 131. cos arctan 7 139. arccsc csc 132. tan arcsec 144. cos (arcsec (5)) = 1 2 136. arcsec sec 119. tan arccos − 143. π 6 = π 3 π 6 =− =− = 2π 3 11π 6 9π 8 3 5 √ 5 5 13 127. sin arccos − = 2 2 √ 2 5 133. arccsc csc = = 114. sin arccot π 6 is undefined. arcsec sec = 115. sin (arccsc (−3)) = − √ 137. arccsc csc 120. arccsc csc 128. cos arcsin − 12 13 2 = 4 1 5 √ =− 3 1 12 √ 3 10 138. arcsec sec 125. tan arcsin − = −2 5 142. arccsc csc = = = 11π 12 π 4 5π 6 5π 3 5π 4 π 2 = = π 4 = = 9π 8 2π 3 5π 6 π 3 π 4 122. arccsc csc 118. tan (arccot (12)) = . arccsc csc − 117. sin arccos 134. sin (arctan (−2)) = − 5 1 3 √ 11π 12 π 8 135. cot arcsin 5 3 12 13 = = 4 3 5 12 141. cot arccos 3 2 √ √ 853 = 3 146. tan arctan(3) + arccos − 159. sec (arctan (x)) = 1 + x2 for all x 166.6 The Inverse Trigonometric Functions √ 145. sin 1 − x2 for −1 ≤ x ≤ 1 1 for all x 1 + x2 x 167. cos (arctan (x)) = √ 169. sin (arccos (x)) = √ = 164. csc arcsin 3 5 5 13 = 5 3 √ 17 2 = 26 3 5 = 1 3 148.10. sec (arctan (10)) = 152. cos 2arcsec arctan(2) 2 25 7 = √ 163. cot arccsc 5 =2 √ 2 3 = 3 √ 101 147. sec arccot − = − 11 10 153. cos (arcsec(3) + arctan(2)) = 158. cos 2arccot − 5 165. sin 2 arcsin − 4 5 4 5 =− 527 625 √ 5− 5 10 =− 24 25 157. sin (2 arctan (x)) = 1 for −1 < x < 1 1 − x2 2x for all x +1 √ 171. cot (arctan (0. sin (2 arctan (2)) = 162. csc (arccos (x)) = √ 170. sin (2 arccos (x)) = 2x 1 − x2 for −1 ≤ x ≤ 1 x2 . csc (arccot (9)) = 2 154. cos 2 arcsin 13 5 3 5 = = 120 169 7 25 2 3 160. tan (arcsin (x)) = √ for −1 < x < 1 1 − x2 √ 168. csc arctan − 3 √ 82 √ =− 13 2 √ √ 5 − 4 10 15 155. sec arccos √ 3 2 149. sin 2arccsc 161. sec arcsin − 150. sin arcsin π + 4 156.25)) = 4 12 13 = 13 5 √ √ 10 151. ∪ − 2 1 − 2x 2 2 2           −  √ x2 + 1 − 1 √ 2 x2 + 1 √ x2 + 1 − 1 √ 2 x2 + 1 for x ≥ 0 for x < 0 √ ∪ 2 . − . sec(arctan(2x)) tan(arctan(2x)) = 2x 1 + 4x2 for all x Foundations of Trigonometry 181. sin arccos = for −5 ≤ x ≤ 5 5 5 √ 4 − x2 x 175. cos(2 arcsin(4x)) = 1 − 32x2 for − ≤ x ≤ 4 4 √ 180.1 2 184. If sin(θ) = for − < θ < . cos arcsin = for −2 ≤ x ≤ 2 2 2 172. but Calculus is required to fully understand what is happening at those x values. we exclude x = ±1 from our answer. sin (arcsin(x) + arccos(x)) = 1 for −1 ≤ x ≤ 1 √ 1 − x2 − x2 √ 182. cos (arcsin(x) + arctan(x)) = for −1 ≤ x ≤ 1 1 + x2 √ √ √ √ 2 2 2 2 10 tan (2 arcsin(x)) = 2x 1 − x for x in 183. You’ll see what we mean when you work through the details of the identity for tan(2t). . then θ − sin(2θ) = arctan − 2 7 2 2 2 2 2 7 x + 49 10 The equivalence for x = ±1 can be verified independently of the derivation of the formula. sin 2 arcsin for − 3 ≤ x ≤ 3 = 3 3 1 1 179. cos (arctan (3x)) = √ 1 + 9x2 √ 1 1 177. then θ + sin(2θ) = arcsin + 2 2 2 2 2 186. sin(arccos(2x)) = 1 − 4x2 for − 1 ≤ x ≤ 1 2 2 √ 25 − x2 x 174. sin(2 arcsin(7x)) = 14x 1 − 49x2 for − ≤ x ≤ 7 7 √ √ √ √ x 3 2x 3 − x2 178. If tan(θ) = x π π 1 1 1 x 7x for − < θ < . sin 1 arctan(x) 2 = √ x π π x x 4 − x2 185. For now. cos (2 arctan (x)) = 1 for all x 176.854 1 − x2 for all x 1 + x2 √ 173. −1. x ≈ 1. in [0. in [0. 5. in [0. x ≈ 1. x = arcsin(0. x = arccos 2 3 1 12 + πk. x ≈ 0. x = arcsin(0. 2π). x = π + arcsin(0. 2π). in [0. then 4 tan(θ) − 4θ = x2 − 16 − 4arcsec 4 2 4 7 11 2 9 + 2πk or x = π − arcsin 7 11 2 9 + 2πk. 5.4535.3502) + 2πk. 2π). x = arccos(−0. in [0. 5. 2π).784 203. 3. x = arccos − + 2πk or x = − arccos − 201.1135 206. 2π).1716 202.5622. 4. 2π). in [0. x ≈ 1.0187 199.117) + 2πk. 2π).7949.8297 192. 5.0300.7038 207.1336 193. x = arccos 359 360 + 2πk or x = 2π − arccos 359 360 + 2πk.6 The Inverse Trigonometric Functions 187. If sec(θ) = 188. in [0.0236. 2π).7348 . x ≈ 3. in [0. x ≈ 2. x = arccos(0.1697. x ≈ 0.2596 200. x ≈ 0.03) + πk.3502) + 2πk or x = π − arcsin(0. 2π).7469. x ≈ 2.9824) + 2πk. x ≈ 0.7572 7 16 + 2πk. x = arctan − 196. 4. 6.10. 2π). 2π).70384 195. 6. x ≈ 0. 2π).721) + 2πk. in [0. 2π). x ≈ 0.5932. x ≈ 2.3578.9468. in [0. 2. x ≈ 3.008) + 2πk. 6. 2. in [0. 4.569) + 2πk. x = arcsin √ x π x for 0 < θ < .8771. x ≈ 3. x ≈ 3. x = arctan(117) + πk. 4. x ≈ 1. in [0. in [0. x = arctan(0.0080. in [0. x = arctan(117) + πk. x ≈ 0. 4.3844.5637) + 2πk. 5. 2π). x = arccos − + 2πk or x = − arccos − 190.117) + 2πk or x = 2π − arccos(0. in [0. 2π). in [0.4780 204. 2. 2π).3316. in [0.2086 194.0932 + 2πk or x = 2π − arccos 17 90 197. x = π + arcsin(0.008) + 2πk or x = π − arcsin(0. x = arctan − 10 + πk.569) + 2πk or x = 2π − arcsin(0.1879. x = arctan(−0. 2π).721) + 2πk or x = 2π − arcsin(0.6898.8411. 2π).4518 + 2πk. in [0. in [0.2000 2 3 + 2πk.4422 17 90 + 2πk. 2π).4883 855 189.6109) + πk. x ≈ 0.0953 205.9824) + 2πk or x = 2π − arccos(0. 2π). x = arccos(0. 4. 3. 6.0746. x ≈ 1. in [0. x = arcsin 3 8 + 2πk or x = π − arcsin 7 16 3 8 + 2πk.5637) + 2πk or x = − arccos(−0.6779 191.0585.56225. x = π + arcsin + 2πk or x = 2π − arcsin √ 198. (−∞. f (x) = 7 sin(10x) − 24 cos(10x) = 25 sin 10x + arcsin − √ 220. ∞) 232.1760) ≈ 5 sin(2x + 0. (−∞.87◦ and 53. −6] ∪ [−4.∞ 2 −∞. ∞) 233. (−∞. f (x) = 3 cos(2x) + 4 sin(2x) = 5 sin 2x + arcsin √ ≈ 13 sin(3x + 1. √ √ √ √ 225. −2] ∪ [2. 5 5 √ √ 2 2 − . 36.1 3 1 3 ≈ 25 sin(10x − 1. 41. 2 2 ≈ √ 5 sin(x − 0.62◦ and 67. f (x) = 2 sin(x) − cos(x) = 1 1 − . [0.∞ 12 12 231. ∞) 229. 1 . ∞) . 223. (−∞.8198) 219. f (x) = − cos(x) − 2 2 sin(x) = 3 sin x + π + arcsin √ √ 5 sin x + arcsin − 5 5 1 − . − 5] ∪ [− 3. 19.13◦ 211. −3) ∪ (−3. 32. 68.9◦ 212. (−∞.81◦ 216. 22. ∞) 227. 3] ∪ [ 5.∞ 2 226.6435) √ 218.4636) 222.5◦ 12 13 3 5 215. f (x) = cos(x) − 3 sin(x) = √ 3 10 10 sin x + arccos − 10 24 25 ≈ 10 sin(x + 2.48◦ 214. 1 . f (x) = 5 sin(3x) + 12 cos(3x) = 13 sin 3x + arcsin 217.856 208.52◦ and 57.4814) 221.2870) ≈ 3 sin(x + 3. 224. ∞) 228. 51◦ Foundations of Trigonometry 210.7◦ 209.38◦ 213. 7. 230. − 1 1 ∪ . 3) ∪ (3. we substitute them into the original equation. • To solve cot(u) = c for c = 0. 18 3 Example 10. π and add 2 2 integer multiples of the period π. there are no real solutions. If c < −1 or of c > 1. sin(2x) = 0. To solve for x. For any integer k we have cos 2 5π 12 √ 5.3 and most recently 10.2. The solutions to cos(u) = − 23 are u = 5π + 2πk or u = 7π + 2πk for integers k.7 Trigonometric Equations and Inequalities In Sections 10. This is the technique employed in the example below.7. • To solve tan(u) = c for any real number c. 1. Since the argument of sine here is 3x. convert to tangent and solve as above.6. 2π) and add integer multiples of the period 2π. Don’t forget to divide the 2πk by 3 as well! . To check these answers 12 12 analytically. there are no real solutions. tan x 2 = −3 6. Strategies for Solving Basic Equations Involving Trigonometric Functions • To solve cos(u) = c or sin(u) = c for −1 ≤ c ≤ 1. List the solutions which lie in the interval [0.1. first solve for u in the interval [0. 10. Since 6 6 the argument of cosine here is 2x. Solving for x gives x = 5π + πk or x = 7π + πk for integers k. • To solve sec(u) = c or csc(u) = c for c ≤ −1 or c ≥ 1.5 for a review of this concept. How do we solve something like sin(3x) = 1 ? Since this equation 6 2 has the form sin(u) = 1 . and obtain x = 18 + 2π k 3 or x = 5π + 2π k for integers k. we solved some basic equations involving the trigonometric functions. respectively. this means 2x = 5π + 2πk or 2x = 7π + 2πk for integers 6 6 k. Solve the following equations and check your answers analytically. we can comfortably solve sin(x) = 1 and find the solution x = π + 2πk 2 6 or x = 5π + 2πk for integers k. first solve for u in the interval − π .7 Trigonometric Equations and Inequalities 857 10. If −1 < c < 1. cos(2x) = − 23 3 4. Note that we use the neutral letter ‘u’ as the argument1 of each circular function for generality. If c = 0. sec2 (x) = 4 Solution.10. we divide both sides2 of these equations by 3. we have 3x = π + 2πk or 3x = 5π + 2πk for 6 6 π integers k.87 + πk = cos = cos = − √ 3 2 5π 6 5π 6 + 2πk (the period of cosine is 2π) 1 2 See the comments at the beginning of Section 10. csc 1 x − π = 2 3. the solution to cot(u) = 0 is u = π + πk for integers k. Below we summarize the techniques we’ve employed thus far. √ √ 2. convert to cosine or sine. 2π) and verify them using a graphing utility. and solve as above. we know the solutions take the form u = π + 2πk or u = 5π + 2πk for 2 6 6 integers k. 2 Using the above guidelines. cot (3x) = 0 1. The solutions we keep come from the values of k = 0 and k = 1 and are x = 5π . 17π and 19π . 7π . 2π). 4 4 π 1 x − π = + 2πk 3 4 1 To solve 3 x − π = π 4 √ or 1 3π x−π = + 2πk 3 4 + 2πk. we first add π to both sides 1 π x = + 2πk + π 3 4 A common error is to treat the ‘2πk’ and ‘π’ terms as ‘like’ terms and try to combine them when they are not.3 We can. To 3 4 4 check the first family of answers. 2π) and examine where these two graphs intersect. 1 x − π = 3π + 2πk produces x = 21π + 6πk for integers k. Since this equation has the form csc(u) = 2. we find cos 2 7π + πk = cos 7π + 2πk = cos 7π = − 23 . we rewrite this as sin(u) = 22 and find 1 u = π + 2πk or u = 3π + 2πk for integers k. To verify this graphically. Using a 12 12 12 12 √ calculator. We see that the x-coordinates of the intersection points correspond to the decimal representations of our exact answers. Despite having infinitely many solutions. 3 Do you see why? . combine the ‘π’ and ‘ π ’ terms to get 4 1 5π x= + 2πk 3 4 We now finish by multiplying both sides by 3 to get x=3 5π + 2πk 4 = 15π + 6πk 4 Solving the other equation. csc 1 3 15π 4 + 6πk − π = csc = csc = csc √ = 2 5π 4 + 2πk π 4 + 2πk π 4 −π (the period of cosecant is 2π) The family x = 21π + 6πk checks similarly. we substitute. however. To determine 12 6 6 which of our solutions lie in [0. we use a√ reciprocal identity to rewrite the cosecant as a sine and we find that y = sin 11x−π and y = 2 do not intersect at ) (3 all over the interval [0.858 Foundations of Trigonometry Similarly. 2π). combine line terms. √ √ 2. we graph y = cos(2x) and y = − 23 over [0. Since the argument of cosecant here is 3 x − π . we substitute integer values for k. and simplify. 2π). we find 4 that none of them lie in [0. The complication in solving an equation like sec2 (x) = 4 comes not from the argument of secant. in this case. 3x = 2 for integers k. As a result. Checking our answers. 3 3 sec π + πk = ± sec π = ±2. we get x = 3 + 2πk or x = 4π + 2πk for 3 integers k. which is just x. we know u = π + πk. However. On many calculators. π . we extract square roots to get sec(x) = ±2. but what happens when you try to find the intersection points? 4 . which lie in [0. there is no function button for cotangent. (Can you show this?) As a result. we see we are finding the measures of all angles with a reference angle of π . To confirm graphically. 2π): x = π . To check the first family of solutions. We choose4 to use the quotient identity cot(3x) = cos(3x) . To get this equation to look like one of the forms listed on page 857. For cos(x) = 2 . we get x = π + 2πk or 3 5π 1 2π x = 3 + 2πk for integers k. Since cot(3x) = 0 has the form cot(u) = 0. 2π) come from 3 the values k = 0 and k = 1.10. The graph on the calculator appears identical. but rather. Solving for x yields x = π + π k. 4π and 5π . we note that. 5π . 1 1 Converting to cosines. To confirm these graphically. namely x = π . we have cos(x) = ± 2 . the fact the secant is being squared. we obtain six answers. If we take a step back and think of these families of solutions geometrically. we use 3 3 3 3 1 The reader is encouraged to see what happens if we had chosen the reciprocal identity cot(3x) = tan(3x) instead. we see sin(3x) sin(3x) that the x-coordinates of the intersection points approximately match our solutions.7 Trigonometric Equations and Inequalities 859 √ y = cos(2x) and y = − 3 2 y= 1 sin( 1 x−π ) 3 and y = √ 2 π 2 3. 4. it is true that for all integers k. depending on the integer k. For cos(x) = − 2 . 3 3 sec2 π 3 + πk = ± sec π 3 2 = (±2)2 = 4 The same holds for the family x = 2π + πk. these 3 solutions can be combined and we may write our solutions as x = π + πk and x = 2π + πk 3 3 for integers k. corresponding to k = 0 through k = 5. 2π . so. 7π . The solutions which lie in [0. 3π and 11π . Graphing y = cos(3x) and y = 0 (the x-axis). we must be 6 2 6 6 2 6 careful. sec π + πk doesn’t always equal sec π . we get 6 3 cot 3 π 6 + πk + πk 3 = cot = cot = 0 π 2 π 2 + πk (the period of cotangent is π) As k runs through the integers. 2π). The x-coordinates of the intersection 1 points of y = (cos(x))2 and y = 4 verify our answers. To check. 6. 2 Hence.87. we first note that it has the form sin(u) = 0. so x = 2 arctan(−3) + 2πk for integers k.27) To determine which of our answers lie in the interval [0. Since this is outside the interval [0. 2π).860 Foundations of Trigonometry a reciprocal identity to rewrite the secant as cosine.87) + πk for integers k. which has the family of solutions u = arcsin(0.87) + 2πk or u = π − arcsin(0. The equation tan x = −3 has the form tan(u) = −3. Multiplying 2 through by 2 gives −π < 2 arctan(−3) < 0. This means x = 2 arctan(−3) + 2π lies in [0. But seriously. x = arctan(−3) + πk. we first need to get an idea of the value of 2 arctan(−3). Starting with the inequality −π < 2 arctan(−3) < 0.7851. To check. we note 2 tan 2 arctan(−3)+2πk 2 = tan (arctan(−3) + πk) = tan (arctan(−3)) = −3 (the period of tangent is π) (See Theorem 10. Graphically. what fun would that be? 5 .5 we proceed analytically. Advancing k to 2 produces x = 2 arctan(−3) + 4π.87) + 2πk or 2x = π − arcsin(0.87) + 2πk for integers k. we get 2x = arcsin(0. We are now in a position to argue which of the solutions x = 2 arctan(−3) + 2πk lie in [0. For k = 0. it follows that − π < arctan(−3) < 0. 2π). To solve sin(2x) = 0. we add 2π and get π < 2 arctan(−3) + 2π < 2π. we discard x = 2 arctan(−3) + 4π and all solutions of the form x = 2 arctan(−3) + 2πk for k > 2. Next. 2 2 Your instructor will let you know if you should abandon the analytic route at this point and use your calculator. Since the argument of sine here is 2x. Once again. we get x = 2 arctan(−3) < 0. 2π) at x = 2 arctan(−3) + 2π ≈ 3.87) + 2πk 1 which gives x = 2 arcsin(0. we see y = tan x and y = −3 intersect only 2 once on [0.87) + πk or x = π − 1 arcsin(0. 2π). so we discard this answer and all answers x = 2 arctan(−3) + 2πk where k < 0. we turn our attention to k = 1 and get x = 2 arctan(−3) + 2π. Since −3 < 0. y= cos(3x) sin(3x) and y = 0 y= 1 cos2 (x) and y = 4 5.87. we get from −π < 2 arctan(−3) < 0 that 3π < 2 arctan(−3) + 4π < 4π. While we could easily find an approximation using a calculator. whose solution is u = arctan(−3)+πk. 87).87.87) < π so that multiplying through by 1 gives us 2 2 0 < 1 arcsin(0.87) + πk 2 To determine which of these solutions lie in [0. we have found four solutions to sin(2x) = 0.87) + 2πk) = sin (arcsin(0.87) + πk = sin (arcsin(0. 2π): 2 2 1 1 x = 2 arcsin(0. Starting with the family of solutions x = 1 arcsin(0. 2π): x = 1 arcsin(0.87) + πk .87). we move to the family x = π − 1 arcsin(0.87) > π .87) < π . or 2 2 2 2 4 π 1 π π 4 < 2 − 2 arcsin(0.87) and 2 x = 3π − 1 arcsin(0.87) and x = 3π − 1 arcsin(0. 0 < arcsin(0. we get 2 = sin (π − arcsin(0.87 in [0.87 (the period of sine is 2π) (See Theorem 10. Here. From the inequality 0 < 1 arcsin(0.87) + πk for integers k.87).26) For the family x = sin 2 π 2 π 2 − 1 arcsin(0. Proceeding with the usual arguments. x = π − 1 arcsin(0.87)) = sin (arcsin(0.87).87 (the period of sine is 2π) (sin(π − t) = sin(t)) (See Theorem 10. 2π).87) and x = 2 arcsin(0.87) + π. we find the only solutions 1 which lie in [0. namely x = π − 2 arcsin(0. All told.26) − 1 arcsin(0. we first need to get an idea of the value of x = 1 arcsin(0.87)) = 0. 2π) correspond to k = 0 and k = 1. Once again.87) < 2 . we confirm our results.87) + π.87).87) < π . but we adopt an analytic 2 route here. y = tan x 2 and y = −3 y = sin(2x) and y = 0. we need to 2 2 get a better estimate of π − 1 arcsin(0. 2 2 2 4 we first multiply through by −1 and then add π to get π > π − 1 arcsin(0. x = 2 arcsin(0.87 . we use 2 4 2 the same kind of arguments as in our solution to number 5 above and find only the solutions 1 corresponding to k = 0 and k = 1 lie in [0.87) + πk.87) + 2πk) = sin (π − arcsin(0. By definition. 2 Next.87)) = 0.7 Trigonometric Equations and Inequalities 861 sin 2 1 2 arcsin(0.10. 2 2 2 2 By graphing y = sin(2x) and y = 0. we could use the calculator. with x = 0 and x = π being the two solutions which lie in [0. it isn’t counted among the solutions in [0. we will need to use identities and Algebra to reduce the equation to the same form as those given on page 857.862 Foundations of Trigonometry Each of the problems in Example 10.1 featured one trigonometric function.7. To check graphically. we get sec2 (x) 1 + tan2 (x) 2 (x) − tan(x) − 2 tan u2 − u − 2 (u + 1)(u − 2) = = = = = tan(x) + 3 tan(x) + 3 (Since sec2 (x) = 1 + tan2 (x). 1. cos(3x) = 2 − cos(x) √ 6. Solve the following equations and list the solutions which lie in the interval [0. 2π) 3 1 1 to be x = arcsin 3 and x = π − arcsin 3 . Analysis of sec2 (x) = tan(x) + 3 reveals two different trigonometric functions. We find the two solutions here which lie in [0. 3 sin3 (x) = sin2 (x) 3. . In the forthcoming solutions. 2π). we use the arcsine function to get x = arcsin 1 +2πk 3 3 or x = π − arcsin 1 + 2πk for integers k. sin2 (x)(3 sin(x) − 1) = 0 1 We get sin2 (x) = 0 or 3 sin(x) − 1 = 0. If an equation involves two different trigonometric functions or if the equation contains the same trigonometric function but with different arguments. The solution to the first equation is x = πk. sin(x) cos Solution. cos(x) − 3 sin(x) = 2 + cos(x) sin x 2 =1 2. 3 sin3 (x) = sin2 (x) 3 sin3 (x) − sin2 (x) = 0 Factor out sin2 (x) from both terms. 2π). cos(2x) = 3 cos(x) − 2 5. Solving for sin(x). so an identity is in order. 1. 0 6 Note that we are not counting the point (2π.2.) 0 0 Let u = tan(x). sec2 (x) = tan(x) + 3 4. 2π). 2π). Since sec2 (x) = 1 + tan2 (x).6 x 2 2.7. 0) in our solution set since x = 2π is not in the interval [0. Some y = (sin(x)) extra zooming is required near x = 0 and x = π to verify that these two curves do in fact intersect four times. 2π) graphically. remember that while x = 2π may be a solution to the equation. We resist the temptation to divide both sides of 3 sin3 (x) = sin2 (x) by sin2 (x) (What goes wrong if you do?) and instead gather all of the terms to one side of the equation and factor. sin(2x) = 3 cos(x) √ 8. cos(3x) = cos(5x) 7. we find sin(x) = 0 or sin(x) = 3 . To solve sin(x) = 1 . Example 10. we plot y = 3(sin(x))3 and 2 and find the x-coordinates of the intersection points of these two curves. Verify your solutions on [0. From Example 10.10. we employ the 4 arctangent function and get x = arctan(2) + πk for integers k. cos(3x) − 3 cos(x) 2 cos3 (x) − 2 cos(x) − 2 4u3 − 2u − 2 4 cos3 (x) = = = = 2 − cos(x) 2 − cos(x) 0 0 Let u = cos(x). 2π). Since u = cos(x). 2π). we rewrite the secant 1 as a cosine and graph y = (cos(x))2 and y = tan(x) + 3 to find the x-values of the points where they intersect. on both sides but the arguments differ. we find. number 4. Using the same sort of argument 4 4 we saw in Example 10. 2π). In the equation cos(2x) = 3 cos(x) − 2. we have the same circular function. Since u = tan(x). π . y = 3(sin(x))3 and y = (sin(x))2 y= 1 (cos(x))2 and y = tan(x) + 3 3. 4.3. This transforms the equation into a polynomial in terms of cos(x). we get x = π + 2πk or x = 5π + 2πk for integers k. Solving 2 2 cos(x) = 1 . we use the same technique as in the previous problem.1. we have tan(x) = −1 or tan(x) = 2. that the curves intersect in three places on [0. From the first set of solutions. From cos(x) = 1. we know that cos(3x) = 4 cos3 (x) − 3 cos(x). and the x-coordinates of these points confirm our results. Using the identity cos(2x) = 2 cos2 (x) − 1. 2π) are x = 0.7 Trigonometric Equations and Inequalities 863 This gives u = −1 or u = 2. To solve tan(x) = 2. we get cos(x) = 1 or cos(x) = 1. we obtain a ‘quadratic in disguise’ and proceed as we have done in the past. we get x = − π + πk for integers k. cos(2x) 2 cos2 (x) − 1 2 cos2 (x) − 3 cos(x) + 1 2u2 − 3u + 1 (2u − 1)(u − 1) = = = = = 3 cos(x) − 2 3 cos(x) − 2 (Since cos(2x) = 2 cos2 (x) − 1. 0 This gives u = 1 or u = 1.4.) 0 0 Let u = cos(x). Using a reciprocal identity. we get x = arctan(2) and x = π + arctan(2) as answers from our second set of solutions which lie in [0. The answers which lie in [0. we get x = 3π and x = 5π as our answers which lie in [0. namely cosine. From tan(x) = −1. after a little extra effort. and 5π . Graphing 3 3 y = cos(2x) and y = 3 cos(x) − 2. To solve cos(3x) = 2 − cos(x). . we get 2 3 3 x = 2πk for integers k.7. 864 Foundations of Trigonometry To solve 4u3 − 2u − 2 = 0, we need the techniques in Chapter 3 to factor 4u3 − 2u − 2 into (u − 1) 4u2 + 4u + 2 . We get either u − 1 = 0 or 4u2 + 2u + 2 = 0, and since the discriminant of the latter is negative, the only real solution to 4u3 − 2u − 2 = 0 is u = 1. Since u = cos(x), we get cos(x) = 1, so x = 2πk for integers k. The only solution which lies in [0, 2π) is x = 0. Graphing y = cos(3x) and y = 2 − cos(x) on the same set of axes over [0, 2π) shows that the graphs intersect at what appears to be (0, 1), as required. y = cos(2x) and y = 3 cos(x) − 2 y = cos(3x) and y = 2 − cos(x) 5. While we could approach cos(3x) = cos(5x) in the same manner as we did the previous two problems, we choose instead to showcase the utility of the Sum to Product Identities. From cos(3x) = cos(5x), we get cos(5x) − cos(3x) = 0, and it is the presence of 0 on the right hand side that indicates a switch to a product would be a good move.7 Using Theorem 10.21, we have that cos(5x) − cos(3x) = −2 sin 5x+3x sin 5x−3x = −2 sin(4x) sin(x). Hence, 2 2 the equation cos(5x) = cos(3x) is equivalent to −2 sin(4x) sin(x) = 0. From this, we get sin(4x) = 0 or sin(x) = 0. Solving sin(4x) = 0 gives x = π k for integers k, and the solution 4 to sin(x) = 0 is x = πk for integers k. The second set of solutions is contained in the first set of solutions,8 so our final solution to cos(5x) = cos(3x) is x = π k for integers k. There are 4 eight of these answers which lie in [0, 2π): x = 0, π , π , 3π , π, 5π , 3π and 7π . Our plot of the 4 2 4 4 2 4 graphs of y = cos(3x) and y = cos(5x) below (after some careful zooming) bears this out. √ 6. In examining the equation sin(2x) = 3 cos(x), not only do we have different circular functions involved, namely sine and cosine, we also have different arguments to contend with, namely 2x and x. Using the identity sin(2x) = 2 sin(x) cos(x) makes all of the arguments the same and we proceed as we would solving any nonlinear equation – gather all of the nonzero terms on one side of the equation and factor. sin(2x) 2 sin(x) cos(x) √ 2 sin(x) cos(x) − 3 cos(x) √ cos(x)(2 sin(x) − 3) √ √ = √3 cos(x) = 3 cos(x) (Since sin(2x) = 2 sin(x) cos(x).) = 0 = 0 √ 3 2 . from which we get cos(x) = 0 or sin(x) = integers k. From sin(x) = 7 8 From cos(x) = 0, we obtain x = 2π 3 π 2 + πk for 3 2 , we get x = π 3 +2πk or x = +2πk for integers k. The answers As always, experience is the greatest teacher here! As always, when in doubt, write it out! 10.7 Trigonometric Equations and Inequalities which lie in [0, 2π) are x = π , 3π , π and 2π . We graph y = sin(2x) and y = 2 2 3 3 after some careful zooming, verify our answers. √ 865 3 cos(x) and, y = cos(3x) and y = cos(5x) y = sin(2x) and y = √ 3 cos(x) 7. Unlike the previous problem, there seems to be no quick way to get the circular functions or their arguments to match in the equation sin(x) cos x + cos(x) sin x = 1. If we stare at 2 2 it long enough, however, we realize that the left hand side is the expanded form of the sum formula for sin x + x . Hence, our original equation is equivalent to sin 3 x = 1. Solving, 2 2 we find x = π + 4π k for integers k. Two of these solutions lie in [0, 2π): x = π and x = 5π . 3 3 3 3 Graphing y = sin(x) cos x + cos(x) sin x and y = 1 validates our solutions. 2 2 8. With the absence of double angles or squares, there doesn’t seem to be much we can do. However, since the arguments of the cosine and sine are the same, we can rewrite the left √ hand side of this equation as a sinusoid.9 To fit f (x) = cos(x) − 3 sin(x) to the form A sin(ωt + φ) + B, we use what we learned in Example 10.5.3 and find A = 2, B = 0, ω = 1 √ and φ = 5π . Hence, we can rewrite the equation cos(x) − 3 sin(x) = 2 as 2 sin x + 5π = 2, 6 6 or sin x + 5π = 1. Solving the latter, we get x = − π + 2πk for integers k. Only one of 6 3 these solutions, x = 5π , which corresponds to k = 1, lies in [0, 2π). Geometrically, we see √ 3 that y = cos(x) − 3 sin(x) and y = 2 intersect just once, supporting our answer. y = sin(x) cos x 2 + cos(x) sin x 2 and y = 1 y = cos(x) − √ 3 sin(x) and y = 2 We repeat here the advice given when solving systems of nonlinear equations in section 8.7 – when it comes to solving equations involving the trigonometric functions, it helps to just try something. We are essentially ‘undoing’ the sum / difference formula for cosine or sine, depending on which form we use, so this problem is actually closely related to the previous one! 9 866 Foundations of Trigonometry Next, we focus on solving inequalities involving the trigonometric functions. Since these functions are continuous on their domains, we may use the sign diagram technique we’ve used in the past to solve the inequalities.10 Example 10.7.3. Solve the following inequalities on [0, 2π). Express your answers using interval notation and verify your answers graphically. 1. 2 sin(x) ≤ 1 Solution. 1. We begin solving 2 sin(x) ≤ 1 by collecting all of the terms on one side of the equation and zero on the other to get 2 sin(x) − 1 ≤ 0. Next, we let f (x) = 2 sin(x) − 1 and note that our original inequality is equivalent to solving f (x) ≤ 0. We now look to see where, if ever, f is undefined and where f (x) = 0. Since the domain of f is all real numbers, we can immediately 1 set about finding the zeros of f . Solving f (x) = 0, we have 2 sin(x) − 1 = 0 or sin(x) = 2 . 5π π The solutions here are x = 6 + 2πk and x = 6 + 2πk for integers k. Since we are restricting our attention to [0, 2π), only x = π and x = 5π are of concern to us. Next, we choose test 6 6 values in [0, 2π) other than the zeros and determine if f is positive or negative there. For x = 0 we have f (0) = −1, for x = π we get f π = 1 and for x = π we get f (π) = −1. 2 2 Since our original inequality is equivalent to f (x) ≤ 0, we are looking for where the function is negative (−) or 0, and we get the intervals 0, π ∪ 5π , 2π . We can confirm our answer 6 6 graphically by seeing where the graph of y = 2 sin(x) crosses or is below the graph of y = 1. 2. sin(2x) > cos(x) 3. tan(x) ≥ 3 (−) 0 (+) 0 (−) 0 π 6 5π 6 2π y = 2 sin(x) and y = 1 2. We first rewrite sin(2x) > cos(x) as sin(2x) − cos(x) > 0 and let f (x) = sin(2x) − cos(x). Our original inequality is thus equivalent to f (x) > 0. The domain of f is all real numbers, so we can advance to finding the zeros of f . Setting f (x) = 0 yields sin(2x) − cos(x) = 0, which, by way of the double angle identity for sine, becomes 2 sin(x) cos(x) − cos(x) = 0 or cos(x)(2 sin(x)−1) = 0. From cos(x) = 0, we get x = π +πk for integers k of which only x = π 2 2 1 and x = 3π lie in [0, 2π). For 2 sin(x) − 1 = 0, we get sin(x) = 2 which gives x = π + 2πk or 2 6 x = 5π + 2πk for integers k. Of those, only x = π and x = 5π lie in [0, 2π). Next, we choose 6 6 6 10 See page 214, Example 3.1.5, page 321, page 399, Example 6.3.2 and Example 6.4.2 for discussion of this technique. 10.7 Trigonometric Equations and Inequalities our test values. For x = 0 we find f (0) = −1; when x = for x = 3π 4 π 4 867 we get f π 4 √ we get f 3π 4 √ = −1 + √ 2 2 x = 7π we get f 7π = −1 − 22 = 4 4 our answer. We can use the calculator to check that the graph of y = sin(2x) is indeed above the graph of y = cos(x) on those intervals. 2−2 ; when x = π we have f (π) = 1, and lastly, for √2 −2− 2 . We see f (x) > 0 on π , π ∪ 5π , 3π , so this is 2 6 2 6 2 √ = 1− 2 2 = √ 2− 2 2 ; = (−) 0 (+) 0 (−) 0 (+) 0 (−) 0 π 6 π 2 5π 6 3π 2 2π y = sin(2x) and y = cos(x) 3. Proceeding as in the last two problems, we rewrite tan(x) ≥ 3 as tan(x) − 3 ≥ 0 and let f (x) = tan(x) − 3. We note that on [0, 2π), f is undefined at x = π and 3π , so those 2 2 values will need the usual disclaimer on the sign diagram.11 Moving along to zeros, solving f (x) = tan(x) − 3 = 0 requires the arctangent function. We find x = arctan(3) + πk for integers k and of these, only x = arctan(3) and x = arctan(3) + π lie in [0, 2π). Since 3 > 0, we know 0 < arctan(3) < π which allows us to position these zeros correctly on the 2 sign diagram. To choose test values, we begin with x = 0 and find f (0) = −3. Finding a convenient test value in the interval arctan(3), π is a bit more challenging. Keep in mind 2 that the arctangent function is increasing and is bounded above by π . This means that 2 the number x = arctan(117) is guaranteed12 to lie between arctan(3) and π . We see that 2 f (arctan(117)) = tan(arctan(117)) − 3 = 114. For our next test value, we take x = π and find f (π) = −3. To find our next test value, we note that since arctan(3) < arctan(117) < π , 2 it follows13 that arctan(3) + π < arctan(117) + π < 3π . Evaluating f at x = arctan(117) + π 2 yields f (arctan(117) + π) = tan(arctan(117) + π) − 3 = tan(arctan(117)) − 3 = 114. We choose our last test value to be x = 7π and find f 7π = −4. Since we want f (x) ≥ 0, we 4 4 see that our answer is arctan(3), π ∪ arctan(3) + π, 3π . Using the graphs of y = tan(x) 2 2 and y = 3, we see when the graph of the former is above (or meets) the graph of the latter. 11 12 See page 321 for a discussion of the non-standard character known as the interrobang. We could have chosen any value arctan(t) where t > 3. 13 . . . by adding π through the inequality . . . 868 Foundations of Trigonometry (−) 0 (+) 0 arctan(3) (−) 0 (+) π 2 (arctan(3) + π) (−) 3π 2 2π y = tan(x) and y = 3 We close this section with an example that puts solving equations and inequalities to good use – finding domains of functions. Example 10.7.4. Express the domain of the following functions using extended interval notation.14 1. f (x) = csc 2x + Solution. 1. To find the domain of f (x) = csc 2x + π 3 π 3 2. f (x) = sin(x) 2 cos(x) − 1 3. f (x) = 1 − cot(x) , we rewrite f in terms of sine as f (x) = Since the sine function is defined everywhere, our only concern comes from zeros in the denominator. Solving sin 2x + π = 0, we get x = − π + π k for integers k. In set-builder notation, 3 6 2 our domain is x : x = − π + π k for integers k . To help visualize the domain, we follow the 6 2 old mantra ‘When in doubt, write it out!’ We get x : x = − π , 2π , − 4π , 5π , − 7π , 8π , . . . , 6 6 6 6 6 6 where we have kept the denominators 6 throughout to help see the pattern. Graphing the situation on a numberline, we have 1 . sin(2x+ π ) 3 − 7π 6 − 4π 6 −π 6 2π 6 5π 6 8π 6 Proceeding as we did on page 756 in Section 10.3.1, we let xk denote the kth number excluded from the domain and we have xk = − π + π k = (3k−1)π for integers k. The intervals which 6 2 6 comprise the domain are of the form (xk , xk + 1 ) = (3k−1)π , (3k+2)π as k runs through the 6 6 integers. Using extended interval notation, we have that the domain is ∞ k=−∞ (3k − 1)π (3k + 2)π , 6 6 We can check our answer by substituting in values of k to see that it matches our diagram. 14 See page 756 for details about this notation. 10.7 Trigonometric Equations and Inequalities 869 2. Since the domains of sin(x) and cos(x) are all real numbers, the only concern when finding sin(x) the domain of f (x) = 2 cos(x)−1 is division by zero so we set the denominator equal to zero and solve. From 2 cos(x)−1 = 0 we get cos(x) = 1 so that x = π +2πk or x = 5π +2πk for integers 2 3 3 k. Using set-builder notation, the domain is x : x = π + 2πk and x = 5π + 2πk for integers k , 3 3 or x : x = ± π , ± 5π , ± 7π , ± 11π , . . . , so we have 3 3 3 3 − 11π 3 − 7π 3 − 5π 3 −π 3 π 3 5π 3 7π 3 11π 3 Unlike the previous example, we have two different families of points to consider, and we present two ways of dealing with this kind of situation. One way is to generalize what we did in the previous example and use the formulas we found in our domain work to describe the intervals. To that end, we let ak = π + 2πk = (6k+1)π and bk = 5π + 2πk = (6k+5)π for 3 3 3 3 integers k. The goal now is to write the domain in terms of the a’s an b’s. We find a0 = π , 3 a1 = 7π , a−1 = − 5π , a2 = 13π , a−2 = − 11π , b0 = 5π , b1 = 11π , b−1 = − π , b2 = 17π and 3 3 3 3 3 3 3 3 b−2 = − 7π . Hence, in terms of the a’s and b’s, our domain is 3 . . . (a−2 , b−2 ) ∪ (b−2 , a−1 ) ∪ (a−1 , b−1 ) ∪ (b−1 , a0 ) ∪ (a0 , b0 ) ∪ (b0 , a1 ) ∪ (a1 , b1 ) ∪ . . . If we group these intervals in pairs, (a−2 , b−2 )∪(b−2 , a−1 ), (a−1 , b−1 )∪(b−1 , a0 ), (a0 , b0 )∪(b0 , a1 ) and so forth, we see a pattern emerge of the form (ak , bk ) ∪ (bk , ak + 1 ) for integers k so that our domain can be written as ∞ ∞ (ak , bk ) ∪ (bk , ak + 1 ) = k=−∞ k=−∞ (6k + 1)π (6k + 5)π , 3 3 ∪ (6k + 5)π (6k + 7)π , 3 3 A second approach to the problem exploits the periodic nature of f . Since cos(x) and sin(x) have period 2π, it’s not too difficult to show the function f repeats itself every 2π units.15 This means if we can find a formula for the domain on an interval of length 2π, we can express the entire domain by translating our answer left and right on the x-axis by adding integer multiples of 2π. One such interval that arises from our domain work is π , 7π . The portion 3 3 of the domain here is π , 5π ∪ 5π , 7π . Adding integer multiples of 2π, we get the family of 3 3 3 3 intervals π + 2πk, 5π + 2πk ∪ 5π + 2πk, 7π + 2πk for integers k. We leave it to the reader 3 3 3 3 to show that getting common denominators leads to our previous answer. This doesn’t necessarily mean the period of f is 2π. The tangent function is comprised of cos(x) and sin(x), but its period is half theirs. The reader is invited to investigate the period of f . 15 870 Foundations of Trigonometry 3. To find the domain of f (x) = 1 − cot(x), we first note that, due to the presence of the cot(x) term, x = πk for integers k. Next, we recall that for the square root to be defined, we need 1−cot(x) ≥ 0. Unlike the inequalities we solved in Example 10.7.3, we are not restricted here to a given interval. Our strategy is to solve this inequality over (0, π) (the same interval which generates a fundamental cycle of cotangent) and then add integer multiples of the period, in this case, π. We let g(x) = 1 − cot(x) and set about making a sign diagram for g over the interval (0, π) to find where g(x) ≥ 0. We note that g is undefined for x = πk for integers k, in particular, at the endpoints of our interval x = 0 and x = π. Next, we look for the zeros of g. Solving g(x) = 0, we get cot(x) = 1 or x = π + πk for integers k and 4 only one of these, x = π , lies in (0, π). Choosing the test values x = π and x = π , we get 4 6 2 √ g π = 1 − 3, and g π = 1. 6 2 (−) 0 (+) 0 We find g(x) ≥ 0 on π 4,π π 4 π . Adding multiples of the period we get our solution to consist of (4k+1)π , (k 4 the intervals π + πk, π + πk = 4 express our final answer as + 1)π . Using extended interval notation, we ∞ k=−∞ (4k + 1)π , (k + 1)π 4 cos (9x) = 9 9. 2 tan(x) = 1 − tan2 (x) . 3 cos(2x) = sin(x) + 2 29. tan2 (x) = 3 18. sec (3x) = 2 √ √ 3 x 2 7. sin = 3 3 2 √ π 1 5π 7π 11. sin(2x) = tan(x) 35. csc3 (x) + csc2 (x) = 4 csc(x) + 4 42. sin (2x) = cos (x) 23. cos(2x) = 5 sin(x) − 2 28. cot2 (x) = 3 csc(x) − 3 32. 2π) 19. cos (2x) = cos (x) 25. 3 cos(2x) + cos(x) + 2 = 0 27.18. tan2 (x) = 1 − sec(x) 31. cos (2x) = sin (x) 24. tan (x) = sec (x) 39. tan3 (x) = 3 tan (x) 38. tan(2x) − 2 cos(x) = 0 41.7. csc(x) = 0 16. cot4 (x) = 4 csc2 (x) − 7 36. 2π). find all of the exact solutions of the equation and then list those solutions which are in the interval [0. cot (2x) = − 8. sin 2x − =− =0 = 3 10. solve the equation. sin (x) = cos (x) 21. cos(2x) = 2 − 5 cos(x) 26. tan2 (x) = 3 sec (x) 2 20. 2 cos x + 3 2 6 4 13.10. sin (5x) = 0 2. csc (4x) = −1 6. tan (6x) = 1 5. √ 1 3 1. cos3 (x) = − cos (x) 40. cos x + 12. sin (2x) = sin (x) 22. sin (−2x) = 2 2 √ 4. 2 sec2 (x) = 3 − tan(x) 30. tan (2x − π) = 1 17. sec2 (x) = 4 3 14. cos2 (x) = 1 2 15. cos(2x) + csc2 (x) = 0 37. cos (3x) = 3. sec(x) = 2 csc(x) 33.7 Trigonometric Equations and Inequalities 871 10. giving the exact solutions which lie in [0. cos(x) csc(x) cot(x) = 6 − cot2 (x) 34.42.1 Exercises In Exercises 1 . sin2 (x) = 3 4 In Exercises 19 . 58. cos2 (x) > 68. tan(x) = cos(x) 53. cos(3x) = cos(5x) 55. restricting your attention to −2π ≤ x ≤ 2π. sin(6x) cos(x) = − cos(6x) sin(x) 45. solve the equation. sec (x) ≤ 2 71. √ 2 cos(x) − √ √ 2 sin(x) = 1 √ 2 44. cos(4x) = cos(2x) 56. cos(x) ≥ sin(x) 4 In Exercises 77 .76. sin(x) + 50. restricting your attention to 0 ≤ x ≤ 2π. cos(x) ≤ 5 3 79. sin(x) + cos(x) = 1 49. sin(3x) cos(x) = cos(3x) sin(x) √ 46. Express the exact answer in interval notation. cot (x) ≥ −1 76. √ √ 3 cos(x) = 1 3 2 3 sin(2x) + cos(2x) = 1 51. Express the exact answer in interval notation. cot(x) ≤ 4 In Exercises 71 . solve the inequality. csc (x) > 1 78. solve the inequality. 3 3 sin(3x) − 3 cos(3x) = 3 3 54. sin (x) ≤ 0 1 2 1 65. sec(x) ≤ √ 2 67. cos(5x) = − cos(2x) 58. sin(5x) ≥ 5 70. 77. tan (x) ≥ √ 3 61.70.872 Foundations of Trigonometry In Exercises 43 . cos (x) > 2 3 3 74. sin(x) > 73. cos (2x) ≤ 0 66. cos(3x) ≤ 1 60. Express the exact answer in interval notation. sin x + π 3 > 1 2 63. giving the exact solutions which lie in [0. solve the inequality. cot(x) ≥ 5 . sin(5x) = sin(3x) 57. 2 cos(x) ≥ 1 69. cos(5x) cos(3x) − sin(5x) sin(3x) = 48. cos(2x) cos(x) + sin(2x) sin(x) = 1 47. sec2 (x) ≤ 4 64. cot2 (x) ≥ 3 62. sin(6x) + sin(x) = 0 In Exercises 59 . √ 3 1 72.82. sin2 (x) < 75. cos(2x) − 3 sin(2x) = √ √ 52. 59. restricting your attention to −π ≤ x ≤ π. 2π) 43. f (x) = csc(2x) 90. f (x) = arcsin(tan(x)) 1 92. What do you find? 2 2 2 2 2 2 2 1 Replace with −1 and repeat the whole exploration. tan2 (x) ≥ 1 81. 2 .) 83. 2π). cos(2x) ≤ sin(x) 873 In Exercises 83 . With the help of your classmates.3. 2π). f (x) = tan2 (x) − 1 sin(x) 2 + cos(x) 87. 1 1 1 Then find the number of solutions to sin(2x) = 2 . Explain how this pattern would help you solve equations like sin(11x) = 1 . Now consider sin x = 1 .10. f (x) = ln (| cos(x)|) 89. f (x) = 86.7 Trigonometric Equations and Inequalities 80. sin 3x = 1 and sin 5x = 1 . express the domain of the function using the extended interval notation. f (x) = cos(x) sin(x) + 1 85.91. sin(2x) ≥ sin(x) 82. (See page 756 in Section 10. f (x) = 3 csc(x) + 4 sec(x) 91. A pattern should emerge. f (x) = 88. f (x) = 1 cos(x) − 1 2 − sec(x) 84.1 for details. sin(3x) = 2 and sin(4x) = 2 in [0. determine the number of solutions to sin(x) = 2 in [0. . x = 3. . . x = − 13. . . x = 4. . . . x = π 5π 9π 13π 5π πk + . x = 4 4 4 π 2π 5π + πk. x = . x= . x = 0. x = . No solution 14. x= . . 12 12 12 12 12. . . . 8 2 8 8 8 8 2πk 7π 2πk π 7π 3π 5π 17π 23π π + or x = + . . . x = − 11. 9 3 9 3 9 9 9 9 9 9 2π 5π 2π 5π 5π 11π + πk or x = + πk. . x = 5. x = 10. 3 3 3 3 3 3 π 5π π 5π 7π 11π + πk or x = + πk. . . x= . 6 6 6 6 6 6 π πk π 3π 5π 7π + . 4 2 4 4 4 4 π 2π π 2π 4π 5π + πk or x = + πk. x = 2. 3 2 3 6 3 6 3π 9π 3π + 6πk or x = + 6πk. x = 15.2 1. . 8 2 8 8 8 8 π 2π π 2π 4π 5π + πk or x = + πk. . x = Answers πk π 2π 3π 4π 6π 7π 8π 9π . x = 6. 12 3 12 3 12 12 4 4 12 12 π πk π 5π 4π 11π + . . x = 16. x = .7. No solution 9. . . x = 18. 24 6 24 24 8 24 24 8 24 24 8 24 24 8 3π πk 3π 7π 11π 15π + . x = 3π 13π π 3π 13π 7π + πk or x = + πk. x= . 3 3 3 3 3 3 . . x = 7. x = .874 Foundations of Trigonometry 10. 4 12 12 4 12 4 19π π π 5π + 2πk or x = + 2πk. . x = . 3 3 3 8. x = . . . . x = . π. . . . x = 17. x= . . x= . . . . . . 3 6 3 6 3 6 π πk π 5π 3π 13π 17π 7π 25π 29π 11π 37π 41π 15π + . . . . 5 5 5 5 5 5 5 5 5 π 2πk 5π 2πk π 5π 7π 11π 13π 17π + or x = + . . x= . . . . π. 34. 6 6 6 6 π π 3π 5π 7π 5π 7π 11π x= . π + arctan 875 1 2 29. 31. x = 25. . 6 2 6 2 π 5π 9π 13π 41. 36. . 37. π. x = 32. . . . . . . . π. . . x = 0. π − arcsin 1 3 1 3 π 5π 20. 40.10. No solution 8 8 8 8 8π 9π 10π 11π 12π 13π π 2π 3π 4π 5π 6π . . 45. . . . 3 3 1 3 1 3 . . 6 6 2 π 7π 5π 11π x= . x = 0 2 2 π 11π 13π 23π 25π 35π 37π 47π 49π 59π 61π 71π 73π 83π 85π 95π 46. . π + arctan(2) π 3π 5π 7π 33. x = 2π 4π . arccos 3 3 7π 11π . x = . arctan 4 4 1 2 . . x = . . . x = . . 2 2 π 5π 7π 3π 11π x= . 6 6 2 π 5π 24. . 3 3 π π 5π 3π 39. . 4 4 π π 5π 3π 21. . x = 0. . 3 3 3 3 π 3π x= . 2 2 6 π 17π π 4π 49. 3 3 π 5π 3π 22. . . 48. 12 12 3 3 17π 41π 23π 47π π 5π 5π 17π 3π 29π 51. . . x = 0. x = . 3 3 19. . . . 3π 7π . x = . π. . 3 3 π 5π 26. x = . x = 0. . x = arctan(2).7 Trigonometric Equations and Inequalities π 5π . . . . . x = . . . x = 0. . x = . x = 0. 43. arcsin 6 6 2π 4π . . 7 7 7 7 7 7 7 7 7 7 7 7 π 3π 44. π. . . x = . x = . . 52. 4 4 4 4 π 3π 35. . 24 24 24 24 6 18 6 18 2 18 π 5π π . x = 0. x = . . . 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48 π 11π π 47. 42. . . 50. . . x = . . x = 30. 6 2 6 2 2π 4π 23. 6 6 6 2 6 . . . π. . . . x = 27. x = . x = 0. . . 6 6 28. 2π − arccos . . 6 4 4 6 6 4 4 6 π 2π 4π 5π x = 0. . 2 2 π 5π 38. . x = 57. ∪ . . . (−2π. 2π] 79. arccot(5) − 2π] ∪ (−π. 8 8 8 8 8 8 8 8 56. x = π π 3π 5π 9π 11π 5π 13π . − 7π 3π . 2π 4π 5π π ∪ .π 2 3 3 2 π 3π ∪ 0. 2π] 70. 4 4 11π . 2π 4 62. 0. . 81. π. . .π 2 2 2 73. 2π) 72. 3 3 1 3 π π − . . 3 3 3 3 . .− 4 2 ∪ − 3π 5π 3π π . ∪ . .− 2 4 4 2 π π π π ∪ − . 4 4 3π 2 ∪ − 3π π π .4754 π π 4π 3π . π. .− ∪ . ∪ . 71. π + arccot(5)] 80. −π ∪ 0. . arcsin − − 1 3 . −π. ∪ π. . π. 2π 3 3 3 3 5π π ∪ . 7 7 7 7 7 7 5 5 5 5 √ √ −1 + 5 −1 + 5 58. ∪ . . 0. − 3π π . π ∪ 4 π 3π . ≈ 2. x = 0. 66. ∪ . 0. 4 4 −2π. ∪ .876 π π 3π 5π 3π 7π 53. 2π 4 68. x = 0. 7 3 7 7 7 7 3 7 Foundations of Trigonometry π 2π 4π 5π 54. 2π π . π 3π 5π 7π . [arccot(4). 4 4 4 4 0. . arccot(5)] ∪ (π. π. 63. 74. . 64. [π. . 65. π. . 77. 3 3 3 3 2π 4π 6π 8π 10π 12π π 3π 7π 9π . . − 75. π − arcsin ∪ π 2π . 4 2 4 4 2 4 π 3π 5π 7π 9π 11π 13π 15π 55.− ∪ − . 0. No solution 69. 2π] 61. 2 2 ∪ 7π . 2 4 4 2 ∪ 3π 7π . − 5π π π 5π ∪ −π. 3 2 3 2 π ∪ 4 π ∪ 2 3π 5π . [0. [−2π.− 3 3 −π. . . . . ∪ . π 2π 4π 5π ∪ . 6 6 π π π π ∪ − .6662. π) ∪ [π + arccot(4). π ∪ π. arccot(5) − π] ∪ (0. . − 78. 2π 6 ∪ 7π . . x = arcsin ≈ 0. . 2π 3 3 67. 2 4 −2π. 76. 0. π − arcsin 2 2 59. . 2 4 4 2 ∪ π 3π 5π 3π . − ∪ 0. . 2π 3 3 3 3 60. . . x = 0. ∞) ∞ 89. 2 2 (4k − 1)π (4k + 1)π . 83. k=−∞ ∞ (6k − 1)π (6k + 1)π ∪ . 3 3 kπ (k + 1)π . k=−∞ (2k − 1)π (2k + 1)π . 4 2 ∪ (2k + 1)π (4k + 3)π . − .− . 2 2 91.10. 6 6 6 6 2 2 ∞ 877 82. 2 2 86. . 4 4 87. 2 2 kπ (k + 1)π .7 Trigonometric Equations and Inequalities π 5π π 3π 11π 7π ∪ ∪. 2 2 85. (2k + 2)π) k=−∞ ∞ 84. (−∞. − ∞ (2kπ. k=−∞ . k=−∞ ∞ 88. 2 4 (4k + 1)π (4k + 3)π . k=−∞ (4k − 1)π (4k + 3)π . k=−∞ ∞ (4k + 1)π (2k + 1)π . k=−∞ ∞ 90. 878 Foundations of Trigonometry . Chapter 11 Applications of Trigonometry 11.1 the cosine and sine functions can be used to model their fair share of natural behaviors.1 Applications of Sinusoids In the same way exponential functions can be used to model a wide variety of phenomena in nature. we remained undecided as to which form we preferred. Properties of the Sinusoid S(t) = A sin(ωt + φ) + B • The amplitude is |A| • The angular frequency is ω and the ordinary frequency is f = • The period is T = 2π 1 = f ω φ ω ω 2π • The phase is φ and the phase shift is − • The vertical shift or baseline is B Along with knowing these formulas. The amplitude measures the maximum displacement of the sine wave from its baseline (determined by the vertical shift). but the time for such indecision is over.5. in the form S(x) = A sin(ωx+φ)+B for ω > 0. since the applications in this section are time-dependent. For clarity of exposition we focus on the sine function2 in this section and switch to the independent variable t.5. At the time. the angular frequency tells how many cycles are completed over an interval of length 2π. The phase indicates what 1 2 See Section 6. it is helpful to remember what these quantities mean in context. we introduced the concept of a sinusoid as a function which can be written either in the form C(x) = A cos(ωx+φ)+B for ω > 0 or equivalently. In section 10. and the ordinary frequency measures how many cycles occur per unit of time. Sine haters can use the co-function identity cos π 2 − θ = sin(θ) to turn all of the sines into cosines. We reintroduce and summarize all of the important facts and definitions about this form of the sinusoid below. the period is the length of time it takes to complete one cycle of the sinusoid. . Example 11.1. we introduced the concept of circular motion and in Section 10.2.1. Solution. It completes two revolutions in 2 minutes and 7 seconds.1. we could just observe the motion of the wheel from the other side. find a sinusoid which describes the height of the passengers above the ground t seconds after they pass the point on the wheel closest to the ground. Recall from Exercise 55 in Section 10.1 that The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height 136 feet. and the phase shift represents how much of a ‘head start’ the sinusoid has over the un-shifted sine function.5. The figure below is repeated from Section 10. we developed formulas for circular motion. Assuming that the riders are at the edge of the circle. .1.3 θ Q h P O 3 Otherwise. Our first foray into sinusoidal motion puts these notions to good use. We sketch the problem situation below and assume a counter-clockwise rotation.1.880 Applications of Trigonometry angle φ corresponds to t = 0. amplitude baseline period In Section 10. assuming it is centered at (0.1. we have that y = 64 sin 127 t describes the y-coordinate on the Giant Wheel after t seconds. Since the base of the Giant Wheel ride is 8 feet above the ground and the Giant Wheel itself has a radius of 64 feet. the 2 2 4π angular frequency is ω = 2π = 127 radians per second. This represents the 8 ‘time delay’ (in seconds) we introduce by starting the motion at the point P as opposed to the point Q.875 seconds to ‘catch up’ to the point Q. Exercise 6b. In our case.5. Said differently.2. To account for this vertical shift upward. 2 4π we know θ = ωt = 127 t. . so the radius is r = 64 feet. From Section 10. note that the amplitude of 64 in our answer corresponds to the radius of the Giant Wheel. 127 .1 Applications of Sinusoids 881 We know from the equations given on page 732 in Section 10. π/2 Second. is in standard position. we need to shift the angle θ in the figure back π radians. 2 y 136 72 8 127 2 t A few remarks about Example 11. In order to find an expression for h. Our next example revisits the daylight data first introduced in Section 2. t = 0 corresponds to the point (r. the phase shift of our answer works out to be 4π/127 = 127 = 15.1 are in order. We 2 2 2 can check the reasonableness of our answer by graphing y = h(t) over the interval 0.1 that the y-coordinate for counterclockwise motion on a circle of radius r centered at the origin with constant angular velocity (frequency) ω is given by y = r sin(ωt). Putting these two pieces of information T 4π together. 0) so that θ. so we (temporarily) write the height in terms of θ as h = 64 sin (θ) + 72.875.11. Since the wheel completes two revolutions in 2 minutes and 7 seconds (which is 127 seconds) the period T = 1 (127) = 127 seconds. 4 We are readjusting our ‘baseline’ from y = 0 to y = 72. 0) with t = 0 corresponding to the point Q. This is where the phase comes into play. This means that passengers on the Giant Wheel never stray more than 64 feet vertically from the center of the Wheel. its center is 72 feet above the ground.1. First. we take the point O in the figure as the origin. Hence. passengers which ‘start’ at P take 15. which makes sense. Next. we need to adjust things so that t = 0 corresponds to the point P instead of the point Q. Here.4 we add 72 to our formula 4π for y to obtain the new formula h = y + 72 = 64 sin 127 t + 72. the diameter of the wheel is 128 feet.2. Geometrically. 4π Subtracting π from θ gives the final answer h(t) = 64 sin θ − π + 72 = 64 sin 127 t − π + 72. the angle measuring the amount of rotation. We do our best to find the constants A. According to the U.8 − 12. it represents H(t) over the interval [0. t = 2 is a sample of H(t) over [1.6 9 12.4 8 15.4 10 9.S. 2009.8 7 19.55 = 12. H 22 20 18 16 14 12 10 8 6 4 2 1 2 3 4 5 6 7 8 9 10 11 12 t The data certainly appear sinusoidal.25. we plot it below. Compare your answer to part 1 to one obtained using the regression feature of a calculator. fitting a sinusoid to data manually is not an exact science. 12]. Find a sinusoid which models these data and use a graphing utility to graph your answer along with the data. which has a length of 11.3+21.6 This Okay. Even though the data collected lies in the interval [1. Similarly.25 sin(ωt + φ) + 12. That is.2.5 but when it comes down to it. we take the period T = 12 months. φ and B so that the function H(t) = A sin(ωt + φ) + B closely matches the data.3 3 12. 6 5 . Next. 2]. At this point. 2.55. we go after the angular frequency ω.8 = 12.8 2 9. we need to think of the data point at t = 1 as a representative sample of the amount of daylight for every day in January.9 5 19. We first go after the vertical shift B whose value determines the baseline. To get a feel for the data.4 6 21. Since the data collected is over the span of a year (12 months). the value of B is the average of the maximum and minimum values. 1.882 Applications of Trigonometry Example 11. Here t = 1 represents January 21. Month Number Hours of Daylight 1 5.55.1 11 5. and so forth. We find A = 21. 1]. the number of hours H of daylight that Fairbanks.55 − 3.4 4 15. and so on. Just humor us.1. Naval Observatory website. So here we take B = 3. t = 2 represents February 21.3 = 9. Solution. we have H(t) = 9. 2009.2. 2 Next is the amplitude A which is the displacement from the baseline to the maximum (and minimum) values. In a typical sinusoid.3 1. Alaska received on the 21st day of the nth month of 2009 is given below. it appears to be the ‘∧’ shape we saw in some of the graphs in Section 2.6 12 3. ω. it is easier in this case to find the phase shift − ω . While both models seem to be reasonable fits to the data. assuming the object isn’t subjected to relativistic speeds . we choose t = 3.1 Applications of Sinusoids 883 means ω = 2π = 2π = π . φ 15. 7 8 See the figure on page 880. from the motion of an object on a spring.5. Before we jump into the Mathematics. there are some Physics terms and concepts we need to discuss. Hence.1 Harmonic Motion One of the major applications of sinusoids in Science and Engineering is the study of harmonic motion. In Physics. 6 2 2.5. we graph the calculator’s regression below. We’ll just have to use our own good judgment when choosing the best sinusoid model.55 (the baseline value). Since we picked A > 0. The reason for this. The calculator does not give us an r2 value like it did for linear regressions in Section 2. is beyond the scope of this course. .55 than the next value.8 while its weight could change. the phase shift corresponds to the first value of t with H(t) = 12.55.4 is closer to 12. We have 6 2 H(t) = 9. The equations for harmonic motion can be used to describe a wide range of phenomena. . Unlike the previous T 12 6 φ example. 11. Using the ‘SinReg’ command. − ω = 3. since its corresponding H value of 12. ‘mass’ is defined as a measure of an object’s resistance to straight-line motion whereas ‘weight’ is the amount of force (pull) gravity exerts on an object. An object’s mass cannot change. we restrict our attention to modeling a simple spring system.9.1. Well.1. so φ = −3ω = −3 π = − π . . cubic and quartic regressions as in Section 3.11. the calculator model is possibly the better fit.25 sin π t − π + 12. Below is a graph of our data with the curve y = H(t). to the response of an electronic circuit. which corresponds to t = 4. nor does it give us an R2 value like it did for quadratic. In this subsection. The last quantity to find is the phase φ.7 Here. much like the reason for the absence of R2 for the logistic model in Section 6. g = 32 second2 . the force involved is weight which is caused by the acceleration due to gravity. Time is always measured in seconds (s). and in the SI meters system. If we let x(t) denote the object’s displacement from the equilibrium position at time t.10 Suppose we attach an object with mass m to a spring as depicted below.12 x(t) = 0 at the equilibrium position x(t) < 0 above the equilibrium position x(t) > 0 below the equilibrium position If we ignore all other influences on the system except gravity and the spring force. 12 To keep units compatible. w is the weight of the object. which comes from Differential Equations. In the English system.) to measure displacement. then Physics tells us that gravity and the spring force will battle each other forever and the object will oscillate indefinitely. but its mass is the same in both places. on Earth a mass of 1 slug weighs 32 lbs. then x(t) = 0 means the object is at the equilibrium position. How far the spring stretches to reach equilibrium depends on the spring’s ‘spring constant’. g = 9.8 N. Here. The function x(t) is called the ‘equation of motion’ of the object.1. 9 . We convert between mass and weight using the formula9 w = mg. gives x(t) as a function of the mass m of the object.3. the initial displacement x0 of the This is a consequence of Newton’s Second Law of Motion F = ma where F is force. If we are in the SI system. In the SI system. second2 second2 11 Look familiar? We saw Hooke’s Law in Section 4. The weight of the object will stretch the spring. and x(t) > 0 means the object is below the equilibrium position. 10 Note that 1 pound = 1 slug foot and 1 Newton = 1 kg meter . The system is said to be in ‘equilibrium’ when the weight of the object is perfectly balanced with the restorative force of the spring. In this case. the object will bounce up and down on the end of the spring until some external force stops it. x(t) < 0 means the object is above the equilibrium position. we describe the motion as ‘free’ (meaning there is no external force causing the motion) and ‘undamped’ (meaning we ignore friction caused by surrounding medium. Hence. the spring constant relates the force F applied to the spring to the amount d the spring stretches in accordance with Hooke’s Law11 F = kd. which in our case is air).) is a measure of force (weight). Usually denoted by the letter k. we measure displacement in meters (m). the spring constant k. we use feet (ft. If the object is released above or below the equilibrium position. or if the object is released with an upward or downward velocity.884 Applications of Trigonometry An object which weighs 6 pounds on the surface of the Earth would weigh 1 pound on the surface of the Moon. ‘pounds’ (lbs. the unit of force is ‘Newtons’ (N) and the associated unit of mass is the ‘kilogram’ (kg). and a mass of 1 kg weighs 9. m is the feet mass and g is the acceleration due to gravity. and the corresponding unit of mass is the ‘slug’. In our present setting. m is mass and a is acceleration. if we are using the English system.8 second2 . In the English system of units. The following theorem. we have A sin(φ) = x0 . 1. As far as the initial velocity v0 is concerned. We are told that the object is released 2 3 feet below the equilibrium position ‘from rest. s Example 11. find the equation of motion of the object. the object would fall towards the ground. . find the equation of motion of the object. What is the longest distance the object travels above the equilibrium position? When does this first happen? Confirm your result using a graphing utility. s2 now proceed to apply Theorem 11. k 1. In order to use the formulas in Theorem 11.1. x(t). depending on which system we choose. x0 < 0 means the object is released above the equilibrium position and x0 > 0 means the object is released below the equilibrium position. v0 = 0 means the object is released ‘from rest.. . As with x(t). or m. any movement upwards is considered ‘negative’. Using F = 64 and d = 8. We can ft.3. Solution.13 Theorem 11. If the object is attached to the spring and released 3 feet below the equilibrium position from rest.1 Applications of Sinusoids 885 object and initial velocity v0 of the object. Only φ = π and angles coterminal to it 2 13 The sign conventions here are carried over from Physics. x0 = 0 means the object is released from the equilibrium position. or k = 8 lbs.’ v0 < 0 means the object is heading upwards and v0 > 0 means the object is heading downwards. To find m. If the object is attached to the spring and released 3 feet below the equilibrium position with an upward velocity of 8 feet per second. Equation for Free Undamped Harmonic Motion: Suppose an object of mass m is suspended from a spring with spring constant k. √ 2 A = = 32 + 02 = 3. x 2 + v0 0 ω which in this case gives 3 sin(φ) = 3 so sin(φ) = 1. To find k. then the displacement x from the equilibrium position at time t is given by x(t) = A sin(ωt + φ) where • ω= k and A = m x2 + 0 v0 ω 2 • A sin(φ) = x0 and Aω cos(φ) = v0 . Suppose an object weighing 64 pounds stretches a spring 8 feet. . Therefore. we get 64 = k · 8. We know the object weighs 64 lbs.11. we use Hooke’s Law F = kd. If the initial displacement from the equilibrium position is x0 and the initial velocity of the object is v0 .1 work out so that ω has units 1 and A has units ft.1. When does the object first pass through the equilibrium position? Is the object heading upwards or downwards at this instant? 2. Since the spring force acts in direct opposition to gravity. we use w = mg with w = 64 lbs. we get ω = m = 8 = 2. and stretches the spring 8 ft. we first need to determine the spring constant k and the mass of the object m. With k = 8 and m = 2. If not for the spring. It is a great exercise in ‘dimensional analysis’ to verify that the formulas given in Theorem 11.1. which is the ‘natural’ or ‘positive’ direction. To determine the phase φ. We get m = 2 slugs.1. x(t).’ This means x0 = 3 and v0 = 0. and g = 32 ft. We still have ω = 2 and x0 = 3. t = 2 arcsin 3 + π ≈ 1. the fact our graph is crossing through the t-axis from positive x to negative x at t = π confirms our answer. but now s we have v0 = −8. Since the amplitude of x(t) is 5. To check this answer. we graph y = 5 sin 2x + π − arcsin 3 on a graphing utility 5 and confirm the coordinates of the first relative minimum to be approximately (1. though beyond the scope of this course. we look for the smallest positive t value which in this case is t = π ≈ 0. To find when the object passes through the equilibrium position we 2 solve x(t) = 3 sin 2t + π = 0. From Aω cos(φ) = v0 . we get 5 sin(φ) = 3 which gives 0 ω 3 sin(φ) = 5 . we graph one cycle of x(t). Going through the usual analysis we find t = − π + π k for 2 4 2 integers k. Going through the usual machinations. and the period of x(t) is T = 2π = 2π = π. Common sense suggests that if we release the object below the equilibrium position. the negative once again signifying that the object is above the equilibrium position. the object will travel 5 at most 5 feet above the equilibrium position. 4 2. Since x(t) is expressed in terms of sine.78 seconds 4 after the start of the motion. we get A = x2 + v0 = 32 + (−4)2 = 5. Since our applied domain in this situation is t ≥ 0. The only difference between this problem and the previous problem is that we now release the object with an upward velocity of 8 ft . x 3 2 1 π 4 π 2 3π 4 2 π −1 −2 −3 t x(t) = 3 sin 2t + π 2 y = 5 sin 2x + π − arcsin 3 5 It is possible. −5). To find when this happens.15 While we may not have the Physics and Calculus background 14 15 For confirmation. we get t = 1 arcsin 3 + π + πk for integers k. the object should be traveling upwards when it first passes through it. we choose to express φ = π − arcsin 3 . to model the effects of friction and other external forces acting on the system.107. or cos(φ) = − 4 . Hence. Hence. The smallest of these values occurs when k = 0. Since we are interested in the first time the object passes through the equilibrium position. To check our 5 4 answer using the calculator. so we pick14 the phase to be φ = π . 5 x(t) = 5 sin 2t + π − arcsin 3 . we get 10 cos(φ) = −8. which in this case reduces to 6 cos(φ) = 0. the equation of motion 2 is x(t) = 3 sin 2t + π . This means 5 that φ is a Quadrant II angle which we can describe in terms of either arcsine or arccosine. 2 5 4 1 that is.886 Applications of Trigonometry satisfy this condition. ω 2 Remembering that x(t) > 0 means the object is below the equilibrium position and x(t) < 0 means the object is above the equilibrium position. we solve the 3 equation x(t) = 5 sin 2t + π − arcsin 5 = −5. From A sin(φ) = x0 . we graph x(t) over the interval [0. the negative indicating the velocity is directed upwards.107 seconds after the start of the motion. Take a good Differential Equations class to see this! . we note that Aω cos(φ) = v0 . π]. Here. In this model. We find √ (cos(2t) + sin(2t)) = 2 sin 2t + π . y = 10e−x/5 sin x + π 3 y = 10e−x/5 sin x + π 3 . Write x(t) = (t + 3) 2 cos(2t) + (t + 3) 2 sin(2t) in the form x(t) = A(t) sin(ωt + φ). Graph x(t) using a graphing utility. The sinusoidal nature 3 continues indefinitely. We convert what’s left in parentheses to the required form using the formulas introduced in Exercise 36 from Section 10. which continues to grow without bound . Solution. Indeed. Graph x(t) using a graphing utility. 3. we see this attenuation taking place.1. We start rewriting x(t) = 5e−t/5 cos(t) + 5e−t/5 3 sin(t) by factoring out 5e−t/5 from both √ terms to get x(t) = 5e−t/5 cos(t) + 3 sin(t) . This equation corresponds to 3 the motion of an object on a spring where there is a slight force which acts to ‘damp’. we can think 3 of the function A(x) = 10e−x/5 as the amplitude.5. so x(t) = 2(t + 3) sin 2t + π . √ 1. the object oscillates forever. Since our 4 amplitude function here is A(x) = 2(x + 3) = 2x + 6. As x → ∞.1 Applications of Sinusoids 887 to derive equations of motion for these scenarios. Graphing this on the 4 4 calculator as y = 2(x + 3) sin 2x + π . we can certainly analyze them. 10e−x/5 → 0 which means the amplitude continues to shrink towards zero. We examine three cases in the following example. Find the period of x(t) = 5 sin(6t) − 5 sin (8t). Proceeding as in the first example. Graphing this on the 3 3 calculator as y = 10e−x/5 sin x + π reveals some interesting behavior. √ √ 2. Example 11. In the case of y = 10e−x/5 sin x + π . or slow the motion. Graph x(t) using a graphing utility. we find the sinusoid’s amplitude growing. but it is being attenuated. In the sinusoid A sin(ωx + φ). but with smaller and smaller amplitude.4. the coefficient A of the sine function is the amplitude. Write x(t) = 5e−t/5 cos(t) + 5e−t/5 3 sin(t) in the form x(t) = A(t) sin(ωt + φ). we factor out (t + 3) 2 from each term in the function √ √ √ x(t) = (t + 3) 2 cos(2t) + (t + 3) 2 sin(2t) to get x(t) = (t + 3) 2(cos(2t) + sin(2t)). √ 1.11. y = ±10e−x/5 √ 2. We find √ cos(t) + 3 sin(t) = 2 sin t + π so that x(t) = 10e−t/5 sin t + π . if we graph y = ±10e−x/5 along with y = 10e−x/5 sin x + π . An example of this kind of force would be the friction of the object against the air. Below we graph y = 5 sin(6x) − 5 sin(8x) and y = ±10 sin(x) over [0. The phenomenon illustrated here is ‘forced’ motion. we take the ratio of their 6 frequencies and reduce to lowest terms: 8 = 3 . π] y = 5 sin(6x) − 5 sin(8x) and y = ±10 sin(x) over [0. −10 sin(t). but not least. That is. To find the period of this function. This is an example of the ‘beat’ phenomena. This equation of motion also results from ‘forced’ motion. we imagine that the entire apparatus on which the spring is attached is oscillating as well. 2π]. A good place to start is this article on beats. 2π] 16 17 The reader is invited to investigate the destructive implications of resonance. we need to determine the length of the smallest interval on which both f (t) = 5 sin(6t) and g(t) = 5 sin(8t) complete a whole number of cycles.888 Applications of Trigonometry as x → ∞. In this case. We graph y = 5 sin(6x) − 5 sin(8x) over [0. Since the sinusoids here have different frequencies. plays an interesting role in the graph of x(t). we come to x(t) = 5 sin(6t) − 5 sin(8t). π] on the calculator to check this. they are ‘out of sync’ and do not amplify each other as in the previous example. we are witnessing a ‘resonance’ effect – the frequency of the external oscillation matches the frequency of the motion of the object on the spring. Last. 4 g makes 4. In other words. this is hardly surprising.16 y = 2(x + 3) sin 2x + π 4 y = 2(x + 3) sin 2x + y = ±2(x + 3) π 4 3. Taking things a step further. the period of x(t) is three times the period of f (t) (which is four times the period of g(t)). To do this. The lower frequency factor in this expression. or π.17 y = 5 sin(6x) − 5 sin(8x) over [0. but here the frequency of the external oscillation is different than that of the object on the spring. and the curious reader is invited to explore this concept as well. we can use a sum to product identity to rewrite x(t) = 5 sin(6t) − 5 sin(8t) as x(t) = −10 sin(t) cos(7t). . This tells us that for every 3 cycles f makes. suppose the yo-yo string is 28 inches and it completes one revolution in 3 seconds. Find a sinsuoid which models the height h of the passenger above the ground in meters t minutes after they board the Eye at ground level.1 Applications of Sinusoids 889 11. England and is one of the largest Ferris Wheels in the world.2 Exercises 1. (a) Find the spring constant k in lbs. Suppose we are in the situation of Exercise 3 above.1. Here we take x(t) > 0 to mean the passenger is to the right of the center. In Exercise 52 in Section 10. Find a sinusoid which models this note. The sounds we hear are made up of mechanical waves. If the closest the yo-yo ever gets to the ground is 2 inches.1. enabling passengers to simply walk on to. It has a diameter of 135 meters and makes one revolution (counterclockwise) every 30 minutes. Assume that the phase is 0. 6. where t = 0 corresponds to the point (r. . The note ‘A’ above the note ‘middle cycles C’ is a sound wave with ordinary frequency f = 440 Hertz = 440 second . ft. 3. and off of. (b) Find the equation of motion of the object if it is released from 1 foot below the equilibrium position from rest. we found the x-coordinate of counter-clockwise motion on a circle of radius r with angular frequency ω to be x = r cos(ωt). 5. Find a sinusoid which models this voltage. 0). find a sinsuoid which models the height h of the yo-yo above the ground in inches t seconds after it leaves its lowest point. The London Eye is a popular tourist attraction in London. Find when the object passes through the equilibrium position heading downwards for the third time. we introduced the yo-yo trick ‘Around the World’ in which a yo-yo is thrown so it sweeps out a vertical circle. It is constructed so that the lowest part of the Eye reaches ground level. Find a sinsusoid which models the horizontal displacement x of the passenger from the center of the Eye in meters t minutes after they board the Eye. while x(t) < 0 means the passenger is to the left of the center. As in that exercise.1.2. assuming that the amplitude is 1 and the phase shift is 0. The voltage V in an alternating current source has amplitude 220 2 and ordinary frequency f = 60 Hertz. and the mass of the object in slugs. √ 2. the ride. When is the first time the object passes through the equilibrium position? In which direction is it heading? (c) Find the equation of motion of the object if it is released from 6 inches above the equilibrium position with a downward velocity of 2 feet per second. Suppose an object weighing 10 pounds is suspended from the ceiling by a spring which stretches 2 feet to its equilibrium position when the object is attached.11. On page 732 in Section 10. 4. noaa. Month Number.erh. θ. in seconds.19 For example. of the pendulum and the period of the motion is given by l g T = 2π where l is the length of the pendulum and g is the acceleration due to gravity.’ Carl remembers the ‘Rule of Thumb’ as being 20◦ or less. the angular displacement of the pendulum from the vertical position. θ0 . (b) In Exercise 40 section 5.3.890 Applications of Trigonometry 7.gov/cle/climate/cle/normals/laketempcle. t. Consider the pendulum below.1. (b) Using a graphing utility. you found the length of the pendulum needed in Jeff’s antique 1 Seth-Thomas clock to ensure the period of the pendulum is 2 of a second. t. 18 Provided θ is kept ‘small. T 1 36 2 33 3 34 4 38 5 47 6 57 7 67 8 74 9 73 10 67 11 56 12 46 (a) Using the techniques discussed in Example 11. find a sinusoid which models the displaceinitial displacement of the pendulum is 15 ment of the pendulum θ as a function of time. Check with your friendly neighborhood physicist to make sure. Ignoring air resistance. fit a sinusoid to these data. . t = 3 represents the average of the temperatures recorded for Lake Erie on every March 1 for the years 1971 through 2000. graph your model along with the data set to judge the reasonableness of the fit. 19 See this website: http://www. 8.2.html. The table below lists the average temperature of Lake Erie as measured in Cleveland. can be modeled as a sinusoid. Ohio on the first of the month for each month during the years 1971 – 2000.18 θ The amplitude of the sinusoid is the same as the initial angular displacement. Arrange things so θ(0) = θ0 . Assuming the ◦ . t Temperature (◦ F). (a) Find a sinusoid which gives the angular displacement θ as a function of time. mil/USNO/astronomical-applications/data-services/frac-moon-ill.navy. 23 The listed fraction is 0.62. 22 You may want to plot the data before you find the phase shift.20 (d) Compare your results to those obtained using a graphing utility. research the phenomena mentioned in Example 11. (c) Use the model you found in part 9a to predict the fraction of the moon illuminated on June 1.03 27 0. graph your model along with the data set to judge the reasonableness of the fit.4. The fraction of the moon illuminated at midnight Eastern Standard Time on the tth day of June. What other things in the world might be roughly sinusoidal? Look to see what models you can find for them and share your results with your class. With the help of your classmates. t Fraction Illuminated.98 9 0. namely resonance and beats. With the help of your classmates.83 15 0. fit a sinusoid to these data. 20 21 The computed average is 41◦ F for April 15th and 71◦ F for September 15th .2.04 24 0.57 18 0.usno.22 (b) Using a graphing utility. 2009 is given in the table below. . 23 (d) Compare your results to those obtained using a graphing utility.1 Applications of Sinusoids 891 (c) Use the model you found in part 8a to predict the average temperature recorded for Lake Erie on April 15th and September 15th during the years 1971–2000.81 6 0. research Amplitude Modulation and Frequency Modulation. 9. 10.98 12 0.1.21 Day of June.1. F 3 0. 2009.11. 12.26 30 0. 11.58 (a) Using the techniques discussed in Example 11. See this website: http://www.27 21 0. 00◦ F and the average temperature on September 15th is approximately T (9. 16 7. and m = ft. x(t) = 67.5 cos π 15 t − π 2 = 67. ≈ (c) x(t) = 22 sin 4t + 7π . the object is heading upwards.34 seconds. h(t) = 67.3 Answers √ 2. (a) k = 5 lbs. .39 seconds after the motion starts. The object first passes through the equilibrium point when t = 2 0.5 sin 5. h(t) = 28 sin π 15 t + 67.5 + 30 5 16 4.1. The sinusoid seems to be shifted to the right of our data.43◦ F and the average temperature on September 15th to be approximately 70. (a) θ(t) = θ0 sin 8.05◦ F.892 Applications of Trigonometry 11.5 (b) Our function and the data set are graphed below. V (t) = 220 2 sin (120πt) − π 2 π 2 1. (c) The average temperature on April 15th is approximately T (4.5) ≈ 39. (a) T (t) = 20.38◦ F.5 sin g l t+ π 2 (b) θ(t) = π 12 sin 4πt + π 2 π 6t − π + 53. This model appears to be more accurate. √ slugs π 8 (b) x(t) = sin 4t + π . (d) Using a graphing calculator. we get the following This model predicts the average temperature for April 15th to be approximately 42. The object passes through the equilibrium point heading down4 wards for the third time when t = 17π ≈ 3.5) ≈ 73.5 sin π 15 t 2π 3 t − 6. At this time. S(t) = sin (880πt) 3. 1 Applications of Sinusoids 893 9.505. This model predicts that the fraction of the moon illuminated on June 1st.505 = 0. 2009 is approximately F (1) ≈ 0.475 sin 15 t − 2π + 0. we either choose A < 0 or we find the second value of t which closely approximates the ‘baseline’ value. (c) The fraction of the moon illuminated on June 1st. .60 (d) Using a graphing calculator. 2009 is approximately 0.59.505 (b) Our function and the data set are graphed below. It’s a pretty good fit.475 sin 15 t + 0. This appears to be a better fit to the data than our first model. we get the following. We choose the latter to π π obtain F (t) = 0. F = 0.11. (a) Based on the shape of the data. First. we label the triangle below.1 are in order.85◦ . c) are called angle-side opposite pairs. In this case.6. Thus.894 Applications of Trigonometry 11. Converting to degrees. we have that sin(β) = 7 so β = arcsin 7 radians and we have β ≈ 34. find the length of the remaining side and the measures of the remaining angles. Second. Given a right triangle with a hypotenuse of length 7 units and one leg of length 4 units. we will strive to solve for quantities using the original data given in the problem whenever possible. b is the side opposite β and c is the side opposite γ. we are more than prepared to do just that. Solution. cos(α) = 4 . a). we adhere to the convention that a lower case Greek letter denotes an angle1 and the corresponding lowercase English letter represents the side2 opposite that angle. To decrease the chances of propagating error.1.2. For definitiveness. we’ve had some experience solving right triangles. A few remarks about Example 11. α = arccos 4 radians. as mentioned earlier. so we want to relate these to α. Taken together. The following example reviews what we know. The main goal of this section and the next is to develop theorems which allow us to ‘solve’ triangles – that is. find the length of each side of a triangle and the measure of each of its angles.3 and 10. a is the side opposite α. it 1 2 as well as the measure of said angle as well as the length of said side 7 . there are several ways we can find α using the inverse trigonometric functions. we stick to using the data given to us in the problem.2. rounded to the nearest hundreth of a degree. 10. In Sections 10. the pairs (α. the lengths 4 and 7 were given. Now 7 that we have the measure of angle α. however. c= β a α b=4 To find the length of the missing side a. (β. Now that all three sides of the triangle are known.2. According to Theorem 10.4.4. we use the Pythagorean Theorem to get a2 + 42 = 72 √ which then yields a = 33 units. we could find the measure of angle β using the fact that α and β are complements so α + β = 90◦ . we find α ≈ 55.15◦ .2 The Law of Sines Trigonometry literally means ‘measuring triangles’ and with Chapter 10 under our belts. Example 11. 7 Since α is an acute angle. b) and (γ. we opt to use the data given to us in the 4 4 problem. Express the angles in decimal degrees. Once again. According to Theorem 10. While this is not always the easiest or fastest way to proceed. Proceeding as before. If we drop an altitude from vertex B. Extending an altitude from vertex A gives two right triangles. After some rearrangement of the c a last equation. If we drop an altitude from vertex A. Theorem 11. consider the triangle ABC below.2. c). b c B B B β h Q γ c α A β γ a c α h γ a c C A C A C Q b b For our next case consider the triangle ABC below with obtuse angle α.10 allow us to easily handle any given right triangle problem. (β. b) and (γ. a b c = = sin(α) sin(β) sin(γ) The proof of the Law of Sines can be broken into three cases. a). we shall adopt this convention for the time being.11. we get sin(α) = sin(γ) . as in the previous case: ABQ and ACQ. The Law of Sines: Given a triangle with angle-side opposite pairs (α. b c B c β a α γ A 3 4 B c b C β Q a h γ A b C Your Science teachers should thank us for this. Don’t worry! Radians will be back before you know it! .3 Third. b) and (γ. For our first case. If we call the length of the altitude h (for height). we divide the triangle into two right triangles: ABQ and BCQ. completing the proof for this case. we can use the Law of Sines to help. but what if the triangle isn’t a right triangle? In certain cases.2 The Law of Sines 895 minimizes the chances of propagated error.4 and 10. the following ratios hold sin(α) sin(β) sin(γ) = = a b c or.4 that sin(α) = h and sin(γ) = h so that h = c sin(α) = a sin(γ). we get from Theorem 10. all of whose angles are acute. c). with angle-side opposite pairs (α. a).4 The Pythagorean Theorem along with Theorems 10. equivalently. since many of the applications which require solving triangles ‘in the wild’ rely on degree measure. we get h = b sin(γ) and h = c sin(β) so that sin(β) = sin(γ) . we can proceed as above a c using the triangles ABQ and ACQ to get sin(β) = sin(γ) . (β. To find γ. we get sin(γ) = h so h = a sin(γ). so in fact. namely α and a. Example 11. Solve the following triangles. we are not immediately given an angle-side opposite pair.72 units. c we have h = c sin(α). α = 85◦ . As in the previous example. 5 2.2. Thus “exact” here means sin(120◦ ) . α = 30◦ . but as we have the measures of α and β. β = 45◦ 3. but that 7 sin(15◦ ) becomes unnecessarily messy for the discussion at hand. a = 2 units. we have no choice but to used the derived value γ = 15◦ . we get sin(γ) = sin(α) . the Law of Sines reduces to the formulas given in Theorem 10. ABQ and BCQ. α = 120◦ . we need at least one angle-side opposite pair. c a B β h c α Q A α b γ C a The remaining case is when ABC is a right triangle. and the pitfalls. Since α = 180◦ − α. c = 4 units Solution. we are forced to use a derived value in our computations since the only The exact value of sin(15◦ ) could be found using the difference identity for sine or a half-angle formula.09 units. In this case. a = 7 units. In order to use the Law of Sines to solve a triangle. In this example. The Law of Sines ◦) c 7 gives us sin(15◦ ) = sin(120◦ ) so that c = 7 sin(15◦ ) ≈ 2. Proceeding to BCQ. α = 30◦ . c = 5. The next example showcases some of the power. it is time to find the third. 1. γ = 180◦ − 120◦ − 45◦ = 15◦ . c = 4 units . Putting this a together with the previous equation. we may proceed in using the Law of √ ◦) 7 b Sines. 1. we use the fact that the sum of the measures of the angles in a triangle is 180◦ . Hence. α = 30◦ . c = 4 units 6.25 units 4. α = 30◦ . Now that sin(120 we have two angle-side pairs. a = 1 units. c = 4 units 5.5 sin(120 2. yet we can minimize the propagation of error here by using the given angle-side opposite pair (α. we use sin(45◦ ) = sin(120◦ ) so b = 7 sin(45◦ ) = 7 3 6 ≈ 5. and we are finished with this case. a).4 and is left to the reader. a = 4 units. we can solve for γ since γ = 180◦ − 85◦ − 30◦ = 65◦ . Since β = 45◦ . We know that sin(α ) = h so that h = c sin(α ).2. To find c. Knowing an angle-side opposite pair. β = 30◦ . sin(α ) = sin(α). of the Law of Sines. a = 3 units.896 Applications of Trigonometry Dropping an altitude from vertex B also generates two right triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. a) and c.25 β = 45◦ c ≈ 2. we see that side a is just too short to make a triangle. Since we are given (α.25 sin(30◦ ) sin(65◦ ) = 5.77 γ = 15◦ b ≈ 2. Now γ is an angle in a triangle which 4 2 also contains α = 30◦ . The next three examples keep the same values for the measure of α and the length of c while varying the length of a. The only angle that satisfies this requirement and has sin(γ) = 1 is γ = 90◦ . Geometrically. we get a = which yields b sin(30◦ ) 5. we have a right triangle.09 α = 120◦ b ≈ 5. We find the measure of β to be β =◦ 180◦ − 30◦ − 90◦ = 60◦ and then determine b using the Law of Sines.46 a=2 α = 30◦ Diagram for number 3 Triangle for number 4 5. To find b we use the angle-side pair (γ. 1].25 sin(65◦ ) .90 units.46 units.72 a=7 α = 85◦ γ = 65◦ a ≈ 5. β = 30◦ c = 5. Proceeding as we have in the previous two examples. Using the Law of Sines. we use the Law of Sines to find γ. In this case. c) 5.25 sin(85◦ ) sin(65◦ ) a sin(85◦ ) 897 = 5. After the usual ≈ 5.77 units. Since the range of the sine function is 4 1 [−1. we use the Law of Sines to find the measure of γ. the side a is precisely long enough to form a unique right triangle. The Law of Sines gives rearrangement. c=4 a=1 c=4 β = 60◦ α = 30◦ b ≈ 3. In this case. we have sin(γ) = sin(30 ) or sin(γ) = 4 sin(30 ) = 2 . we get sin(γ) = sin(30 ) so sin(γ) = 2 sin (30◦ ) = 1. there is no real number with sin(γ) = 2. We will discuss this case in more detail after we see what happens in those examples. Since γ lies in a triangle with α = 30◦ .25 sin(65◦ ) hence b = ≈ 2.11. 4 3 3 3 . We start ◦ with sin(γ) = sin(30 ) and get sin(γ) = 4 sin (30◦ ) = 2. This means that γ must measure between 0◦ and 150◦ in order to fit inside the triangle with α.2 The Law of Sines angle-side pair available is (γ. we have◦ the measure of α = 30◦ . In other words. c). We find √ sin(60 b = 2sin(30◦ )) = 2 3 ≈ 3.90 Triangle for number 1 Triangle for number 2 3. 4. In this ◦ ◦ case. a = 2 and c = 4. In both cases. using the 30 √ 4 sin(120 ) Law of Sines one last time.19 triangles are drawn below. it must be true that γ. We first note that if we are given the measures of two of the angles in a triangle.81 . there is just one angle which satisfies both required conditions.19◦ a=3 α = 30◦ b ≈ 1. say α and β. 3 6 6 2 3 . both candidates for γ are ‘compatible’ with the given angle-side pair (α. we 6 β ≈ 180◦ − 30◦ − 41.81◦ = 108. a) = (30◦ .19 ) ≈ 5.81◦ b ≈ 5. In the case γ = π − arcsin 2 radians 3 sin(30◦ ) ◦ .70 units. has greater measure than α which is opposite a. a) and β. ◦ c=4 α = 30◦ β = 120◦ a=4 γ = 30◦ b ≈ 6. Using the Law of Sines with the angle-side opposite find ◦ pair (α. which is opposite c.70 6. In the case γ = arcsin 3 radians ≈ 41.19◦ . Since c > a. namely γ = ◦ ◦ .23 units. Hence. we pause to see if it makes sense that we actually have two viable cases to consider.81◦ and γ = π − arcsin 2 radians ≈ 138.93 Some remarks about Example 11. 2 Since the measure of γ must be strictly less than 150◦ . the measure of the third angle γ is uniquely To find an exact expression for β.898 Applications of Trigonometry we must have that 0◦ < γ < 150◦ . 6 3 6 3 7 ◦ An exact answer for β in this case is β = arcsin 2 − π radians ≈ 11. we find b ≈ 3 sin(108. we convert everything back to radians: α = 30◦ = π radians.7 Both ≈ 138.23 c=4 a=3 γ ≈ 138.2 are in order. At 3 3 this point. γ > α. The only other given piece of information is that c = 4 units. b = sin(30◦ ) = 4 3 ≈ 6.19◦ . Thus have two triangles on our hands. we repeat the exact same steps and find β ≈ 11. we repeat the usual Law of Sines routine to find that sin(γ) = sin(30 ) so 4 4 that sin(γ) = 1 .93 units. Since γ must inhabit a triangle with α = 30◦ .81◦ c=4 α = 30◦ β ≈ 108. As we have discussed.19◦ . γ = arcsin 6 radians and 180◦ = π radians.2. So β = 180◦ − 30◦ − 30◦ = 120◦ and. so both candidates for γ are compatible with this last piece of given information as 2 well.81◦ and b ≈ 1. For this last problem. β ≈ 11. 3) in that both choices for γ can fit in a triangle with α and both have a sine of 2 3 .81◦ . β = π − π − arcsin 2 = 5π − arcsin 2 radians ≈ 108.19◦ γ ≈ 41. we must have 0◦ < γ < 150◦ . There are two angles γ that fall in this range and have 2 sin(γ) = 3 : γ = arcsin 2 radians ≈ 41. which is c a a a impossible. the length of the one given side a was too short to even form a triangle. no triangle a = h. From high school Geometry. a and c are given. and the remaining parts of the theorem If this sounds familiar. This means if a ≥ h. the length of a was just long enough to form a right triangle. then a < c sin(α). the given side is adjacent to both angles which means we are in the so-called ‘Angle-Side-Angle’ (ASA) case. this is called the ‘Side-Side-Angle’ or SSA case. a) and (γ. Such is the case in numbers 1 and 2 above. we are always guaranteed to have at least one triangle. In the figure below.11. c) are intended to be angle-side pairs in a triangle where α. Let h = c sin(α) • If a < h. we are given the measure of just one of the angles in the triangle along with the length of two sides. 9 In more reputable books. the given side is adjacent to just one of the angles – this is called the ‘Angle-Angle-Side’ (AAS) case. Side-Angle-Side (SAS) and Side-Side-Side (SSS).3. but not too long. • If a ≥ c. These four cases exemplify all of the possibilities in the Angle-Side-Side case which are summarized in the following theorem. In number 1. Knowing the measures of all three angles of a triangle completely determines its shape. only one of which is adjacent to the given angle. in 5. If a < h. • If a = h. then two distinct triangles exist which satisfy the given criteria. if a < h the side a is too short to connect to form a triangle.8 In number 2. If a triangle were to exist. If in addition we are given the length of one of the sides of the triangle. If. If we are given information about a triangle that meets one of these four criteria. then γ is acute and exactly one triangle exists which satisfies the given criteria Theorem 11. 8 . side a was long enough to form a triangle but too long to swing back and form two. then no triangle exists which satisfies the given criteria. in number 4. Theorem 11. then γ = 90◦ so exactly one (right) triangle exists which satisfies the criteria. we see geometrically why this is the case. it should. so that two triangles were possible. we know there are four congruence conditions for triangles: Angle-Angle-Side (AAS). and in number 6. c a h = c sin(α) c a = h = c sin(α) α α a < h. on the other hand. γ = 90◦ Simply put. Suppose (α. a was long enough.2 The Law of Sines 899 determined using the equation γ = 180◦ − α − β. we are in the ‘Angle-Side-Side’ (ASS) case. Angle-Side-Angle (ASA). • If h < a < c.3 is proved on a case-by-case basis.9 In number 3. we can then use the Law of Sines to find the lengths of the remaining two sides to determine the size of the triangle. then we are guaranteed that exactly one triangle exists which satisfies the given criteria. the Law of Sines would have sin(γ) = sin(α) so that sin(γ) = c sin(α) > a = 1. Here. Think about this before reading further. Sasquatch Island lies off the coast of Ippizuti Lake. completing the proof. Since a c sin(α) < 1 which means there are two solutions to sin(γ) = c sin(α) : an h < a. Since γ0 is acute.2. then a = c sin(α) and the Law of Sines gives sin(α) = sin(γ) so that sin(γ) = c sin(α) = a = 1. Assuming a straight coastline. which forces γ to be an acute angle. we assume a ≥ c. are made to the island. The angle between the shore and the island at the first observation point is 30◦ and at the second point the angle is 45◦ . we note that the angles γ and 45◦ are supplemental. and its supplement. we need to find the perpendicular distance from the island to the coast. We need to argue that each of these angles ‘fit’ into a triangle with α. giving us the second triangle predicted in the theorem. As before. Hence. Since (α. now suppose h < a < c.66 miles. 180◦ − γ0 . To that end.3. the Law of Sines10 gives sin(γ) = c sin(α) . the assumption c > a in this case gives us γ0 > α. if you are given an obtuse angle to begin with then it is impossible to have the two triangle case. we first need to find the measure of β which is the angle opposite the side of length 5 miles. find the distance from the second observation point to the island. one triangle One last comment before we use the Law of Sines to solve an application problem. so that γ = 180◦ − 45◦ = 135◦ . This proves a triangle can contain both of the angles α and (180◦ − γ0 ). If a = h. we have already argued that a triangle exists in this case! Do you see why C must lie to the right of Q? . taken 5 miles apart.11 10 11 ◦ Remember. we have 180◦ −γ0 < 180◦ −α which gives (180◦ − γ0 ) + α < 180◦ . γ = 90◦ as required. In order to use the Law of Sines to find the distance d from Q to the island. We can now d 5 find β = 180◦ − 30◦ − γ = 180◦ − 30◦ − 135◦ = 15◦ . we get only one triangle in this case. a c a a Moving along. to find the point on the coast closest to the island. we have sin(30◦ ) = sin(15◦ ) sin(30 which gives d = 5sin(15◦ )) ≈ 9. we must have that α is acute as well. Example 11. To prove the last case in the theorem. c a h a c a γ h α α γ0 γ0 h < a < c. By the Law of Sines. What point on the shore is closest to the island? How far is the island from this point? Solution. giving us one of the triangles promised in the theorem. Next. Then α ≥ γ. which we’ve labeled as C. a) and (γ0 . Two sightings. If we manipulate the inequality γ0 > α a bit. c) are angle-side opposite pairs. We sketch the problem below with the first observation point labeled as P and the second as Q. two triangles a ≥ c. In the AngleSide-Side case.900 Applications of Trigonometry tell us what kind and how many triangles to expect in each case. This means one triangle can contain both α and γ0 . c sin(α) < a or a a acute angle which we’ll call γ0 . Find the area of the triangle in Example 11.2.66 22 ≈ 6.2 number 1.83 miles from the coast.2 The Law of Sines 901 Let x denote the distance from the second observation point Q to the point C and let y denote the distance from C to the island.12 we get x = y ≈ 6. . the point on the shore closest to the island is approximately 6. Hence. . we choose A = 2 ac sin(β) from Theorem 11. Its proof uses the same cases and diagrams as the proof of the Law of Sines and is left as an exercise. a).2 number 1.4. we have all three angles and all three sides 1 to work with.2. Then the area A enclosed by the triangle is given by 1 1 1 A = bc sin(α) = ac sin(β) = ab sin(γ) 2 2 2 Example 11.66 miles β y miles 30◦ P 5 miles γ Q 45◦ Shoreline Q 45◦ C x miles We close this section with a new formula to compute the area enclosed by a triangle. Suppose (α.4.4.2. we find A = 1 (7) 7 sin(15◦ ) sin (45◦ ) =≈ 5. to minimize propagated error. we note that β = 180◦ − 90◦ − 45◦ = 45◦ so by symmetry. the island is approximately 6.4 again . 12 Or by Theorem 10.4 because it uses the most pieces of given information.11. b) and (γ. To find the distance from Q to C.4. After some rearranging.66 miles d ≈ 9. Sasquatch Island Sasquatch Island √ β d ≈ 9. Solution.83 miles. From our work in Example 11. . Theorem 11. We are given a = 7 and β = 45◦ .83 miles.18 square 2 sin(120 sin(120 units. However. we get sin (45◦ ) = y .83 miles down the coast from the second observation point. The reader is encouraged to check this answer against the results obtained using the other formulas in Theorem 11. Using these values. Hence. Using Theorem 10. and we ◦) ◦) calculated c = 7 sin(15◦ ) . (β. c) are the angle-side opposite pairs of a triangle. d we find y = d sin (45◦ ) ≈ 9. 20. α = 68. β = 17◦ .) 23. (α.24. However.1 Exercises In Exercises 1 . CA and their road is actually this steep. b = 16.902 Applications of Trigonometry 11.2. 12 and 20 above. γ = 83. α = 73.13◦ .5 10. What grade is given by a 9.6◦ .75 19. γ = 35◦ . a = 33. a = 57. b = 42 8. γ = 120◦ .6) in that it expresses the ratio of rise/run.5 13. β = 102◦ . γ = 53◦ .05 16. β = 61◦ .65◦ angle made by the road and the horizontal?13 13 I have friends who live in Pacifica.95◦ . .5 21. c = 28. b = 14 12. b = 92 9. (β. a = 45. a = 7. α = 13◦ . c) are angle-side opposite pairs.7◦ . b = 16. a = 117 4. β = 62◦ . α = 53◦ . a = 25. c = 4 2. a = 35.75. b = 12.75. solve for the remaining side(s) and angle(s) if possible. α = 42◦ . a = 88.2◦ . c = 3. It’s not a nice road to drive. a = 17. if we want to apply any Trigonometry to a story problem involving roads going uphill or downhill. α = 117◦ . α = 30◦ . c = 18 18. α = 95◦ . α = 117◦ . In Exercises 22 . show that a 7% grade means that the road (hypotenuse) makes about a 4◦ angle with the horizontal. b = 90 11. b = 42 7. a = 70. α = 42◦ . α = 50◦ . b) and (γ. b = 23. we need to view the grade as an angle with respect to the horizontal. a = 3. a = 33. β = 54.33 6. (It will not be exactly 4◦ . In the case of a road.6. c = 13 17. γ = 74.15 20. a = 39. a road which rises 7 feet for every 100 feet of (horizontal) forward progress is said to have a 7% grade. For example. b = 314. Find the area of the triangles given in Exercises 1. this ratio is always positive because it is measured going uphill and it is usually given as a percentage. β = 29. Using a right triangle with a horizontal leg of length 100 and vertical leg with length 7. b = 23. (Another Classic Application: Grade of a Road) The grade of a road is much like the pitch of a roof (See Example 10. As in the text. α = 95◦ . b = 16. we first have you change road grades into angles and then use the Law of Sines in an application. a).1◦ . β = 102◦ . β = 85◦ . α = 6◦ . α = 68. 22.7◦ . but it’s pretty close.33 5. β = 102◦ .01 14. b = 100 15. 1. a = 5 3. For example.5◦ E (d) due south 14 The word ‘plumb’ here means that the tree is perpendicular to the horizontal. Find the angle θ in standard position with 0◦ ≤ θ < 360◦ which corresponds to each of the bearings given below. but rather. (Hint: Look at the diagram above. .14 From a point 500 feet downhill from the tree.11. and it is assumed that you know which quadrantal angle goes with each cardinal direction. a bearing is the direction you are heading according to a compass. These four bearings are drawn in the plane below.) 25. Simply put. Use the Law of Sines to find the height of the tree. you see a tall tree standing perfectly plumb alongside the road. S50◦ W would point into Quadrant III along the terminal side of θ = 220◦ because we started out pointing due south (along θ = 270◦ ) and rotated clockwise 50◦ back to 220◦ . respectively. however. (a) due west (b) S83◦ E (c) N5. Counter-clockwise rotations would be found in the bearings N60◦ W (which is on the terminal side of θ = 150◦ ) and S27◦ E (which lies along the terminal side of θ = 297◦ ). ‘due east’ and ‘due west’. east and west are usually not given as bearings in the fashion described above. Along a long. (Hint: First show that the tree makes a 94◦ angle with the road. N40◦ E (read “40◦ east of north”) is a bearing which is rotated clockwise 40◦ from due north. the angle of inclination from the road to the top of the tree is 6◦ . Similarly. one just refers to them as ‘due north’. The classic nomenclature for bearings.2 The Law of Sines 903 24. straight stretch of mountain road with a 7% grade. A bearing is given as an acute angle of rotation (to the east or to the west) away from the north-south (up and down) line of a compass rose. this bearing would have us heading into Quadrant I along the terminal side of θ = 50◦ . is not given as an angle in standard position. N N40◦ E N60◦ W 60◦ 40◦ W 50◦ 27◦ E S50◦ W S S27◦ E The cardinal directions north. If we imagine standing at the origin in the Cartesian Plane. so we must first understand the notation.) (Another Classic Application: Bearings) In the next several exercises we introduce and work with the navigation tool known as bearings. ‘due south’. south. rounded to the nearest foot. the captain of the HMS Sasquatch finds the signal flare to be at a bearing of N75◦ W. If Jeff and Carl are 500 feet apart.1 for a review of the DMS system. find the distances from the flare to each vessel. How far will she have to walk to get from the Muffin Ridge Observatory to Sasquach Point? What is the distance between Sasquatch Point and the Chupacabra Trailhead? 28. 30. If the SS Bigfoot is 5 miles from the HMS Sasquatch and the bearing from the SS Bigfoot to the HMS Sasquatch is N50◦ E. they position themselves 2 miles apart on an abandoned stretch of desert runway. 29. A watchtower spots a ship off shore at a bearing of N70◦ E. . 31. How far is she from the lodge at this point? Round your answer to the nearest hundredth of a mile. Skippy spies a UFO hovering over a spot on the runway directly between him and Sally. who is at a bearing of N50◦ E from Carl’s position. determines the bearing to the ship to be N25◦ W. How high off the ground is the UFO at this point? Round your answer to the nearest foot. 27. A hiker determines the bearing to a lodge from her current position is S40◦ W. 32. reckons the bearing to the fire to be N20◦ W from his current position. From there. she knows a bearing of S65◦ E will take her straight back to Sasquatch Point. He records the angle of inclination from the ground to the craft to be 75◦ and radios Sally immediately to find the angle of inclination from her position to the craft is 50◦ .) 15 See Example 10. rounded to the nearest tenth of a mile. which is 50 miles from the first at a bearing of S80◦ E from the first tower. One night. An hour into their investigation.25◦ W (f) S72◦ 41 12 W15 Applications of Trigonometry (g) N45◦ E (h) S45◦ W 26. Determine the distance from the campfire to each man. Sarge.1. how far is Jeff from the Sasquatch nest? Round your answer to the nearest foot. (Recall: 1 mile is 5280 feet. The Colonel spots a campfire at a of bearing N42◦ E from his current position. the nest is at a bearing of S70◦ W. Skippy and Sally decide to hunt UFOs. How far is the boat from the second tower? Round your answer to the nearest tenth of a mile. The captain of the SS Bigfoot sees a signal flare at a bearing of N15◦ E from her current location. who is positioned 3000 feet due east of the Colonel. From Jeff’s position.1 in Section 10. From the Chupacabra Trailhead she hikes for 5 miles along a bearing of N53◦ W which brings her to the Muffin Ridge Observatory. From his position. A hiker starts walking due west from Sasquatch Point and gets to the Chupacabra Trailhead before she realizes that she hasn’t reset her pedometer.904 (e) N31. A second tower. Carl spies a potential Sasquatch nest at a bearing of N10◦ E and radios Jeff. She proceeds to hike 2 miles at a bearing of S20◦ E at which point she determines the bearing to the lodge is S75◦ W. 38. The angle of depression from an observer in an apartment complex to a gargoyle on the building next door is 55◦ . the angle of inclination to the gargoyle is 20◦ . b = 10 and your choice of a yield only one triangle where that unique triangle has three acute angles.) 34. Prove that the Law of Sines holds when ABC is a right triangle.4. Why do those formulas yield square units when four quantities are being multiplied together? . (Said another way.11. choose four different values for a so that (a) the information yields no triangle (b) the information yields exactly one right triangle (c) the information yields two distinct triangles (d) the information yields exactly one obtuse triangle Explain why you cannot choose a in such a way as to have α = 30◦ . Round your answers to the nearest foot. 35.) 37. Use the cases and diagrams in the proof of the Law of Sines (Theorem 11.2 The Law of Sines 905 33. Discuss with your classmates why knowing only the three angles of a triangle is not enough to determine any of the sides. Discuss with your classmates why the Law of Sines cannot be used to find the angles in the triangle when only the three sides are given. Given α = 30◦ and b = 10. From a point five stories below the original observer. explain why the Law of Sines cannot be used in the SSS and SAS cases. Find the distance from each observer to the gargoyle and the distance from the gargoyle to the apartment complex.2) to prove the area formulas given in Theorem 11. (Use the rule of thumb that one story of a building is 9 feet. Also discuss what happens if only two sides and the angle between them are given. 36. arctan 7 100 ≈ 0. The area of the triangle from Exercise 1 is about 8.33 b ≈ 29. About 53 feet . 7. The area of the triangle from Exercise 20 is about 149 square units. 22. α = 30◦ β = 90◦ γ = 60◦ √ a=7 b = 14 c=7 3 α = 53◦ β = 74◦ γ = 53◦ a = 28.54 c ≈ 13.61◦ β = 102◦ γ ≈ 49.39◦ a ≈ 8.7◦ a = 45 b = 42 c ≈ 5. 14.7◦ β ≈ 103.1◦ γ = 52.36 α = 68.699 radians.5 c ≈ 31.48◦ a = 25 b = 12.01 b ≈ 33.217 c = 3 16. α ≈ 28.05 b ≈ 0. 5.5 c ≈ 11.6◦ a = 3.1◦ γ ≈ 8.9◦ γ ≈ 34.6◦ a = 3.5 c ≈ 53.50 c ≈ 11. The area of the triangle from Exercise 12 is about 377.00 c ≈ 97.1 square units.81◦ γ = 74.51 11.75 c ≈ 9.7◦ a = 117 b ≈ 99.43◦ γ ≈ 4.15 α = 6◦ β ≈ 169.41◦ β ≈ 3. 19.15 c ≈ 641.43◦ a = 57 b = 100 c ≈ 155.2◦ β = 54.5 c ≈ 23.95◦ a ≈ 593. 8.2◦ a = 88 b = 92 c ≈ 13. 12.05 b ≈ 1.11 Information does not produce a triangle Information does not produce a triangle α = 68.93 α = 42◦ β ≈ 112. which is equivalent to 4. 3. 4.82 Information does not produce a triangle 21. 15. About 17% 24. Information does not produce a triangle α = 42◦ β ≈ 23.68 b = 16. 18.75 α = 50◦ β ≈ 22.34◦ γ ≈ 25.22◦ a = 39 b = 23. α = 73.01 α ≈ 78.52◦ γ ≈ 107.47 9.89 α = 42◦ β ≈ 67. 17.906 Applications of Trigonometry 11.66◦ a = 17 b = 23.99◦ γ = 74.07 α = 117◦ β ≈ 56.40 c=3 α ≈ 101.20 b = 16.69 b = 314.78◦ γ ≈ 114.57◦ a = 57 b = 100 c ≈ 43.66◦ γ ≈ 70. 6.71 c = 28.3◦ γ ≈ 6.22 α = 95◦ β = 62◦ γ = 23◦ a = 33.57◦ γ ≈ 163.00 10.1 square units. Information does not produce a triangle α = 66. 13.7◦ β ≈ 76.75 c = 13 α = 43◦ β = 102◦ γ = 35◦ a ≈ 11.2.34◦ a = 17 b = 23.004◦ 23.92◦ β = 29. Answers 2. 20.2 1.45 α = 6◦ β ≈ 10.13 α = 13◦ β = 17◦ γ = 150◦ a=5 b ≈ 6.59◦ β ≈ 26.13◦ γ = 83.4◦ a = 88 b = 92 c ≈ 53. The Colonel is about 3193 feet from the campfire. 27.5◦ (g) θ = 45◦ (d) θ = 270◦ (h) θ = 225◦ 907 26. The gargoyle is about 27 feet from the observer on the lower floor. 33. 30. . The UFO is hovering about 9539 feet above the ground. 28.25◦ (b) θ = 353◦ (f) θ = 197◦ 18 48 (c) θ = 84.12 miles.46 miles. The distance from the Muffin Ridge Observatory to Sasquach Point is about 7. The SS Bigfoot is about 4.2 The Law of Sines 25. (a) θ = 180◦ (e) θ = 121. Jeff is about 342 feet from the nest. The boat is about 25. The gargoyle is about 25 feet from the other building.1 miles from the flare.02 miles from the lodge 31. 29. The gargoyle is about 44 feet from the observer on the upper floor. The HMS Sasquatch is about 2.1 miles from the second tower. 32.11. She is about 3.9 miles from the flare. Sarge is about 2525 feet to the campfire. The distance from Sasquatch Point to the Chupacabra Trailhead is about 2. Law of Cosines: Given a triangle with angle-side opposite pairs (α. the given angle is adjacent to both of the given sides. b) and (γ.5. we developed the Law of Sines (Theorem 11. ‘Side-Angle-Side’ means that we are given two sides and the ‘included’ angle .908 Applications of Trigonometry 11. From Theorem 10. In this section.3 The Law of Cosines In Section 11. c sin(α)) c α A = (0. we have cos(α) = b2 + c2 − a2 2bc cos(β) = a2 + c2 − b2 2ac cos(γ) = a2 + b2 − c2 2ab To prove the theorem.1 We state and prove the theorem below. B = (c cos(α).3.2. the coordinates Here. (β. a). we consider a generic triangle with the vertex of angle α at the origin with side b positioned along the positive x-axis.2) to enable us to solve triangles in the ‘Angle-Angle-Side’ (AAS). the ‘Angle-Side-Angle’ (ASA) and the ambiguous ‘Angle-Side-Side’ (ASS) cases. y) lies on a circle of radius c. 0) b a C = (b. we immediately find that the coordinates of A and C are A(0. 0) and C(b. we develop the Law of Cosines which handles solving triangles in the ‘Side-Angle-Side’ (SAS) and ‘Side-Side-Side’ (SSS) cases. 1 . Theorem 11. 0). solving for the cosine in each equation. c). 0) From this set-up. we know that since the point B(x.that is. the following equations hold a2 = b2 + c2 − 2bc cos(α) b2 = a2 + c2 − 2ac cos(β) c2 = a2 + b2 − 2ab cos(γ) or. a) is an angle-side opposite pair and b and c are the sides adjacent to α – the same can be said of any other angle-side opposite pair in the triangle.3. c = 2 units Solution. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. The cosine of an acute is positive. The advantage to using the Law of Cosines over the Law of Sines in cases like this is that unlike the sine function. Using the distance formula. With no angle-side opposite pair to use. the Pythagorean Theorem. That being said. the cosine function distinguishes between acute and obtuse angles.3 The Law of Cosines 909 of B are B(x.) We note that the distance between the points B and C is none other than the length of side a. we get a = a2 = (c cos(α) − b)2 + (c sin(α) − 0)2 (c cos(α) − b)2 + c2 sin2 (α) 2 a2 = (c cos(α) − b)2 + c2 sin2 (α) a2 = c2 cos2 (α) − 2bc cos(α) + b2 + c2 sin2 (α) a2 = c2 cos2 (α) + sin2 (α) + b2 − 2bc cos(α) a2 = c2 (1) + b2 − 2bc cos(α) a2 = c2 + b2 − 2bc cos(α) The remaining formulas given in Theorem 11. Equation 1.2 Example 11. or. the sine of an angle alone is not This shouldn’t come as too much of a shock. We could use the Law of Cosines again. 2 Since cos2 (α) + sin2 (α) = 1 2. If we have a triangle in which γ = 90◦ . 1. β = 50◦ .1. and the measure of the included angle. c = 5 units . Since the sine of both acute and obtuse angles are positive. What’s important about a and α in the above proof is that (α. since we have the angle-side opposite pair (β. All of the theorems in Trigonometry can ultimately be traced back to the definition of the circular functions along with the distance formula and hence. whereas the cosine of an obtuse angle is negative. 1. a = 4 units. the following computations hold for any angle α drawn in standard position where 0 < α < 180◦ . Solve the following triangles. the Law of Cosines and the Pythagorean Theorem amount to pretty much the same thing.5 can be shown by simply reorienting the triangle to place a different vertex at the origin. b) we could use the Law of Sines. the Law of Cosines can be thought of as a generalization of the Pythagorean Theorem. b = 7 units.11. a = 7 and c = 2. β = 50◦ . We leave these details to the reader. In order to determine the measures of the remaining angles α and γ. (This would be true even if α were an obtuse or right angle so although we have drawn the case when α is acute. What this means is that in the larger mathematical sense. y) = B(c cos(α).1. we apply the Law of Cosines. We get b2 = 72 + 22 − 2(7)(2) cos (50◦ ) which yields b = 53 − 28 cos (50◦ ) ≈ 5. There are two ways to proceed at this point. We are given the lengths of two sides. c sin(α)). we are forced to used the derived value for b. a = 7 units.92 units. then cos(γ) = cos (90◦ ) = 0 so we get the familiar relationship c2 = a2 + b2 . Notice that the proof of the Law of Cosines relies on the distance formula which has its roots in the Pythagorean Theorem. however. since we are guaranteed it will be acute. Using the angle-side opposite pair (β. 2. . α. we get sin(γ) = √ sin(50 ) ◦ . we know the radian measure of α must lie between 0 and π radians. since it is opposite the longest side.01◦ b ≈ 5. We sketch the triangle below. When using the Law of Cosines. To that end. we proceed with this method first. we use the formula cos(α) = b +c −a 2bc and substitute a = 7. that is if we trust our approximation for α. Your instructor will let you know which procedure to use. since this will give us the obtuse angle of the triangle if there is one. Plugging in a = 7.99◦ − 50◦ = 15.01◦ . As in 2ac 5 5 3 4 after simplifying .910 Applications of Trigonometry enough to determine if the angle in question is acute or obtuse.01◦ and 2 53−28 cos(50 ) α = 180◦ − β − γ ≈ 180◦ − 50◦ − 15. and if there is one. it must be the largest. Since all three sides and no angles are given. β = 50◦ c=2 a=7 α ≈ 114. b = 53 − 28 cos (50◦ ) and c = 2.54◦ . To minimize propagation of error. we are forced to use the Law of Cosines.5 In this case. Since both authors of the textbook prefer the Law of Cosines.92 As we mentioned earlier. we must proceed with caution as we are in the ambiguous (ASS) case. Since the largest angle is opposite 2 2 2 the longest side. b). 2 2 2 b.99◦ At this point. Here. we 2 2 2 could use the Law of Cosines again. We get cos(β) = a +c −b = − 1 .99◦ . The usual calculations produces γ ≈ 15. We get3 cos(α) = 2 − 7 cos (50◦ ) 53 − 28 cos (50◦ ) Since α is an angle in a triangle. so we have α = arccos 2 − 7 cos (50◦ ) 53 − 28 cos (50◦ ) radians ≈ 114. 5 There can only be one obtuse angle in the triangle. it’s always best to find the measure of the largest unknown angle first. we could find γ using γ = 180◦ − α − β ≈ 180◦ − 114. once we’ve determined b it is possible to use the Law of Sines to find the remaining angles. It all boils down to how much you trust your calculator. . we get γ = arccos √7−2 cos(50 ◦) 53−28 cos(50◦ ) radians ≈ 15. It is advisable to first find the smallest of the unknown angles. This matches the range of the arccosine function. we choose to find α first.4 in this case using cos(γ) = a +b −c . 2ab b= 53 − 28 cos (50◦ ) and c = 2. . however. so we get β = arccos − 1 radians ≈ 101. Following our discussion in the previous problem.01◦ .01◦ = 114. we find β first.99◦ γ ≈ 15. we would find γ since the side opposite γ is smaller than the side opposite the◦ other unknown angle. 3. which is geometrically impossible. 7 β ≈ 101. find the width of the pond. so we may apply the Law of Cosines to find the length of the missing side opposite the given angle. A great example of this is number 2 in Example 11. The Law of Cosines.54◦ c=5 a=4 α ≈ 34. 1000 feet 60◦ 950 feet P Solution. we could proceed using the Law of Sines. From a point P . 29 Using this. offers us a rare opportunity to find the remaining angles using only the data given to us in the statement of the problem.05◦ b=7 γ ≈ 44. where the approximate values we record for the measures of the angles sum to 180.42◦ We note that.3. Next. while the distance to the western-most point of the pond from P is 1000 feet.11.01◦ . however.3 The Law of Cosines 911 the previous problem. we get w2 = 9502 + 10002 − 2(950)(1000) cos (60◦ ) = 952500 from which √ we get w = 952500 ≈ 976 feet. b).42◦ and α = arccos 35 radians ≈ 34. we have an application of the Law of Cosines. A researcher wishes to determine the width of a vernal pond as drawn below. depending on how many decimal places are carried through successive calculations. . now that we have obtained an angle-side opposite pair (β. If the angle between the two lines of sight is 60◦ . he finds the distance to the eastern-most point of the pond to be 950 feet. we get γ = arccos 5 radians ≈ 44. the approximate answers you obtain may differ slightly from those the authors obtain in the Examples and the Exercises.2. Example 11. Calling this length w (for width). We are given the lengths of two sides and the measure of an included angle.1. and depending on which approach is used to solve the problem.05◦ . let s = 1 (a + b + c). we have A = 2 ab sin(γ) from Theorem 11. we start by manipulating the expression for A2 . 1 ab sin(γ) 2 2 A2 = = = 1 2 2 2 a b sin (γ) 4 a2 b2 1 − cos2 (γ) 4 a2 +b2 −c2 .2. Let s be the semiperimeter of the triangle. .Heron’s Formula. so substituting this into our equation for A2 gives The Law of Cosines tells us cos(γ) = A2 = = = = = = = = a2 b2 1 − cos2 (γ) 4 a2 b2 1− 4 a2 + b2 − c2 2ab 2 2 a2 + b2 − c2 a2 b2 1− 4 4a2 b2 a2 b2 4 4a2 b2 − a2 + b2 − c2 4a2 b2 2 2 4a2 b2 − a2 + b2 − c2 16 (2ab)2 − a2 + b2 − c2 16 2 + b2 − c2 2ab − a 2ab + a2 + b2 − c2 16 2 − a2 + 2ab − b2 c a2 + 2ab + b2 − c2 16 2 difference of squares.4. In this section.912 Applications of Trigonometry In Section 11.4. 2ab since sin2 (γ) = 1 − cos2 (γ). that is. Theorem 11. we used the proof of the Law of Sines to develop Theorem 11. In order to simplify computations. b and c denote the lengths of the three sides of a triangle.6.4 as an alternate formula for the area enclosed by a triangle.6 using Theorem 11. Using the convention that the angle γ is opposite the 1 side c. Heron’s Formula: Suppose a. we use the Law of Cosines to derive another such formula . Then the 2 area A enclosed by the triangle is given by A= s(s − a)(s − b)(s − c) We prove Theorem 11. 1 Solution. we note that (s − a) = Similarly. difference of squares.3. (s − b) = 8 − 7 = 1 and (s √ c) = 8√ 5 = 3.3 The Law of Cosines 913 A2 = = = = = c2 − a2 − 2ab + b2 c2 − (a − b)2 a2 + 2ab + b2 − c2 16 (a + b)2 − c2 16 (c − (a − b))(c + (a − b))((a + b) − c)((a + b) + c) 16 (b + c − a)(a + c − b)(a + b − c)(a + b + c) 16 (b + c − a) (a + c − b) (a + b − c) (a + b + c) · · · 2 2 2 2 perfect square trinomials.80 square units. we find (s − b) = A2 = 1 2 (a + b + c) = a+b+c 2 . . we find s = 2 (4 + 7 + 5) = 8. Using these values.11.3. a+c−b 2 Hence. we recognize the last factor as the semiperimeter. we get (b + c − a) (a + c − b) (a + b − c) (a + b + c) · · · 2 2 2 2 = (s − a)(s − b)(s − c)s so that A = s(s − a)(s − b)(s − c) as required. s = complete the proof. (s − a) = 8 − 4 = 4.3. b = 7 and c = 5. At this stage. Find the area enclosed of the triangle in Example 11. Example 11. We are given a = 4. we get − − A = s(s − a)(s − b)(s − c) = (8)(4)(1)(3) = 96 = 4 6 ≈ 9. Using Heron’s Formula.1 number 2. To a+b+c a + b + c − 2a b+c−a −a= = 2 2 2 and (s − c) = a+b−c 2 . We close with an example of Heron’s Formula. a = 153. α = 42◦ . b = 25. α = 63◦ . . β = 7◦ . a = 7. b = 302. a = 5.5◦ E for 207 miles. 6 Please refer to Page 903 in Section 11. c = 88 12.6 17. b = 117. From her camp. b = 20 13. Round your answer to the nearest hundredth of an inch. c = 153 5. From the Pedimaxus International Airport a tour helicopter can fly to Cliffs of Insanity Point by following a bearing of N8. what is the diameter of the crater? Round your answer to the nearest hundredth of a mile.6 Find the distance between Cliffs of Insanity Point and Bigfoot Falls. c = 26 14.1 Exercises In Exercises 1 .5 inches long. b = 12. 8 and 10 above. α = 104◦ . α = 63◦ .2◦ . a = 18. c = 13 In Exercises 11 . c = 98.2◦ E for 192 miles and it can fly to Bigfoot Falls by following a bearing of S68. γ = 90◦ 6.10. a = 1. using any appropriate technique. Cliffs of Insanity Point and Bigfoot Falls from Exericse 20 above both lie on a straight stretch of the Great Sasquatch Canyon.3. 21. solve for the remaining side(s) and angle(s). a = 3. a = 5. a = 22. c = 4 7. What bearing would the tour helicopter need to follow to go directly from Bigfoot Falls to Cliffs of Insanity Point? Round your angle to the nearest tenth of a degree. if possible. use the Law of Cosines to find the remaining side(s) and angle(s) if possible. A geologist wants to measure the diameter of a crater. b = 10. Round your answer to the nearest mile.16.914 Applications of Trigonometry 11. The hour hand on my antique Seth Thomas schoolhouse clock in 4 inches long and the minute hand is 5. . b = 5. b = 4. b = 12. β = 8. α = 120◦ .2 for an introduction to bearings. 19. c = 48 10. Find the area of the triangles given in Exercises 6. γ = 59. 18. b = 3. a = 300. Find the distance between the ends of the hands when the clock reads four o’clock. 20. b = 20 15. a = 37. If the angle between the two lines of sight is 117◦ . b = 2. a = 7. a = 16. b = 20 16. c = 13 8. α = 63◦ . c = 37 4. 11. c = 5 9. it is 4 miles to the northern-most point of the crater and 2 miles to the southern-most point. γ = 170◦ . 1.3◦ 3. c = 5 2. b = 45. 5 miles.3◦ . If his bearing to the harbor is now S70◦ W. you will first need to use right angle Trigonometry to find the lengths of the lines of sight. 23. The angle of depression7 made by the line of sight from the ranger to the first fire is 2. After 1. If you apply the Law of Cosines to the ambiguous Angle-Side-Side (ASS) case. The SS Bigfoot leaves a harbor bound for Nessie Island which is 300 miles away at a bearing of N32◦ E. 25. How far is it from port? Round your answer to the nearest hundredth of a mile. she changes her bearing to S17◦ W and continues hiking for 3 miles.5◦ and the angle of depression made by line of sight from the ranger to the second fire is 1. What course should the captain set to head to the island? Round your angle to the nearest tenth of a degree. The angle formed by the two lines of sight is 117◦ . Discuss with your classmates why Heron’s Formula yields an area in square units even though four lengths are being multiplied together. Round your answer to the nearest foot. If the equation has only one positive real zero then exactly one triangle is formed and if the equation has two distinct positive real zeros then two distinct triangles are formed.) fire 117◦ firetower 26. A naturalist sets off on a hike from a lodge on a bearing of S80◦ W. If the equation has no positive real zeros then the information given does not yield a triangle. (Hint: In order to use the 117◦ angle between the lines of sight. Apply the Law of Cosines to Exercises 11. how far is the SS Bigfoot from Nessie Island? Round your answer to the nearest hundredth of a mile.3 for the definition of this angle. What bearing should she follow to return to the lodge? Round your angle to the nearest degree. From a point 300 feet above level ground in a firetower. 27.3 The Law of Cosines 915 22. fire 7 See Exercise 78 in Section 10. . the result is a quadratic equation whose variable is that of the missing side. a ranger spots two fires in the Yeti National Forest. Find the distance between the two fires.11. A storm moves in and after 100 miles. the captain of the Bigfoot finds he has drifted off course. 24. This will give you a Side-Angle-Side case in which to apply the Law of Cosines. The HMS Sasquatch leaves port on a bearing of N23◦ E and travels for 5 miles. Find her distance from the lodge at this point. Round your answer to the nearest hundredth of a mile. What is its bearing to port? Round your angle to the nearest degree. 13 and 14 above in order to demonstrate this result. It then changes course and follows a heading of S41◦ E for 2 miles. 90◦ a = 153 b ≈ 21.30 b = 117 c = 88 14.40◦ γ ≈ 46.89◦ γ ≈ 35. (Try to avoid rounding errors. 23. It is about 229. 25.16◦ γ = 59.77◦ a ≈ 78.916 Applications of Trigonometry 11.22 miles.41 b = 25 c = 37 α ≈ 36.30◦ a = 37 b = 45 c = 26 α ≈ 35.72 b ≈ 69. Answers 2.4◦ E to reach the island.98 α ≈ 3◦ β = 7◦ γ = 170◦ a ≈ 29.61 miles from the island and the captain should set a course of N16.26 inches.30◦ β ≈ 89. 20. She is about 3.50 miles from port and its heading to port is S47◦ W. About 313 miles 21.92 miles from the lodge and her bearing to the lodge is N37◦ E. 16. 12.2 1.88 c = 153 α = √ ◦ β ≈ 25. 5. It is about 4.14◦ a = 300 b = 302 c = 48 α ≈ 22.23◦ γ ≈ 48.38◦ γ = 90◦ a=5 b = 12 c = 13 α ≈ 55. The area of the triangle given in Exercise 8 is 51764375 ≈ 7194. 10.2 c = 98.58◦ γ ≈ 98.) .05◦ β ≈ 87.54 α = 63◦ β ≈ 81.6 √ √ 17. The area of the triangle given in Exercise 10 is exactly 30 square units.3◦ a=7 b = 12 c ≈ 10.81◦ γ ≈ 9.87◦ β ≈ 53.36 α ≈ 85. 15.8◦ W 22.3.31◦ β ≈ 49. α = 63◦ β ≈ 54. The distance between the ends of the hands at four o’clock is about 8.72◦ 120 c=4 a = 37 b = 3 Information does not produce a triangle α = 60◦ β = 60◦ γ = 60◦ a=5 b=5 c=5 α = 63◦ β ≈ 98. The area of the triangle given in Exercise 6 is √1200 = 20 3 ≈ 34.11◦ a = 18 b = 20 c ≈ 11. 3.64 square units.90◦ β = 8.54◦ β ≈ 85. N31.1◦ γ ≈ 62.62◦ β ≈ 67.11◦ γ ≈ 18. 18.89◦ a = 18 b = 20 c ≈ 6.2◦ γ ≈ 85. α = 104◦ β ≈ 29. 24.21◦ a=7 b = 10 c = 13 α ≈ 83.75 square units.62 13.60◦ a ≈ 49.13◦ γ = 90◦ a=3 b=4 c=5 α ≈ 32. 11. The diameter of the crater is about 5. The fires are about 17456 feet apart. Information does not produce a triangle α = 42◦ β ≈ 89.9◦ a = 22 b = 20 c ≈ 21. 9. 7. 8. 6. 4. 19.40◦ γ ≈ 35.28◦ γ ≈ 34. 5π this way. 6 move out along the polar axis 4 units. the ‘motions’ of the Cartesian system (over and up) describe a rectangle. Roughly speaking. the Cartesian coordinates of a point are often called ‘rectangular’ coordinates. of course. For the most part.4 Polar Coordinates 917 11. 4). We defined the Cartesian coordinate plane using two number lines – one horizontal and one vertical – which intersect at right angles at a point we called the ‘origin’. The authors encourage the reader to take time to think about both approaches to plotting points given in polar coordinates. θ) For example. In this section. We then locate a point P using two coordinates. if we wished to plot the point P with polar coordinates 4. the more ways you have to look at something. the polar coordinates (r. travel up 4 units. θ). where r represents a directed distance from the pole2 and θ is a measure of rotation from the polar axis.3 To plot P 4. and a ray called the polar axis. θ) of a point measure ‘how far out’ the point is from the pole (that’s r). then up 4 units.1 For this reason. then move outwards from the pole 4 units. 6 P 4. we introduce a new system for assigning coordinates to points in the plane – polar coordinates. Alternatively.11. we introduced the Cartesian coordinates of a point in the plane as a means of assigning ordered pairs of numbers to points in the plane. then to the left 3 units and arrive at the same location. To plot a point.4 Polar Coordinates In Section 1. say P (−3. We start with an origin point. we could start at the origin. travel horizontally to the left 3 units. called the pole. the better. θ= r=4 Pole Pole Pole 5π 6 5π 6 We may also visualize this process by thinking of the rotation first.1. 2 1 . the points in which one or both coordinates are 0. we’d start at the pole. 4) 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x Pole r Polar Axis r θ P (r. y P (−3. we start at the origin. We will explain more about this momentarily. 6 we rotate 5π counter-clockwise from the polar axis. (r. 5π . 3 As with anything in Mathematics. 6 Excluding. then rotate 5π radians counter-clockwise. and most points can be thought of as the corner diagonally across the rectangle from the origin. and ‘how far to rotate’ from the polar axis. (that’s θ). − 3π we have the following.5. not π . P 4. 4 Pole θ = − 3π 4 Pole θ = − 3π 4 R 3. Hence. θ= 5π 6 5π 6 θ= 5π 6 Pole Pole Pole If r < 0. we begin by moving in the opposite direction on the polar axis from the pole.5. π 4 If we interpret the angle first. we rotate − 3π then move out 3. we rotate π radians. For example. to plot Q −3.5.5 units away from the pole on the terminal side of 5π .5. − 3π 4 Pole .5 units from the pole.918 Essentially we are locating a point on the terminal side of 5π 6 Applications of Trigonometry which is 4 units away from the pole. − 3π 4 Pole From an ‘angles first’ approach. We see that 4 R is the point on the terminal side of θ = − 3π which is 3.5 units from the pole. θ < 0 means the rotation away from the polar axis is clockwise instead of counter-clockwise. 4 Here we are locating a point 3. 4 r = 3.5 Pole Pole θ = − 3π 4 R 3.5 units. π 4 As you may have guessed.5. then move back through the pole 3.5 Pole θ= π 4 Pole Pole Q −3. 4 4 θ= Pole π 4 θ= Pole π 4 Pole Q −3.5. to plot R 3. π we have 4 r = −3. We plot −4. the same point despite the fact that their polar coordinate representations are different. we need only to rotate θ = 60◦ to arrive at location coterminal with 240◦ . Example 11. P −4. We explore this notion more in the following example. Next. we plot P (2. 240◦ ) below.4. our answer here is (−2. 7π 6 3. one of which has r > 0 and the other with r < 0. The given representation for P is (2. θ = 240◦ Pole Pole 2. P −3. (Can you see why?) One such angle is θ = −120◦ so one answer for this case is (2. we visualize our rotation starting 2 units to the left of the pole. − 5π 2 4. a point can be represented by infinitely many polar coordinate pairs. P (2.4 Polar Coordinates 919 The points Q and R above are. −120◦ ) P (−2. 7π 6 Pole Pole θ= 7π 6 . P 117. From this position. Since P is 2 units from the pole. 1. 60◦ ).1. For the case r = −2. Hence. Whether we move 2 units along the polar axis and then rotate 240◦ or rotate 240◦ then move out 2 units from the pole. we find our point lies 4 units from the pole on the terminal side of π . b) and (c. 240◦ ) Solution.11. 240◦ ) We now set about finding alternate descriptions (r. in fact. For each point in polar coordinates given below plot the point and then give two additional expressions for the point. r = ±2. We check our answers by plotting them. Pole θ = −120◦ θ = 60◦ Pole P (2. 1. d) represent the same point if and only if a = c and b = d. θ) for the point P . 60◦ ) 2. 6 6 Since r = −4 < 0. −120◦ ). 240◦ ) so the angle θ we choose for the r = 2 case must be coterminal with 240◦ . we choose angles θ for each of the r values. 6 P −4. − π 4 P (2. Unlike Cartesian coordinates where (a. 7π by first moving 4 units to the left of the pole and then rotating 7π radians. 6 6 6 P 4. − 5π as our second answer. 3π as one answer. and get 117. gives us 4. θ) for P satisfies r = ±117.920 Applications of Trigonometry To find alternate descriptions for P . To find a different 6 6 representation for P with r = −4. 2 Pole θ = − 5π 2 Pole P 117. 3π 2 P −117. P lies on the terminal side of π . coupled with r = 4. any representation (r. π as one of our answers. we visualize moving left 117 2 2 units from the pole and then rotating through an angle θ to reach P . we can take θ to be any angle coterminal with − 5π . π 2 . we may choose any angle coterminal with the angle in the original representation of P −4. we note that the distance from P to the pole is 4 units. 2 Pole θ= 3π 2 Pole θ= π 2 P 117. − 5π . To plot P 117. we move along the polar axis 117 units from the pole and rotate 2 clockwise 5π radians as illustrated below. In this case. − 5π 2 Since P is 117 units from the pole. θ= π 6 π 6 θ = − 5π 6 P −4. π . 7π . so this. we choose 2 θ = 3π . so our second answer is −117. For the r = −117 case. As we noted above. For the r = 117 case. We find that θ = π 2 satisfies this requirement. so any representation (r. θ) for P must have r = ±4. − 5π 6 Pole Pole 3. We pick − 5π and get −4. To 4 4 find a different representation for P with r = −3. If r = 0. Then (r.4 Polar Coordinates 921 π 4 4. 4 4 4 P 3. r = 0 and the angles are measured in radians. θ) means (directed distance from pole. It could be considered as a definition or a theorem. θ) and (r . angle of rotation). is to keep in mind that (r. θ) represent the pole regardless of the value of θ. we may choose any angle coterminal with − π . and indeed the whole polar coordinate system. − π 4 θ = −π 4 Pole Pole Since P lies on the terminal side of 3π . depending on your point of view. We state it as a property of the polar coordinate system. θ= 7π 4 7π 4 θ= Pole Pole Now that we have had some practice with plotting points in polar coordinates. Thus (0. 3π 4 3π 4 P −3. 3π . The following result characterizes when two sets of polar coordinates determine the same point in the plane. Equivalent Representations of Points in Polar Coordinates Suppose (r. We choose θ = 7π for our final answer −3. θ ) determine the same point P if and only if one of the following is true: • r = r and θ = θ + 2πk for some integer k • r = −r and θ = θ + (2k + 1)π for some integer k All polar coordinates of the form (0. θ) is the pole for all . The key to understanding this result. θ ) are polar coordinates where r = 0. 7π . θ) and (r . 4 4 P −3. one alternative representation for P is 3. the point never leaves the pole.11. We move three units to the left of the pole and follow up with a clockwise rotation of radians to plot P −3. We see that P lies on the terminal side of 3π . then no matter how much rotation is performed. it should come as no surprise that any given point expressed in polar coordinates has infinitely many other representations in polar coordinates. − π . The remaining case is r = 0. Q(−3. θ + π). and hence are the same point. Conversely. x2 + y 2 = (−r)2 = r2 . If r = r. θ ). Since the pole is identified with the origin (0. θ ) determine the same point. r = −r. θ = π + θ + 2πk which we rewrite as θ = θ + (2k + 1)π for some integer k. θ) is the pole. too. Convert each point in rectangular coordinates given below into polar coordinates with r ≥ 0 and 0 ≤ θ < 2π. so the theorem is true in this case. our first step is to move the same distance from the pole as P . √ 1. Theorem 11.7 is an immediate consequence of Theorem 10. we have that either r = r or r = −r. then when plotting (r. θ ) lie the same (directed) distance from the pole on the terminal sides of coterminal angles.7. θ). the theorem in this case amounts to checking ‘0 = 0. We know that this means θ = θ + 2πk for some integer k. Since this distance is controlled by the first coordinate. we identify the pole and polar axis in the polar system to the origin and positive x-axis. applying the theorem to (−r. Theorem 11. θ) and (r . θ) and (r . then when plotting (r.2. Hence. to plot P . Conversion Between Rectangular and Polar Coordinates: Suppose P is represented in rectangular coordinates as (x. θ) is (−r. Now let’s assume that neither r nor r is zero. S(−3. in which case (r.922 Applications of Trigonometry values of θ. y) and in polar coordinates as (r. Since θ = θ + (2k + 1)π = (θ + π) + 2πk for some integer k. but in the opposite direction.3 along with the sin(θ) quotient identity tan(θ) = cos(θ) . Next. Use exact values if possible and round any approximate values to two decimal places. θ) = (0. Since cos(θ + π) = − cos(θ) and sin(θ + π) = − sin(θ). To plot P . as required. if r = r and θ = θ + 2πk for some integer k. R(0. we have two points equidistant from the pole rotated exactly π radians apart. Hence. P 2. θ) and P (r . then we know an alternate representation for (r. θ ) determine the same point P then the (non-zero) distance from P to the pole in each case must be the same.’ The following example puts Theorem 11. the initial side of θ is rotated π radians away from the initial side of θ. we marry the polar coordinate system with the Cartesian (rectangular) coordinate system. If. We get the following result. Then • x = r cos(θ) and y = r sin(θ) • x2 + y 2 = r2 and tan(θ) = y (provided x = 0) x In the case r > 0. θ + π) gives x = (−r) cos(θ + π) = (−r)(− cos(θ)) = r cos(θ) and y = (−r) sin(θ + π) = y (−r)(− sin(θ)) = r sin(θ). 4) . At this intermediate stage. we see that θ is coterminal to (θ + π) and it is this extra π radians of rotation which aligns the points P and P . If (r. −3) 3. then the points P (r. −3) 4. respectively. In this case. in the rectangular system. If r < 0. the angles θ and θ have the same initial side.7 to good use. θ ). on the other hand. Check your answer by converting them back to rectangular coordinates. −2 3 2. 0) in rectangular coordinates. θ) and (r . θ) and (r . we must have that θ is coterminal with θ.4. we first move a directed distance r from the pole. To do so. Example 11. if (r. θ must be coterminal with π + θ. Now suppose r = −r and θ = θ + (2k + 1)π for some integer k. Moreover. and x = tan(θ + π) = tan(θ). −3) lies in Quadrant III. 5π 4 3. 4 we choose θ = 5π . Since we are asked for r ≥ 0. −2 3 shows that √ √ 2 it lies in Quadrant IV. θ) = 4. We find −3 tan(θ) = −3 = 1. 5π . we convert (r. With x = 2 and y = −2 3.4 Polar Coordinates Solution. Since we require r ≥ 0. we choose r = 4. −2 3) P has polar coordinates 4. . we know θ is a Quadrant IV angle. so we are done. as required. we find x = r cos(θ) = (3 2) cos 5π = (3 2) − 22 = −3 4 4 √ √ √ 2 5π and y = r sin(θ) = (3 2) sin 4 = (3 2) − 2 = −3. which satisfies the requirement that 0 ≤ θ < 2π. 923 1. the angle θ = 3π satisfies 0 ≤ θ < 2π 2 4 Since x = 0. we would have to determine θ geometrically. Even though we are not explicitly told to do so. Concerning θ. Our final answer is 4 √ √ √ √ (r. we get r2 = x2 + y 2 = (2)2 + −2 3 = 4 + 12 = 16 so r = ±4. we can avoid many common mistakes by taking √ the time to plot the points before we do any calculations. θ) = 3 2. in this case we can find the polar coordinates of R using the definition. To check. We are asked to have 0 ≤ θ < 2π. 5π 3 3 3 back to rectangular coordinates and we find x = r cos(θ) = 4 cos 5π = 4 1 = 2 and 3 2 √ √ y = r sin(θ) = 4 sin 5π = 4 − 23 = −2 3. we can easily tell the point R is 3 units from the pole. θ) of R we know r = ±3. 3 2. While we could go through the usual computations4 to find the polar form of R. we have √ √ y that tan(θ) = x = −22 3 = − 3. 5π . 5π 3 Q has rectangular coordinates (−3. which means in the polar representation (r. and since P 3 lies in Quadrant IV. so we choose θ = 5π . which means θ has a reference angle of π . we get r2 = (−3)2√ (−3)2 = 18 + √ so r = ± 18 = ±3 2. Since Q lies in Quadrant III. −3) √ Q has polar coordinates 3 2. To check. The point √ Q(−3. we choose r = 3. Since the pole is identified with the origin.11. Plotting P 2. our answer is 4. y y θ= 5π 3 θ= x 5π 4 x Q P √ P has rectangular coordinates (2. Hence. Using x = y = −3. −3) lies along the negative y-axis. This tells us θ has a reference angle of π . The point R(0. we choose r = 3 2. Since we are asked for r ≥ 0. To find θ. we note 2 x = r cos(θ) = 3 cos 3π = (3)(0) = 0 and y = r sin(θ) = 3 sin 3π = 3(−1) = −3. Convert each equation in polar coordinates into an equation in rectangular coordinates. With x = −3 and y = 4. Since θ lies in Quadrant II and must satisfy 0 ≤ θ < 2π. so our answer is 3. These are good review exercises and are hence 3 3 4 4 left to the reader. −3) R has polar coordinates 3. (a) r = −3 (b) θ = 4π 3 (c) r = 1 − cos(θ) . we choose 4 θ = π − arctan 3 radians. We find cos π − arctan 4 = − 3 and sin π − arctan 3 = 5 . equations in the variables r and θ represent relations in polar coordinates. To 3 check our answers requires a bit of tenacity since we need to simplify expressions of the form: cos π − arctan 4 and sin π − arctan 4 . 4) lies in Quadrant II. As usual. and since this isn’t the tangent of one the common angles. 1. Example 11. 2 2 4.924 Applications of Trigonometry with its terminal side along the negative y-axis.3. θ) = 5. we choose r = 5 ≥ 0 and proceed to determine θ. (a) (x − 3)2 + y 2 = 9 (b) y = −x (c) y = x2 2. π − arctan 4 ≈ (5. We convert equations between the two systems using Theorem 11. our answer is (r. 3π . we now set about converting equations from one system to another. Hence. 4) 4 S has polar coordinates 5. 2. 3π 2 S has rectangular coordinates (−3.4. we get r2 = (−3)2 +(4)2 = 25 so r = ±5.21). The point S(−3. To check. π − arctan 3 Now that we’ve had practice converting representations of points between the rectangular and polar coordinate systems. We have tan(θ) = y 4 4 x = −3 = − 3 .7 as the next example illustrates. Just as we’ve used equations in x and y to represent relations in rectangular coordinates. Convert each equation in rectangular coordinates into an equation in polar coordinates. 5 y S y θ= 3π 2 θ = π − arctan x 4 3 x R R has rectangular coordinates (0. we resort to using the arctangent function. so that 3 5 4 x = r cos(θ) = (5) − 3 = −3 and y = r sin(θ) = (5) 5 = 4 which confirms our answer. we get r cos(θ) + r sin(θ) = 0 or r(cos(θ) + sin(θ)) = 0. One strategy to convert an equation from rectangular to polar coordinates is to replace every occurrence of x with r cos(θ) and every occurrence of y with r sin(θ) and use identities to simplify. and necessity is the mother of invention. we 4 are essentially drawing the line containing the terminal side of θ = − π which is none 4 other than y = −x. we follow our mathematical instincts and see where they take us. Hence. Factoring. but as a general rule. we are not losing any Experience is the mother of all instinct. we get r(r cos2 (θ) − sin(θ)) = 0 so that either r = 0 or r cos2 (θ) = sin(θ). or r2 cos2 (θ) − r sin(θ) = 0. 4 5 .1. Consider the equation θ = − π . We can solve the latter equation for r by dividing both sides of the equation by cos2 (θ). If we imagine plotting points (r.8 4 (c) We substitute x = r cos(θ) and y = r sin(θ) into y = x2 and get r sin(θ) = (r cos(θ))2 . 925 1. We know y = −x describes a line through the origin. We get r = 0 or r = 6 cos(θ). In this particular case. 6 Note that when we substitute θ = π into r = 6 cos(θ). the variable 4 r is free. we get θ = − π + πk for integers 4 k. This is the technique we employ below.11. Solving the latter equation for θ. then sin(θ) would also have to be 0. r = 0 describes the origin. 8 We could take it to be any of θ = − π + πk for integers k.7 meaning it can assume any and all values including r = 0. then cos(θ) = 0. 7 See Section 8. Since there are no angles with both cos(θ) = 0 and sin(θ) = 0. so we aren’t losing anything 2 by disregarding r = 0. we take a step back and think geometrically. Study this example and see what techniques are employed. In this equation. (a) We start by substituting x = r cos(θ) and y = sin(θ) into (x−3)2 +y 2 = 9 and simplifying.6 (b) Substituting x = r cos(θ) and y = r sin(θ) into y = −x gives r cos(θ) = −r sin(θ). we are safe since if cos2 (θ) = 0.5 (r cos(θ) − 3)2 + (r sin(θ))2 = 9 r2 cos2 (θ) − 6r cos(θ) + 9 + r2 sin2 (θ) = 9 r2 cos2 (θ) + sin2 (θ) − 6r cos(θ) = 0 Subtract 9 from both sides. but nothing else. Rearranging. negative and zero). r2 − 6r cos(θ) = 0 r(r − 6 cos(θ)) = 0 Since cos2 (θ) + sin2 (θ) = 1 Factor. then try your best to get your answers in the homework to match Jeff’s. As we did in the previous example. With no real direction in which to proceed. This gives r = 0 or cos(θ) + sin(θ) = 0. we choose r = 6 cos(θ) for our final answer. − π ) for all conceivable values of r (positive. and since r = 0 describes just a point (namely the pole/origin). we can take as our final answer θ = − π here.2 we know the equation (x − 3)2 + y 2 = 9 describes a circle.4 Polar Coordinates Solution. we recover the point r = 0. we never divide through by a quantity that may be 0. From Section 7. and for the equation r cos2 (θ) = sin(θ) to hold. As before. not just r = −3. As before. we need to manipulate r = 1 − cos(θ) a bit before using the conversion formulas given in Theorem 11.7. Once again. Doing so. converting equations from polar to rectangular coordinates isn’t as straight forward as the reverse process. so we opt for a second strategy – rearrange the given polar equation so that the expressions r2 = x2 + y 2 . Here.3. (c) Once again. We could square both sides of this equation like we did in part 2a above to obtain an r2 on the left hand side. θ) can be represented as (−3. We now have an r2 and an r cos(θ) in the equation. As a general rule. for instance . the r = 0 case is recovered in the solution r = sec(θ) tan(θ) (let θ = 0). y r cos(θ) = x. 3 3 √ √ y y Since tan(θ) = x . geometrically. to y = x 3 in which we assume x can be 0. but when interpreted as polar coordinates. θ + π).926 Applications of Trigonometry information by dividing both sides of r cos2 (θ) = sin(θ) by cos2 (θ). we pause a moment to wonder √ if. √ 11 In addition to taking the tangent of both sides of an equation (There are infinitely many solutions to tan(θ) = 3. we have performed some Exercise 5. they correspond to the same point in the plane.2. The concern here is that the equation r2 = 9 might be satisfied by more points than r = −3. Back then. this appears to be the case since r2 = 9 is equivalent to r = ±3. √ (b) We take the tangent of both sides the equation θ = 4π to get tan(θ) = tan 4π = 3. we get sin(θ) r = cos2 (θ) . which means any point (r. we can square both sides to get r2 = (−3)2 or r2 = 9. we concern ourselves only with ensuring that the sets of points in the plane generated by two equations are the same. Since polar coordinates were defined geometrically to describe the location of points in the plane. a point in the plane was identified with a unique ordered pair given by its Cartesian coordinates. we also went from x = 3. any point with polar coordinates (3. ‘equivalent’ means they represent the same point in the plane. however. the equations θ = 4π and y = x 3 generate the same set of points. r sin(θ) = y and/or tan(θ) = x present themselves. we multiply both sides by r to obtain r2 = r − r cos(θ). when we first defined relations as sets of points in the plane in Section 1. or r = sec(θ) tan(θ). but that does nothing helpful for the right hand side. in which x cannot be 0. √ √ y 4π and θ = 3 is only one of them!). Mathematically speaking.3. produce an equivalent equation. As we have seen. Substituting r2 = x2 + y 2 and 2 r cos(θ) = x gives x2 + y 2 = x2 + y 2 + x . in general. (3. π) are different. Rewriting the equation as r = r2 + r cos(θ) 2 and squaring both sides yields r2 = r2 + r cos(θ) . relations are sets of ordered pairs. Instead. We may now substitute r2 = x2 + y 2 to get the equation x2 + y 2 = 9. As ordered pairs. but we also have another r to deal with. θ) whose polar coordinates satisfy the relation r = ±3 has an equivalent10 representation which satisfies r = −3. Of course. 0) and (−3.1 in Section 5. On the surface. so we state the latter as our final answer. This was not an issue. .9 squaring an equation does not. However. which we can easily handle. We could solve r2 = x2 + y 2 for r to get r = ± x2 + y 2 y y and solving tan(θ) = x requires the arctangent function to get θ = arctan x + πk for integers k.11 3 The same argument presented in number 1b applies equally well here so we are done. we get x = 3 or y = x 3. . 10 9 . (a) Starting with r = −3. so the equations r2 = 9 and r = −3 represent different relations since they correspond to different sets of ordered pairs. Neither of these expressions for r and θ are especially user-friendly. 2. by the way. We substitute r = −r and θ = θ + π into r = r2 + r cos(θ) to see if we get a true statement. squaring equations like r = −3 to arrive at r2 = 9 happens without a second thought. In practice. much of the pedantic verification of the equivalence of equations in Example 11. so the multiplication by r doesn’t introduce any new points. number 2c above shows the price we pay if we insist on always converting to back to the more familiar rectangular coordinate system. Indeed. if any. Then r = ± (r )2 + r cos(θ ) . Your instructor will ultimately decide how much. θ + π). we multiplied both sides by r. which determine the same point as (r . r = 1 − cos(θ). we devote our attention to graphing equations like the ones given in Example 11.4. can turn ugly in polar coordinates. The squaring of both sides of this equation is also a reason to pause. (−r .3 number 2 on the Cartesian coordinate plane without converting back to rectangular coordinates.3 is left unsaid. In the next section. θ ) satisfies r2 = r2 + r cos(θ) . θ ).4 Polar Coordinates 927 algebraic maneuvers which may have altered the set of points described by the original equation. in most textbooks. Are there points with coordinates 2 (r. it should be that relatively nice things in rectangular coordinates. θ + π) satisfies r = r2 + r cos(θ) which 2 means that any point (r. and we are done. we are done. This means that now r = 0 is a viable solution to the equation.11. If nothing else. we see that θ = 0 gives r = 0. such as y = x2 . . θ) which satisfies r2 = r2 + r cos(θ) has a representation which satisfies r = r2 + r cos(θ). First.4. justification is warranted. In the original equation. θ) which satisfy r2 = r2 + r cos(θ) but do not satisfy r = r2 + r cos(θ)? Suppose 2 (r . What if r = − (r )2 + r cos(θ ) = −(r )2 −r cos(θ )? We claim that the coordinates (−r . If we have that r = (r )2 +r cos(θ ).3. and vice-versa. satisfy r = r2 + r cos(θ). = (−r )2 + (−r cos(θ + π)) = (r )2 − r cos(θ + π) ? ? ? −r − −(r )2 − r cos(θ ) Since r = −(r )2 − r cos(θ ) Since cos(θ + π) = − cos(θ ) (r )2 + r cos(θ ) = (r )2 − r (− cos(θ )) (r )2 + r cos(θ ) = (r )2 + r cos(θ ) Since both sides worked out to be equal. If you take anything away from Example 11.4. 36. 34.− 4 4 √ √ 42. −7) √ 43. 3 1 . convert the point from polar coordinates into rectangular coordinates. 7π 2 7π 6 20. 5. 5. 14. (−117. 8. √ 7. − 2. π − arctan (5) 2 √ 2 . 5 2 42. 33. (7. (10. (6. 2 . π + arctan 1 − . 29. 2π 3 9. −5. − 9π 4 3 π .928 Applications of Trigonometry 11. 1 3π . arctan(π)) In Exercises 37 . −3. (−20. plot the point given in polar coordinates and then give three different expressions for the point such that (a) r < 0 and 0 ≤ θ ≤ 2π. −1. convert the point from rectangular coordinates into polar coordinates with r ≥ 0 and 0 ≤ θ < 2π. 2 6 7 13π .36. 2. arctan − 4 3 28. (−3.4. 117π) 27.56. √ 3) 39. π 3 7π 6 2. 5 5π . 11. −4 3 √ 40. 15. − √ 4π − 5. 32. −π 11. 3 2 4. −4. 5) 41. π − arctan −1. 31. − 3) √ 44. 2. 5. (π. − 9. arctan 2. − −4. 2 2. 7π 4 5π 4 5π 4 π 4 3. 5. (−3. arctan(2)) 30. (0. 23. 7π 4 18. 37. −π) In Exercises 17 . π 2 12. arctan(3)) −3. π 3 5π 6 19. 26. (−20. 3π) 13.− 2 6 −3. 6. (−π. arctan 12 5 35. 3π) 24. 10. 17. 13π 6 4 3 1 2 3 4 −4. 0) 38.5. 22. − 11π 6 −2. 25. − 3 16. − 3. 12. π + arctan 2 2 3 13.16.1 Exercises In Exercises 1 . 21. (3. (b) r > 0 and θ ≤ 0 (c) r > 0 and θ ≥ 2π 1. With the help of your classmates. 1) 51. x2 + (y − 3)2 = 9 58. Solve for r in all but #60 through #63. r = 7 81. 77. y = 4x − 19 70. x2 + y 2 = 25 68. θ = π 4 3π 2 84. In Exercises 57 . −7) 54. 2 5) 929 45.96. r = 2 sec(θ) tan(θ) 96. 0) into polar coordinates in four different ways. 5r = cos(θ) 89. r2 = sin(2θ) 79. (−2 10. y = −3x2 72. x2 + y 2 − 2y = 0 59.11. In Exercises 60 . 5 5 53. r = − 5 csc(θ) 95. x2 + y 2 = 117 69. 56. x2 + y 2 = x =1 In Exercises 77 . −12) √ √ 2 6 . x = 3y + 1 71. (x + 2)2 + y 2 = 4 76. θ = √ 2 80. Convert the origin (0. (12. convert the equation from rectangular coordinates into polar coordinates. y = −x 65. 98. 47. y = x 3 66. x = 6 61. 4 4 52. θ = 2π 3 78. y = 7 63. x = −3 √ 62. r = 7 sec(θ) 92. r = 3 sin(θ) 90. (6. 12r = csc(θ) 93. y 2 = 7y − x2 75. r = 4 cos(θ) 88. r = 1 + sin(θ) 85. use the Law of Cosines to develop a formula for the distance between two points in polar coordinates.63. 6 10) √ √ 48. r = −2 sec(θ) 94. 4x = y 2 73. r = 1 − 2 cos(θ) 97. .− 10 10 √ √ 46.76. you need to solve for θ 57. r = −2 sin(θ) √ 91. r = − csc(θ) cot(θ) 87. − 5. y = 2x 67. (−5. ( 5. − 5 √ √ 50.4 Polar Coordinates √ 3 3 3 − . (−8. y = 0 64. 8) 49. 4x2 + 4 y − 1 2 2 60. x2 − 4x + y 2 = 0 74. r = 83. r = −3 82. convert the equation from polar coordinates into rectangular coordinates.− 15 15 √ √ 65 2 65 − . √ √ 5 2 5 − . −9) 55. θ = π 86. (24. − .− . 3 3 7π 5π . − . 1 π 1 3π . 3 2 3 2 1 π 1 7π . 7π . −2. 2 6 2 6 5 7π 5 17π . 2 6 2 6 y 3 2 1 x −3 −2 −1 −1 −2 −3 1 2 3 . 2.− . 4 −5. 3π 4 15π 5. − . 5 5π 5 11π . 4 π 5. 2.2 1. 3 2 3 2 y 1 1 x 4. . −1 −1 −2 −3 1 2 3 x 3. 2. . . − 3 3 y 2 1 −1 1 y 1 2 x 2. 4 5. Answers π 4π . .4.930 Applications of Trigonometry 11. −12. 4 4 y 3 2 1 −3 −2 −1 −1 −2 −3 1 2 3 x √ √ 7. 3 −12 −9 −6 −3 x 6. −3. . 2 6 2 6 7 π 7 23π . 0 √ √ 2 2. 2 2. 4 4 13π 11π 3.11. 2 2. − y 6 931 5. 12. − . −3π . 3.4 Polar Coordinates 11π 7π .− . 7 13π 7 5π . − . 3. −π . 2 6 2 6 y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x . 3π y 3 2 1 −3 −2 −1 −1 −2 −3 1 2 3 x 8. 12. − 7π 5π . − 6 6 12. −2 2.− . 6 6 17π 13π . 5π 4 7π 4. 3 3 π 11π 1. . (−20. −4. (20. − . −1. − . 1. 2 −3. y 3 2 1 −3 −2 −1 −1 −2 −3 1 2 3 x . 5π 2 7π 3. 3 3 −1. 3π). − 4 . 2π 8π . π . 13π 4 9π 4. 2 π 3. y 2 1 −2 −1 −1 −2 1 2 x 12. π) (20. (−20. 4π) −20 Applications of Trigonometry y 1 −10 10 20 x −1 10. −2π).932 9. 2 −3. 4 −4. y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 11. 14.11. y 2 1 −2 −1 −1 −2 1 2 x 15. − . − . − . 3. π) (π.5. √ √ 2π 4π − 5. (π. − y 3 2 1 −3 −2 −1 −1 −2 −3 1 2 3 x 933 13. 4 5π 2.5.5.5. 3. 4 7π 4 11π 2. − π . 3 3 y 2 1 −2 −1 −1 −2 1 2 x 16. 6 6 19π 5π . −2. 2π) y 3 2 1 −3 −2 −1 −1 −2 −3 1 2 3 x . −π) . (−π. 5. (−π. 3 3 √ √ 11π π 5. − 6 6 −3.4 Polar Coordinates π 11π . 4 −2. − 5. −2π) . −3. π 2 26. 5 5 π2 √ π . 28. π + arctan (2) 3 15. 4π 3 4 3 40. θ = 66. 33. 53. 3 √ 19. r = π 3 19 4 cos(θ)−sin(θ) 13. (0. 6 2. x = 1 cos(θ)−3 sin(θ) √ 117 69. 22. 47. 41. 5 5 5. 46. 36. 58. arctan (2)) 50. r = 7 csc(θ) 63. r = cos(θ) . 0) 27. 2 2 √ √ 10. r = 5 68. θ = 0 64. 43. 44. 21 3.− 9 9 √ π 2 3. r = 6 sec(θ) 61. 42.934 √ √ 5 2 5 2 . √ 25. π − arctan 10. 2π − arctan √ 3 4 13.− 5 5 4 3 . π − arctan(3)) 52. √ 7π 7 2. r = 4 csc(θ) cot(θ) 70. √1+π2 1+π 2 24. 34. r = − sec(θ) tan(θ) 3 57. (117. −9) √ √ 6 5 12 5 . π) 45. −2 23. −4) √ 2 4 2 − .− 2 2 3 0. √ 3π 4 10. π + arctan 25. 54. 5 3 √ 65. 29. 0) √ √ 5 2 5 2 − . r = 3π 4 59. r = −3 sec(θ) 62. 2 3. − 2 2 17. 3 10 √ 26 5 26 . r = 2 sin(θ) 71. 5 5 √ √ 4 5 2 5 − . 6 1 11π . θ = 65. 49. 21 9 12 − . 20. 4 8. 5 √ Applications of Trigonometry √ 11 3 11 . (3. 35. π − arctan(2) 60. (5. arctan 48. (3. 56. 18.− 52 52 21. 1. 51. 2 3 55. 2π − arctan √ 2 π . (20. 1 . 30. (20. θ = arctan(2) 67. 1 8 12 5 7 24 39. √ 32. 5π 4 31. 3 2π . r = 4 cos(θ) 72. 12) √ 7π 2 3. 38. 2 6 37. (5. (0. 5x2 + 5y 2 = x or x− 1 10 2 935 75. y = − 3x 83. r = −4 cos(θ) 78. x2 = 2y 94. x2 + 2x + y 2 2 = 2xy 2 = x2 + y 2 96.11. θ) will work. y = 93. x2 + y 2 = 4x or (x − 2)2 + y 2 = 4 + y2 = 1 100 86. r = 7 sin(θ) 77. −117). Any point of the form (0. y = − 5 89. r = 6 sin(θ) 79. r = 1 4 sin(θ) 80. π). y = 0 84. (0. x2 + y 2 = 3y or x2 + y − 88. x2 + y 2 + y = x2 + y 2 and (0. y 2 = −x 95. x = 7 90. x2 + y 2 2 74. 0). 23π 4 . e. 0. x2 + y 2 = −2y or x2 + (y + 1)2 = 1 1 12 √ 91. 97. x = −2 92. y = x 3 2 2 = 9 4 87.4 Polar Coordinates 73.g. x2 + y 2 = 9 76. x2 + y 2 = 2 82. x = 0 85. x2 + y 2 = 49 √ 81. The Fundamental Graphing Principle for Polar Equations The graph of an equation in polar coordinates is the set of points which satisfy the equation. θ) is on the graph of an equation if and only if there is a representation of P . Example 11. θ).1 This makes these graphs easier to visualize than others. such that r and θ satisfy the equation. Since any given point in the plane has infinitely many different representations in polar coordinates. only one of the variables r and θ is present making the other variable free.3 number 2a. θ is free The graph of r = 4 √ 2.5. r = −3 2 3.936 Applications of Trigonometry 11.4.1. we discuss how to graph equations in polar coordinates on the rectangular coordinate plane. centered at the origin. Plotting all of the points of the √ √ form (−3 2. 1. θ) gives us a circle of radius 3 2 centered at the origin. In each of these equations. That is. θ = 5π 4 4. Once again we have θ being free in the equation r = −3 2. our ‘Fundamental Graphing Principle’ in this section is not as clean as it was for graphs of rectangular equations on page 23. for any choice of θ. r = 4 √ 2. We state it below for completeness. all points which have a polar coordinate representation (4.5 Graphs of Polar Equations In this section. 1. The graph of this equation is. a point P (r. . Graph the following polar equations. y 4 y θ>0 x θ<0 −4 4 x −4 In r = 4. with a radius of 4. say (r . θ ). 1 See the discussion in Example 11. This is exactly the definition of circle. Graphically this translates into tracing out all of the points 4 units away from the origin. θ = − 3π 2 Solution. therefore. Our first example focuses on the some of the more structurally simple polar equations. θ is free. In the equation r = 4. r is free. 4 4 What we find is that we are tracing out the line which contains the terminal side of θ = 5π 4 when plotted in standard position. r is free The graph of θ = 5π 4 4. 5π . so we plot all of the points with polar representation r. As in the previous example. In the equation θ = 5π . . Plotting r.11. θ is free √ The graph of r = −3 2 3. − 3π 2 2 for various values of r shows us that we are tracing out the y-axis. y 4 r<0 θ= 5π 4 y r=0 x −4 4 x r>0 −4 In θ = 5π 4 . the variable r is free in the equation θ = − 3π .5 Graphs of Polar Equations y 4 y 937 θ<0 x θ>0 −4 4 x −4 √ In r = −3 2. Theorem 11. r as the dependent variable. 3π 2 √ 7π 3 2. • The graph of the polar equation r = a on the Cartesian plane is a circle centered at the origin of radius |a|. Suppose we wish to graph r = 6 cos(θ). 4 (−6. 4 0. π 2 √ 3π −3 2.5. A reasonable way to start is to treat θ as the independent variable.2 We generate the table below.938 y r>0 Applications of Trigonometry y 4 r=0 x θ = − 3π 2 −4 4 x r<0 −4 In θ = − 3π . see Sections 1. evaluate r = f (θ) at some ‘friendly’ values of θ and plot the resulting points.1 makes the following result clear. • The graph of the polar equation θ = α on the Cartesian plane is the line containing the terminal side of α when plotted in standard position. 0) √ π 3 2.6. 2π) y 3 π 5π 4 3π 2 7π 4 3 6 x −3 2π 2 For a review of these concepts and this process. θ 0 π 4 π 2 3π 4 r = 6 cos(θ) 6 √ 3 2 0 √ −3 2 −6 √ −3 2 0 √ 3 2 6 (r. 4 0. our experience in Example 11.8. a = 0. r is free 2 The graph of θ = − 3π 2 Hopefully. Graphs of Constant r and θ: Suppose a and α are constants.4 and 1. 4 (6. . π) √ 5π −3 2. θ) (6. this means that the curve starts 6 units from the origin on the positive x-axis (θ = 0) and gradually returns to the origin by the time the curve reaches the y-axis (θ = π ). At this stage. with all of the angle rotations starting from the negative x-axis. we employ a slightly different strategy. It is a less-precise way to generate the graph than computing the actual function values. Since the |r| for these values of θ match the r values for θ in The graph looks exactly like y = 6 cos(x) in the xy-plane.5 Graphs of Polar Equations 939 Despite having nine ordered pairs. θ) on the xy-plane. r ranges from 6 2 to 0. Here. r 6 θ runs from 3 π 2 y to π π 2 π 3π 2 2π θ x −3 r < 0 so we plot here −6 As θ ranges from π to 3π . We see that as θ ranges from 0 to π . the r values are still negative. r 6 θ runs from 0 to 3 y π 2 π 2 π 3π 2 2π θ x −3 −6 Next. we get only four distinct points on the graph. the arrows are drawn from the θ-axis to the curve r = 6 cos(θ). In the θr-plane. which means the graph is traced out in 2 Quadrant I instead of Quadrant III. The 2 arrows drawn in the figure below are meant to help you visualize this process. and for good reason. we repeat the process as θ ranges from π to π. each of these arrows starts at the origin and is rotated through the corresponding angle θ. In the xy-plane. instead of graphing in Quadrant II. we graph in Quadrant IV. the r values are all negative. In the xy-plane. 3 .11. For this reason. We graph one cycle of r = 6 cos(θ) on the θr-plane3 and use it to help graph the equation on the xy-plane. but it is markedly faster. we are just graphing the relationship between r and θ before we interpret them as polar coordinates (r. in accordance with how we plot polar coordinates. This 2 means that in the xy-plane. 4 We present the final graph below. we find 2 that when 3π ≤ θ ≤ 2π. r = 2 + 4 cos(θ) r = 6 cos(θ) in the xy-plane 3. r 6 y 3 3 π 2 π θ −3 3 6 x −3 −6 r = 6 cos(θ) in the θr-plane Example 11. The reader is invited to verify that plotting any range of θ outside the interval 2 [0. π . 3π and 3π . r = 5 sin(2θ) 4. π . r 6 y θ runs from 0 to π 2 4 x 2 π 2 π 3π 2 2π θ 4 The graph of r = 6 cos(θ) looks suspiciously like a circle. we have that the curve begins to retrace itself at this point. To help us visualize what is going on graphically.5. . As θ ranges from 0 to π . π] results in retracting some portion of the curve. 2. for good reason.2.940 Applications of Trigonometry 0. This means that the curve in the xy-plane starts 4 units from the origin on the positive x-axis and gradually pulls in towards the origin as it moves towards the positive y-axis. Proceeding further. 1. π . 2 2 π. we divide up [0. and proceed as we did above. r2 = 16 cos(2θ) 1. r decreases from 2 2 2 4 to 2. 2π . See number 1a in Example 11. We first plot the fundamental cycle of r = 4 − 2 sin(θ) on the θr-axes. π . 2π] into the usual four subintervals 0. we retrace the portion of the curve in Quadrant IV that we first traced 2 out as π ≤ θ ≤ π. Graph the following polar equations.3. r = 4 − 2 sin(θ) Solution.4. as θ takes on values from 3π to 2π. we see that r increases from 2 to 4. so we are finished. we see that r increases from 4 to 6. On the xy-plane. the curve 2 sweeps out away from the origin as it travels from the negative x-axis to the negative y-axis.11. 2π] results in retracing portions of the curve. Picking up where we left 2 off. The graph on the 2 xy-plane pulls in from the negative y-axis to finish where we started. 3π . as θ runs from π to π. r 6 x y 4 2 π 2 π 3π 2 2π θ runs from θ 3π 2 to 2π We leave it to the reader to verify that plotting points corresponding to values of θ outside the interval [0.5 Graphs of Polar Equations 941 Next. . r decreases from 6 back to 4. we gradually pull the graph away from the origin until we reach the negative x-axis. r 6 θ runs from π 2 y to π 4 x 2 π 2 π 3π 2 2π θ Over the interval π. r 6 x y 4 2 π 2 π 3π 2 2π θ θ runs from π to 3π 2 Finally. Setting r = 0 we get 2 + 4 cos(θ) = 0. 2π . Plotting this on the xy-plane. As 2 2 3 3 3 3 2 2 θ ranges from 0 to π . 3 5 The ‘tangents at the pole’ theorem from second semester Calculus. we break the interval [0. π . Since these values of θ are important geometrically. 2π] is that the graph crosses through the θ-axis. r decreases from 2 to 0. 4π . The first thing to note when graphing r = 2 + 4 cos(θ) on the θr-plane over the interval [0. Solving for θ in [0. r decreases from 6 to 2. r 6 y θ runs from 0 to π 2 4 2 2π 3 π 2 4π 3 x π 3π 2 2π θ −2 On the interval π . .942 r 6 Applications of Trigonometry y 2 x 4 −4 4 2 π 2 π 3π 2 2π −6 θ r = 4 − 2 sin(θ) in the θr-plane r = 4 − 2 sin(θ) in the xy-plane. π. 4π . a 3 theorem from Calculus5 states that the curve hugs the line θ = 2π as it approaches the origin. This corresponds to the graph of the curve passing through the origin in the xy-plane. 2π] 2π 4π gives θ = 3 and θ = 3 . 2π] into six subintervals: 0. 2π . we start 6 units 2 out from the origin on the positive x-axis and slowly pull in towards the positive y-axis. or cos(θ) = − 2 . 2π . and our first task is to determine when this 1 happens. which means the graph is heading into (and 2 3 will eventually cross through) the origin. 2π . π . 3π and 3π . Not only do we reach the origin when θ = 2π . 2. π . 11. 3 6 Recall that one way to visualize plotting polar coordinates (r. we continue our 3 graph in the first quadrant. heading into the origin along the line θ = 4π . Rotating between 2π and π radians from the negative x-axis in 3 this case determines the region between the line θ = 2π and the x-axis in Quadrant IV. Since r ≤ 0. r ranges from 0 to −2.5 Graphs of Polar Equations r 6 θ= 4 2π 3 943 y 2 2π 3 π 2 4π 3 x π 3π 2 2π θ −2 On the interval 2π . π . Since r ≤ 0. the curve passes through the 3 origin in the xy-plane. θ) with r < 0 is to start the rotation from the left side of the pole . 3 .6 Since |r| is increasing from 0 to 2. r 6 θ= 2π 3 y 4 2 2π 3 π 2 4π 3 x π 3π 2 2π θ −2 Next.in this case. the curve pulls away from the origin to finish at a point on the positive x-axis. the negative x-axis. following the line θ = 2π and continues upwards through Quadrant IV 3 towards the positive x-axis. as θ progresses from π to 4π . r ranges from −2 to 0. We hug the 3 2 line θ = 4π as we move through the origin and head towards the negative y-axis. r increases from 2 out 2 to 6. r returns to positive values and increases from 0 to 2.944 r 6 y Applications of Trigonometry 4 θ= 4π 3 2 2π 3 π 2 4π 3 x π 3π 2 2π θ −2 On the interval 4π . 3π . 3 r 6 θ= 4π 3 y 4 2 2π 3 π 2 4π 3 x π 3π 2 2π θ −2 As we round out the interval. we find that as θ runs through 3π to 2π. 6 units from the origin on the positive x-axis. r 6 y 4 2 2π 3 π 2 4π 3 x π 3π 2 2π θ θ runs from −2 3π 2 to 2π . and we end up back where we started. We partition our interval into subintervals to help us with the graphing. π . namely 0. r is 4 2 heading negative as θ crosses π . r 4 4 2 2 4 4 4 increases from 0 to 5. occurs as θ ranges from 0 to π. π . 4 r 5 y π 4 π 2 3π 4 π x θ −5 Next.5 Graphs of Polar Equations 945 Again. 3π and 3π . we see that r decreases from 5 to 0 as θ runs through π . r 6 y 4 θ= 2π 3 2 2 2π 3 π 2 4π 3 2 3π 2 6 x π 2π θ θ= 4π 3 −2 −2 r = 2 + 4 cos(θ) in the θr-plane r = 2 + 4 cos(θ) in the xy-plane 3. As θ ranges from 0 to π . we invite the reader to show that plotting the curve for values of θ outside [0. Hence. . π . As usual. 2π] results in retracing a portion of the curve already traced. which in this case. π . we draw the curve hugging the line θ = π (the y-axis) 2 2 as the curve heads to the origin. we start by graphing a fundamental cycle of r = 5 sin(2θ) in the θr-plane. π . π .11. Our final graph is below. This means that the graph of r = 5 sin(2θ) in the xy-plane starts at the origin and gradually sweeps out so it is 5 units away from the origin on the line θ = π . and furthermore. r 5 y π 4 π 2 3π 4 π x θ −5 For 3π 4 ≤ θ ≤ π. the curve 2 4 pulls away from the negative y-axis into Quadrant IV. r becomes negative and ranges from 0 to −5. r 5 y π 4 π 2 3π 4 π x θ −5 . r increases from −5 to 0.946 r 5 Applications of Trigonometry y π 4 π 2 3π 4 π x θ −5 As θ runs from π to 3π . Since r ≤ 0. so the curve pulls back to the origin. inclusive. we sketch a fundamental period of r = cos(2θ) which we have dotted in the figure below. if we continue plotting beyond θ = π. so we don’t have any values on the interval π . 3π . How do we sketch such a curve? First off.11. Graphing r2 = 16 cos(2θ) is complicated by the r2 .5 Graphs of Polar Equations 947 Even though we have finished with one complete cycle of r = 5 sin(2θ). so we solve to get r = ± 16 cos(2θ) = ±4 cos(2θ). On the intervals which remain.7 From this. respectively. When cos(2θ) < 0. Hence. we find that the curve continues into the third quadrant! Below we present a graph of a second cycle of r = 5 sin(2θ) which continues on from the first. cos(2θ) is undefined. As we have seen in earlier Owing to the relationship between y = x and y = former is defined. r 5 3 y 4 π 1 5π 4 2 3π 2 3 7π 4 4 2π θ x 1 2 −5 We have the final graph below. The boxed labels on the θ-axis correspond to the portions with matching labels on the curve in the xy-plane. 1]. r 5 5 y π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π θ −5 5 x −5 −5 r = 5 sin(2θ) in the θr-plane r = 5 sin(2θ) in the xy-plane 4. cos(2θ) ranges from 0 to 1 as well. 7 √ x over [0. Below we graph both r = 4 cos(2θ) and r = −4 cos(2θ) on the θr plane and use them to sketch the corresponding pieces of the curve r2 = 16 cos(2θ) in the xy-plane. we also know cos(2θ) ≥ cos(2θ) wherever the . 4 4 cos(2θ) ranges from 0 to 1. we know r = ±4 cos(2θ) ranges continuously from 0 to ±4. 5.9 the authors truly feel that taking the time to work through each graph in the manner presented here is the best way to not only understand the polar 8 In this case. π]. While many of the ‘common’ polar graphs can be grouped into families.5. 2π] in the θr-plane. In Example 11. in general. the lines θ = π and θ = 3π . despite the fact that the period of f (θ) = 6 cos(θ) is 2π. but in order to obtain the complete graph of r = 5 sin(2θ).2 are examples of ‘lima¸ons. θ= 3π 4 3 θ= 1 π 4 π 4 π 2 3π 4 π 3 θ 2 4 x 2 −4 r = 4 cos(2θ) and r = −4 cos(2θ) As we plot points corresponding to values of θ outside of the interval [0.948 Applications of Trigonometry examples. 9 Numbers 1 and 2 in Example 11. there is no relation.’ number 3 is an example of a ‘polar rose. but graphed over the interval [0. we needed to run θ from 0 to 2π.8 so our final answer is below. As we saw on page 939. r 4 1 4 y cos(2θ). we find ourselves retracing parts of the curve. between the period of the function f (θ) and the length of the interval required to sketch the complete graph of r = f (θ) in the xyplane. we sketched the complete graph of r = 6 cos(θ) in the xy-plane just using the values of θ as θ ranged from 0 to π.2. r 4 y θ= 3π 4 4 θ= π 4 π 4 π 2 3π 4 π θ −4 4 x −4 −4 r = ±4 cos(2θ) in the θr-plane r2 = 16 cos(2θ) in the xy-plane A few remarks are in order.’ and c number 4 is the famous ‘Lemniscate of Bernoulli. we could have generated the entire graph by using just the plot r = 4 cos(2θ).’ . which are the zeros of the functions r = ±4 4 4 serve as guides for us to draw the curve as is passes through the origin. First. the period of f (θ) = 5 sin(2θ) is π. We leave the details to the reader. number 3. if a point P is on the graph of two different polar equations. 1. The graph of r = 2 − 2 sin(θ) is a special kind of lima¸on called a ‘cardioid.11. r = 2 sin(θ) and r = 2 − 2 sin(θ) 3.3. r = 3 sin θ 2 and r = 3 cos θ 2 −2 2 x −4 r = 2 − 2 sin(θ) and r = 2 sin(θ) It appears as if there are three intersection points: one in the first quadrant. . it must have a representation P (r. in order for a point P to be on the graph of a polar equation. In addition to the usual kinds of symmetry discussed up to this point in the text (symmetry about each axis and the origin).5. it is possible to talk about rotational symmetry. We leave the discussion of symmetry to the Exercises. it is entirely possible that the representation P (r. According to the Fundamental Graphing Principle for Polar Equations on page 936. the name is derived from its resemblance to a stylized human heart. r = 2 and r = 3 cos(θ) 4. Our next task is to find polar representations of these points. and the origin. In 10 Presumably. one in the second quadrant. 1. Here. more than ever. r = 3 and r = 6 cos(2θ) Solution. θ) which satisfies the equation. 1) with a radius of 1. Following the procedure in Example 11. Find the points of intersection of the graphs of the following polar equations. Second. we need to rely on the Geometry as much as the Algebra to find our solutions.2. What complicates matters in polar coordinates is that any given point has infinitely many representations. the symmetry seen in the examples is also a common occurrence when graphing polar equations. we graph r = 2 sin(θ) and find it to be a circle centered at the point with rectangular coordinates (0. but also prepare you for what is needed in Calculus.5 Graphs of Polar Equations 949 coordinate system. Example 11.’10 c y 2 2. θ) which satisfies one of the equations does not satisfy the other equation. As a result. we are given the task of finding the intersection points of polar curves. In our next example.5. For the point of intersection in the second quadrant. 1. On the curve r = 2 − 2 sin(θ). then P has a (possibly different) representation P (r . 6 6 2. We get 3 11 We are really using the technique of substitution to solve the system of equations r r = = 2 sin(θ) 2 − 2 sin(θ) . 1. θ) for any angle θ. What about the origin? We know from Section 11. then equating11 the expressions for r 1 gives 2 sin(θ) = 2 − 2 sin(θ) or sin(θ) = 2 . and as the reader can verify. we try θ = 5π . we have determined the three points of intersection to be 1. θ) which satisfies both r = 2 and r = 3 cos(θ). we first determine if any of the intersection points P have a representation (r. Plugging θ = 6 into r = 2 sin(θ). To solve this equation. On the graph of r = 2 sin(θ). π is one 6 representation for the point of intersection in the first quadrant. θ) that satisfies both r = 2 sin(θ) and r = 2 − 2 sin(θ). From this. In any case. The graph of r = 3 cos(θ) is also a circle . we are at the origin exactly when θ = πk for integers k. when θ = π + 2πk for integers k. we get r = 2 sin π = 2 1 = 1. it must have a representation P (r. one in Quadrant I and one in Quadrant IV. 5π .4 that the pole may be represented as (0. however.but this one is centered at the point 3 with rectangular coordinates 3 . As before. and more generally. so this is 6 6 our answer here. θ ) which satisfies r = 2 sin(θ ). There is no integer value of k 2 2 for which πk = π + 2πk which means while the origin is on both graphs. If P is also on the graph of r = 2−2 sin(θ). π . Equating these two expressions for r. we get cos(θ) = 2 . 0 and has a radius of 2 . we need the arccosine function. Assuming such a pair (r. we get θ = π + 2πk or θ = 5π + 2πk 6 6 π for integers k. θ) exists. θ) which satisfies r = 2 sin(θ). the point is never 2 reached simultaneously.950 Applications of Trigonometry order for a point P to be on the graph of r = 2 sin(θ). we reach the origin when θ = π . The graph of r = 2 is a circle. Proceeding as above. we start at the origin when θ = 0 and return to it at θ = π. 5π and the origin. Hence. with a radius of 2. centered at the origin. 2 y 2 −2 2 3 x −2 r = 2 and r = 3 cos(θ) We have two intersection points to find. Both equations give us the point 1. we make a quick sketch of r = 2 and r = 3 cos(θ) to get feel for the number and location of the intersection points. We first try to see if we can find any points which have a single representation P (r. which 6 2 is also the value we obtain when we substitute it into r = 2 − 2 sin(θ). we get any more solutions for θ. From these solutions. At this point. Since cos(2(θ + 2πk)) = cos(2θ + 4πk) = cos(2θ) for every integer k. we obtain just four distinct points represented by 3. this case would reduce to the previous case instantly. we get 3 2 2. then either r = r or r = −r . Solving.5. θ) with a representation that satisfies both r = 3 and r = 6 cos(2θ). θ ) represent the same point and r = 0.13 f we replace every occurrence of θ in the equation r = 6 cos(2θ) with (θ+2πk) and then see if. The reader is encouraged to check these results algebraically and geometrically. we have to consider how the representations of the points of intersection can differ. we first graph r = 3 and r = 6 cos(2θ) to get an idea of how many intersection points to expect and where they lie. however. we have that θ = θ + (2k + 1)π for integers k. The graph of r = 3 is a circle centered at the origin with a radius of 3 and the graph of r = 6 cos(2θ) is another four-leafed rose. by equating the resulting expressions for r. π . See Example 11. the equation r = 6 cos(2(θ + 2πk)) reduces to the same equation we had before. Since there is no θ in r = 3. 13 12 . θ) and (r . and 2. Moving on to the case where r = −r . We look to see if we can find points P which have a representation (r. 5π . For these points.4 that if (r. so one possibility is that an intersection point P has a representation (r. 3. We could have just as easily chosen to do this substitution in the equation r = 3. Out of all of these solutions.2 number 3. 6 cos(2θ) = 3 or cos(2θ) = 1 . r = 6 cos(2θ).11. 2π − arccos 2 3 as one representation for our answer in Quadrant IV. we get θ = π + πk or θ = 5π + πk 2 6 6 for integers k. 3.two in each quadrant. We know from Section 11. θ+2πk) for some integer. The authors have chosen to replace θ with θ + 2πk in the equation r = 6 cos(2θ) for illustration purposes only. Proceeding as above. however.12 y 6 3 −6 −3 −3 3 6 x −6 r = 3 and r = 6 cos(2θ) It appears as if there are eight points of intersection . then θ = θ+2πk. θ) that satisfies r = 3 and another. which means we get no additional solutions. 3. k which satisfies r = 6 cos(2θ). 7π and 3. If r = r . To determine the coordinates of the remaining four 6 6 6 6 points. arccos 3 as one representation for our answer in Quadrant I. We first look to see if there any points P (r.5 Graphs of Polar Equations 951 2 θ = arccos 2 + 2πk or θ = 2π − arccos 3 + 2πk for integers k. The reader is encouraged to follow this latter procedure in the interests of efficiency. 11π . θ) which satisfies r = 3 and another representation (r. however. 3 16 A quick sketch of r = 3 sin θ and r = 3 cos θ in the θr-plane will convince you that. then for our equation θ to hold. 3 3 From these solutions. Since cos(2(θ + (2k + 1)π)) = cos(2θ + (2k + 1)(2π)) = cos(2θ) for all integers k. we find that we need to plot both functions as θ ranges from 0 to 4π to obtain the complete graph.16 we need to show that. 5π . we could ‘plug’ these values for θ into r = 3 (where there is no θ) and get the list of points: 3. 15 14 . Coupling this 1 equation with r = 3 gives −6 cos(2θ) = 3 or cos(2θ) = − 2 . θ) which θ θ satisfies both r = 3 sin 2 and r = 3 cos 2 . that satisfies r = 6 cos(2θ). once again. 5π . for illustration purposes. From these solutions. While normally we discourage dividing by a variable θ expression (in case it could be 0). To our surprise and/or delight. the equation −r = 6 cos(2(θ + (2k + 1)π)) reduces to −r = 6 cos(2θ). Since no angles have both cosine and sine equal to zero. −3. we obtain15 the remaining four intersection points with representations −3. in the interests of efficiency. 3 sin 2 = 0 as well. viewed as functions of r. the graphs of these two equations intersect at all points on the plane. we could have easily chosen to substitute these into r = 3 which would give −r = 3. 4π and −3. or r = −3. we begin by graphing r = 3 sin 2 and r = 3 cos 2 . we note here that if 3 cos 2 = 0. π . To do this. it appears as if these two equations describe the same curve! y 3 −3 3 x −3 θ r = 3 sin 2 and r = 3 cos θ 2 appear to determine the same curve in the xy-plane To verify this incredible claim. We get θ = π + πk or θ = 2π + πk. which we can readily check graphically. 2 2 these are two different animals. As usual.5.2. we 2 Again. We obtain these representations by substituting the values for θ into r = 6 cos(2θ). π represents the same point 3 3 3 3 3 as −3. 3. 3 3 3 3 θ θ 4. 3. −3. Using the techniques presented in Example 11. in fact. we substitute14 (−r) for r and (θ + (2k + 1)π) for θ in the equation r = 6 cos(2θ) and get −r = 6 cos(2(θ + (2k + 1)π)). While it is not true that 3. we still get the same set of solutions. Again. π . 2π . π . 4π and 3. θ θ θ we are safe to divide both sides of the equation 3 sin 2 = 3 cos 2 by 3 cos 2 to get θ tan 2 = 1 which gives θ = π + 2πk for integers k. Suppose P has a representation (r. θ + (2k + 1)π). Equating these two expressions for r gives θ θ the equation 3 sin 2 = 3 cos 2 . 2π .952 Applications of Trigonometry (−r. or r = −6 cos(2θ). We now investigate 2 other representations for the intersection points. Suppose P is an intersection point with θ a representation (r. If we choose k = 0.11. . since sin(πk) = 0 for all integers k. θ) which satisfy both equations. we expand 3 cos 2 + πk = 3 cos 2 cos(πk) − 3 sin 2 sin (πk) = ±3 cos 2 . which gives the intersection point 2 √ 3 2 π 2 . θ) which satisfies r = 3 sin 2 and a representation θ (−r. π . we get r = −3 cos 2 . θ θ we get the same equation r = 3 cos 2 as before. θ + π) for P automatically satisfies r = 3 cos 2 . If k is an even integer.5 Graphs of Polar Equations √ 953 get only one intersection point which can be represented by 3 2 2 . θ + (2k + 1)π) which satisfies r = 3 cos 2 for some integer k. This θ θ θ latter expression for r leads to the equation 3 sin 2 = −3 cos 2 . • Substitute (−r) for r and (θ + (2k + 1)π) for θ in either one of E1 or E2 (but not both) and solve for pairs (r. θ we assume P has a representation (r. θ) for the point P satisfies r = 3 sin( 2 ). Our work in Example 11.−2 . or r = 3 sin 2 which is the other equation under consideration! θ What this means is that if a polar representation (r. • Solve for pairs (r. Substituting (−r) for θ 1 r and (θ + (2k + 1)π) in for θ into r = 3 cos 2 gives −r = 3 cos 2 [θ + (2k + 1)π] . and cos (πk) = ±1 for all integers k. Using the sum formula for 2 θ θ θ θ cosine. Solving. = 3 cos 1 [θ + (2k + 1)π] in this case 2 θ θ reduces to −r = −3 sin 2 . Keep in mind that k is an integer. θ + 2πk) for some integer k which satisfies r = 3 cos 2 . θ) which satisfy both E1 and E2 . Next. we use the sum formula for cosine to get 1 2 θ 2 θ 2 (2k+1)π 2 cos (2k+1)π 2 θ 2 cos [θ + (2k + 1)π] = cos = cos + − sin θ 2 sin (2k+1)π 2 = ± sin where the last equality is true since cos Hence. Once again. Guidelines for Finding Points of Intersection of Graphs of Polar Equations To find the points of intersection of the graphs of two polar equations E1 and E2 : • Sketch the graphs of E1 and E2 . θ) which satisfies r = 3 sin 2 and the same point P has a different θ representation (r. and the equation −r (2k+1)π = ±1 for integers k. or tan 2 = −1. θ) which satisfy both equations. −r = 3 cos then sin (2k+1)π 2 1 2 (2k+1)π 2 = 0 and sin [θ + (2k + 1)π] can be rewritten π 2 = sin = 1.3 justifies the following. Check to see if the curves intersect at the origin (pole). Hence the θ θ equations r = 3 sin( 2 ) and r = 3 cos( 2 ) determine the same set of points in the plane. θ then the representation (−r. 2 θ as r = ±3 sin 2 .5. • Substitute (θ + 2πk) for θ in either one of E1 or E2 (but not both) and solve for pairs (r. Keep in mind that k is an integer. we get r = 3 cos 1 [θ + 2πk] = 3 cos 2 + πk . we get θ = − π + 2πk for integers k. If k is odd. Substituting θ into the latter. 0 ≤ θ ≤ (r. In other words. 4.4. Sketch the region in the xy-plane described by the following sets. θ) | 3 ≤ r ≤ 6 cos(2θ). Our first step in these problems is to sketch the graphs of the polar equations involved to get a sense of the geometric situation. We know from Example 11. Moreover. Hence. 0 ≤ θ ≤ π 6 . 0 ≤ θ ≤ ∪ (r. θ) | 0 ≤ r ≤ 5 sin(2θ). y θ= π 6 y x x r = 3 and r = 6 cos(2θ) (r.5. the region we seek is the leaf itself. 0 ≤ θ ≤ (r.954 Applications of Trigonometry Our last example ties together graphing and points of intersection to describe regions in the plane. 2. θ) | 3 ≤ r ≤ 6 cos(2θ). θ) | 0 ≤ r ≤ 5 sin(2θ). 2π 3 π 2 π 6 4π 3 π 6 ≤θ≤ π 6 (r. Since all of the equations in this example are found in either Example 11. 2 r 5 y π 4 π 2 3π 4 π θ x −5 (r.5. Example 11. θ) | 0 ≤ r ≤ 2 sin(θ). (r.3. we are 6 looking at the points outside or on the circle (since r ≥ 3) but inside or on the rose (since r ≤ 6 cos(2θ)).5. ≤θ≤ π 2 Solution. 1. so 6 the region that is being described here is the set of points whose directed distance r from the origin is at least 3 but no more than 6 cos(2θ) as θ runs from 0 to π . θ) | 0 ≤ r ≤ 2 − 2 sin(θ). We know from Example 11. We shade the region below. we are tracing out the ‘leaf’ of the rose 2 which lies in the first quadrant. θ) | 2 + 4 cos(θ) ≤ r ≤ 0. 0 ≤ θ ≤ π 2 2.3 number 3 that r = 3 and r = 6 cos(2θ) intersect at θ = π . 3. we know from our work there that as 0 ≤ θ ≤ π .5. most of the work is done for us. 1.2 number 3 that the graph of r = 5 sin(2θ) is a rose.2 or Example 11. The inequality 0 ≤ r ≤ 5 sin(2θ) means we want to capture all of the points between the origin (r = 0) and the curve r = 5 sin(2θ) as θ runs through 0.5. π . 4π . θ) | 0 ≤ r ≤ 2 − 2 sin(θ).5. we are shading the region between the origin 6 (r = 0) out to the circle (r = 2 sin(θ)) as θ ranges from 0 to π . θ picks up 6 2 where it left off at π and continues to π . we are shading from the origin 6 2 (r = 0) out to the cardioid r = 2 − 2 sin(θ) which pulls into the origin at θ = π .5 Graphs of Polar Equations 955 3.5.3. (r.11. for the 6 first region. we know that the graph of r = 2 + 4 cos(θ) is a lima¸on c whose ‘inner loop’ is traced out as θ runs through the given values 2π to 4π . Hence. In this case. π ≤ θ ≤ π 6 2 . From Example 11. y 1 1 y θ= π 6 1 x 1 x r = 2 − 2 sin(θ) and r = 2 sin(θ) (r. 0 ≤ θ ≤ π . θ) | 0 ≤ r ≤ 2 sin(θ). θ) | 0 ≤ r ≤ 2 sin(θ). we found that the curves r = 2 sin(θ) and r = 2 − 2 sin(θ) intersect when θ = π . and we are looking for all of the points between the pole r = 0 and the lima¸on as θ ranges c over the interval 2π . In Example 11. however. In other words.’ We shade each in turn and find our final answer by combining the two. θ) | 2 + 4 cos(θ) ≤ r ≤ 0. 0 ≤ θ ≤ π ∪ 6 (r. θ) | 0 ≤ r ≤ 2 − 2 sin(θ). the inequality 2 + 4 cos(θ) ≤ r ≤ 0 makes sense. π ≤ θ ≤ π .2 number 2. number 1. For the second region. c 3 3 r 6 y 4 θ= 2π 3 2 x 2π 3 π 2 4π 3 π 3π 2 2π θ θ= 4π 3 −2 (r. which is the angle of intersection 6 of the two curves. (r. Since the values 3 3 r takes on in this interval are non-positive. Putting 2 these two regions together gives us our final answer. we shade in the inner loop of the lima¸on. 2π ≤ θ ≤ 3 4π 3 4. We have two regions described here connected with the union symbol ‘∪. Carefully label your graphs. r = 1 + sin(θ) and r = 1 − cos(θ) 24. 1. find the exact polar coordinates of the points of intersection of graphs of the polar equations. 0 ≤ θ ≤ π} 34. θ) | 0 ≤ r ≤ 3 cos(θ). 0 ≤ θ ≤ π 2 . (r. θ) | 0 ≤ r ≤ 3. r = 3 cos(θ) and r = 1 + cos(θ) 23. Rose: r = 2 sin(2θ) 5. sketch the region in the xy-plane described by the given set.5. Rose: r = 5 sin(3θ) 7. θ) | 0 ≤ r ≤ 4 sin(θ).40. Lima¸on: r = 1 − 2 cos(θ) c √ 15. Remember to check for intersection at the pole (origin). plot the graph of the polar equation by hand. Lima¸on: r = 3 − 5 cos(θ) c 18.20. r2 = 2 sin(2θ) and r = 1 30. 21. Cardioid: r = 2 + 2 cos(θ) 13. Cardioid: r = 1 − sin(θ) 14. r = 1 − 2 sin(θ) and r = 2 √ 25. Lemniscate: r2 = sin(2θ) 2. θ) | 0 ≤ r ≤ 2 sin(2θ). (r. r = 4 cos(2θ) and r = 2 In Exercises 31 . Rose: r = sin(4θ) 9. Lima¸on: r = 3 − 5 sin(θ) c 19.956 Applications of Trigonometry 11. Rose: r = 4 cos(2θ) 6. Rose: r = cos(5θ) 8. Lima¸on: r = 2 + 7 sin(θ) c 20. 0 ≤ θ ≤ 2π} 33. r2 = 4 cos(2θ) and r = √ 2 22. Cardioid: r = 5 + 5 sin(θ) 12.1 Exercises In Exercises 1 . r = 2 sin(2θ) and r = 1 29. r = 2 cos(θ) and r = 2 3 sin(θ) 27. {(r. Cardioid: r = 3 − 3 cos(θ) 11. Circle: r = 6 sin(θ) 3. Circle: r = 2 cos(θ) 4. r = 1 − 2 cos(θ) and r = 1 26. {(r. − π ≤ θ ≤ 2 π 2 32. Rose: r = 3 cos(4θ) 10. Lima¸on: r = 2 3 + 4 cos(θ) c 17. Lemniscate: r2 = 4 cos(2θ) In Exercises 21 . Lima¸on: r = 1 − 2 sin(θ) c 16. r = 3 cos(θ) and r = sin(θ) 28.30. 31. θ) | 1 ≤ r ≤ 2 sin(2θ). θ) | 1 + cos(θ) ≤ r ≤ 3 cos(θ). 40.11. ∪ (r. 0 ≤ θ ≤ (r. θ) | 0 ≤ r ≤ 4 cos(2θ). π 2 ≤θ≤ 3π 2 (r. The inside of the petal of the rose r = 3 cos(4θ) which lies on the positive x-axis 48. which stands for “polar”. The region in Quadrant I which lies inside both the circle r = 3 as well as the rose r = 6 sin(2θ) While the authors truly believe that graphing polar curves by hand is fundamental to your understanding of the polar coordinate system. The first thing you must do is switch the MODE of your calculator to POL. We now give a brief demonstration of how to use the graphing calculator to plot polar curves. Science and Engineering. The region inside the circle r = 4 cos(θ) which lies in Quadrant IV. − π ≤ θ ≤ 4 π 4 π 3 957 36. The region which lies inside of the circle r = 3 cos(θ) but outside of the circle r = sin(θ) 50. θ) | 0 ≤ r ≤ 2 sin(2θ). we would be derelict in our duties if we totally ignored the graphing calculator. The region inside the circle r = 5. 41. θ) | 0 ≤ r ≤ 1. 44. θ) | 0 ≤ r ≤ 2 cos(θ). 0 ≤ θ ≤ π 12 ∪ (r. θ) | 1 ≤ r ≤ 1 − 2 cos(θ). The region inside the left half of the circle r = 6 sin(θ). The region inside the cardioid r = 2 − 2 sin(θ) which lies in Quadrants I and IV. The region inside the circle r = 5 but outside the circle r = 3. there are some important polar curves which are simply too difficult to graph by hand and that makes the calculator an important tool for your further studies in Mathematics. − π ≤ θ ≤ 3 (r. 42. use set-builder notation to describe the polar region. π 12 π 6 ≤θ≤ π 4 π 2 ≤θ≤ In Exercises 41 . 38. 43. Indeed. (r. The region inside the circle r = 5 which lies in Quadrant III. Assume that the region contains its bounding curves. θ) | 0 ≤ r ≤ 2 3 sin(θ). 47. 39.50. (r. 49. . 45. 37. The region inside the top half of the cardioid r = 3 − 3 cos(θ) 46.5 Graphs of Polar Equations 35. 13π 12 ≤θ≤ π 6 17π 12 √ (r. The issue here is that the calculator screen is 96 pixels wide but only 64 pixels tall. however. a θstep of 0. so we enter the data you see above. but we also what values of θ to use. To get a true geometric perspective.958 Applications of Trigonometry This changes the “Y=” menu as seen above in the middle. θmax] using the θstep. Xmax] into 96 equal subintervals. In polar mode. Let’s plot the polar rose given by r = 3 cos(4θ) from Exercise 8 above. We type the function into the “r=” menu as seen above on the right. In function mode. You will need to experiment with the settings in order to get a nice graph. we find three extra lines. From our previous work. It you make it too big.1 is fine. If you make it too small then the calculator takes a long time to graph. For most graphs. we need to hit ZOOM SQUARE (seen below in the middle) to produce a more accurate graph which we present below on the right. you get chunky garbage like this. Notice that some of them have explicit bounds on θ and others do not. We need to set the viewing window so that the curve displays properly. we need to tell it not only the dimensions which x and y will assume. we know that we need 0 ≤ θ ≤ 2π. the calculator automatically divided the interval [Xmin.60 give you some curves to graph using your calculator.) Hitting GRAPH yields the curve below on the left which doesn’t look quite right. In order for the calculator to be able to plot r = 3 cos(4θ) in the xy-plane. Exercises 51 . but when we look at the WINDOW menu. (I’ll say more about the θ-step in just a moment. . we must specify how to split up the interval [θmin. r = θ. (a) Show that f (θ) = 2 + 4 cos(θ) is even and verify that the graph of r = 2 + 4 cos(θ) is indeed symmetric about the x-axis.5. r = sin(5θ) − 3 cos(θ) 57. (See Example 11. r = 60. we have you and your classmates explore some of the more basic forms of symmetry seen in common polar curves.1θ .5. r = ln(θ). r = θ3 − θ. (See Example 11.) θ (b) Show that f (θ) = 3 cos 2 is not odd. (See Example 11.64. (See Example 11.) θ 2 is symmetric about 63. classifying symmetry for polar curves is not as straight-forward as it was for equations back on page 26.11. make a conjecture as to how many petals the polar rose r = sin(nθ) has for any natural number n. Replace sine with cosine and repeat the investigation. r = e. 1 ≤ θ ≤ 12π 54.3 number 4. Show that if f is odd18 then the graph of r = f (θ) is symmetric about the origin. However. How many petals does the polar rose r = sin(2θ) have? What about r = sin(3θ). r = sin3 58. show that f (π − θ) = f (θ) and the graph of r = 4 − 2 sin(θ) is symmetric about the y-axis.2 56.5 Graphs of Polar Equations 51. it’s clear that many polar curves enjoy various forms of symmetry. as required. Show that if f is even17 then the graph of r = f (θ) is symmetric about the x-axis. r = 1 2 − cos(θ) 52. r = arctan(θ). −1.2 number 2. 0 ≤ θ ≤ 12π 53.3 number 4. 0 ≤ θ ≤ 12π 55. r = θ 2 959 + cos2 θ 3 1 1 − cos(θ) 1 2 − 3 cos(θ) 61. How many petals does r = cos(nθ) have for each natural number n? Looking back through the graphs in the section. In Exercises 62 . yet the graph of r = 3 cos the origin. (a) Show that f (θ) = 5 sin(2θ) is odd and verify that the graph of r = 5 sin(2θ) is indeed symmetric about the origin. (See Example 11.) 17 18 Recall that this means f (−θ) = f (θ) for θ in the domain of f .2 number 1. 62.2 number 3.5.5.2 ≤ θ ≤ 1. yet the graph of r = 3 sin the x-axis. (a) For f (θ) = 4 − 2 sin(θ).) θ 2 is symmetric about 64.) θ (b) Show that f (θ) = 3 sin 2 is not even.5. Show that if f (π − θ) = f (θ) for all θ in the domain of f then the graph of r = f (θ) is symmetric about the y-axis. −π ≤ θ ≤ π 59. . r = sin(4θ) and r = sin(5θ)? With the help of your classmates. Recall that this means f (−θ) = −f (θ) for θ in the domain of f . 5. let f (θ) = cos(θ) and g(θ) = 2 − sin(θ). Some of these curves have polar representations which we invite you and your classmates to research. in the paragraph before Exercise 53. show that f π − π = f π . Repeat this process for g(θ). r = 2 f (θ). compare the graph of r = f (θ) to each of the graphs of r = f θ + π . r = f θ − π and r = f θ − 3π . how would the graph of r = f (−θ) compare with the graph of r = f (θ) for a generic function f ? What about the graphs of r = −f (θ) and r = f (θ)? What about r = f (θ) and r = f (π − θ)? Test out your conjectures using a variety of polar functions found in this section with the help of a graphing utility. In general.960 Applications of Trigonometry (b) For f (θ) = 5 sin(2θ). Repeat this process 4 4 4 4 for g(θ). (See Example 11. With the help of your classmates. yet the graph of r = 5 sin(2θ) is 4 4 symmetric about the y-axis.64. (a) Using your graphing calculator. r = f θ + 3π . r = −f (θ) and r = −3f (θ). how do you think the graph of r = f (θ + α) compares with the graph of r = f (θ)? (b) Using your graphing calculator. Back in Section 1.) In Section 1. we gave you this link to a fascinating list of curves. compare the graph of r = f (θ) to each of the graphs of 1 r = 2f (θ). . how do you think the graph of r = k · f (θ) compares with the graph of r = f (θ)? (Does it matter if k > 0 or k < 0?) 66. 67.2 number 3. 65. In light of Exercises 62 . In Exercise 65 we have you and your classmates explore transformations of polar graphs.7. we discussed transformations of graphs. In general.2. research cardioid microphones. For Exercises 65a and 65b below. 68. 2 Answers 2. Rose: r = cos(5θ) y θ= π 3 5 1 θ= 7π 10 θ= 3π 10 θ= −5 5 9π 10 θ= π 10 x −1 1 x −5 −1 .5 Graphs of Polar Equations 961 11.5. Rose: r = 5 sin(3θ) y θ= 2π 3 6. Circle: r = 6 sin(θ) y 6 −6 6 x −2 2 x −6 −2 3.11. Circle: r = 2 cos(θ) y 2 1. Rose: r = 2 sin(2θ) y 2 4. Rose: r = 4 cos(2θ) y 4 θ= 3π 4 θ= π 4 −2 2 x −4 4 x −2 −4 5. Cardioid: r = 5 + 5 sin(θ) y 10 3 5 −6 −3 −3 3 6 x −10 −5 −5 5 10 x −6 −10 11. Cardioid: r = 3 − 3 cos(θ) y 6 10.962 7. Cardioid: r = 2 + 2 cos(θ) y 4 12. Rose: r = sin(4θ) y 1 θ= 3π 4 Applications of Trigonometry 8. Cardioid: r = 1 − sin(θ) y 2 2 1 −4 −2 −2 2 4 x −2 −1 −1 1 2 x −4 −2 . Rose: r = 3 cos(4θ) y θ= θ= π 4 5π 8 3 θ= 3π 8 θ= 7π 8 θ= π 8 −1 1 x −3 3 x −1 −3 9. Lima¸on: r = 1 − 2 sin(θ) c y 3 π 6 θ= 1 −3 −1 −1 1 3 5π 6 θ= 1 x −3 −1 −1 1 3 x θ= −3 5π 3 −3 √ 15. Lima¸on: r = 2 + 7 sin(θ) c y 9 3 5 θ = arcsin 5 −8 −3 −2 3 8 x −9 θ = π + arcsin −2 2 7 2 9 2 7 x θ = 2π − arcsin −8 −9 .11. Lima¸on: r = 2 3 + 4 cos(θ) c y √ 2 3+4 5π 6 16. Lima¸on: r = 3 − 5 sin(θ) c y 8 θ = π − arcsin 3 5 18. Lima¸on: r = 1 − 2 cos(θ) c y 3 θ= π 3 963 14. Lima¸on: r = 3 − 5 cos(θ) c y 8 θ = arccos 3 5 θ= √ 2 3 3 x √ −2 3 − 4 θ= 7π 6 √ 2 3+4 √ −2 3 −8 −2 −3 8 x √ −2 3 − 4 θ = 2π − arccos 3 5 −8 17.5 Graphs of Polar Equations 13. . . 2 4 √ 2 − 2 7π . .964 19. r = 3 cos(θ) and r = 1 + cos(θ) y 3 2 1 −3 −2 −1 −1 −2 −3 1 2 3 3 π . pole 2 4 −2 −1 −1 −2 1 2 x . Lemniscate: r2 = sin(2θ) y 1 Applications of Trigonometry 20. r = 1 + sin(θ) and r = 1 − cos(θ) y 2 1 √ 2 + 2 3π . pole 2 3 x 22. . 2 3 3 5π . Lemniscate: r2 = 4 cos(2θ) y 2 θ= 3π 4 θ= π 4 −1 1 x −2 2 x −1 −2 21. 1 −3 −1 −1 1 3 x −3 24.11. 0) 2 1 −3 −1 −1 1 3 x −3 √ 25. r = 1 − 2 sin(θ) and r = 2 y 3 965 7π .5 Graphs of Polar Equations 23. pole 6 x . π . 2 1. 6 11π 6 2. (−1. 3π . r = 2 cos(θ) and r = 2 3 sin(θ) y 4 3 2 1 −3 −2 −1 −1 −2 −3 −4 1 2 3 √ 3. π . 2. r = 1 − 2 cos(θ) and r = 1 y 3 1. 6 √ 2. 12 1. r = 3 cos(θ) and r = sin(θ) y 3 2 1 −3 −2 −1 −1 −2 −3 1 2 3 Applications of Trigonometry √ 3 10 . 6 √ 2. 7π . 5π . π . r2 = 4 cos(2θ) and r = y 2 √ 2 √ 2. 12 1. π .966 26. 17π 12 1 √ − 2 −1 1 √ 2 x −1 √ − 2 . 11π 6 −2 2 x −2 28. pole 10 x 27. arctan(3) . 13π . 5π . r2 = 2 sin(2θ) and r = 1 y √ 2 1. 6 √ 2. 12 1. −4 4 x −4 30. 5π . −2 2 19π . 1. 11π . 2. −2. 2.5 Graphs of Polar Equations 29.11. π 11π . r = 2 sin(2θ) and r = 1 y 2 1. 12 −1. 6 3 4π . 12 −1. 12 1. 6 5π . 6 −2. 6 7π . x −2 . 12 7π . 2π . 12 23π 12 −1. 12 17π . 5π 3 −2. 3 −2. 1. 3 2. . 13π . 12 −1. π . r = 4 cos(2θ) and r = 2 y 4 967 π . 2. 0 ≤ θ ≤ y 2 π 2 x −2 2 x −2 35. θ) | 0 ≤ r ≤ 3 cos(θ). {(r. − π ≤ θ ≤ 2 y 3 2 1 −3 −2 −1 −1 −2 −3 1 2 3 π 2 34. (r. θ) | 0 ≤ r ≤ 2 sin(2θ). (r. {(r. (r. y 3 π 2 ≤θ≤ 3π 2 1 −4 4 x −3 −1 −1 1 3 x −4 −3 . θ) | 0 ≤ r ≤ 4 cos(2θ). 0 ≤ θ ≤ 2π} y 3 2 1 −3 −2 −1 −1 −2 −3 1 2 3 Applications of Trigonometry 32. 0 ≤ θ ≤ π} y 4 3 2 1 x −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 33. θ) | 1 ≤ r ≤ 1 − 2 cos(θ). (r. θ) | 0 ≤ r ≤ 3. θ) | 0 ≤ r ≤ 4 sin(θ).968 31. − π ≤ θ ≤ 4 y 4 π 4 36. θ) | 0 ≤ r ≤ 2 3 sin(θ). √ (r. (r. π 6 ≤θ≤ π 2 x .5 Graphs of Polar Equations 37. 0 ≤ θ ≤ y 4 3 2 1 −3 −2 −1 −1 −2 −3 −4 1 2 3 π 6 ∪ (r. − π ≤ θ ≤ 3 y 3 2 1 −3 −2 −1 −1 −2 −3 1 2 3 π 3 969 x 38.11. (r. 13π 12 ≤θ≤ 17π 12 1 √ − 2 −1 1 √ 2 x −1 √ − 2 39. θ) | 1 ≤ r ≤ y √ 2 2 sin(2θ). θ) | 1 + cos(θ) ≤ r ≤ 3 cos(θ). θ) | 0 ≤ r ≤ 2 cos(θ). (r. θ) | 0 ≤ r ≤ 2 − 2 sin(θ). {(r. 0 ≤ θ ≤ y 2 π 12 Applications of Trigonometry ∪ (r. θ) | 0 ≤ r ≤ 3 cos(4θ). θ) | 0 ≤ r ≤ 2 sin(2θ). 47. 0 ≤ θ ≤ 2π} 49. 5π ≤ θ ≤ π 12 2 π 12 ≤θ≤ 5π 12 ∪ . 0 ≤ θ ≤ 12 ∪ (r. θ) | 0 ≤ r ≤ 2 − 2 sin(θ). θ) | 0 ≤ r ≤ 3 cos(θ). θ) | 0 ≤ r ≤ 3 cos(4θ). 0 ≤ θ ≤ or (r. (r. θ) | 0 ≤ r ≤ 3 − 3 cos(θ). π 2 π 2 3π 2 ≤θ≤π ≤θ≤π 45. (r. θ) | 0 ≤ r ≤ 3 cos(4θ). θ) | 0 ≤ r ≤ 3. θ) | 3 ≤ r ≤ 5. π 8 ≤ θ ≤ 2π or (r. 3π 2 π 2 ∪ (r. 0 ≤ θ ≤ π} 46. θ) | 0 ≤ r ≤ 5.970 40. 0 ≤ θ ≤ 2π} 42. {(r. (r. 43. − π ≤ θ ≤ 0 ∪ {(r. 5π 2 15π 8 3π 2 ≤ θ ≤ 2π ≤θ≤ π 8 (r. θ) | 0 ≤ r ≤ 6 sin(2θ). θ) | 0 ≤ r ≤ 5. θ) | 0 ≤ r ≤ 6 sin(θ). θ) | sin(θ) ≤ r ≤ 3 cos(θ). (r. {(r. π 12 ≤θ≤ π 4 −2 2 x −2 41. 44. 50. θ) | 0 ≤ r ≤ 2 − 2 sin(θ). − π ≤ θ ≤ 8 48. θ) | 0 ≤ r ≤ 1. 0 ≤ θ ≤ arctan(3)} 2 π (r. (r. θ) | 4 cos(θ) ≤ r ≤ 0. θ) | 0 ≤ r ≤ 6 sin(2θ). 0 ≤ θ ≤ ∪ (r. π ≤ θ ≤ (r. That is. Suppose we wished to find rectangular coordinates based on the x . y ) into polar coordinates.and y-axes. we saw that the graph of y = x is actually a hyperbola. we wish to determine P (x . y ). Consider the angle φ whose initial side is the positive x -axis and whose terminal side contains the point P .3 in Section 8. y ) by converting them to polar coordinates. More specifically. Using the sum formulas for sine and cosine. it is the hyperbola obtained by rotating the graph of x2 − y 2 = 4 counter-clockwise through a 45◦ angle.1 Rotation of Axes Consider the x. choose the same r > 0 (since the origin is the same in both systems) and get x = r cos(φ) and y = r sin(φ). we have x = r cos(θ + φ) = r cos(θ) cos(φ) − r sin(θ) sin(φ) = (r cos(φ)) cos(θ) − (r sin(φ)) sin(θ) = x cos(θ) − y sin(θ) Since x = r cos(φ) and y = r sin(φ) Sum formula for cosine .6 Hooked on Conics Again In this section. y) to polar coordinates with r > 0 yields x = r cos(θ + φ) and y = r sin(θ + φ). The coordinates (x.11. 11.6 Hooked on Conics Again 971 11. we revisit our friends the Conic Sections which we began studying in Chapter 7. y y P (x. we can generalize the process of rotating axes as shown below. y) are rectangular coordinates and are based on the x. y) and P (x .3. To convert the point P (x .3.and y -axes.and y -axes obtained by rotating the xand y-axes counter-clockwise through an angle θ and consider the point P (x. Converting P (x. Armed with polar coordinates. we first match the polar axis with the positive x -axis.6. y) = P (x . Our first task is to formalize the notion of rotating axes so this subsection is actually a follow-up 2 to Example 8. In that example. y ) θ φ θ x x We relate P (x.and y-axes below along with the dashed x . it is nearly trivial if we use polar coordinates. While this seems like a formidable challenge. y). The matrix A introduced here is revisited in the Exercises. we find A−1 = so that X x y x y = A−1 X = = cos(θ) sin(θ) − sin(θ) cos(θ) x cos(θ) + y sin(θ) −x sin(θ) + y cos(θ) x y cos(θ) sin(θ) − sin(θ) cos(θ) From which we get x = x cos(θ) + y sin(θ) and y = −x sin(θ) + y cos(θ). Sound familiar? In Section 11. we need to solve the system x cos(θ) − y sin(θ) = x x sin(θ) + y cos(θ) = y for x and y .9. 1 . Using the formula given in Equation 8. These equations enable us to easily convert points with x y -coordinates back into xy-coordinates. y ) are related by the following systems of equations x = x cos(θ) − y sin(θ) y = x sin(θ) + y cos(θ) and x y = x cos(θ) + y sin(θ) = −x sin(θ) + y cos(θ) We put the formulas in Theorem 11. Rotation of Axes: Suppose the positive x and y axes are rotated counterclockwise through an angle θ to produce the axes x and y . the determinant of A is not zero so A is invertible and X = A−1 X. Then the coordinates P (x. using the sum formula for sine we get y = x sin(θ) + y cos(θ). we write the above system as the matrix equation AX = X where A= cos(θ) − sin(θ) sin(θ) cos(θ) . Using the machinery developed in Section 8. of course. They also enable us to easily convert equations in the variables x and y into equations in the variables in terms of x and y .2 with det(A) = 1.4. and they make it easy to convert equations from rectangular coordinates into polar coordinates. interchange the roles of x and x . 2 We could.4. Theorem 11. y and y and replace φ with −φ to get x and y in terms of x and y.972 Applications of Trigonometry Similarly. the equations x = r cos(θ) and y = r sin(θ) make it easy to convert points from polar coordinates into rectangular coordinates. but that seems like cheating.1 If we want equations which enable us to convert points with xy-coordinates into x y -coordinates. respectively. X= x y Since det(A) = (cos(θ))(cos(θ)) − (− sin(θ))(sin(θ)) = cos2 (θ) + sin2 (θ) = 1. Perhaps the cleanest way2 to solve this system is to write it as a matrix equation. y) and P (x .9 to good use in the following example. X = x y . To summarize. −2 − 3 back into x and y coordinates. To check our answer graphically.6.9 to convert P (x . y ). we obtain y = −4 as required. We get x = x cos(θ) − y sin(θ) √ √ = (1 − 2 3) cos π − (−2 − 3) sin 3 √ √ 3 = 1 − 3 − − 3− 2 2 = 2 Similarly. To check our answer √ √ algebraically.46. we √ sketch in the x -axis and y -axis to see if the new coordinates P (x . Theorem 11.9 gives x = x cos(θ) + y sin(θ) = 2 cos π + (−4) sin π which simplifies 3 3 3 √ to x = 1 − 2 3. Using these values for x and y along with θ = π . −4) P (x . Solution. 144 and y 3 x 3 π π x π π we substitute x = x cos 3 − y sin 3 = 2 − 2 and y = x sin 3 + y cos 3 = 2 + y 2 . 1. −2 − 3 ≈ (−2. Let P (x.11.and y . y) = (2.73) seem reasonable. −4) and find P (x . −2 − 3 . y ) = 1 − 2 3. −4) then x = 2 and y = −4.46. y x y π 3 π 3 π 3 x P (x. using y = x sin(θ) + y cos(θ). y) = (2. Convert the equation 21x2 + 10xy 3 + 31y 2 = 144 to an equation in x and y and graph. Our graph is below. y ) = 1 − 2 3. y ) ≈ (−2.1. y = −x sin(θ) + y cos(θ) = (−2) sin π + (−4) cos π which 3 √ √ √ √3 gives y = − 3 − 2 = −2 − 3. 3 1.and y. −3. y ) = √ 1 − 2 3.axes. √ 2. Similarly.6 Hooked on Conics Again 973 Example 11. If P (x. respectively. Hence P (x .axes are both rotated counter-clockwise through the angle θ = π to produce the x . Suppose the x. y) = (2. Check your answer algebraically and graphically.73) √ 2. To convert the equation 21x2 +10xy 3+31y 2 = √ to an equation in the variables x √ y . we use the formulas in Theorem 11. −3. 0). We leave it to the reader to verify that x2 = (x )2 cos2 (θ) − 2x y cos(θ) sin(θ) + (y )2 sin(θ) xy = (x )2 cos(θ) sin(θ) + x y cos2 (θ) − sin2 (θ) − (y )2 cos(θ) sin(θ) y 2 = (x )2 sin2 (θ) + 2x y cos(θ) sin(θ) + (y )2 cos2 (θ) . the equation in the rotated variables x and y contains no x y term? To explore this conjecture. We will not go through the entire computation. Bxy and Cy 2 . That is. it is nothing more than a hefty dose of Beginning Algebra. y = + + 4 2 4 4 2 4 √ To our surprise and delight. 0) with vertices along the y -axis with (x y -coordinates) (0. ±3) and whose minor axis has endpoints with (x y -coordinates) (±2. we make the usual substitutions x = x cos(θ) − y sin(θ) and y = x sin(θ) + y cos(θ) into the equation Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0 and set the coefficient of the x y term equal to 0. We graph it below. It is natural to wonder if we can always do this. The latter is an ellipse centered at (0. While this is by no means a trivial task. is there an angle θ so that if we rotate the x and yaxes counter-clockwise through that angle θ. the equation 21x2 + 10xy 3 + 31y 2 = 144 in xy-coordinates 2 2 reduces to 36(x )2 + 16(y )2 = 144. given an equation of the form Ax2 +Bxy +Cy 2 +Dx+Ey +F = 0. Terms containing x y in this expression will come from the first three terms of the equation: Ax2 . or (x4) + (y9) = 1 in x y -coordinates.1 number 2 allowed us to graph the equation by hand using what we learned in Chapter 7.974 Applications of Trigonometry and simplify.6. with B = 0. y π 3 x y π 3 x √ 21x2 + 10xy 3 + 31y 2 = 144 √ The elimination of the troublesome ‘xy’ term from the equation 21x2 + 10xy 3 + 31y 2 = 144 in Example 11. the reader should take the time to do it. but rather. 4 2 4 2 √ √ √ (x )2 3 x y (y )2 3 3(x )2 x y 3 (y )2 2 xy = − − . Start by verifying that √ (x )2 x y 3 3(y )2 x = − + . 10. 16x2 + 24xy + 9y 2 + 15x − 20y = 0 Solution. 2) (in x y -coordinates) and asymptotes y = ± 2 x + 2. we identify A = 5.10 to good use in the following example. Since no angle θ can have both sin(2θ) = 0 and cos(2θ) = 0. 2) opening in the x direction with vertices 3 (±2. We graph it below. 2 2 x √ 2 2 − y √ 2 2 and y = x √ 2 2 + y √ 2 2 . from Bxy it is B cos2 (θ) − sin2 (θ) .2. The equation Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0 with B = 0 can be transformed into an equation in variables x and y without any x y terms by rotating the xand y. we can safely assume3 sin(2θ) = 0. this would force cos(2θ) = 0. and since B = 0. we get B cos(2θ) = (A − C) sin(2θ). B = 26 and C = 5 so that cot(2θ) = A−C = 5−5 = 0. and our goal is to solve for θ in terms of the coefficients A. B We put Theorem 11. Graph the following equations. We get cos(2θ) = A−C . Example 11.11. The latter is the equation 9 of a hyperbola centered at the x y -coordinates (0. Since we are assuming B = 0.6. (x )2 (y )2 +xy + 2 2 √ √ Making the other substitutions. 5x2 + 26xy + 5y 2 − 16x 2 + 16y 2 − 104 = 0 2. If sin(2θ) = 0. B and C. or (x4) − (y −2) = 1. 2 2 xy = (x )2 (y )2 − . To solve for θ we would like to divide both sides of the equation by sin(2θ). What happens to the axes in this case? . y2 = 3 The reader is invited to think about the case sin(2θ) = 0 geometrically. we get that 5x2 + 26xy + 5y 2 − 16x 2 + 16y 2 − 104 = 0 2 2 reduces to 18(x )2 − 8(y )2 + 32y − 104 = 0. or cot(2θ) = A−C . and from Cy 2 it is 2C cos(θ) sin(θ). we get −2A cos(θ) sin(θ) + B cos2 (θ) − sin2 (θ) + 2C cos(θ) sin(θ) = 0 −A sin(2θ) + B cos(2θ) + C sin(2θ) = 0 Double Angle Identities From this. We have B B sin(2θ) just proved the following theorem. This means cot(2θ) = 0 which gives θ = π + π k for integers k.10.axes counter-clockwise through an angle θ which satisfies cot(2θ) = A−C . then we would have B cos(2θ) = 0. B 26 4 2 We choose θ = π so that our rotation equations are x = 4 The reader should verify that x2 = (x )2 (y )2 −xy + .6 Hooked on Conics Again 975 The contribution to the x y -term from Ax2 is −2A cos(θ) sin(θ). Theorem 11. √ √ 1. √ √ 1. we can divide both sides of this equation by B. provided of course that we have assurances that sin(2θ) = 0. Equating the x y -term to 0. Since the equation 5x2 + 26xy + 5y 2 − 16x 2 + 16y 2 − 104 = 0 is already given to us in the form required by Theorem 11. and 7 then go after cos(θ) and sin(θ). or y = (x )2 . Ultimately. we would need to use half angle 2 7 identities. (Hint: 3 − 4 = −1. whose graph is a parabola opening along the positive y -axis with vertex (0. there are infinitely many solutions to tan(θ) = 3 . While either of these values of tan(θ) satisfies the equation 4 7 cot(2θ) = 24 . 0). we can start with cot(2θ) = 24 . B = 24 and C = 9 so that 7 cot(2θ) = 24 . after the usual calculations we 7 get θ = 1 arccot 24 . 25 25 25 12(x )2 7x y 12(y )2 + − . From cot(2θ) = 24 . Factoring. Using the techniques developed in Section 10. which 7 7 gives 24 tan2 (θ) + 14 tan(θ) − 24 = 0. since this produces an acute angle. Since this isn’t one of the values of the common angles. we get 2(3 tan(θ) + 4)(4 tan(θ) − 3) = 0 which 4 gives tan(θ) = − 3 or tan(θ) = 3 . We choose the acute angle θ = arctan 3 . We adopt the second approach. 5 5 5 5 As usual. we need cos(θ) = cos arctan 4 and sin(θ) = sin arctan 4 . we will need to use inverse functions. we need to find cos(θ) and sin(θ).) 3 4 3 . If we use the arccotangent function immediately. we have 1−tan2 (θ) = 24 . As a first step. we get A = 16. The 4 4 reader is encouraged to think about why there is always at least one acute answer to cot(2θ) = A−C and what this B means geometrically in terms of what we are trying to accomplish by rotating the axes.976 Applications of Trigonometry 2. We graph this equation below. y y y x x y θ = arctan θ= π 4 3 4 x x √ √ 5x2 + 26xy + 5y 2 − 16x 2 + 16y 2 − 104 = 0 16x2 + 24xy + 9y 2 + 15x − 20y = 0 4 As usual. 25 25 25 9(x )2 24x y 16(y )2 + + 25 25 25 x2 = xy = y2 = Once the dust settles. we have 2 tan(θ) tan(2θ) = 24 .6 we get cos(θ) = 4 and sin(θ) = 3 . we choose tan(θ) = 3 . we get 25(x )2 − 25y = 0. The reader is also encouraged to keep a sharp lookout for the angles which satisfy tan(θ) = − 4 in our final graph. the reader can verify 16(x )2 24x y 9(y )2 − + . From 16x2 + 24xy + 9y 2 + 15x − 20y = 0. Using the double angle identity for tangent. we now substitute these quantities into 16x2 + 24xy + 9y 2 + 15x − 20y = 0 and simplify. use a double angle identity. To get cos(θ) and sin(θ) from this.4 θ = arctan 3 . which means we have two options. Our rotation 5 5 equations are x = x cos(θ) − y sin(θ) = 4x − 3y and y = x sin(θ) + y cos(θ) = 3x + 4y . To 4 4 3 3 find the rotation equations. Alternatively. See page 497. so we proceed here under the assumption that B = 0.11. the coefficient A on (x )2 and the coefficient C on (y )2 are A = A cos2 (θ) + B cos(θ) sin(θ) + C sin2 (θ) C = A sin2 (θ) − B cos(θ) sin(θ) + C cos2 (θ) In order to make use of the condition cot(2θ) = A−C . it is possible to determine which conic section we have by looking at a special. we try to make sense of the product (2A )(2C ) = {[(A + C) + (A − C) cos(2θ)] + B sin(2θ)} {[(A + C) − (A − C) cos(2θ)] − B sin(2θ)} 5 We hope that someday you get to see why this works the way it does.5 First note that if the coefficient B = 0 in the equation Ax2 +Bxy +Cy 2 +Dx+Ey +F = 0. a Recall that this means its graph is either a circle. after gathering like terms. we rewrite our formulas for A and C using B the power reduction formulas. we make the usual substitutions x = x cos(θ) − y sin(θ) y = x sin(θ) + y cos(θ) into Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0. We leave it to the reader to show that. Our goal is to find the product A C in terms of the coefficients A. the graph in part 1 turned out to be a hyperbola and the graph in part 2 worked out to be a parabola.a • If B 2 − 4AC > 0 then the graph of the equation is a hyperbola.11 reduces to the result presented in Exercise 34 in Section 7. the quantity B 2 −4AC mentioned in Theorem 11. We rotate the xy-axes counter-clockwise through an angle θ which satisfies cot(2θ) = A−C to produce an equation with no x y -term in accordance with B Theorem 11.6.2. After some regrouping. • If B 2 − 4AC < 0 then the graph of the equation is an ellipse or circle. Nevertheless. Theorem 11. we could easily pick out which conic section we were dealing with based on the presence (or absence) of quadratic terms and their coefficients. we can invoke Exercise 34 in Section 7. In this form.11 is called the discriminant of the conic section.2 demonstrates that all bets are off when it comes to conics with an xy term which require rotation of axes to put them into a more standard form.11 mechanically to show that it is true. parabola. Theorem 11. Suppose the equation Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0 describes a non-degenerate conic section. Example 11. we will at least work through the proof of Theorem 11. To that end.5 once more using the product A C . B and C in the original equation. . • If B 2 − 4AC = 0 then the graph of the equation is a parabola.11. While we will not attempt to explain the deep Mathematics which produces this ‘coincidence’.10: A (x )2 + C(y )2 + Dx + Ey + F = 0. We have the following theorem. we get 2A 2C = [(A + C) + (A − C) cos(2θ)] + B sin(2θ) = [(A + C) − (A − C) cos(2θ)] − B sin(2θ) Next.6 Hooked on Conics Again 977 We note that even though the coefficients of x2 and y 2 were both positive numbers in parts 1 and 2 of Example 11. As you may expect.6. Whereas in Chapter 7. ellipse or hyperbola. familiar combination of the coefficients of the quadratic terms.5. We use this substitution B B sin(2θ) twice along with the Pythagorean Identity cos2 (2θ) = 1 − sin2 (2θ) to get 4A C = (A + C)2 − (A − C)2 cos2 (2θ) − 2B(A − C) cos(2θ) sin(2θ) − B 2 sin2 (2θ) = (A + C)2 − (A − C)2 1 − sin2 (2θ) − 2B cos(2θ)B cos(2θ) − B 2 sin2 (2θ) = (A + C)2 − (A − C)2 + (A − C)2 sin2 (2θ) − 2B 2 cos2 (2θ) − B 2 sin2 (2θ) = (A + C)2 − (A − C)2 + [(A − C) sin(2θ)]2 − 2B 2 cos2 (2θ) − B 2 sin2 (2θ) = (A + C)2 − (A − C)2 + [B cos(2θ)]2 − 2B 2 cos2 (2θ) − B 2 sin2 (2θ) = (A + C)2 − (A − C)2 + B 2 cos2 (2θ) − 2B 2 cos2 (2θ) − B 2 sin2 (2θ) = (A + C)2 − (A − C)2 − B 2 cos2 (2θ) − B 2 sin2 (2θ) = (A + C)2 − (A − C)2 − B 2 cos2 (2θ) + sin2 (2θ) = (A + C)2 − (A − C)2 − B 2 = A2 + 2AC + C 2 − A2 − 2AC + C 2 − B 2 = 4AC − B 2 Hence.6. 5x2 + 26xy + 5y 2 − 16x 2 + 16y 2 − 104 = 0 3. so the quantity B 2 − 4AC has the opposite sign of A C . Putting all of this together yields 4A C = (A + C)2 − (A − C)2 cos2 (2θ) − 2B(A − C) cos(2θ) sin(2θ) − B 2 sin2 (2θ) From cot(2θ) = A−C . 16x2 + 24xy + 9y 2 + 15x − 20y = 0 Solution. This is a straightforward application of Theorem 11. Use Theorem 11.978 Applications of Trigonometry We break this product into pieces. or (A−C) sin(2θ) = B cos(2θ). . 21x2 + 10xy 3 + 31y 2 = 144 √ √ 2. we add the product of the ‘outer’ and ‘inner’ quantities in each factor to get −B sin(2θ) [(A + C) + (A − C) cos(2θ)] +B sin(2θ) [(A + C) − (A − C) cos(2θ)] = −2B(A − C) cos(2θ) sin(2θ) The product of the ‘last’ quantity in each factor is (B sin(2θ))(−B sin(2θ)) = −B 2 sin2 (2θ).5. we get cos(2θ) = A−C . First. we use the difference of squares to multiply the ‘first’ quantities in each factor to get [(A + C) + (A − C) cos(2θ)] [(A + C) − (A − C) cos(2θ)] = (A + C)2 − (A − C)2 cos2 (2θ) Next.3.11 to classify the graphs of the following non-degenerate conics. Example 11. The result now follows by applying Exercise 34 in Section 7. √ 1. B 2 − 4AC = −4A C .11. 2 number 2. e was also defined as a ratio of distances.6. We have our ‘new’ definition below.2 number 1. a conic section is the set of all points P such that the distance from P to F =e the distance from P to L The line L is called the directrix of the conic section. directrix and eccentricity with the ‘new’ ones presented in Definition 11. Here. Theorem 11.5 in Section 7.11. B = 10 3 and C = 31 so B 2 − 4AC = (10 3)2 − 4(21)(31) = −2304 < 0. One way to reconcile the ‘old’ ideas of focus. y d r cos(θ) P (r.11 tells us that the graph is a parabola. and later extended to hyperbolas in Exercise 31 in Section 7. which matches our answer to Example 11. There. Definition 11.3.1 number 2. Theorem 11.1 and compare these parameters with what we know from Chapter 7. we have A = 16. There.6 Hooked on Conics Again 979 √ √ 1. so B 2 − 4AC = 262 − 4(5)(5) = 576 > 0. We have seen the notions of focus and directrix before in the definition of a parabola.11 classifies the graph as a hyperbola. Definition 7.2 The Polar Form of Conics In this subsection. B = 24 and C = 9 which gives 242 − 4(16)(9) = 0. the point F is called a focus of the conic section. and the constant e is called the eccentricity of the conic section.11 predicts the graph is an ellipse. which checks with our work from Example 11. It was introduced for ellipses in Definition 7. Given a fixed line L. Theorem 11. a focus F at the origin and that the directrix is the vertical line x = −d as in the figure below.1 is to derive equations for the conic sections using Definition 11. We have also seen the concept of eccentricity before.6. θ) r θ O=F x = −d x . Finally.6. and a positive number e.5. We have A = 21. giving an eccentricity e = 1 according to Definition 11.1. A = 5. B = 26 and C = 5.1. a parabola is defined as the set of points equidistant from the focus and directrix. 2. 11.6. we start from scratch to reintroduce the conic sections from a more unified perspective. matching our result from Example 11. though in these cases the distances involved were measurements from the center to a focus and from the center to a vertex.4. We begin by assuming the conic section has eccentricity e. a point F not on L. 3. We get7 a2 = e2 d2 = e2 d2 and to take the opposite reciprocal of the coefficient of y 2 )2 (1−e (e2 −1)2 e d e d b2 = − 1−e2 = e2 −1 . and the formula given in Exercise 31 in Section 7. we need a2 b 2 to find b2 . our ‘new’ understanding of ‘conic section’. We could have just as cos(θ) easily chosen the directrix to be x = d. we note that in order to arrive at our general equation of a conic r = 1−eed . ‘eccentricity’ and ‘directrix’ as presented in Definition 11. Hence. 0) and working through the formula given in Definition 7. but e = 1. as required. respectively. 6 −d. we convert the equation r = e(d + r cos(θ)) back into a rectangular equation in the variables x and y. Hence. complete the square on x and clean things up. 2 −1 e2 −1 Additionally. The key thing to remember is that in any of these cases. we get e= the distance from P to F r = the distance from P to L d + r cos(θ) so that r = e(d + r cos(θ)). 0 whose 2 2 transverse axis lies along the x-axis. θ) for a point on the conic with r > 0. Turn r = e(d + r cos(θ)) into r = e(d + x) and square both sides to get r2 = e2 (d + x)2 . In the language of Section 7. 0). If e > 0. 4p = 2d so p = d . We have established the following theorem. Using the notation from Section 7. the usual conversion process outlined in Section 11. we have shown that in all cases.3. 1 − e2 < 0. 0 2 . For parabolas. Replace r2 with x2 + y 2 .4. As the reader can verify. 0) and the directrix also tells us which way the parabola opens. we verify that one focus is at (0. 2 the focal diameter is 2d and the directrix is x = −d. 0). yields ed r= 1 − e cos(θ) At this point. 2 2 7 Since e > 1 in this case. we rewrite 1 − e2 = e2 − 1 to help simplify things later on.4 gives6 1 − e2 e2 d2 2 x− e2 d 1 − e2 2 + 1 − e2 e2 d2 y2 = 1 2 e d We leave it to the reader to show if 0 < e < 1. the equation r = 1−eed cos(θ) reduces to 2 2 2 2 2 2 r= d 1−cos(θ) which gives the rectangular equation y 2 = 2d x + d 2 .1 correspond with the ‘old’ definitions given in Chapter 7. we assumed that the directrix was the line x = −d for d > 0. then the equation generates a hyperbola with center 1−e2 . the directrix is always perpendicular to the major axis of an ellipse and it is always perpendicular to the transverse axis of the hyperbola. ‘focus’. the focus is (0. 0 2 = e d2 with major axis along the x-axis. in these cases ed we obtain the forms r = 1+eed . this is the equation of an ellipse centered at 2 1−e2 . If e = 1. y = −d or y = d. we find 1−e that one focus is (0. we have a and (1−e2 )2 2ed e d b2 = 1−e2 . so the transverse axis has length e2ed and the conjugate axis has length √2ed . Solving this equation for r. This is a parabola with vertex opening to the right. combine like terms. knowing the focus is (0.5 gives the eccentricity e2 d to be e. r = 1−eed cos(θ) sin(θ) and r = 1+e sin(θ) .980 Applications of Trigonometry Using a polar coordinate representation P (r. as required. expand (d + x)2 . Before we summarize our findings. Moreover. so the major axis has length 1−e2 and the minor axis has length √2ed 2 . Since such hyperbolas have the form (x−h) − y2 = 1. If e > 1.5 gives the eccentricity is e in this case as well. r = 12 3 − cos(θ) 3. r = 4 1 − sin(θ) 2. is the graph of a conic section with directrix y = −d. Suppose e and d are positive numbers. 0) and (6. 0). 0 and this allows us to find the 2 2 8 As a quick check.4. 0 . is the graph of a conic section with directrix x = d. Then • the graph of r = • the graph of r = • the graph of r = • the graph of r = ed 1−e cos(θ) ed 1+e cos(θ) ed 1−e sin(θ) ed 1+e sin(θ) 981 is the graph of a conic section with directrix x = −d. we have d = 4 and considering the form of the equation. which in this case is 3 . Sketch the graphs of the following equations. which is 3 from the center 3 . based on the form of the equation. The center of the ellipse is the midpoint of the vertices. 0) is a focus of the conic and the number e is the eccentricity of the conic. we know that the graph of this equation is an ellipse. We first rewrite r = 3−cos(θ) in the form found in Theorem 11. 4 1. With d = 4. Since the focus is at (0. 1 Since e = 3 satisfies 0 < e < 1. . 1.6 Hooked on Conics Again Theorem 11. 0). we have d = 12 and. the graph is a parabola whose focal diameter is 2d. Since ed = 4. 0). Example 11. This means that the ellipse has its major axis along the x-axis. we have from Theorem 11. the graph is an ellipse whose major axis has length axis has length √2ed 2 1−e • If e = 1. • If e > 1.12 in the next example. the graph is a hyperbola whose transverse axis has length conjugate axis has length √2ed .12. this puts the directrix at y = −4. 12 4 2.11. so the parabola contains the points (±4. r = 6 1 + 2 sin(θ) 2ed e2 −1 2ed 1−e2 and whose minor and whose Solution. (0.12 the major axis should have length 2ed 1−e2 = (2)(4) 1−(1/3)2 = 9. From r = 1−sin(θ) . e2 −1 We test out Theorem 11. We find r(0) = 6 and r(π) = 3 which correspond to the rectangular points (−3. We graph r = 1−sin(θ) below. In each case above. we first note e = 1 which means we have a parabola on our hands. so these are our vertices. is the graph of a conic section with directrix y = d. • If 0 < e < 1. namely r = 1−(1/3) cos(θ) . the directrix is x = −12. we have a focal 4 diameter of 2d = 8.12.8 2 We know one focus is (0. We can find the vertices of the ellipse by finding the points of the ellipse which lie on the x-axis. Since ed = 4.6. 0). −2) and must open upwards. we know that the vertex is located at the point (in rectangular coordinates) (0. we know from Theorem √ 4 11. y 8 7 6 5 4 y=3 2 1 −5 −4 −3 −2 −1 1 2 3 4 5 x r= 6 1+2 sin(θ) . We graph the hyperbola below. According to Theorem 11. 6) which puts the center of the hyperbola at (0. We now have everything we need to graph r = 3−cos(θ) . Since the center of the hyperbola is (0. Since ed = 6. This means the transverse axis of the hyperbola lies along the y-axis. so we can find the vertices by looking where the hyperbola intersects the y-axis. we get d = 3.12 that the length of the minor axis is √2ed 2 = √1−(1/3)2 = 6 3 which means the endpoints 1−e √ 12 of the minor axis are 3 . From r = 1+2 sin(θ) we get e = 2 > 1 so the graph is a hyperbola. and from the form of the equation.982 Applications of Trigonometry other focus (3. 2 y 4 3 3 2 1 −4 −3 −2 −1 −1 −2 −3 y = −4 x = −12 1 2 3 4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 5 6 x 2 1 r= 4 1−sin(θ) r= 12 3−cos(θ) 6 3. Putting this together with the location of the vertices. Finally. 4).12. the asymptotes are y = ± 3 3 x + 4. ±3 2 . 2) and (0. We find r π = 2 and r 3π = −6. we get that e2 −1 2 the asymptotes of the hyperbola have slopes ± 2√3 = ± √ √ 3 3 . which is 4 units away from the center. 4). the conjugate axis has a length √ (2)(6) of √2ed = √22 −1 = 4 3. 0). we know the directrix is y = 3. These two 2 2 points correspond to the rectangular points (0. 0). 8). Since one focus is at (0. we know the other focus is at (0. even though we are not asked to do so. 12 2 remain intact. 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 1 2 3 4 r= 4 1−sin(θ− π ) 4 Using rotations. the graph of the equation r= 1 − e cos(θ − φ) is a conic section with eccentricity e and one focus at (0. 0) and the directrix contains the point with polar coordinates (−d. we can greatly simplify the form of the conic sections presented in Theorem 11.6 Hooked on Conics Again 983 In light of Section 11. 4 to graph r = 1−sin 4θ− π all we need to do is rotate the graph of r = 1−sin(θ) . as shown below. the reader may wonder what the rotated form of the conic sections would look like in polar form. all of the formulas for lengths Theorem 11. In the theorem below. • If e = 0. Since rotations do not affect lengths.5 that replacing θ with (θ − φ) in an expression r = f (θ) rotates the graph of r = f (θ) counter-clockwise by an angle φ. 2ed – If 0 < e < 1.12.4 number 1. For instance. e2 −1 2ed e2 −1 and whose . the graph is a parabola whose focal diameter is 2d.6.13. since any three of the forms given there can be obtained from the fourth by rotating through some multiple of π . φ) where d = e . • If e = 0. counter-clockwise by π radians. 0). 0) with radius . We conclude this section with the statement of the following theorem. the graph is an ellipse whose major axis has length 1−e2 and whose minor axis has length √2ed 2 1−e – If e = 1. which we obtained in ( 4) Example 11. then the conic has a focus at (0. Given constants > 0. We know from Exercise 65 in Section 11. the graph is a hyperbola whose transverse axis has length conjugate axis has length √2ed . e ≥ 0 and φ. we also generalize our formula for conic sections to include circles centered at the origin by extending the concept of eccentricity to include e = 0.1.6. – If e > 1.11. the graph is a circle centered at (0. Theorem 11. x2 + 2xy + y 2 − x 2 + y 2 − 6 = 0 √ √ 3. √ √ 1. 18. 19.9. x2 + 2 3xy + 3y 2 + 2 3x − 2y − 16 = 0 √ 6. Show the matrix from Example 8. 9. We’ve seen this matrix most sin(θ) cos(θ) recently in the proof of used in the proof of Theorem 11. r = 3 2 + sin(θ) 2 1 + sin(θ) 2 1 − 2 sin(θ) 6 3 − cos θ + π 4 cos(θ) − sin(θ) is called a rotation matrix. show A(θ)−1 = A(−θ). . 13x2 − 34xy 3 + 47y 2 − 64 = 0 √ √ 7. 7x2 − 4xy 3 + 3y 2 − 2x − 2y 3 − 5 = 0 √ √ 4. r = 12. Discuss with your classmates how to use A(θ) to rotate points in the plane.3 is none other than A π 4 . r = 14.3. x2 − 2 3xy − y 2 + 8 = 0 8.3 Exercises √ √ 2. x2 − 4xy + 4y 2 − 2x 5 − y 5 = 0 Graph the following equations. r = 16. Using the even / odd identities for cosine and sine. r = 11. r = 13. r = 15. Interpret this geometrically. The matrix A(θ) = 17.6.984 Applications of Trigonometry 11.3 in Section 8. r = 2 1 − cos(θ) 3 2 − cos(θ) 4 1 + 3 cos(θ) 2 1 + sin(θ − π ) 3 10. 8x2 + 12xy + 17y 2 − 20 = 0 Graph the following equations. 5x2 + 6xy + 5y 2 − 4 2x + 4 2y = 0 √ 5. 5x2 + 6xy + 5y 2 − 4 2x + 4 2y = 0 2 becomes (x )2 + (y +2) = 1 after rotating 4 counter-clockwise through θ = π .6 Hooked on Conics Again 985 11. x2 + 2xy + y 2 − x 2 + y 2 − 6 = 0 becomes (x )2 = −(y − 3) after rotating counter-clockwise through θ = π . 4 y y x θ= π 3 x x √ √ x2 + 2xy + y 2 − x 2 + y 2 − 6 = 0 √ √ 3.6.4 Answers √ √ 2. 4 y y x √ √ 7x2 − 4xy 3 + 3y 2 − 2x − 2y 3 − 5 = 0 √ √ 4. 7x2 − 4xy 3 + 3y 2 − 2x − 2y 3 − 5 = 0 2 becomes (x −2) + (y )2 = 1 after rotating 9 counter-clockwise through θ = π 3 y x y θ= π 4 √ √ 1.11. x2 + 2 3xy + 3y 2 + 2 3x − 2y − 16 = 0 becomes(x )2 = y + 4 after rotating counter-clockwise through θ = π 3 y x y θ= π 4 θ= π 3 x x √ √ 5x2 + 6xy + 5y 2 − 4 2x + 4 2y = 0 √ √ x2 + 2 3xy + 3y 2 + 2 3x − 2y − 16 = 0 . x2 − 2 3xy − y 2 + 8 = 0 2 2 becomes (x4) − (y4) = 1 after rotating counter-clockwise through θ = π 3 y x y x θ= π 3 x x √ 13x2 − 34xy 3 + 47y 2 − 64 = 0 √ √ 7. 13x2 − 34xy 3 + 47y 2 − 64 = 0 2 becomes (y )2 − (x ) = 1 after rotating 16 counter-clockwise through θ = π .986 √ 5. θ = arctan x 1 2 θ = arctan(2) x √ √ x2 − 4xy + 4y 2 − 2x 5 − y 5 = 0 8x2 + 12xy + 17y 2 − 20 = 0 . 6 y y θ= π 6 Applications of Trigonometry √ 6. x2 − 4xy + 4y 2 − 2x 5 − y 5 = 0 becomes (y )2 = x after rotating counter-clockwise through θ = arctan y y √ x2 − 2 3xy − y 2 + 8 = 0 8. 8x2 + 12xy + 17y 2 − 20 = 0 2 becomes (x )2 + (y4) = 1 after rotating counter-clockwise through θ = arctan(2) y x y x 1 2 . 0). vertices (−1. 1) focus (0. r = 1−cos(θ) is a parabola directrix x = −2 . −3) center (0. (2. (3. 0) √ minor axis length 2 3 y 4 3 2 1 −4 −3 −2 −1 −1 −2 1 2 3 4 x 1− 1 cos(θ) 2 3 2 is an ellipse 2 12. 0) . 0) focus (0. −2) (0. r = 1+sin(θ) is a parabola directrix y = 2 . 0). r = 3 2+sin(θ) = directrix y = 3 . 0). r = 3 2−cos(θ) = directrix x = −3 . foci (0. foci√ 0).11. 0) center (1. minor axis length 2 3 y 1+ 1 2 3 2 sin(θ) is an ellipse y 4 4 3 3 2 2 1 1 −4 −3 −2 −1 −1 −2 −3 −4 −4 1 2 3 4 x −4 −3 −2 −1 −1 −2 1 2 3 4 x 11. vertex (0. vertex (−1. focal diameter 4 987 10. 0). focal diameter 4 y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x −4 . (0.6 Hooked on Conics Again 2 9. −2) . 1). vertices (0. (0. − 8 3 conjugate axis length y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 √ 2 3 3 1 2 3 4 x 15. 0). 0) √ conjugate axis length 2 2 y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x Applications of Trigonometry 2 14. (3. 0. foci (0. 0) 3 center 2 . r = 6 is the ellipse 3−cos(θ+ π ) 4 6 r = 3−cos(θ) = 1− 1 2 cos(θ) 3 rotated through φ = − π 4 y 4 y φ= π 3 3 2 1 x −4 −3 −2 −1 −1 −2 −3 1 2 3 y 4 x φ = −π 4 x −4 . r = 1−2 sin(θ) is a hyperbola directrix y = −1. vertices (1. (2. foci (0. (0. − 3 . vertices 0. − 2 . 0). r = 1+3 cos(θ) is a hyperbola 4 directrix x = 3 . −2) 3 4 center 0.988 4 13. 0 . r = 2 the parabola r = 1+sin(θ) rotated through φ = π 3 y x 2 1+sin(θ− π ) 3 is 16. 0). we can still give r and θ special names in relation to z. denoted Im(z). the number Re(z) is always the same. Of course. Although it is not a straightforward as the definitions of Re(z) and Im(z). denoted Re(z). is defined by |z| = r. then θ is the principal argument of z. then a = c and b = d. which corresponds to the real number line as usual. The set of all arguments of z is denoted arg(z). the expression z = a + bi is called the rectangular form of z.4. In this case.7 Polar Form of Complex Numbers In this section. we associate each complex number z = a + bi with the point (a. • The modulus of z. −3) ←→ z = −3i (3. Imaginary Axis 4i (−4. which means Re(z) and Im(z) are well-defined. Recall that a complex number is a number √ the form z = a + bi where a and b are real of numbers and i is the imaginary unit defined by i = −1. we return to our study of complex numbers which were first introduced in Section 3. Definition 11. b) gives the rectangular coordinates associated with the complex number z = a + bi. The plane determined by these two axes is called the complex plane. 0) ←→ z = 3 0 1 2 3 4 Real Axis The Complex Plane Since the ordered pair (a. written θ = Arg(z). and the y-axis is relabeled as the imaginary axis. we could just as easily associate z with a pair of polar coordinates (r. θ). b.11. and the number Im(z) is always the same. which is demarcated in increments of the imaginary unit i. denoted |z|. we know that if z = a + bi = c + di where a. Re and Im are functions of complex numbers. the x-axis is relabeled as the real axis. θ) be a polar representation of the point with rectangular coordinates (a.2. b) where r ≥ 0. c and d are real numbers. • If z = 0 and −π < θ ≤ π. 2) ←→ z = −4 + 2i 3i 2i i −4 −3 −2 −1 −i −2i −3i −4i (0. • The angle θ is an argument of z.7 Polar Form of Complex Numbers 989 11. From Intermediate Algebra. ‘Well-defined’ means that no matter how we express z. Let (r. while the real number b is called the imaginary part of z. 1 . b) on the coordinate plane. In other words. The Modulus and Argument of Complex Numbers: Let z = a + bi be a complex number with a = Re(z) and b = Im(z). The number a is called the real part of z.1 To start off this section. we choose r = 2. the set arg(z) of all arguments of z can be described using set-builder notation as arg(z) = {Arg(z) + 2πk | k is an integer}. 4 If we had Calculus. Hence for z = 0. |z| = 0 is well-defined. z = −2 + 4i 3. we would regard Arg(0) as an ‘indeterminate form. In fact. so |z| = 2. we find a corresponding angle θ. the use of the absolute value notation |z| for modulus will be explained shortly. If z = 0 then the point in question is the origin. too. |z|. Since coterminal angles are exactly 2π radians apart. we have arg(0) = (−∞. To find θ. √ 1. π]. 6 2. ∞) and since there is no one value of θ which lies (−π. For each of the following complex numbers find Re(z). Thus the modulus is well-defined in this case. z = 3 − i 2. more succinctly θ = π − arctan(2) + 2πk for integers k. there are infinitely many angles θ which can be used in a polar representation of a point (r. In this case. hence Arg(z) = − π . We know tan(θ) = √3 = − 33 . Im(z). Next. so |z| = 2 5. and this angle is what we call the principal argument of z. θ is a Quadrant IV angle. θ) for any angle θ. we will write ‘θ ∈ arg(z)’ to mean ‘θ is in3 the set of arguments of z’. Since we require r ≥ 0. To find |z|. we stipulated r ≥ 0 in our definition so this pins down the value of |z| to one and only one number. Concerning the modulus. and since r > 0 and P lies in Quadrant II. Hence arg(z) = {π − arctan(2) + 2πk | k is an integer}.’ But we don’t. z = −117 Solution. we know θ is a Quadrant II angle. √ √ √ 1. The complex number z = −2 + 4i has Re(z) = −2. We know r2 = ( 3)2 + (−1)2 = 4. θ) which means it’s worth our time to make sure the quantities ‘modulus’.990 Applications of Trigonometry Some remarks about Definition 11. We find θ = π + arctan(−2) + 2πk. Only θ = π − arctan(2) satisfies the requirement −π < θ ≤ π. so θ = − π + 2πk 6 for integers k.1. we get tan(θ) = −2. Our next task is to find a polar representation (r. Note that since arg(z) is a set. Of these values. and there are two possibilities for r: one positive and one negative. Recall the symbol being used here. which we know can be represented in polar coordinates as (0. ‘∈.4 It is time for an example. However. if z = 0 then the point associated with z is the origin. we have Re(z) = 3 and Im(z) = −1. so Arg(z) = π − arctan(2). arg(z) and Arg(z). so we won’t. θ).2 are in order. In this case.7. 2 3 In case you’re wondering.2 Even with the requirement r ≥ 0. Plot z in the complex plane. only θ = − π 6 6 satisfies the requirement that −π < θ ≤ π. ‘argument’ and ‘principal argument’ are well-defined. we need to find a polar √ representation (r. the only r-value which can be used here is r = 0. For z = 3 − i = 3 + (−1)i. so all of these angles θ are coterminal. we are guaranteed that only one of them lies in the interval (−π.4 that every point in the plane has infinitely many polar coordinate representations (r. Since r > 0 and P lies √ −1 in Quadrant IV. and is associated with the point P (−2. −1) associated with z. z = 3i 4. arg(z) = − π + 2πk | k is an integer . We know from Section 11.√ for P where r ≥ 0. or. Hence. If z = 0. If z = 0 then the point in question is not the origin. so r = ±2. 4). Im(z) = 4. arg(z) √ and Arg(z).’ is the mathematical symbol which denotes membership in a set. Example 11. θ) √ Running through the usual calculations gives r = 2 5. then the point associated with z is not the origin. . Arg(z). π]. θ) with r ≥ 0 for the point P ( 3. we leave Arg(0) undefined. We plot z along with the other numbers in this example below. provided w = 0 w |w| To prove the first three properties in Theorem 11. Since we require r ≥ 0. The point (−117. To determine |z|. We get arg(z) = π + 2πk | k is an integer and Arg(z) = π . 0). From √ Section 11. n • Quotient Rule: z |z| = . we find a polar representation (r. and this is another instance where we can determine the polar form ‘by eye’. We rewrite z = 3i as z = 0 + 3i to find Re(z) = 0 and Im(z) = 3. • |z| is the distance from z to 0 in the complex plane • |z| ≥ 0 and |z| = 0 if and only if z = 0 • |z| = Re(z)2 + Im(z)2 • Product Rule: |zw| = |z||w| • Power Rule: |z n | = |z|n for all natural numbers. we find the distance . we can almost ‘see’ the answer. θ = π. The point (0.14. Hence. As in the previous problem. r = |z| = 3 and θ = π + 2πk for 2 integers k. 0) is 117 units away from the origin along the negative x-axis. Hence. The number z = −117 corresponds to the point (−117. The point in the plane which corresponds to z is (0. We have the following theorem. We have arg(z) = {(2k + 1)π | k is an integer}. 3) and while we could go through the usual calculations to find the required polar form of this point.14. Using the distance formula. then it must be √ √ that r = a2 + b2 . Properties of the Modulus: Let z and w be complex numbers. it is time to explore their properties. just barely lies in the interval (−π. b). Only one of these values.11.7 Polar Form of Complex Numbers 991 3.4. Theorem 11. which means |z| = a2 + b2 . Imaginary Axis z = −2 + 4i 4i 3i 2i z = −117 −117 i −2 −1 −i 1 3 4 √ z = 3−i 2 Real Axis z = 3i Now that we’ve had some practice computing the modulus and argument of some complex numbers. 2 2 4. suppose z = a + bi where a and b are real numbers. θ) with r ≥ 0 for the point (a. 3) lies 3 units away from the origin on the positive y-axis. r = |z| = 117 and θ = π + 2π = (2k + 1)πk for integers k. π] which means and Arg(z) = π. we know r2 = a2 + b2 so that r = ± a2 + b2 . we write z = −117 = −117 + 0i so Re(z) = −117 and Im(z) = 0. That is. b. Like the Power Rule. 8 See Section 9. |zw| = = = = = = √ √ √ (ac − bd)2 + (ad + bc)2 a2 c2 − 2abcd + b2 d2 + a2 d2 + 2abcd + b2 c2 Expand a2 c2 + a2 d2 + b2 c2 + b2 d2 a2 (c2 + d2 ) + b2 (c2 + d2 ) (a2 + b2 ) (c2 + d2 ) √ a2 + b2 c2 + d2 Rearrange terms Factor Factor Product Rule for Radicals Definition of |z| and |w| = |z||w| Hence |zw| = |z||w| as required. 6 This may be considered by some √ be a bit of a cheat. which is what√ were asked to show. We know |z| = 0 if and only if a2 + b2 = 0 if and only if a2 + b2 = 0.3 for a review of this technique.4 for a review of complex number arithmetic. which means = |z|n is true for all natural numbers n. which is true if and only if a = b = 0. note that since |z| is a distance. and the latter happens if and only if z = 0.992 Applications of Trigonometry √ from (0. . 0) to (a. so we work through the underlying Algebra to see this is to true. this first property justifies the notation |z| for modulus. then the definition of modulus coincides with absolute value so the notation |z| is unambiguous. Then P (1) is true since z 1 = |z| = |z|1 . namely z k+1 = |z|k+1 . assume P (k) is true. Furthermore. There. |z| ≥ 0. Next. we first try to reduce the problem in such a way as to use the Induction Hypothesis. = |z| 5 Since the absolute value |x| of a real number x can be viewed as the distance from x to 0 on the number line. z k+1 = = = zk z zk |z|k |z| |z n | Properties of Exponents |z| Product Rule Induction Hypothesis Properties of Exponents = |z|k+1 Hence. Now that the Product Rule has been established. Therefore. the Quotient Rule can also be established with the help of the Product Rule.5 For the second property. Then zw = (a + bi)(c + di).6 For the third we property. we use it and the Principle of Mathematical Induction8 to prove the power rule. Our job is to show that P (k + 1) is true. Let P (n) be the statement |z n | = |z|n . 7 See Example 3. c and d. establishing the first property. |z| = 0 if and only if the distance from z to 0 is 0. After the usual arithmetic7 we get zw = (ac − bd) + (ad + bc)i. assume z k = |z|k for some k ≥ 1. We assume w = 0 (so |w| = 0) and we get z w = (z) 1 w 1 w Product Rule. As is customary with induction proofs. suppose z = a + bi and w = c + di for real numbers a. To prove the product rule. We leave it to the reader to show that if z is real. we note that since a = Re(z) and b = Im(z).1 in Section 3. b) is also a2 + b2 .4. P (k + 1) is true. z = a2 + b2 = Re(z)2 + Im(z)2 . The latter happens if and only if z = a + bi = 0. we characterize the argument of a complex number in terms of its real and imaginary parts. then arg(z) = − π + 2πk | k is an integer .7 that a = r cos(θ) and b = r sin(θ). θ) is a polar representation for (Re(z). provided Re(z) = 0. 2 • If Re(z) = Im(z) = 0. then z lies 2 on the negative imaginary axis. Next. Properties of the Argument: Let z be a complex number. To prove Theorem 11.15. • If Re(z) = 0 and Im(z) < 0. b) = (Re(z).2. The expression ‘cos(θ) + i sin(θ)’ is abbreviated cis(θ) so we can write z = rcis(θ). Im(z)). A Polar Form of a Complex Number: Suppose z is a complex number and θ ∈ arg(z). then z = 0 and arg(z) = (−∞. This is left as an exercise. (r. Our next goal is to completely marry the Geometry and the Algebra of the complex numbers. The expression: |z|cis(θ) = |z| [cos(θ) + i sin(θ)] is called a polar form for z. we know that if (r. . We know from Theorem 11. From Section 11. Imaginary Axis (a. consider the figure below.11.15. The last 2 property in the theorem was already discussed in the remarks following Definition 11. By definition. so the point associated with z is (a. θ) bi √ a2 + = | b2 = r |z θ ∈ arg(z) 0 a Real Axis Polar coordinates. π 2 + 2πk | k is an integer . Im(z)).7 Polar Form of Complex Numbers 993 1 1 Hence. If Re(z) = 0 and Im(z) < 0. we get Definition 11. and the result follows. the proof really boils down to showing w = |w| . Theorem 11. To that end.3. suppose z = a + bi for real numbers a and b. ∞). • If Re(z) = 0 and θ ∈ arg(z). b) ←→ z = a + bi ←→ (r. Making these substitutions for a and b gives z = a + bi = r cos(θ) + r sin(θ)i = r [cos(θ) + i sin(θ)]. then arg(z) = Im(z) Re(z) . a = Re(z) and b = Im(z). then z lies on the positive imaginary axis. Re(z) If Re(z) = 0 and Im(z) > 0. Since r = |z| and θ ∈ arg(z). then tan(θ) = • If Re(z) = 0 and Im(z) > 0. θ) associated with z = a + bi with r ≥ 0.4. Since we take r > 0. we have that θ is coterminal with π . then tan(θ) = Im(z) . and a similar argument shows θ is coterminal with − π . Find Re(z) and Im(z).1 to find what we need.2. we get Re(z) = 3 and Im(z) = 0. |z| = 2 and θ = − π . Rotating π radians 2 counter-clockwise lands you exactly 3 units above 0 on the imaginary axis at z = 3i. Use the results from Example 11. 1. . (a) z = Solution. . there infinitely many polar forms for z. Head out 3 units from 0 along the positive real axis. It is a good exercise to actually show that this polar form reduces to z = −2 + 4i.3. It is time for an example.994 Applications of Trigonometry Since there are infinitely many choices for θ ∈ arg(z). 1. √ √ (b) For z = −2 + 4i. We can check our answer by 6 6 √ converting it back to rectangular form to see that it simplifies to z = 3 − i. we find √ 3−i (b) z = −2 + 4i (c) z = 3i (d) z = −117 (c) We get z = 3cis(0) = 3 [cos(0) + i sin(0)] = 3. This can be checked 2 2 geometrically. z = 2 5cis(π − arctan(2)). |z| = 2 5 and θ = π − arctan(2). (b) Expanding.7. so z = 2cis − π . we get Re(z) = 0 2 2 2 and Im(z) = 1. (d) Lastly. As with the previous problem. so Re(z) = − 2 = Im(z). our answer is easily checked geometrically. z = 4cis 2π = 4 cos 2π + i sin 2π .’ these answers make perfect sense. From this. Example 11. 2. To write a polar form of a complex number z. Since i is called the ‘imaginary unit. Find the rectangular form of the following complex numbers. |z| = 3 and θ = π . we get 3 3 3 √ √ z = −2 + 2i 3. We shamelessly mine our solution to Example 11. We get z = 117cis(π). The key to this problem is to write out cis(θ) as cos(θ) + i sin(θ). so that Re(z) = −2 and Im(z) = 2 3. z = 3cis π . (a) z = 4cis 2π 3 (b) z = 2cis − 3π 4 (c) z = 3cis(0) (d) z = cis π 2 2. Hence. for z = −117. Writing 3 = 3 + 0i. After some simplifying.1 to find a polar form of the following complex numbers. Since i = 0 + 1i. (d) Last but not least. so we used the indefinite article ‘a’ in Definition 11.7. (a) By definition. we need two pieces of information: the modulus |z| and an argument (not necessarily the principal argument) of z. we get z = 2cis − 3π = 2 cos − 3π + i sin − 3π 4 4 √ √ √4 z = − 2 − i 2. which makes sense seeing as 3 is a real number.7. √ (a) For z = 3 − i. In this case. |z| = 117 and θ = π. (c) For z = 3i. we have z = cis π = cos π + i sin π = i. we next take aim at the Power Rule. We have z k+1 = z k z = = = 9 Properties of Exponents (|z|cis(θ)) Induction Hypothesis Product Rule + 1)θ) cis(kθ + θ) |z|k cis(kθ) |z|k |z| |z|k+1 cis((k Compare this proof with the proof of the Power Rule in Theorem 11. Products. as required. [cos(α) + i sin(α)] [cos(β) + i sin(β)] = cos(α) cos(β) + i cos(α) sin(β) + i sin(α) cos(β) + i2 sin(α) sin(β) = cos(α) cos(β) + i2 sin(α) sin(β) + i sin(α) cos(β) + i cos(α) sin(β) Rearranging terms = (cos(α) cos(β) − sin(α) sin(β)) Since i2 = −1 + i (sin(α) cos(β) + cos(α) sin(β)) Factor out i = cos(α + β) + i sin(α + β) = cis(α + β) Sum identities Definition of ‘cis’ Putting this together with our earlier work. arithmetic and identities. Then • Product Rule: zw = |z||w|cis(α + β) • Power Rule (DeMoivre’s Theorem) : z n = |z|n cis(nθ) for every natural number n • Quotient Rule: z |z| = cis(α − β).16 requires a healthy mix of definition. that is. We now assume P (k) is true.14. zw = [|z|cis(α)] [|w|cis(β)] = |z||w| [cos(α) + i sin(α)] [cos(β) + i sin(β)] We now focus on the quantity in brackets on the right hand side of the equation. Our goal is to show that P (k + 1) is true. Moving right along. Let P (n) be the sentence z n = |z|n cis(nθ). better known as DeMoivre’s Theorem. Theorem 11. . Then P (1) is true. We first start with the product rule. Powers and Quotients Complex Numbers in Polar Form: Suppose z and w are complex numbers with polar forms z = |z|cis(α) and w = |w|cis(β).9 We proceed by induction on n. we assume z k = |z|k cis(kθ) for some k ≥ 1. since z 1 = z = |z|cis(θ) = |z|1 cis(1 · θ).11. or that z k+1 = |z|k+1 cis((k + 1)θ). we get zw = |z||w|cis(α + β).7 Polar Form of Complex Numbers 995 The following theorem summarizes the advantages of working with complex numbers in polar form.16. provided |w| = 0 w |w| The proof of Theorem 11. The next example makes good use of Theorem 11.996 Applications of Trigonometry Hence. we multiply both the numerator and denominator of the right hand side by (cos(β) − i sin(β)) which is the complex conjugate of (cos(β) + i sin(β)) to get z w = |z| |w| cos(α) + i sin(α) cos(β) − i sin(β) · cos(β) + i sin(β) cos(β) − i sin(β) If we let the numerator be N = [cos(α) + i sin(α)] [cos(β) − i sin(β)] and simplify we get N = [cos(α) + i sin(α)] [cos(β) − i sin(β)] = cos(α) cos(β) − i cos(α) sin(β) + i sin(α) cos(β) − i2 sin(α) sin(β) Expand = [cos(α) cos(β) + sin(α) sin(β)] + i [sin(α) cos(β) − cos(α) sin(β)] = cos(α − β) + i sin(α − β) = cis(α − β) Rearrange and Factor Difference Identities Definition of ‘cis’ If we call the denominator D then we get D = [cos(β) + i sin(β)] [cos(β) − i sin(β)] = cos2 (β) − i cos(β) sin(β) + i cos(β) sin(β) − i2 sin2 (β) Expand = cos2 (β) − i2 sin2 (β) = cos2 (β) + sin2 (β) = 1 Putting it all together. assuming P (k) is true. . z n = |z|n cis(nθ) for all natural numbers n. Assuming |w| = 0 we have z w = = |z|cis(α) |w|cis(β) |z| cos(α) + i sin(α) |w| cos(β) + i sin(β) Next. we have that P (k + 1) is true. i2 = −1 Pythagorean Identity and we are done.16.16 to prove is the quotient rule. we get z w = = = |z| cos(α) + i sin(α) cos(β) − i sin(β) · |w| cos(β) + i sin(β) cos(β) − i sin(β) |z| cis(α − β) |w| 1 |z| cis(α − β) |w| Simplify Again. so by the Principle of Mathematical Induction. The last property in Theorem 11. for instance. The long and short of it is that w = −2i. we get w5 = 32 cos 4π + i sin 4π = −16 − 16i 3. a lot of work was needed to convert the numbers z and w in Example 11. to accomplish the same feat by to expanding (−1 + i 3)5 .7. z w Write your final answers in rectangular form. Last. the formula zw = |z||w|cis(α + β) can be viewed geometrically as a two step process. 6 6 √ √ √ 3 we have |w| = (−1)2 + ( 3)2 = 2. 3 3 1. we know tan(θ) = Im(z) = 2√3 = 33 . If z = |z|cis(α) and w = |w|cis(β). we need to write z and w in polar form. For w = −1 + i 3. If θ ∈ arg(z). We use DeMoivre’s Theorem which yields w5 = 2cis 2π 3 3 √ Since 10π is coterminal with 4π . Theorem 11. . √ Solution.16 pays huge dividends when computing powers of complex numbers. There is geometric reason for studying these polar forms and we would be derelict in our duties if we did not mention the Geometry hidden in Theorem 11. by the factor |w|. Adding the argument of w to the argument of z can be interpreted geometrically as a rotation of β radians counter-clockwise. However. z = 4cis π . 1. Use Theorem 11. then rotating π radians clockwise to arrive at the point 2 units below 0 on the 2 z imaginary axis. Since w lies in Quadrant II. Consider how we computed w5 above and compare that √ using the Binomial Theorem.11 Focusing on z and w from Example 10 11 Assuming |w| > 1.7. we have 10π 3 z 4cis( π ) 6 = 2cis 2π = 4 cis π − 2π = 2cis − π . we have tan(θ) = −1 = − 3.3 into polar form. We get zw = 4cis π 2cis 2π = 8cis 6 3 √ After simplifying. Since − π is a 2 6 3 2 2 (3) w quadrantal angle. π 6 + 2π 3 = 8cis 5 5π 6 = 8 cos 5π 6 + i sin 5π 6 .11. Take the product rule. For z = 2 3+2i. but not least. Let z = 2 3 + 2i and w = −1 + i 3.16. Indeed. and convert back√ rectangular form – certainly more work than is required to √ to multiply out zw = (2 3 + 2i)(−1 + i 3) the old-fashioned way. zw 2. We can now proceed. √ √ √ 2 we find |z| = (2 3)2 + (2)2 = 16 = 4. compute their product. Some remarks are in order. −1+i 3 rationalizing the denominator. In order to use Theorem 11.7 Polar Form of Complex Numbers 997 √ √ Example 11. and so forth. 3 3 3 3 3. the reader may not be sold on using the polar form of complex numbers to multiply complex numbers – especially if they aren’t given in polar form to begin with.4. For an argument θ of w. w5 3. First. Since Re(z) √ z lies in Quadrant I. we have θ = π + 2πk for integers k.16 to find the following. Assuming β > 0.3. The multiplication of |z| by |w| can be interpreted as magnifying10 the distance |z| from z to 0. . θ = 2π + 2πk for integers k and w = 2cis 2π .16 to find and simplify w using their polar forms as opposed to starting with √ 2 3+2i √ . we get zw = −4 3 + 4i.16. we can ‘see’ the rectangular form by moving out 2 units along the positive real axis. Hence. = 25 cis 5 · 2π = 32cis 2. Division is tricky in the best of times. Theorem 9. and we saved ourselves a lot of z time and effort using Theorem 11. followed up by a clockwise 13 rotation of β z radians.7. both 4 and −4 are 12 13 Again. the formula w = |w| cis(α − β) may be interpreted as shrinking12 the distance from 0 to z by the factor |w|. doubling its distance from 0 (since |w| = 2). Visualizing zw for z = 4cis and w = 2cis 2π 3 . In the case of z and w from Example 11. Rotating counter-clockwise by Arg(w) = π 6 radians. we can arrive at the product zw by plotting z. . Definition 11. Using this definition. we arrive at w by first halving the distance from 0 to z. |z| z We may also visualize division similarly. we do not specify one particular prinicpal nth root. then w is an nth root of z. 3 Imaginary Axis 3i 2i i 1 |w| π 6 Imaginary Axis i 1 |w| z = 2cis π 6 z = 4cis z = 2cis π 6 0 −i −2i 1 2 3 Real Axis 0 zw = 2cis 1 2 3 Real Axis π 2π 6 3 2π 3 Dividing z by |w| = 2. assuming |w| > 1.3. Unlike Definition 5.7.4 in Section 5. and rotating 2π radians counter-clockwise. The sequence of diagrams below attempts to describe 3 this process geometrically.3. Here. Again. Our last goal of the section is to reverse DeMoivre’s Theorem to extract roots of complex numbers. assuming β > 0. If there is a natural number n such that wn = z. then rotating clockwise 2π radians. hence the use of the indefinite article ‘an’ as in ‘an nth root of z’.3. Imaginary Axis 6i 5i 4i 3i 2i i 0 1 2 3 4 5 6 7 Real Axis −7 −6 −5 −4 −3 −2 −1 z = 4cis π 6 Imaginary Axis 6i zw = 8cis z|w| = 8cis π 6 π 6 + 2π 3 5i 4i 3i 2i i 0 1 2 3 4 5 6 2π 3 z|w| = 8cis π 6 7 Real Axis Multiplying z by |w| = 2.998 Applications of Trigonometry 11. and w = 2cis . Let z and w be complex numbers. Visualizing z for z = 4cis w π 6 Rotating clockwise by Arg(w) = 2π 3 radians.4. . The nth roots of a Complex Number: Let z = 0 be a complex number with polar form z = rcis(θ). this procedure can be generalized to find. w1 . 2π and 2. while 16 means the principal square root of 16 as in 16 = 4.4. Theorem 11. namely w = 3 8 = 2. If we let w = |w|cis(α) be a polar form of w. . there are three complex zeros. This produces three distinct points with polar coordinates corresponding to 3 k = 0. we get we solve |w| α = 2πk for integers k. . and they are given by the formula wk = √ n rcis θ 2π + k n n The proof of Theorem 11. These correspond to the complex numbers 3 3 w0 = 2cis(0). For each natural number n.14. z has n distinct nth roots. Writing these out in rectangular form 3 √ 3 √ yields w0 = 2. the equation w3 = 8 becomes w3 = 8 (|w|cis(α))3 = 8cis(0) |w|3 cis(3α) = 8cis(0) DeMoivre’s Theorem The complex number on the left hand side of the equation corresponds to the point with polar coordinates |w|3 . 1 and 2: specifically (2. 3 = 8 by extracting the principal cube root to get |w| = 3 8 = 2. 2. counting multiplicity. we know |w|3 = 8 and 3α = 0 + 2πk for integers k. which we denote by w0 . To do so.11. so there are exactly three distinct cube roots of 8. . we get (w − 2) w2 + 2w + 4 = 0. 0). we get z = 8cis(0). while the complex number on the right hand side corresponds to the point with polar coordinates (8. . which means |w|3 . we arrive at the following theorem. we express z = 8 in polar form. Let us now solve this same problem using the machinery developed in this section. w1 = −1 + i 3 and w2 = −1 − i 3. While this process seems a tad more involved than our previous factoring approach. and second. Algebraically. we know these are all of the zeros. . Since z = 8 lies 8 units away on the positive real axis. Since we have found three distinct zeros. both with positive r values. all of the fifth roots of 32.17 breaks into to two parts: first. To show wk is an nth root of z. showing that the set {wk | k = 0. showing that each wk is an nth root. Suppose we wish to find all complex third (cube) roots of 8. 3α . (Try using Chapter 3 techniques on that!) If we start with a generic complex number in polar form z = |z|cis(θ) and solve wn = z in the same manner as above.17. for a total of three cube roots of 8. 1. √ Since |w| is a real number. respectively. As for α. w1 = 2cis 2π and w2 = 2cis 4π . Since |w| ≥ 0. so is |w|3 . 4π .7 Polar Form of Complex Numbers 999 √ √ square roots of 16. 0). we are trying to solve w3 = 8. . for example. since the degree of p(w) = w3 − 8 is three. . wn − 1 . (n − 1)} consists of n different complex numbers. . In accordance with Theorem 3. 3α and (8. The √ quadratic factor gives two more cube roots w = −1 ± i 3. we use DeMoivre’s Theorem to show (wk )n = z. 0) are two polar representations corresponding to the same complex number. From Section 11. We √ know that there is only one real solution to this equation. but if we take the time to rewrite this equation as w3 − 8 = 0 and factor. so it cannot be a multiple of n. Use Theorem 11. both square roots of z = −2 + 2i 3 2. We know that z has two square roots. w1 = 4 16cis π + 2π (1) = 4 4 4 4 4 √ √ 2cis 3π . and in keeping with the notation 3 √ in Theorem 11. Solution. We get w0 = 4cis (2π/3) + 2π (0) = 2cis π 2 2 3 √ (2π/3) 2π 4π and w1 = 4cis + 2 (1) = 2cis 3 . Therefore.17 in the next example. In rectangular form. the three cube roots of z = 2 + i 2 4. w2 = − 2−i 2 to √ and w3 = 2 − i 2. n n n (k − j) must be a multiple of n. w1 = − 2+i 2. √ 1. (k − j) is a positive number less than n. it follows that cis(θ + 2πk) = cis(θ). .) Hence.7.17 to find the following: √ 1. with k = j. But because of the restrictions on k and j. we’ll call them w0 and w1 . θ = π and n = 4. let’s assume for the sake of argument θ θ that k > j. For this to be an integer multiple of 2π. we get z = −16 = 16cis(π). 0 < k − j ≤ n − 1. wk and wj are different complex numbers. Example 11. 3 θ = 2π and n = 2.17 generates n distinct numbers. and we are done. cos(θ + 2πk) = cos(θ) and sin(θ + 2πk) = sin(θ). and we have just found all of them. we know there at most n distinct solutions to wn = z. Hence. 4 4 4 4 √ 4 √ √ 4 √4 √ √ Converting these√ rectangular form gives w0 = 2+i 2. we get the √ √ four fourth roots of z to be w0 = 4 16cis π + 2π (0) = 2cis π .17. (Think this through.1000 √ n Applications of Trigonometry (wk )n = √ n = ( n r) cis n rcis θ n + n 2π n k θ · n + 2π k n DeMoivre’s Theorem = rcis (θ + 2πk) Since k is a whole number. we assume n ≥ 2 (or else there is nothing to prove) and note √ that the modulus of each of the wk is the same.17. so (wk )n = rcis(θ) = z. w2 = 4 16cis π + 2π (2) = 2cis 5π and w3 = 4 16cis π + 2π (3) = 2cis 7π . Since k and j are different. the five fifth roots of z = 1. the two square roots of √ 2 √ z are w0 = 1 + i 3 and w1 = −1 − i 3. We can check our answers by squaring them and √ showing that we get z = −2 + 2i 3. With r = 16. Suppose k and j are whole numbers between 0 and (n − 1). the only way any two of these polar forms correspond to the same number is if their arguments are coterminal – that is. By Theorem 3. We start by writing z = −2 + 2i 3 = 4cis 2π . Then n + 2π k − n + 2π j = 2π k−j . if the arguments differ by an integer multiple of 2π.4. To show that the formula in Theorem 11. Proceeding as above. as required. We illustrate Theorem 11. inclusive. we identify r = 4. namely n r.14. As a result. To use Theorem 11. 2. the four fourth roots of z = −16 √ √ 3. . w1 = cis 2π . The complex plane is without a doubt one of the most important mathematical constructs ever devised. we first take the nth root of the modulus and divide the argument by n. √ Since 5 1 = 1. w1 = 3 2cis 9π = 3 2cis 3π and w2 = 3 2cis 17π .4 found on page 293. The situation here is even graver than in the previous example. we have z = 2cis π . 14 For more on this. spaced equally around the complex plane. To find the five fifth roots of 1. 4. but messy. As n an example of this. Coupled with Calculus.7. see the beautifully written epilogue to Section 3.11. We have r = 1. we would need to use either the Sum and Difference Identities in Theorem 10. Imaginary Axis 2i w1 i w0 −2 −1 −i 0 1 2 Real Axis w2 −2i w3 The four fourth roots of z = −16 equally spaced 2π 4 = π 2 around the plane. Each succeessive root is found by adding 2π to the argument. This gives the first root w0 . We have only glimpsed at the beauty of the complex numbers in this section. the following exercises will have to suffice. w2 = cis 4π . exercise.19 to evaluate w0 and w2 . it is the venue for incredibly important Science and Engineering applications. At this stage.4 below.17 says that to find the nth roots of a complex number.16 or the Half-Angle Identities in Theorem 10. w3 = cis 6π and 5 5 5 w4 = cis 8π . For z = 2+i 2. Since we are not explicitly told to do so. Essentially. we write 1 = 1cis(0). and we leave this as an exercise. we 5 could approximate our answers using a calculator. we take a step back to look at things geometrically. This results in n roots. Now that we have done some computations using Theorem 11. θ = π and n = 3 the usual computations 4 4 √ √ √ √ π yield w0 = 3 2cis 12 . we leave this as a good. which amounts to n rotating w0 by 2π radians. θ = 0 and n = 5. since we have 5 not developed any identities to help us determine the cosine or sine of 2π . Theorem 11.14 For now. we plot our answers to number 2 in Example 11. If we were 12 4 12 to convert these to rectangular form.17. With r = 2. the roots are w0 = cis(0) = 1.7 Polar Form of Complex Numbers 1001 √ √ 3. z = 6i √ 7. z = 3 + 4i 17.1 Exercises In Exercises 1 . z = −5i 13. z = 2cis 7π 4 7π 8 24. z = −3 − 3i √ 12. z = 38. find the rectangular form of the given complex number. z = 5 + 5i 3 6. z = −5 − 2i In Exercises 21 . z = 50cis π − arctan √ √ 3 arctan − 2 5 12 1 40. z = cis π + arctan 2 √ √ √ 3 3 3 For Exercises 41 . z w 43. z = −12 − 5i √ √ 2. z = 9cis (π) 1 31. z 4 48. find a polar representation for the complex number z and then identify Re(z). z = cis 2 34. z = 3cis π 2 4π 3 π 3 6cis 29. 21. z = 9 + 9i 5. z = −2 + 6i 20. z = − 11.1002 Applications of Trigonometry 11. z = 3cis 28. z = √ 2+i 3.52. 1. z = 7 2cis 4 27. z = √ √ π 6 3π 4 3π 2 √ π 23. z = 6 15. z = 2cis 26. z = 2 2 − 2i 2 14. z = i 3 7 16. z = 12cis − π 12 4 3 35. z 5 w2 47.20. Express 2 2 your answers in polar form using the principal argument. zw 45. arg(z) and Arg(z). Im(z). z = 8cis 13cis 32. z = 5cis arctan 36. w3 42.7. z = −3 2+3i 2 8. z = −6 3 + 6i 9. 41. z = 7 24 √ 10cis arctan 1 3 37. |z|. w z 44. z = 7cis − 33. z = 15cis (arctan (−2)) 39. z = 1 − 3i 18. z = 6cis(0) 25. use z = − + i and w = 3 2 − 3i 2 to compute the quantity. z = 30.40. Use whatever identities are necessary to find the exact values. z = 4cis 2π 3 3π 4 22. z 3 w2 . z2 w 46. z = −7 + 24i 19. z = 4 − 2i 3 1 − i 2 2 √ √ 4. z = −2 √ √ 10. According to Theorem 3.4. (2 + 2i)5 √ 63. 65.16 or √ Half Angle Identities in the √ Theorem 10. the three cube roots of z = i 73.7.76. 52. the four fourth roots of z = 16 75. In Exercise 28 in Section 8. the four fourth roots of z = −81 76. 3 3 − i 2 2 3 √ 60. (See Example 11.4.4. the two square roots of z = 4i √ 67.7 Polar Form of Complex Numbers w z2 z3 w2 w2 z3 w z 6 1003 49. 3 58. (−3 + 3i)4 56. 5 5 + i 2 2 √ 61.11. we showed you how to factor this polynomial into the product of two irreducible quadratic factors using a system of non-linear equations. 50. 4 √ 1 3 − − i 2 2 6 59. −2 + 2i 3 57. Use the 12 complex 12th roots of 4096 to factor p(x) = x12 − 4096 into a product of linear and irreducible quadratic factors.7. By multiplying the first two factors together and then the second two factors together.15 told us we’d get. √ √ √ 3 54. (− 3 − i)3 55. Express your answers in polar form and then convert them into rectangular form. (See Example 11. use DeMoivre’s Theorem to find the indicated power of the given complex number. the polynomial p(x) = x4 + 4 can be factored into the product linear and irreducible quadratic factors.16 in Section 3. number 3. find the indicated complex roots. we can simply apply the Complex Factorization Theorem. ( 3 + i)4 53. 3 1 − i 3 3 4 √ 2 2 + i 2 2 62. (1 − i)8 In Exercises 65 .64. 51. Now that we can compute the complex fourth roots of −4 directly. thus pairing up the complex conjugate pairs of zeros Theorem 3. to obtain the linear factorization p(x) = (x − (1 + i))(x − (1 − i))(x − (−1 + i))(x − (−1 − i)). the six sixth roots of z = −729 77. Use the Sum and Difference Identities in Theorem 10.7. the six sixth roots of z = 64 66. ( 3 − i)5 64. Use a calculator to approximate the five fifth roots of 1. Express your final answers in rectangular form.14. the three cube roots of z = 64 71. the two square roots of 5 2 − √ 5 3 2 i 70. Theorem 3. In Exercises 53 . the two square roots of z = 1 + i 3 69.19 to express the three cube roots of z = 2 + i 2 in rectangular form. the three cube roots of z = −125 72. the two square roots of z = −25i 68. the three cube roots of z = −8i 74.) 79. number 4.) 78. we have that p(x) = (x2 − 2x + 2)(x2 + 2x + 2). . 16 to show that eix = ei(nx) for any real number x and any natural number n. assume that n is a fixed.16 to show that iy = ei(x−y) for all real numbers x and y. (e) Is it always true that Arg (z) = −Arg(z)? 82. Given any natural number n ≥ 2. is given by z = a − bi. Hint: If wj = cis(θ) let wj = cis(2π − θ).) (f) Show that cos(t) = eit + e−it eit − e−it and that sin(t) = for all real numbers t.16 to show that eix eiy = ei(x+y) for all real numbers x and y. natural number such that n ≥ 2. You’ll need to verify that wj = cis(2π − θ) is indeed an nth root of unity.1004 Applications of Trigonometry 1 w 80. show that z = reit where θ = t radians. the n complex nth roots of the number z = 1 are called the nth Roots of Unity. Euler’s Formula defines eit = cos(t) + i sin(t). (a) Use Theorem 11. (b) Show that if both wj and wk are nth roots of unity then so is their product wj wk . 2 2i n . eix (c) Use Theorem 11. Another way to express the polar form of a complex number is to use the exponential function. (This famous equation relates the five most important constants in all of Mathematics with the three most fundamental operations in Mathematics. Interpret this result geometrically. (a) Show that w = 1 is an nth root of unity. In the following exercises.14 by showing that if w = 0 than = 1 |w| . 81. Recall from Section 3. e (d) If z = rcis(θ) is the polar form of z. Complete the proof of Theorem 11. (b) Use Theorem 11. (c) Show that if wj is an nth root of unity then there exists another nth root of unity wj such that wj wj = 1. but arbitrary.4 that given a complex number z = a+bi its complex conjugate. For real numbers t. (e) Show that eiπ + 1 = 0. 83. (a) Prove that |z| = |z|. √ (b) Prove that |z| = zz z+z z−z (c) Show that Re(z) = and Im(z) = 2 2i (d) Show that if θ ∈ arg(z) then −θ ∈ arg (z). denoted z. Im(z) = 4. 2 √ √ √ √ 10. Re(z) = 5. z = i 3 7 = 3 7cis π . 4 √ √ 2. Im(z) = 3 7. Im(z) = 6. |z| = 5 4 3 + 2πk | k is an integer and Arg(z) = arctan . |z| = 2 arg(z) = {(2k + 1)π | k is an integer} and Arg(z) = π. 4 4 √ √ 5. Re(z) = − 3 2 . |z| = 3 2 4 arg(z) = arg(z) = 5π 4 + 2πk | k is an integer and Arg(z) = − 3π . Im(z) = 6. z = 6i = 6cis . z = 9 + 9i = 9 2cis arg(z) = π 4 √ . |z| = 6 arg(z) = π + 2πk | k is an integer and Arg(z) = π . |z| = 10 3 arg(z) = π 3 + 2πk | k is an integer and Arg(z) = π . Re(z) = −3 2. 6 √ √ 8. 4 11. z = −2 = 2cis (π). |z| = 3 7 2 arg(z) = π 2 + 2πk | k is an integer and Arg(z) = π . √ √ 7. Re(z) = 9. Im(z) = −2 2. |z| = 6 4 arg(z) = 3π + 2πk | k is an integer and Arg(z) = 3π . Im(z) = 9. Im(z) = −5. Re(z) = 0. Re(z) = −2. 4 3π 2 9. z = − 3 2 − 1 i = cis 2 7π 6 7π 6 . Im(z) = 5 3. √ √ √ √ 12. Re(z) = 2 2. 3 π 2 3. |z| = 5 arg(z) = + 2πk | k is an integer and Arg(z) = − π . Re(z) = 6. z = −6 3 + 6i = 12cis 5π . Re(z) = 3. Im(z) = 3 2. z = −3 − 3i = 3 2cis 5π . z = −5i = 5cis 3π 2 . z = 6 = 6cis (0). z = 3 + 4i = 5cis arctan arg(z) = arctan 4 3 . .7. Re(z) = −3. |z| = 12 6 arg(z) = 5π 6 + 2πk | k is an integer and Arg(z) = 5π 6 . Re(z) = 0. Im(z) = − 1 . Re(z) = −6 3. |z| = 1 2 + 2πk | k is an integer and Arg(z) = − 5π . Im(z) = 0. |z| = 6 arg(z) = {2πk | k is an integer} and Arg(z) = 0. Re(z) = 0. z = 2 2 − 2i 2 = 4cis 7π . Im(z) = 0.2 Answers π 4 √ 1. 6. 2 4 3 13. z = 5 + 5i 3 = 10cis π .11. 2 2 √ √ √ √ 4. |z| = 4 4 arg(z) = 7π 4 + 2πk | k is an integer and Arg(z) = − π .7 Polar Form of Complex Numbers 1005 11. Im(z) = −3. |z| = 9 2 + 2πk | k is an integer and Arg(z) = π . z = −3 2 + 3i 2 = 6cis 3π . Im(z) = −2. z = 2 cis π + arctan . z = 1 cis 2 33. z = 2cis = 7 + 7i 24. z = π 6 π 2 = 3+i = 3i 3π 4 √ = −2 + 2i 3 √ 6cis √ √ =− 3+i 3 = −3 − 2 √ 3i 3 2 27. |z| = 2 5 arg(z) = arctan − 1 + 2πk | k is an integer and Arg(z) = arctan − 1 = − arctan 2 2 √ √ 20. Im(z) = −2. z = 7 2cis 25. Re(z) = −7. Re(z) = −2. z = 15cis (arctan (−2)) = 3 5 − 6i 5 39. z = 1 − 3i = 10cis (arctan (−3)). 17. |z| = 10 √ 1 2 . arg(z) = {arctan (−3) + 2πk | k is an integer} and Arg(z) = arctan (−3) = − arctan(3). z = −12 − 5i = 13cis π + arctan 5 12 . 7 7 √ √ 16. z = 8cis 7π 4 π 12 √ √ √ 7 2 2 i 28. |z| = 2 10 arg(z) = {π − arctan (3) + 2πk | k is an integer} and Arg(z) = π − arctan (3). Re(z) = √ 2. Im(z) = 1. Re(z) = 4. |z| = 13 5 5 arg(z) = π + arctan 12 + 2πk | k is an integer and Arg(z) = arctan 12 − π. z = 6cis(0) = 6 √ 23. z = 7cis − 3π = − 7 2 2 − 4 31. Re(z) = 1. Im(z) = −3. √ √ 18. Re(z) = −12. |z| = 25 arg(z) = π − arctan 24 + 2πk | k is an integer and Arg(z) = π − arctan 24 . z = √ √ 7π 8 =4 2+ 4 3 √ 3 + 4i =− 2+ 1 3 √ 2+i 2− √ 2 35. |z| = √ 3 . z = √ 2+i= √ √ Applications of Trigonometry 3cis arctan √ 2 2 2 2 . z = 4π 3 √ 13cis 3π 2 √ = −i 13 = 2 4 √ −i 2 4 √ 32. z = −5 − 2i = 29cis π + arctan 2 . z = −2 + 6i = 2 10cis (π − arctan (3)). √ arg(z) = arctan + 2πk | k is an integer and Arg(z) = arctan 24 7 2 2 15. z = −7 + 24i = 25cis π − arctan .1006 14. Im(z) = 6. z = 3cis 26. z = 38. Im(z) = 24. z = 50cis π − arctan 7 24 √ 3cis arctan − 2 5 12 = −48 + 14i 1 40. 5 √ √ 1 19. 21. z = 4cis 2π 3 π 4 22. z = 9cis (π) = −9 29. z = 3cis 30. z = 4 − 2i = 2 5cis arctan − 2 . z = 5cis arctan = 3 + 4i 10cis arctan =3+i √ =1−i 2 6 = − 13 − 5i 26 √ √ 37. Im(z) = −5. z = 12cis − π = 6 − 6i 3 3 2− √ 3 34. z = 2cis 36. |z| = 29 5 2 arg(z) = π + arctan 2 + 2πk | k is an integer and Arg(z) = arctan 5 − π. Re(z) = −5. Since z = 64 = 64cis (0) we have w0 = 4cis (0) = 4 w1 = 4cis 2π 3 √ = −2 + 2i 3 w2 = 4cis 4π 3 √ = −2 − 2i 3 . Since z = w0 = √ − = 5cis √ 15 2 5cis 5π 6 =− + = 15 2 √ − 5 2 i 69. (−3 + 3i)4 = −324 125 4 i √ √ √ 56.52. w z2 2 = 3 cis π 12 = 3 cis(π) 4 3 = 4 cis(π) 3 w 6 z = 64cis − π 2 √ 53. (− 3 − i)3 = −8i 57. Since z = −25i = 25cis w0 = 5cis 3π 4 √ = −522 + √ 5 2 2 i π 3 = √ 5 2 2 − √ 5 2 2 i √ 67. z3 w2 45. z w = 1 cis − 11π 2 12 43. Since z = 1 + i 3 = 2cis w0 = √ 2cis 5 2 π 6 √ we have w1 = we have w1 = √ 5cis 11π 6 √ = √ 5 3 2 i 6 2 √ + 2 2 i 5π 3 √ 5 2 i √ 2cis 7π 6 √ =− 6 2 √ − 2 2 i 68. = 2cis 44. w3 = 216cis − 3π 4 48. zw = 18cis 7π 12 42. −2 + 2i 3 = 64 √ 54. z 3 w2 = 972cis(0) 50. z2 w w2 z3 π = 3 cis − 12 2 46. 51. z 4 = 81cis − 2π 3 47.11. −1 − 2 √ 2 2 √ 6 i 3 2 √ 2 2 i 4 =1 = −1 − 3i 2 3 = − 27 − 4 27 4 i √ − 1i 3 8 = − 81 − 8i81 3 + 62. 5 2 55. 61. 52. (1 − i)8 = 16 = √ √ 2+i 2 3π 2 √ √ =− 2−i 2 66.7 Polar Form of Complex Numbers 3 In Exercises 41 . 3 2 + 5i 2 3 3 3 = − 125 + 4 4 58. (2 + 2i)5 − 128 − 128i 65. Since z = 4i = 4cis w0 = 2cis π 4 π 2 √ √ 63. √ 5π 6 1007 √ √ and w = 3 2 − 3i 2 = 6cis − π so 4 w z 11π 12 41. 60. ( 3 + i)4 = −8 + 8i 3 59. we have that z = − 3 2 3 + 2 i = 3cis we get the following. ( 3 − i)5 = −16 3 − 16i we have w1 = 2cis we have w1 = 5cis 7π 4 5π 4 64. z 5 w2 = 8748cis − π 3 49. Since z = −125 = 125cis (π) we have w0 = 5cis π 3 Applications of Trigonometry = 5 2 + π 2 √ 5 3 2 i w1 = 5cis (π) = −5 w2 = 5cis 5π 3 = 5 2 − √ 5 3 2 i 71. Note: In the answers for w0 and w2 the first rectangular form comes from applying the π appropriate Sum or Difference Identity ( 12 = π − π and 17π = 2π + 3π . respectively) and the 3 4 12 3 4 second comes from using the Half-Angle Identities. Since z = −8i = 8cis w0 = 2cis π 2 = 2i √ =− 3−i w2 = cis 11π 6 = √ 3−i 73. Since z = 64 = 64cis(0) we have w0 = 2cis(0) = 2 w3 = 2cis (π) = −2 w1 = 2cis π 3 =1+ √ 3i √ 3i w2 = 2cis 2π 3 w4 = 2cis − 2π = −1 − 3 w5 = 2cis − π 3 √ = −1 + 3i √ = 1 − 3i 76.1008 70. Since z = −729 = 729cis(π) we have w0 = 3cis w3 = 3cis π 6 7π 6 = √ 3 3 2 3 + 2i √ w1 = 3cis π 2 = 3i w2 = 3cis 5π 6 = −323 + 3i 2 √ 3 3 2 √ = −323 − 3i 2 w4 = 3cis − 3π = −3i 2 w5 = 3cis − 11π = 6 − 3i 2 77. Since z = 16 = 16cis (0) we have w0 = 2cis (0) = 2 w2 = 2cis (π) = −2 74. Since z = −81 = 81cis (π) we have w0 = 3cis w2 = 3cis π 4 5π 4 w1 = 2cis w3 = 2cis π 2 3π 2 = 2i = −2i = √ 3 2 2 = √ 3 2 2 i √ √ −322 − 322i + w1 = 3cis w3 = 3cis 3π 4 7π 4 = −322 + = √ 3 2 2 √ − √ 3 2 2 i √ 3 2 2 i 75. Since z = i = cis w0 = cis π 6 √ we have w1 = cis we have w1 = 2cis 7π 6 5π 6 √ = 3 2 1 + 2i 3π 2 =− 3 2 + 1i 2 w2 = cis 3π 2 = −i 72. √ √ √ √ √ √ √ √ √6+√2 √ 2+ 3 3 3 3 6− 2 π +i = 2 + i 2− 3 w0 = 2cis 12 = 2 4 4 2 2 w1 = w2 = √ 3 √ 3 2cis 2cis 3π 4 17π 12 = √ 3 √ 2 − √ 2 2 √ + 2 2 i √ √ − 2− 6 4 = √ 3 2 √ 2− 6 4 +i = √ 3 √ 2 √ 2− 3 2 √ +i √ 2+ 3 2 . 951i ≈ −0.951i √ √ 79.11.309 − 0.309 + 0. p(x) = x12 −4096 = (x−2)(x+2)(x2 +4)(x2 −2x+4)(x2 +2x+4)(x2 −2 3x+4)(x2 +2 3+4) . w0 = cis(0) = 1 w1 = cis w2 = cis w3 = cis w4 = cis 2π 5 4π 5 6π 5 8π 5 1009 ≈ 0.7 Polar Form of Complex Numbers 78.588i ≈ 0.809 + 0.588i ≈ −0.809 − 0. we shall adopt2 the ‘arrow’ notation. regardless of its initial point. 4 . where the first number. any directed line segment with the same length and direction as v is considered to be the same vector as v. how close the nearest Sasquatch nest is as well as the direction in which it lies. 4 with initial point P (−2. we use the mathematical objects called vectors. that is.) Q (4. 3 2 1 . If we wanted to reconstruct v = 3. it should. Mathematics can be used to model and solve real-world problems. is called the y-component of v. Below is a typical vector v with endpoints P (1. 3).8 Vectors As we have seen numerous times in this book. 3. along with a direction. When referring to vectors in this text. so we have chosen the ‘arrow’ notation. Since we can reconstruct − − → v completely from P and Q. any vector which moves three units to the right and four up3 from its initial point to arrive at its terminal point is considered the same vector as v. 6). magnitude and direction. 3 If this idea of ‘over’ and ‘up’ seems familiar. For many applications. v = 3. (Think about this before moving on. is called the x-component of v and the second number. Perhaps it is important to know. 6) P (1. so the symbol v is read as ‘the vector v’. (Foreshadowing the use of bearings in the exercises. 2) and Q (4. The point P is called the initial point or tail of v and the point Q is called the terminal point or head of v. for instance. it is important to note that since vectors are defined in terms of their two characteristics. when these kinds of quantities do not suffice.1 A vector is represented geometrically as a directed line segment where the magnitude of the vector is taken to be the length of the line segment and the direction is made clear with the use of an arrow at one endpoint of the segment. The notation we use to capture this idea is the component form of the vector. 7).’ Other textbook authors use bold vectors such as v. real numbers suffice. perhaps?) To answer questions like these which involve both a quantitative answer. then we would find the terminal point of v by adding 3 to the x-coordinate and adding 4 to the y-coordinate to obtain the terminal point Q (1. In the case of our vector v above. 2) − − → v = PQ While it is true that P and Q completely determine v. where the order of points P (initial point) and Q (terminal point) is important.1010 Applications of Trigonometry 11. We find that writing in bold font on the chalkboard is inconvenient at best. or magnitude. The slope of the line segment containing v is 4 . we write v = P Q. as seen below. 4. real numbers with the appropriate units attached can be used to answer questions like “How close is the nearest Sasquatch nest?” There are other times though. The word ‘vector’ comes from the Latin vehere meaning ‘to convey’ or ‘to carry. y0 ) and terminal point Q (x1 . the speed of the plane relative to the air around it. y1 − y0 Using the language of components. A plane leaves an airport with an airspeed5 of 175 miles per hour at a bearing of N40◦ E. v2 if and only if v1 = v1 and v2 = v2 . Definition 11. 3).8. How does wind affect this? Keep reading! 5 4 . rounded to the nearest degree. think about this before reading on. The component form of a vector is what ties these very geometric objects back to Algebra and ultimately Trigonometry. Find the true speed of the plane. That is. y1 ). 4 with initial point P (−2. plot v. The sum. review page 903 and Section 11. plot w so that its initial point is the terminal point of v. We generalize our example in our definition below. Suppose v is represented by a directed line segment with initial point P (x0 . The component form of v is given by − − → v = P Q = x1 − x0 . w.11. Suppose we are given two vectors v and w. If necessary. rounded to the nearest mile per hour.4 Example 11. or resultant vector v + w is obtained as follows.1. Next. A 35 mile per hour wind is blowing at a bearing of S60◦ E. we have that two vectors are equal if and only if their corresponding components are equal.3. and the true bearing of the plane. First. That is.8 Vectors Q (1. 3) over 3 v = 3. To plot the vector v + w we begin at the initial point of v and end at the terminal point of w. v2 = v1 .) We now set about defining operations on vectors. 7) 1011 up 4 P (−2. If there were no wind. plane’s airspeed would be the same as its speed as observed from the ground. w v+w v v. and v + w Our next example makes good use of resultant vectors and reviews bearings and the Law of Cosines. v1 . (Again.5. It is helpful to think of the vector v + w as the ‘net result’ of moving along v then moving along w. Coupling speed (as a magnitude) with direction is the concept of velocity which we’ve seen a few times before in this textbook. is obtuse. for instance. we could use the Law of Sines without any ambiguity here. vectors are defined as 1 × n or n × 1 matrices. N v 35 v+w 40◦ 175 α 40◦ c 100◦ N E 60◦ w 60◦ E From the Law of Cosines. we are given their speeds and their directions.3. which we’ll call c. The vector v + w is defined by v + w = v1 + w1 . the magnitude of w. Given the geometry of the situation. In fact. we need to determine the angle α. which is 175. we need to write w in component form. the lengths of whose sides are the magnitude of v. we get a triangle. Using the Law of Cosines once more. Suppose v = v1 . Using Definition 11.8 Definition 11. Before can add the vectors using Definition 11. and the magnitude of v + w. we determine c = 31850 − 12250 cos(100◦ ) ≈ 184. w2 . which means the true speed of the plane is (approximately) 184 miles per hour. we get w = −2 − (−3).5.6 We let v denote the plane’s velocity and w denote the wind’s velocity in the diagram below.6. v2 and w = w1 . 5). The ‘true’ speed and bearing is found by analyzing the resultant vector. 7 6 . From the vector diagram. since our given angle. Find v + w and interpret this sum geometrically. which is 35. 8 Adding vectors ‘component-wise’ should seem hauntingly familiar. in more advanced courses such as Linear Algebra. we go through the usual geometry to determine that the angle between the sides of length 35 and 175 measures 100◦ .1.6. 5 − 7 = 1. v + w.1. we add α to the given 40◦ 350c and find the true bearing of the plane to be (approximately) N51◦ E. Let v = 3. To determine the true bearing of the plane. depending on the situation.8. 4 and suppose w = P Q where P (−3. 100◦ . v2 + w2 − − → Example 11. Or. From the given bearing information. Thus See Section 10.2. Solution. 7) and Q(−2. Compare this with how matrix addition was defined in section 8.7 we find 2 cos(α) = c +29400 so that α ≈ 11◦ . −2 .1012 Applications of Trigonometry Solution: For both the plane and the wind. Our next step is to define addition of vectors component-wise to match the geometric action. we find the terminal point of w to be (4. 0 = 0. We have the following theorem. Theorem 11. vectors are supposed to have both a magnitude (length) and a direction. but this minor hiccup in the natural flow of things is worth the benefits we reap by including 0 in our discussions. 4 + 1. 2 To visualize this sum. −2 3 + 1. y 4 3 v 2 1 v+w 1 2 3 4 x w In order for vector addition to enjoy the same kinds of properties as real number addition. 0) and terminal point (4.8 Vectors 1013 v+w = = = 3. we graph w with its initial point at (3. The reader may well object to the inclusion of 0. as required. v + 0 = 0 + v = v. 2) so its component form is 4. • Inverse Property: Every vector v has a unique additive inverse. That is. for every vector v. there is a vector −v so that v + (−v) = (−v) + v = 0. Geometrically. (u + v) + w = u + (v + w). since after all. v + w = w + v. . it is not clear what its direction is. 2 . We see that the vector v + w has initial point (0. • Identity Property: The vector 0 acts as the additive identity for vector addition. 2). the direction of 0 is in fact undefined. Next. Properties of Vector Addition • Commutative Property: For all vectors v and w.11. As we shall see. denoted −v. 0 . 4). we draw v with its initial point at (0. it is necessary to extend our definition of vectors to include a ‘zero vector’. which we can think of as a directed line segment with the same initial and terminal points. While it seems clear that the magnitude of 0 should be 0. v and w. • Associative Property: For all vectors u. Moving one to the right and two down. That is. 0 represents a point. 0) (for convenience) so that its terminal point is (3. 4 + (−2) 4. 4).18. for all vectors v. v − w = v + (−w). and moreover. we can ‘see’ the commutative property by realizing that the sums v + w and w + v are the same directed diagonal determined by the parallelogram below. but opposite directions. when adding the vectors geometrically. or in other words. v2 + w2 = w1 + v1 . v2 . −v2 . 0 . Geometrically.9 For the commutative property. and the reader is encouraged to verify them and provide accompanying diagrams.3 and the discussion there. v2 and w = w1 . v2 and −v = −v1 . As a result. we can define the difference of two vectors.18 and the ensuing discussion with Theorem 8. We get w1 = −v1 and w2 = −v2 so that w = −v1 . v has an additive inverse. w2 then The interested reader is encouraged to compare Theorem 11. v w w + v+ v w w v Demonstrating the commutative property of vector addition. v2 + w2 = 0. Given a vector v = v1 . w2 so that v + w = 0. w2 + v2 = w+v Geometrically. suppose we wish to find a vector w = w1 . and hence. The proofs of the associative and identity properties proceed similarly. v −v Using the additive inverse of a vector. 9 . it is unique and can be obtained by the formula −v = −v1 . the sum v + (−v) results in starting at the initial point of v and ending back at the initial point of v. we have v1 + w1 . w2 = v1 + w1 . −v2 . we note that if v = v1 .1014 Applications of Trigonometry The properties in Theorem 11. the net result of moving v then −v is not moving at all. By the definition of vector addition. v2 + w1 . If v = v1 . v1 + w1 = 0 and v2 + w2 = 0.3 in Section 8. −v2 have the same length. Hence. w2 then v + w = v1 . the vectors v = v1 . v2 and w = w1 . The existence and uniqueness of the additive inverse is yet another property inherited from the real numbers.18 are easily verified using the definition of vector addition. the vector v − w is one of the diagonals – the other being v + w. v w v−w v−w w w v v Next. v2 − w2 In other words. We define scalar multiplication for vectors in the same way we defined it for matrices in Section 8. v2 = kv1 . we discuss scalar multiplication – that is. we define kv by kv = k v1 . taking a real number times a vector. To interpret the vector v − w geometrically. we see that v − w may be interpreted as the vector whose initial point is the terminal point of w and whose terminal point is the terminal point of v as depicted below. v2 . v2 + (−w2 ) = v1 − w1 . It is also worth mentioning that in the parallelogram determined by the vectors v and w. kv2 Scalar multiplication by k in vectors can be understood geometrically as scaling the vector (if k > 0) or scaling the vector and reversing its direction (if k < 0) as demonstrated below. vector subtraction works component-wise. like vector addition.3.7.11. Definition 11. −w2 = v1 + (−w1 ) . If k is a real number and v = v1 . we note w + (v − w) = = = = = w + (v + (−w)) w + ((−w) + v) (w + (−w)) + v 0+v v Definition of Vector Subtraction Commutativity of Vector Addition Associativity of Vector Addition Definition of Additive Inverse Definition of Additive Identity This means that the ‘net result’ of moving along w then moving along v − w is just v itself. v2 + −w1 . .8 Vectors 1015 v − w = v + (−w) = v1 . From the diagram below. we let v = v1 . −v = (−1)v. If k and r are scalars then (kr)v = (kr) v1 . 1v = v. (−1)v = (−1) v1 . rv2 = k (r v1 . v2 ) = k(rv) . (kr)v2 k(rv1 ). • Identity Property: For all vectors v. then kv = 0 if and only if k = 0 or v = 0 The proof of Theorem 11. Properties of Scalar Multiplication • Associative Property: For every vector v and scalars k and r. ultimately boils down to the definition of scalar multiplication and properties of real numbers. k(rv2 ) Definition of Scalar Multiplication Associative Property of Real Number Multiplication Definition of Scalar Multiplication Definition of Scalar Multiplication = k rv1 .19. v2 . • Distributive Property of Scalar Multiplication over Scalar Addition: For every vector v and scalars k and r.18. like the proof of Theorem 11. k(v + w) = kv + k w • Zero Product Property: If v is vector and k is a scalar. This. v2 = (−1)v1 . (kr)v = k(rv). • Additive Inverse Property: For all vectors v. v2 = = (kr)v1 . (k + r)v = kv + rv • Distributive Property of Scalar Multiplication over Vector Addition: For all vectors v and w and scalars k. to prove the associative property. Theorem 11. by definition 11. For example. and other properties of scalar multiplication are summarized below.7.1016 Applications of Trigonometry 2v v 1 v 2 −2v Note that. (−1)v2 = −v1 . −v2 = −v.19. 4 ) 3v + [ −2. −4 ) 1 3 (3) v = (2). . −2 )] 5v + [(−1)(2)] (v + 1. Solve 5v − 2 (v + 1. −2 ) = 0 for v. If v = v1 . −3 2 4 3. Solution. 4 ) 0 + (−1) −2.3. 4 0 + (−1)(−2). −2 ) 5v + (−2) (v + 1. (−1)(4) 2. v2 ). 4 )] 3v + 0 3v 1 3 1 3 = = = = = = = = = = = = = 0 0 0 0 0 0 0 0 0 0 + (− −2. −2 ) 5v + [(−2)v + (−2) 1. we spell out the solution in excruciating detail to encourage the reader to think carefully about why each step is justified.1 in Section 8. 1017 Our next example demonstrates how Theorem 11. As in that example.19 allows us to do the same kind of algebraic manipulations with vectors as we do with variables – multiplication and division of vectors notwithstanding. 0) is said to be in standard position. Example 11. −2 ) 5v + (−1) [2 (v + 1.8 Vectors The remaining properties are proved similarly and are left as exercises. then its terminal point is necessarily (v1 . 4 + (− −2. −4 1 3 (3v) = 1v = v = ( 2. This is the same treatment we gave Example 8.3. (−2)(−2) ] [5v + (−2)v] + −2. v2 ) x v = v1 .) y (v1 . (Once more. If the pedantry seems familiar.8. it should. v2 is plotted in standard position. −2 ] 5v + [(−2)v + (−2)(1). 4 ) + (− −2.11.3. think about this before reading on. −3 A vector whose initial point is (0. 1 3 (−4) 2 4 3. 4 (3v + −2. 4 3v + −2. v2 in standard position. 5v − 2 (v + 1. 4 (5 + (−2))v + −2. Definition 11. v2 and k is a scalar then If this all looks familiar.20. as required. v2 v cos(θ). The second property is a result of the definition of magnitude and scalar multiplication along with a propery of radicals.1018 Applications of Trigonometry Plotting a vector in standard position enables us to more easily quantify the concepts of magnitude and direction of the vector. The following theorem summarizes the 0 important facts about the magnitude and direction of a vector. sin(θ ) making v is well-defined. A few remarks are in order.2 in Section 11. so that v = ˆ ˆ 1 v v. 2 2 2 2 v1 + v2 = 0 if and only of v1 + v2 = 0 if and only if v1 = v2 = 0. • The magnitude of v. Moreover. θ) in polar coordinates where r ≥ 0. v2 ) in rectangular coordinates to a pair (r. v = 0 if and only if v = 0. then v = v1 + v2 which is by definition greater than or equal to 0. • v ≥ 0 and v = 0 if and only if v = 0 • For all scalars k. which we said earlier was length 2 2 of the directed line segment. if θ and θ both satisfy the conditions of Definition 11. v sin(θ) v cos(θ). v2 .7. then v = 0. 0 = 0. The proof of the first property in Theorem 11. we know v1 = r cos(θ) = v cos(θ) and v2 = r sin(θ) = v sin(θ).20 is a direct consequence of the definition of v . We can convert the point (v1 .8. then cos(θ) = cos(θ ) and sin(θ) = sin(θ ). and as such. 2 2 If v = v1 . Properties of Magnitude and Direction: Suppose v is a vector. v sin(θ) . • If v = 0 then v = v v . For this reason. 0 . 10 . v2 ) with r ≥ 0. Theorem 11. θ) is a polar representation for the origin for any angle θ. is given by v = r = 2 2 v1 + v2 v1 . we get v = v cos(θ). sin(θ) ˆ ˆ Taken together. Suppose v is a vector with component form v = v1 . we get v = = = This motivates the following definition. the (vector) direction of v. the stipulation r > 0 means that all of the angles are coterminal.4.8. First. it should. Hence. From the definition of scalar multiplication and vector equality. sin(θ) = cos(θ ). The interested reader is invited to compare Definition 11. From Section 11. denoted v is given by v = cos(θ). If v = v1 . k v = |k| v .10 If ˆ v = 0. sin(θ) • If v = 0. ˆ is undefined.4 that (0. Hence. cos(θ). we note that if v = 0 then even though there are infinitely many angles θ which satisfy Definition 11. denoted v .8 to Definition 11. The magnitude of v. Let (r. and we know from Section 11.8. is r = v1 + v2 and is denoted by v . θ) be a polar representation of the point with rectangular coordinates (v1 . v2 . For v = 3. 0 ≤ θ < 2π so that v = v cos(θ). we will usually use degree measure for the angle when giving the vector’s direction. so we can use the formula v = v v to get the component form of v. 12 Due to the utility of vectors in ‘real-world’ applications. 1. 2 1 v √ Of course. In words. kv2 Definition of scalar multiplication (kv1 )2 + (kv2 )2 Definition of magnitude 2 2 k 2 v1 + k 2 v2 2 2 k 2 (v1 + v2 ) √ 2 2 = k 2 v1 + v2 = |k| = |k| v v2 1 + v2 2 Product Rule for Radicals √ Since k 2 = |k| The equation v = v v in Theorem 11. so v = v v = ˆ 2 5 −1. We are told that v lies in Quadrant II and makes a 60◦ angle with the negative x-axis. . when plotted in standard position is (5.19. it lies in Quadrant II and makes a 60◦ angle12 with the negative x-axis.8 on page 1018. 120◦ ). Example 11. The equation v = v v is a result of ˆ solving v = v v for v by multiplying11 both sides of the equation by ˆ ˆ of Theorem 11. To determine v .8. √ ˆ (See the diagram below. find the following. we appeal to Definition ˆ ˆ 11. 4 and w = 1. 2 11 1 v and using the properties (b) v −2 w (c) v − 2w (d) w ˆ √ 3 2 = −5. 1. However. −2 . sin (120◦ ) = − 1 . however.11. 3. Find the component form of the vector v with v = 5 so that when v is plotted in standard position.20 is a consequence of the definitions of v and v and was ˆ ˆ worked out in the discussion just prior to Definition 11. The authors encourage the reader. since Carl doesn’t want you to forget about radians. (a) v ˆ Solution. We are overdue for an example. √ 2. we are essentially ‘dividing both sides’ of the equation by the scalar v . he’s made sure there are examples and exercises which use them. −3 3 . to work out the details carefully to gain an appreciation of the properties in play.) Thus v = cos (120◦ ) . to go from v = v v to v = ˆ ˆ v. 23 . the equation v = v v says that any given vector is the product of its magnitude and its direction – an important ˆ 1 concept to keep in mind when studying and using vectors. find v and θ. v2 kv1 .4. sin(θ) . We are told that v = 5 and are given information about its direction. For the vectors v = 3. 523 .8 Vectors 1019 kv = = = = = k v1 . so the polar form of the terminal point of v.8. −3 3 . and the true bearing of the plane. rounded to the nearest mile per hour.1 and let v denote the plane’s velocity and w denote the wind’s velocity. we get w = 12 + (−2)2 = 5.1020 y 5 4 v 3 2 60◦ −3 −2 −1 1 Applications of Trigonometry θ = 120◦ x 1 2 3 √ 2. we get v = √ (3)2 + (−3 3)2 = 6. −3 3 = 6 cos 5π .5. 8 . 4 = 5 .4. θ) where r = v > 0. −2 = = √ 1 = 1. We 3 √ may check our answer by verifying v = 3. As an application of this process. we get w = ˆ √ √ 5 . v − 2 w = 5 − 2 5. 2 1 √ . If we regard the airport as being . we pick θ = 5π . √ 1 (d) To find w . A 35 mile per hour wind is blowing at a bearing of S60◦ E. and set about determining v + w.1 below. In light of Definition 11. Hence. For v = 3. notice that the arithmetic on the vectors comes first. 8 = 12 + 82 = 65. −2 . −3 3) is a point in Quadrant IV.8. so to find v −2 w . we need only find w . ˆ 5 5 v. −255 5 √ 1 √ 5 20 25 1. Since √ (3.8. rounded to the nearest degree. (c) In the expression v −2w . Hence. We √ √ get v − 2w = 3. ˆ ˆ ˆ which we found the in the previous problem. Hence. (a) Since we are given the component form of v. we first need w. we have v = 32 + 42 = 25 = 5. −3 3) to polar √ √ form (r. 4 − 2 1. then the magnitude. v − 2w = 1. we have tan(θ) = −33 3 = − 3. Find the true speed of the plane. 4 . v = 1 3. −√ 5 5 = . Hence.4 above by which we take information about the magnitude and direction of a vector and find the component form of a vector is called resolving a vector into its components. 4 . Using the formula w = w w along with w = 5. we can √ find the θ we’re after by converting the point with rectangular coordinates (3. For (b) We know from our work above that v = 5. −2 = 1.8. sin 5π . we’ll use the formula v = ˆ √ √ 3 v = 3.8. Hence. Example 11. we revisit Example 11. w = ˆ 5 5 2 + −255 √ 2 = 5 25 + The process exemplified by number 1 in Example 11. 3 3 1 v 3. Hence. √ √ Since w = 1. From Section 11. θ is a Quadrant IV angle. Solution: We proceed as we did in Example 11.8. our first step is to find the component form of the vector v − 2w. A plane leaves an airport with an airspeed of 175 miles per hour with bearing N40◦ E. a calculator approximation is the quickest way to see this. but you can also use good old-fashioned inequalities and the fact that 45◦ ≤ 50◦ ≤ 60◦ . If v = 1. Since. we compute the magnitude of this resultant vector v+w = (175 cos(50◦ ) + 35 cos(−30◦ ))2 + (175 sin(50◦ ) + 35 sin(−30◦ ))2 ≈ 184 Hence. y) = (175 cos(50◦ ) + 35 cos(−30◦ ). r > 0. Hence.11. To find the true bearing. we leave both vectors in terms of cosines and sines. Since we have no convenient way to express the exact values of cosine and sine of 50◦ . y (N) v y (N) v v+w 40◦ 50◦ θ x (E) w x (E) 60◦ −30◦ w In part 3d of Example 11. Since both of these coordinates are positive.8 Vectors 1021 at the origin. 175 sin(50◦ ) + 35 sin(−30◦ )). Unit Vectors: Let v be a vector. 175 sin(50◦ ) and the component form of w = 35 cos(−30◦ ). tan(θ) = y 175 sin(50◦ ) + 35 sin(−30◦ ) = . we see that the vectors v and w are in standard position and their directions correspond to the angles 50◦ and −30◦ . we say that v is a unit vector. respectively.14 we know θ is a Quadrant I angle. we take the complement of θ and find the ‘true’ bearing of the plane to be approximately N51◦ E.4. the component form of v = 175 cos(50◦ ). of the point (x. 35 sin(−30◦ ) . Furthermore. we need to find the angle θ which corresponds to the polar form (r. we get θ ≈ 39◦ . the ‘true’ speed of the plane is approximately 184 miles per hour. sin(50◦ ) = 175 cos(50◦ ). 175 sin(50◦ ) + 35 sin(−30◦ ) . for once! Yes. θ). Vectors with length 1 have a special name and ˆ are important in our further study of vectors.13 Adding corresponding components.8. we find the resultant vector v + w = 175 cos(50◦ ) + 35 cos(−30◦ ). Keeping things ‘calculator’ friendly. 14 13 . Definition 11. To find the ‘true’ speed of the plane. x 75 cos(50◦ ) + 35 cos(−30◦ ) so using the arctangent function.9. we saw that w = 1. as depicted below. we need the angle between v + w and the positive y-axis. for the purposes of bearing. the positive y-axis acting as due north and the positive x-axis acting as due east. 1 = v1 . If v = 0. (You should take the time to show this. then necessarily. In practice. Principal Vector Decomposition Theorem: Let v be a vector with component form v = v1 . . lie on the Unit Circle.21 in Section 11. 1 . 0 + 0.17 Theorem 11. when plotted in standard position.’ and the resulting vector v is called the ‘unit vector in the direction of ˆ v’. 0 ı ı • The vector  is defined by ˆ = 0.9. 17 We will see a generalization of Theorem 11. . Stay tuned! . v2 = v ı ˆ 15 One proof uses the properties of scalar multiplication and magnitude. consider v = ˆ 1 v v . back to the Unit Circle. 16 . Conversely.10.21 is straightforward.1022 Applications of Trigonometry If v is a unit vector. Since ˆ = 1. v2 = v1 . if v > 1 . ı ˆ The proof of Theorem 11. The terminal points of unit vectors. when plotted in standard position. v2 . Then v = v1ˆ + v2 . v = v v = 1 · v = v . two deserve special mention. . while  represents the positive ı ˆ y-direction.21. . we have from the ı ˆ definition of scalar multiplication and vector addition that v1ˆ + v2  = v1 1. if v is a unit v vector we write it as v as opposed to v because we have reserved the ‘ˆ’ notation for unit vectors. ˆ 1 The process of multiplying a nonzero vector by the factor v to produce a unit vector is called ‘normalizing the vector. We have the following ‘decomposition’ theorem. we visualize normalizing a nonzero vector v as shrinking16 its terminal point. Definition 11. 1 ˆ ı 1 v v We can think of the vector ˆ as representing the positive x-direction.) As a result. 0 + v2 0. y v 1 v ˆ −1 1 x −1 Visualizing vector normalization v = ˆ Of all of the unit vectors. Use the fact that v ≥ 0 is a scalar and consider factoring. we leave it as an exercise15 ˆ ˆ ˆ 1 to show that v = ˆ v is a unit vector for any nonzero vector v. The Principal Unit Vectors: • The vector ˆ is defined by ˆ = 1. 0 and  = 0. and is measured in Netwons (N) in the SI system or pounds (lbs.6. A ‘force’ is defined as a ‘push’ or a ‘pull. which we’ll call T1 and T2 (for ‘tensions’) acting upward at angles 60◦ and 30◦ .’ The intensity of the push or pull is the magnitude of the force. pulling the speaker directly downward.8. We represent the problem schematically below and then provide the corresponding vector diagram.18 The following example uses all of the concepts in this section. A 50 pound speaker is suspended from the ceiling by two support braces. respectively. .8 Vectors Geometrically.1. and should be studied in great detail. ı ˆ We conclude this section with a classic example which demonstrates how vectors are used to model forces. This is the criteria for ‘static equilbrium’. which are the magnitudes T1 and T2 . Example 11. v 2 1023  ˆ x ˆ ı v1ˆ ı v = v1 . which we’ll call w. what are the tensions on each of the supports? Solution.11.1.) in the English system. We are looking for the tensions on the support. v2 = v1ˆ + v2 . w We have three forces acting on the speaker: the weight of the speaker.19 we require w + T1 + T2 = 0. and the forces on the support rods. 30◦ 60◦ 30◦ T2 30◦ 60◦ T1 60◦ 50 lbs. In order for the speaker to remain stationary. the situation looks like this: y v2  ˆ v = v1 . Viewing the common initial point of these vectors as the 18 19 See also Section 11. If one of them makes a 60◦ angle with the ceiling and the other makes a 30◦ angle with the ceiling. Substituting that into (E2) gives ( T2 2 3) 3 + T2 − 50 = 0 2 √ √ which yields 2 T2 − 50 = 0. For the force in the first support. so the angle needed to write T2 in component form is 150◦ . we get T1 = T2 3. −1 . we note that the angle 30◦ is measured from the negative x-axis. + − . −1 = 0.  √ T2 3 T1    (E1) − = 0 2 2 √  T1 3 T2   (E2) + − 50 = 0 2 2 √ √ √ From (E1). We can model the weight of the speaker as a vector pointing directly downwards with a magnitude of 50 pounds.20 to get component representations for the three vectors involved. Hence. we get T1 = = T1 cos (60◦ ) . w + T1 + T2 = 0 √ T1 T1 3 T2 3 T2 . we get a system of linear equations in the variables T1 and T2 . 2 2 For the second support. Hence T2 = = T2 cos (150◦ ) . 2 2 The requirement w + T1 + T2 = 0 gives us this vector equation. . sin (60◦ ) √ T1 T1 3 . −50 .1024 Applications of Trigonometry origin and the dashed line as the x-axis. 0 − . = 0. Hence. sin (150◦ ) √ T2 3 T2 − . That is. −50 + Equating the corresponding components of the vectors on each side. w = 50 and w = −ˆ = 0. ˆ  w = 50 0. 0 2 2 2 2 √ √ T1 T2 3 T1 3 T2 = 0. we use Theorem 11. + − 50 2 2 2 2 √ 0. T2 = 25 pounds and T1 = T2 3 = 25 3 pounds. v = − 3. 3 5 5 √ . when drawn in standard position v lies in Quadrant I and makes a 60◦ angle with the positive y-axis v = 12. − 3 1 2 7. when drawn in standard position v lies along the negative x-axis √ 18. v = 2. when drawn in standard position v lies in Quadrant II and makes a 30◦ angle with the negative x-axis √ 16. 12. v = 10. when drawn in standard position v lies in Quadrant I and makes a 45◦ angle with the positive x-axis 2 v = 3 . 1 . 11.8 Vectors 1025 11. w = −5. w = 3. w = ı ˆ (ˆ − ) ı ˆ In Exercises 11 . − 22 2 √ . v = 5 6. when drawn in standard position v lies along the positive y-axis v = 4. v = −7. 13. 5 6. 5 . v = 6. w = −4. Give exact values. v = 12. 4 √ √ 5. w = −2. v = 8. v = 10. w = −2ˆ ı   (ˆ + ). v = 1 2 3 4 5. w = 2 3. w = −2.11.w= − √ 2 .25. 4 3.1 Exercises In Exercises 1 . 4 . when drawn in standard position v lies in Quadrant III and makes a 45◦ angle with the negative x-axis 19. 7 v = 2 . verify that the vectors satisfy the Parallelogram Law v 1. −12 4. −5 . 14. when drawn in standard position v lies in Quadrant II and makes a 30◦ angle with the positive y-axis 17. 23 2 9. 15. 24 . v = 3ˆ + 4ˆ. w = −1. 2 √ 2 + w 2 = 1 2 v+w 2 + v−w 2 2. •v + w • w − 2v • v+w • v + w • v w− w v • w v ˆ Finally. use the given pair of vectors v and w to find the following quantities.8. −1 . find the component form of the vector v using the information given about its magnitude and direction.10.25. when drawn in standard position v lies along the negative y-axis . v = 2 3. State whether the result is a vector or a scalar. v = 6. when drawn in standard position v lies in Quadrant I and makes a 60◦ angle with the positive x-axis v = 3. 22 2 √ 1 . v = √ 2 . find the magnitude v and an angle θ with 0 ≤ θ < 360◦ so that v = v cos(θ). 29. v = 5. 5 √ √ 34. when drawn in standard position v makes a 252◦ angle with the positive x-axis v = 26. v = 5 2.52. 4 43. 0 41. 5 .92. 26.3◦ angle with the positive x-axis v = 5280. v = −2. 31. v = 0.1026 20.) Round approximations to two decimal places. when drawn in standard position v makes a 78. √ 7 √ 3 2 36. v = −2 3.31. when drawn in standard position v makes a 304. − 2 40. √ 32. v = 12. for the given vector v. v = 392. 2 37.5◦ angle with the positive x-axis In Exercises 32 . v = − √ 2 2 2 . when drawn in standard position v makes a 210. v = 6. 30.7. when drawn in standard position v lies in Quadrant I and makes an measuring arctan(2) with the positive x-axis √ 23.− 2 39. sin(θ) (See Definition 11. v = − 1 . v = − 2. Applications of Trigonometry angle angle angle angle √ v = 4 3. when drawn in standard position v lies in Quadrant IV and makes a 45◦ with the negative y-axis √ 22. approximate the component form of the vector v using the information given about its magnitude and direction. v = 3. v = 10.8. v = 1. v = 5.5. when drawn in standard position v makes a 117◦ angle with the positive x-axis v = 63. Round your approximations to two decimal places.75◦ angle with the positive x-axis v = 168. 2 38. when drawn in standard position v lies in Quadrant III and makes an angle measuring arctan 4 with the negative x-axis 3 v = 26. 28. v = 2 5. 0 42. 3 √ √ 35. 27. when drawn in standard position v lies in Quadrant IV and makes an angle 5 measuring arctan 12 with the positive x-axis In Exercises 26 . when drawn in standard position v lies in Quadrant II and makes an measuring arctan(3) with the negative x-axis 24. when drawn in standard position v makes a 12◦ angle with the positive x-axis v = 450. when drawn in standard position v lies in Quadrant IV and makes a 30◦ with the positive x-axis √ 21. v = −10ˆ  33. 25. find the tension on each cable.05 45. v = 123. −6 50.1. what speed and heading should she set to take into account the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing. If the cables are each to make a 42◦ angle with the ceiling. If the captain of the HMS Sasquatch in Exercise 54 wishes to reach Chupacabra Cove.6 46. 3 47. Two cables are to support an object hanging from a ceiling. The SS Bigfoot leaves Yeti Bay on a course of N37◦ W at a speed of 50 miles per hour.3 1027 53. v = 965. The river is flowing due east at 8 miles per hour. 57. rounded to the nearest tenth of a degree. The HMS Sasquatch leaves port with bearing S20◦ E maintaining a speed of 42 miles per hour (that is. (The distance between the airport and the cliffs is 192 miles. v = −7. After traveling half an hour. v = −2.4. with respect to the water). What is the boat’s true speed and heading? Round the speed to the nearest mile per hour and express the heading as a bearing. rounded to the nearest tenth of a degree. v = ˆ − 4ˆ ı  52. 42. If each cable makes a 60◦ angle with the ceiling. rounded to the nearest tenth of a degree. If the ocean current is 5 miles per hour with a bearing of N60◦ E. what is the heaviest object that can be supported? Round your answer down to the nearest pound. 24 48. 58. . v = ˆ +  ı ˆ 51. a plane flying from the Pedimaxus International Airport can reach Cliffs of Insanity Point in two hours by following a bearing of N8. A small boat leaves the dock at Camp DuNuthin and heads across the Nessie River at 17 miles per hour (that is. 54. 55. Round your answer to the nearest pound. and each cable is rated to withstand a maximum tension of 100 pounds.8 Vectors 44. Round the speed to the nearest mile per hour and express the heading as a bearing. HINT: If v denotes the velocity of the HMS Sasquatch and w denotes the velocity of the current. v = −4. 59. an island 100 miles away at a bearing of S20◦ E from port. in three hours. v = −114.) If the wind is blowing from the southeast at 25 miles per hour. −77. the captain determines he is 30 miles from the bay and his bearing back to the bay is S40◦ E. what does v + w need to be to reach Chupacabra Cove in three hours? 56. A 600 pound Sasquatch statue is suspended by two cables from a gymnasium ceiling.15. find the HMS Sasquatch’s true speed and bearing. what speed and bearing should the pilot take so that she makes the trip in two hours along the original heading? Round the speed to the nearest hundredth of a mile per hour and your angle to the nearest tenth of a degree. 831. −1 49.2◦ E at 96 miles an hour. v = −2. rounded to the nearest tenth of a degree. with respect to the water) at a bearing of S68◦ W. In calm air.11. What is the speed and bearing of the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing. consider the line y = 2x − 4. 2 . when drawn in standard position. They need the resultant force to be at least 300 pounds otherwise the couch won’t move. Let v = v1 . b and let s = 1. loading the couch with all but one of their friends and pulling it due west down the street. 65. v2 be any non-zero vector. has its terminal point on the line y = 2x − 4. v 64. Show that this means ˆ ˆ ˆ ˆ v = k w for some non-zero scalar k and that k > 0 if the vectors have the same direction and k < 0 if they point in opposite directions. (Hint: Show that v0 + ts = t. Prove the associative and identity properties of vector addition in Theorem 11. Let v0 = 0. the second points due west and the third points S80◦ W. Show that 1 v has length 1. b (the position vector of the y-intercept) and a scalar multiple of the slope vector s = 1. 62. the force applied to the second rope is 40 pounds and the force applied (by the non-riding friend) to the third rope is 160 pounds. A 300 pound metal star is hanging on two cables which are attached to the ceiling. The goal of this exercise is to use vectors to describe non-vertical lines in the plane. Prove the properties of scalar multiplication in Theorem 11. 2t − 4 for any real number t. What is the tension on each of the cables? Round your answers to three decimal places. is it heading due west? 63. our intrepid young scholars have decided to pay homage to the chariot race scene from the movie ‘Ben-Hur’ by tying three ropes to a couch. Two drunken college students have filled an empty beer keg with rocks and tied ropes to it in order to drag it down the street in the middle of the night. −4 and let s = 1. The first rope points N80◦ W. The force applied to the first rope is 100 pounds. 61. To that end. Show that the vector defined by v = v0 + ts. Let t be any real number. m .1028 Applications of Trigonometry 60. v = 0 and w = 0 are parallel if either v = w or v = −w.) Now consider the non-vertical line y = mx + b. . m . Does it move? If so. What force should the weaker student apply to his rope so that the keg of rocks heads due east? What resultant force is applied to the keg? Round your answer to the nearest pound. Repeat the previous analysis with v0 = 0. Thus any non-vertical line can be thought of as a collection of terminal points of the vector sum of 0.18. The stronger of the two students pulls with a force of 100 pounds at a heading of N77◦ E and the other pulls at a heading of S68◦ E. That is. The left hand cable makes a 72◦ angle with the ceiling while the right hand cable makes a 18◦ angle with the ceiling. Emboldened by the success of their late night keg pull in Exercise 61 above.19. 66. We say that two non-zero vectors v and w are parallel if they have same or opposite directions. 67. scalar v w − w v = −34. 3 . scalar 3 4 5. scalar √ √ v w − w v = −14 29. vector • v + w = 2. vector • v + w = 2. vector • w − 2v = 9. 0 . −1 . vector √ 3. 12 . vector • • v + w = 0. 2 .8 Vectors 1029 11. scalar 4. 2 . • v + w = −12. 6 . 2 . • v + w = 8. scalar • w v = − 91 . vector • √ v + w = 3 5. −612 . scalar √ √ v w − w v = −6 5. • v+w = • • √ v + w = 2 3. vector • √ v + w = 3 29. 7 • v + w = − 1 . vector . vector • • v+w = √ 145. vector • w v = 5. 5 6. 3 2 2 . vector ˆ 25 25 • w − 2v = −6. − 5 .11. vector • • √ v + w = 12 2. 6 29 . vector √ √ • v + w = 0. 9 . vector • w v = 4. scalar 3. vector • • v+w = √ 226. vector 5 • w v= ˆ . vector ˆ • w − 2v = −22. −1 . vector • • v + w = 3. vector • v + w = 18. vector • w v= ˆ . scalar 5. 6 5 . − 22 2 • w v= ˆ . vector • w − 2v = − 3 2 2 . − 13 v w − w v = −21. −60 . vector √ • w v = −2 3. Answers • v + w = 15. scalar √ √ v w − w v = − 2. vector ˆ • w − 2v = −2. 312 .8. scalar √ √ 2 . 0 .2 1. scalar √ v w − w v = 8 3. vector ˆ √ • w − 2v = 4 3. scalar 1 v w − w v = − 7 . • v + w = 0. vector • v + w = 6. −2 . 0 . scalar 60 25 13 . 5 . 3 . vector 2. scalar • w − 2v = −21. vector • v + w = 38. vector 5 • • 7. v+w = √ 2. 14 . −3 . 77 . v = 3. 35. scalar √ 3 .77 31.57◦ . 3 3 14. 38.3 √ 11.13◦ v = 25. v ≈ −386. • v + w = 1. vector • w v = 1. v ≈ −177. −230. 3 26. 43. v ≈ 12.13◦ 13. θ ≈ 143. ˆ • v + w = 3. θ = 0◦ v = 10. v = −3. 5 v w − w v = −6. v = 0. vector • • v+w = √ 13. scalar √ v w − w v = −2. vector 12. scalar v w − w v = 0. θ = 270◦ v = 5. −2 3 . 39. • v + w = −1.43 34. v ≈ −52. −2 3 23. 45. θ ≈ 53. v = −2 3. v = 5. v = 6.08 32. −160. v = √ √ 3 2 3 2 .73. −6. θ = 225◦ v = 2. scalar 6 8 5. vector • v + w = 7.1030 8. θ = 90◦ v = 13. θ = 150◦ v = 1. 4 25. −18 . v = 24. √ 3 2 Applications of Trigonometry . −2 3 .73. −4 27. v = 2. 349. vector √ • w − 2v = −2. v = − 7 .59 30. 62. 37. 44. 0 2 √ 20. scalar √ • w − 2v = − 1 . 2 √ 2. vector 10. − 3 .44 33. θ = 240◦ v = √ 7. θ = 60◦ v = 2. 46.27 29. 1097. 2 .96. vector • w v= ˆ . scalar • w − 2v = −6. 2 √ √ 18.5. −10 .13. √ v = 5 2.25 22. 40. vector • v + w = 3. θ = 135◦ v = 6. vector v + w = 1. − . 41. −10 28.62◦ v = √ 5. vector √ 3 1 3 . v = 0. v = 2. vector • • v + w = 1.96. v ≈ 5164. 0 . −5 24. θ ≈ 106. 36. v = 4. θ = 180◦ v = 5.62. v = −1. vector 2 2 • 2 2 v + w = 1 1 2. 12 17. θ = 45◦ v = 1. v = − 3. 42.26◦ √ 16. − 2 • • 9. v ≈ 14. θ ≈ 206. 3 19. v = √ 15. v = −5 3. 2 2 • w v= ˆ . −21. −5 3 21. θ ≈ 22. 62.66◦ 53.6◦ W. v ≈ 145. 55. She should maintain a speed of about 35 miles per hour at a heading of S11.02◦ 51. The maximum weight that can be held by the cables in that configuration is about 133 pounds. 58.00. v = √ 1031 17.46 miles per hour with a heading of N22. The tension on the left hand cable is 285.48. She should fly at 83. The boat’s true speed is about 10 miles per hour at a heading of S50.1◦ E 57. θ ≈ 251.8 Vectors 47. The current is moving at about 10 miles per hour bearing N54.11.317 lbs. 59. θ ≈ 45◦ 49. and on the right hand cable is 92. √ v = 2 10. θ ≈ 284. . the stronger force on the third rope would have made the couch drift slightly to the south as it traveled down the street. v ≈ 1274.75◦ 52. 61. θ ≈ 40.57◦ 48. Even if it did move.8◦ E. v = √ 2. θ ≈ 159. The tension on each of the cables is about 346 pounds. The resultant force is only about 296 pounds so the couch doesn’t budge.04◦ 50. v ≈ 121. 60. 54. 56. The net force on the keg is about 153 pounds.8◦ E.69.705 lbs. The weaker student should pull about 60 pounds. θ ≈ 328.6◦ W. The HMS Sasquatch’s true speed is about 41 miles per hour at a heading of S26. v · v = v 2 . • Relation to Magnitude: For all vectors v. w2 . • Distributive Property: For all vectors u. v2 and w = w1 . u · (v + w) = u · v + u · w. v2 and w = w1 . To show the commutative property for instance. v · w = w · v. 4 · 1. v2 = w·v The distributive property is proved similarly and is left as an exercise. kv2 · w1 . Theorem 11. −2 = (3)(1) + (4)(−2) = −5. 4 and w = 1. Note that the dot product takes two vectors and produces a scalar. For that reason. the proof of Theorem 11. w2 = k(v · w) We leave the proof of k(v · w) = v · (k w) as an exercise. the quantity v· w is often called the scalar product of v and w. v2 ) · w1 . assume that v = v1 . we define a product of vectors. Definition 11. let v = 3.8.22. w2 Definition of Dot Product Commutativity of Real Number Multiplication Definition of Dot Product = v 1 w1 + v 2 w2 = w1 v 1 + w2 v 2 = w1 . v and w. w2 Definition of Scalar Multiplication Associativity of Real Number Multiplication Distributive Law of Real Numbers Definition of Dot Product = (kv1 )(w1 ) + (kv2 )(w2 ) Definition of Dot Product = k(v1 w1 ) + k(v2 w2 ) = k(v1 w1 + v2 w2 ) = k v1 . v2 · w1 . . v2 · w1 . let v = v1 . In this section. For the scalar property. Then v·w = v1 . Suppose v and w are vectors whose component forms are v = v1 . The dot product of v and w is given by v · w = v1 . w2 . Properties of the Dot Product • Commutative Property: For all vectors v and w. • Scalar Property: For all vectors v and w and scalars k. w2 = kv1 . w2 and k is a scalar.11. v2 and w = w1 . we learned how add and subtract vectors and how to multiply vectors by scalars. w2 = v1 w1 + v2 w2 For example. Then (kv) · w = (k v1 .9 The Dot Product and Projection In Section 11. (kv) · w = k(v · w) = v · (k w). −2 . The dot product enjoys the following properties. Like most of the theorems involving vectors. Then v · w = 3.22 amounts to using the definition of the dot product and properties of real number arithmetic. We begin with the following definition. v2 · w1 .1032 Applications of Trigonometry 11. w2 · v1 . 9. v2 = v1 + v2 = v 2 .1.9 The Dot Product and Projection 1033 2 2 For the last property. Theorem 11. If we take a step back from the pedantry in Example 11.8. it should. Solution.9.) v w θ w θ=0 0<θ<π θ=π The following theorem gives us some insight into the geometric role the dot product plays.23. w v v . We require 0 ≤ θ ≤ π. The following example puts Theorem 11.8.11. (Think about why this is needed in the definition. Geometric Interpretation of Dot Product: If v and w are nonzero vectors then v · w = v w cos(θ).22 to good use.9. where θ is the angle between v and w. v−w 2 = (v − w) · (v − w) = (v + [−w]) · (v + [−w]) = (v + [−w]) · v + (v + [−w]) · [−w] = v · (v + [−w]) + [−w] · (v + [−w]) = v · v + v · [−w] + [−w] · v + [−w] · [−w] = v · v + v · [(−1)w] + [(−1)w] · v + [(−1)w] · [(−1)w] = v · v + (−1)(v · w) + (−1)(w · v) + [(−1)(−1)](w · w) = v · v + (−1)(v · w) + (−1)(v · w) + w · w = v · v − 2(v · w) + w · w = v 2 − 2(v · w) + w 2 2 Hence. If we draw v and w with the same initial point. we work out the problem in great detail and encourage the reader to supply the justification for each step. If this looks familiar. we see that the bulk of the work is needed to show that (v − w)·(v − w) = v ·v −2(v · w)+ w· w. and hence we get similar looking results. Suppose v and w are two nonzero vectors. v − w 2 = v 2 − 2(v · w) + w as required.1 plays a large role in the development of the geometric properties of the dot product. we note that if v = v1 . we define the angle between v and w to be the angle θ determined by the rays containing the vectors v and w. The identity verified in Example 11.1. the machinations required to expand (v − w) · (v − w) for vectors v and w match those required to expand (v − w)(v − w) for real numbers v and w. as illustrated below.22.3. Since the dot product enjoys many of the same properties enjoyed by real numbers. As in Example 11. We begin by rewriting v − w 2 in terms of the dot product using Theorem 11. which we now explore. where the last equality comes courtesy of Definition 11. Example 11. v2 · v1 . then v · v = v1 . v2 . Prove the identity: v − w 2 = v 2 − 2(v · w) + w 2 . proving that the formula holds for θ = 0. If θ = 0.23 in cases. An immediate consequence of Theorem 11. v · w = v · (kv) = k(v · v) = k v 2 = k v v . Next. Since v and w are nonzero.2. if v = w then w = w v = ˆ ˆ ˆ ˆ ˆ ( v v) = ˆ w v v. Then θ = arccos v·w v w = arccos(ˆ · w) v ˆ v−w We obtain the formula in Theorem 11. k = w v > 0.9. It follows1 that there is a real number k > 0 so that w = kv. Hence. ˆ ˆ v ˆ We are overdue for an example. √ √ 1. If θ = π. 2 . giving us the alternative formula θ = arccos(ˆ · w). k v v = v (k v ) = v kv = v w . and w = − 3. or v · w = v w cos(θ). . We use the formula θ = arccos 1 v·w v w from Theorem 11. the values of θ exactly match the w range of the arccosine function. we can rewrite = 1 v v · 1 w w = v · w. Theorem 11. Since 0 ≤ θ ≤ π by definition. Since cos(0) = 1. as required. 1 Solution.1034 Applications of Trigonometry We prove Theorem 11. we know v − w 2 = v 2 − 2(v · w) + w 2 .1. w θ v−w v w θ v The Law of Cosines yields v − w 2 = v 2 + w 2 − 2 v w cos(θ).23 is the following. θ = arccos v·w v w v·w v w . Hence. so are v and w . and w = 5. we may divide both sides of v · w = v w cos(θ) by v w to get cos(θ) = vv·w . Find the angle between the following pairs of vectors. Equating these two expressions for v − w 2 gives v 2 + w 2 −2 v w cos(θ) = v 2 −2(v· w)+ w 2 which reduces to −2 v w cos(θ) = −2(v· w). as required. we get v · w = − v w = v w cos(π).24 by solving the equation given in Theorem 11.9. In this case.23 for θ. we repeat the argument with the difference being w = kv where k < 0. −4 . 1 2.22. −5 3. |k| = −k. then v and w have the same direction. v = 3. Hence. if 0 < θ < π. Using Theorem 11. In this case. the vectors v. w v Since v = v v and w = w w. so k v = |k| v = kv by Theorem 11. we get v · w = k v v = v w = v w cos(0). Since k > 0. −3 3 . w and v − w . Since cos(π) = −1. From Example 11. Let v and w be nonzero vectors and let θ the angle between v and w. k = |k|.24. v = 2. so k v = −|k| v = − kv = − w .24 in each case below. Example 11. w and v − w determine a triangle with side lengths v . as seen below. Hence. respectively. v = 3. and w = 2.20. yet v · w = 0. which has the same direction as L2 and show v1 ⊥ v2 if and only if m1 m2 = −1. For v = 2. and v2 . 1 = −3 3−3 3 = −6 3.23. 2 and w = 5. when orthogonal vectors 2 are sketched with the same initial point.1. 2 2 if v and w are nonzero vectors and v · w = 0.11. −4 · 2. We can use Theorem 11. The vectors v = 2. 1 = 6 − 4 = 2. we find v · w = 2. the angle between v and w is π . we first assume v and w are nonzero vectors with v ⊥ w.3. By definition. We find v · w = 3. Prove that L1 is perpendicular to L2 if and only if m1 · m2 = −1. Since 2255 isn’t the cosine of one of the common angles. 2 · 5. Theorem 11. −3 3 · − 3.1. We have v· w = 3.25.3 Example 11. . matter what v and w are. −5 in Example 11. Our strategy is to find two vectors: v1 . v ⊥ w We state the relationship between orthogonal vectors and their dot product in the following theorem.9.2 θ = arccos vv·w 2 w √ 3. then Theorem 11. Also v = 32 + (−4)2 = 25 = 5 and √ √ √ √ 2 w = 22 + 12 = 5.2 are called orthogonal and we write v ⊥ w. θ = arccos −6 3 = arccos − 23 = 5π . Geometrically. Conversely. To prove Theorem 11. Then v ⊥ w if and only if v · w = 0.24 gives θ = arccos arccos 0 v w v·w v w = = arccos(0) = π 2. Since v = 32 + (−3 3)2 = √ √ √ √ √ 36 = 6 and w = (− 3)2 + 12 = 4 = 2. −5 . v · w = v w cos π = 0. we leave our answer as θ = arccos √ 2 5 25 . because the angle between them is π radians = 90◦ . which has the same direction as L1 . By Theorem 11. 12 6 2. The Dot Product Detects Orthogonality: Let v and w be nonzero vectors. 2 . −5 = 10 − 10 = 0. namely P (0. Solution. so v ⊥ w.25 in the following example to provide a different proof about the relationship between the slopes of perpendicular lines.9 The Dot Product and Projection 1035 √ √ √ √ √ √ 1. w v v and w are orthogonal. b1 ) 2 3 Note that there is no ‘zero product property’ for the dot product since neither v nor w is 0. See Exercise 2. it doesn’t = arccos(0) = π . the lines containing the vectors are perpendicular. Let L1 be the line y = m1 x + b1 and let L2 be the line y = m2 x + b2 . and w = 5. To that end.25. Hence.9.1 in Section 2. so θ = arccos 5√5 = arccos 2255 . we substitute x = 0 and x = 1 into y = m1 x + b1 to find two points which lie on L1 . m1 .22. Putting this together with ˆ p = −w. Similarly. we get p = p p = −(v · w)(−w) = (v · w)w in this case as well. L1 and L2 are perpendicular if and only if v1 ⊥ v2 . we find p = v cos(θ ). The vector p is then defined as p = OR. ˆ ˆ ˆ ˆ ˆ ˆ ˆ .25 certainly gives us some insight into what the dot product means geometrically. Using Theorem 1 ˆ ˆ ˆ ˆ 11. To determine p . v1 ⊥ v2 if and only if v1 · v2 = 0. Since θ +θ = π. we now have a w formula for p completely in terms of v and w. Notice that v1 · v2 = 1. which we’ll denote θ. assume that the angle between v and w. m2 which has the same direction as the line L2 . and note that since v1 is determined by two points on L1 . Hence. we see that p = −w and using the triangle ORT . we get the vector v2 = 1. we rewrite v·w = v · w w = v · w. indicated below. We wish to develop a formula for the vector p. which means p = v cos(θ ) = − v cos(θ). as required. and since p = w. and consider the diagram below. m1 · 1. which is true if and only if m1 m2 = −1. m2 = 1 + m1 m2 . p = v · w. Hence. (m1 + b1 ) − b1 = 1. p is determined by its magnitude p and its direction p according to the formula p = p p. v1 · v2 = 0 if and only if 1 + m1 m2 = 0.25. we make use of Theorem 10. While Theorem 11. To get things in terms of just v and w. For the moment. According to Theorem 11. Since we want p to have the ˆ ˆ ˆ same direction as w. we get p = −(v · w). or p = v cos(θ). Consider the two nonzero vectors v and w drawn with a common initial point O below. Rewriting this last equation in terms of v and w as before. we use Theorem 11. we have p = w.1036 Applications of Trigonometry − − → and Q(1. Like any vector. ˆ ˆ ˆ v v T T v w θ θ − − → p = OR O O O R w θ p R Now suppose that the angle θ between v and w is obtuse. ˆ ˆ it follows that cos(θ ) = − cos(θ). m1 + b1 ). The vector p is obtained geometrically as follows: drop a perpendicular from the terminal point T of v to the vector w and call the point − − → of intersection R. which is called the orthogonal projection of v onto w. is acute. there is more to the story of the dot product. We let v1 = P Q = 1 − 0. We find cos(θ) = v . it may be viewed as lying on L1 .4 as applied ˆ ˆ p to the right triangle ORT . Hence. In this case.23 to get p = v cos(θ) = v w cos(θ) w = v·w w . namely p = p p = (v · w)w. Hence it has the same direction as L1 . 2 v · w = 0. because of the presence of the normalized unit vector w. which we leave to the reader as an exercise. w and p below. v· Hence. It is time for an example. 2 = −3. 2 . too. It follows that v · w = 0 and p = 0 = 0w = (v · w)w in this case. 2 = (−1) + 16 = 15 and w · w = −1. computing the projection using the formula ˆ ˆ ˆ projw (v) = (v · w)w can be messy.12 is theoretically appealing. 2 · −1. 8 · −1. Alternate Formulas for Vector Projections: If v and w are nonzero vectors then ˆ ˆ projw (v) = (v · w)w = v·w w 2 w= v·w w·w w The proof of Theorem 11. ˆ Example 11. Let v = 1. Since v ⊥ w in this case. Let v and w be nonzero vectors. We plot v. so p = OR = OO = 0. the point R coincides with the point O. p = w·w w = 15 −1. The scalar v · w is a measure ˆ of how much of the vector v is in the direction of the vector w and is thus called the scalar projection of v onto w. w 5 − → − − → In this case.9. We find v · w = 1.9 The Dot Product and Projection T v θ θ 1037 w O R − − → p = OR If the angle between v and w is π then it is easy to show4 that p = 0.12. amounts to using the 1 formula w = w w and properties of the dot product.4. Solution. denoted projw (v) is given by projw (v) = (v · w)w. 6 . 2 = 1 + 4 = 5. and plot v. While the formula given in Definition 11.26. 4 . We present two other formulas that are often used in practice. Theorem 11. 8 and w = −1.26. This gives us ˆ ˆ ˆ ˆ Definition 11. w and p in standard position. ˆ ˆ Definition 11. The orthogonal projection of v onto w.12 gives us a good idea what the dot product does.11. Find p = projw (v). 4. 8 − −3. 2 . Next. Theorem 11.27 are nonzero. we compute q · w. In the case of Example 11. by definition. q p 8 7 6 5 4 3 2 v w −3 −2 −1 1 From the definition of vector arithmetic.9. the vector p is sometimes called the ‘vector component of v parallel to w’ and q is called the ‘vector component of v orthogonal to w.’ To prove Theorem 11. it has the correct direction.9. Consider the vector q whose initial point is the terminal point of p and whose terminal point is the terminal point of v. so that q = v − p. We first note that since p is a scalar multiple of w. so q = 1.1038 Applications of Trigonometry 8 7 p 6 5 4 3 w 2 v −3 −2 −1 1 Suppose we wanted to verify that our answer p in Example 11. 6 = 4. There are unique vectors p and q such that v = p + q where p = k w for some scalar k. Then q· w = 4.4 is indeed the orthogonal projection of v onto w. Note that if the vectors p and q in Theorem 11. 5 See Exercise 64 in Section 11. p + q = v. so what remains to check is the orthogonality condition. Then p is. Generalized Decomposition Theorem: Let v and w be nonzero vectors. 8 and p = −3. 6 . v = 1. 2 = (−4)+4 = 0. . a scalar multiple of w. 2 · −1.27. we take p = projw (v) and q = v − p. which shows q ⊥ w. This result is generalized in the following theorem.8. then we can say p is parallel 5 to w and q is orthogonal to w. as required. In this case. and q · w = 0.27. we get W = (F · P Q) P Q . it must be that q = q as well. . we have shown there is only one way to write v as a sum of vectors as described in Theorem 11.27 exist. we can use the dot product to find the work done. so w · (p − p ) = w · (q − q) = w · q − w · q = 0 − 0 = 0. − − → − − → − − → − − → W = (F · P Q) P Q = F · ( P Q P Q) = F · P Q = F P Q cos(θ). the work W done by the force is given by W = F d. w·w = v·w−v·w = 0 Hence. which means the only way w · (p − p ) = (k − k ) w 2 = 0 is for k − k = 0. Now we need to show that they are unique. With q − q = p − p = p − p = 0. w 2 = 0. If the force applied is not in the direction of the motion. where θ is the angle between − − → the applied force F and the trajectory of the motion P Q. q · w = 0. This is precisely what the dot product F · P Q represents. or k = k .9 The Dot Product and Projection 1039 q · w = (v − p) · w = v·w−p·w = v·w− Definition of q.27. we assume the force is being applied in the direction of the motion. F θ P Q θ F To find the work W done in this scenario. we have shown that the vectors p and q guaranteed by Theorem 11.27 as p and q. w · (p − p ) = 0. We close this section with an application of the dot product.11. Since w = 0. Hence. Now there are scalars k and k so that p = k w and p = k w. w·w v·w = v·w− (w · w) Properties of Dot Product. Here. Properties of Dot Product v·w w ·w Since p = projw (v). Hence. Consider the scenario below where the constant force F is applied to move an object from the point P to the point Q. Since P Q = P Q P Q. Since − − → − − → − − → − − → the distance the object travels is P Q . This means p = k w = k w = p . At this point. In Physics. as required. Then p − p = q − q. This means w · (p − p ) = w · (k w − k w) = w · ([k − k ]w) = (k − k )(w · w) = (k − k ) w 2 . we need to find how much of the force F is in the − − → direction of the motion P Q. We have proved the following. if a constant force F is exerted over a distance d. Suppose v = p + q = p + q where the vectors p and q satisfy the same properties described in Theorem 11. Example 11. we have F = 10. the displacement vector is P Q = 50ˆ = 50 1.5. Alternatively. There are two ways to attack this problem. One way is to find the vectors F and P Q − − → mentioned in Theorem 11. Work as a Dot Product: Suppose a constant force F is applied along the − − → vector P Q. we get W = 250 3 foot-pounds. 0 . 30◦ − − → Solution. we assume the origin is at the point where the handle of the wagon meets the wagon and the positive x-axis lies along the dashed line in the figure above. Since the force applied is a constant 10 pounds. Since it is being applied at a constant angle of θ = √ ◦ with respect to the positive x-axis. Since force is measured in pounds and distance is √ measured in feet. If the handle of the wagon makes a 30◦ angle with the horizontal. 0 = 50.5 and compute W = F · P Q. how much work did Taylor do pulling the wagon? Assume Taylor exerts the force of 10 pounds at a 30◦ angle for the duration of the 50 feet. To do this. .1040 Applications of Trigonometry Theorem 11. We get ı √ √ − − → W = F · P Q = 5 3. Definition 30 ◦ .28. The work W done by F is given by − − → W = F · PQ = F − − → where θ is the angle between F and P Q. 5 · 50.9. − − → P Q cos(θ). 0 = 250 3.9. Taylor exerts a force of 10 pounds to pull her wagon a distance of 50 feet over level ground. 5 . we can use the formulation √ − − → W = F P Q cos(θ) to get W = (10 pounds)(50 feet) cos (30◦ ) = 250 3 foot-pounds of work. Since the wagon is being pulled along 50 11. sin(30◦ ) = 5 3.8 gives us F = 10 cos(30 − − → feet in the positive direction. 3 and w = − 3.2 √ • projw (v) • q = v − projw (v) (Show that q · w = 0. 23 2 √ 3 1 2 . the keg was being pulled due east and the student’s heading was N77◦ E. 1 14. 4 and w = −6. Find the work done towing the trailer along a flat stretch of road 300 feet. v = 3 ˆ + 3  and w = ˆ −  ı ˆ 2ı 2ˆ 17. Assume the force is applied in the direction of the motion. In Exercise 61 in Section 11. two drunken college students have filled an empty beer keg with rocks which they drag down the street by pulling on two attached ropes. A force of 1500 pounds is required to tow a trailer. Find the work done lifting a 10 pound book 3 feet straight up into the air. v = √ 2 . Round your answer to two decimal places. −2 and w = 1. v = 19.1 Exercises In Exercises 1 . −5 and w = 10. v = 1. −91 and w = 0. − 23 2 √ √ 2 2 2 . v = 5ˆ + 12ˆ and w = −3ˆ + 4ˆ ı  ı  √ 2 . • v·w • The angle θ (in degrees) between v and w 1. v = 3. v = 20. v = −8.11. v = 3. − 2 21.− 2 √ and w = − and w = − 18. Suppose Taylor fills her wagon with rocks and must exert a force of 13 pounds to pull her wagon across the yard. v = −6. 24. v = −3 3. −5 11.9 The Dot Product and Projection 1041 11. use the pair of vectors v and w to find the following quantities.9. −7 12. compute how much work Taylor does pulling her wagon 25 feet. v = −24ˆ + 7ˆ and w = 2ˆ ı  ı 16. 22 2 √ 2 2 2 . −12 4. Assume the force of gravity is acting straight downwards. v = −2. 22 2 and w = and w = √ 1 . − 3 5.8. v = −5.) 2. 23. . −8 √ √ 6. The stronger of the two students pulls with a force of 100 pounds on a rope which makes a 13◦ angle with the direction of motion. −1 8. −7 and w = 5.20. If she maintains a 15◦ angle between the handle of the wagon and the horizontal. v = 1. 3 and w = 2. −9 √ √ 3. v = √ 1 . 6 and w = 4. v = −2. 6 7. 12 10. v = 3ˆ −  and w = 4ˆ ı ˆ  15. 4 and w = 5. v = −4.) Find the work done by this student if the keg is dragged 42 feet. 6 13. v = 34. (In this case. 3 and w = 1. 0 9. 22. 17 and w = −1. 1 and w = 3.− 2 √ √ 3 1 2. Round your answer to two decimal places. Prove the distributive property of the dot product in Theorem 11. We can now establish a Triangle Inequality for vectors. d and just follow your nose! 6 It is also known by other names. 28. Check out this site for details. the force being applied acts directly upwards. (a) (Step 1) Show that u + v 2 = u 2 + 2u · v + v 2 . we prove that u + v ≤ u + v for all pairs of vectors u and v.1042 Applications of Trigonometry 25. Use the identity in Example 11. we can now show that the Triangle Inequality |z + w| ≤ |z| + |w| holds for all complex numbers z and w as well. 27. (e) As an added bonus. Finish the proof of the scalar property of the dot product in Theorem 11.22. This is the celebrated Cauchy-Schwarz Inequality.1 to prove the Parallelogram Law v 2 + w 2 = 1 2 v+w 2 + v−w 2 29.) (c) (Step 3) Show that u + v 2 = u 2 u v + v 2 = ( u + v )2 .5◦ incline. This means that the angle between the applied force in this case and the motion of the object is not the 12. b and identify the complex number w = c + di with the vector v = c.6 (Hint: To show this inequality. (b) (Step 2) Show that |u · v| ≤ u v . 2 + 2u · v + v 2 ≤ u 2 + 2|u · v| + v 2 ≤ u 2 + (d) (Step 4) Use Step 3 to show that u + v ≤ u + v for all pairs of vectors u and v. Identify the complex number z = a + bi with the vector u = a. . In this exercise. HINT: Since you are working to overcome gravity only. start with the fact that |u · v| = | u v cos(θ) | and use the fact that | cos(θ)| ≤ 1 for all θ. Ignore all forces acting on the barrel except gravity.2. Find the work done pushing a 200 pound barrel 10 feet up a 12. which acts downwards.5◦ of the incline! 26. We know that |x + y| ≤ |x| + |y| for all real numbers x and y by the Triangle Inequality established in Exercise 36 in Section 2.22.9. 4 and w = −6. −5 v·w =6 θ≈ q= 74. 9 2 8. v = −4.2 Answers 2.94◦ projw (v) = − 248 . 6 v·w =0 θ = 90◦ projw (v) = 0. 1 and w = 3. − 169 √ 7. 3 and w = 1. −7 v · w = −62 θ ≈ 169. 3 and w = − 3. 0 v · w = −1 θ≈ 93. −2 1. −5 4. 434 65 65 44 q = − 77 . 0 √ √ 6. −7 and w = 5. 0 q = −6. 0 q = −2. v = −6. 4 q = 0. 0 q = 0. −2 and w = 1. −8 v · w = −50 θ = 180◦ projw (v) = 3. − 65 65 projw (v) = .74◦ 3 15 13 . 169 192 80 169 . 1 projw (v) = − 3 2 3 .11. 4 and w = 5. v = 1. 6 and w = 4. v = −2.9 The Dot Product and Projection 1043 11. − 3 v · w = −2 θ= 120◦ √ 1 projw (v) = − 2 . −5 2 2 projw (v) = 0. −12 v·w =0 θ = 90◦ 9 5 2. v = 3. v = 3. − 13 − 55 . 2 5. − 11 13 13 10.9. v = −3 3.25◦ projw (v) = q= 315 756 169 . v = −5. √ 3 2 q= 3 3 2. 17 and w = −1. 17 9. − 3 2 q = −323. 12 v · w = 63 θ ≈ 14. v = −2. −1 v·w =6 θ = 60◦ √ √ √ 3. v = 1. −9 v · w = 53 θ = 45◦ projw (v) = q= −9. −5 and w = 10.37◦ projw (v) = 1. (1500 pounds)(300 feet) cos (0◦ ) = 450. 1+4 3 √ √ √ 2 . 25 √ √ √ 2 . v = v·w =− θ = 165◦ projw (v) = q= 3+1 4 . 2 projw (v) = 0. − 3+1 4 4 √ √ q = 1−4 3 . v = −8. −91 q = 34. 1−4 3 projw (v) = 13. v = 5ˆ + 12ˆ and w = −3ˆ + 4ˆ ı  ı  v · w = 33 θ ≈ 59. −1 q = 3. v = 90◦ 3 3 2.1044 11. − 23 2 2 √ √ v · w = 2− 6 4 ◦ θ = 105 √ √ √ √ 2− 6 3 2− 6 projw (v) = . 3−1 4 √ √ q = 1+4 3 . 4 3+1 4 21. (10 pounds)(3 feet) cos (0◦ ) = 30 foot-pounds . 22 2 √ √ v · w = 6− 2 4 ◦ θ = 75 √ √ projw (v) = 1−4 3 . −91 and w = 0. 8 8 √ √ √ √ q = 3 2+ 6 . 0 √ √ √ 1 . 000 foot-pounds 22. − 22 2 √ √ v · w = 6+ 2 4 ◦ θ = 15 √ √ 3+1 projw (v) = .74◦ projw (v) = −24. − 22 2 19. 6 v·w =2 θ≈ q= 87. v = 34.43◦ projw (v) = 0. 2+ 6 8 8 √ √ √ 1 . √ √ 3−1 1− 3 4 .49◦ projw (v) = − 99 . − 23 and w = 22 . v = −24ˆ + 7ˆ and w = 2ˆ ı  ı v · w = −48 θ ≈ 163. 23 and w = − 22 . 0 ı ˆ 15. 10 27 − 81 . 10 10 Applications of Trigonometry 12. v = 3ˆ −  and w = 4ˆ ı ˆ  v · w = −4 θ≈ 108. 0 14. 1 v · w = −91 θ ≈ 159. v = 224 168 25 . v = 3 ˆ + 3  and w = ˆ −  2ı 2ˆ v·w =0 θ= q= 17.51◦ projw (v) = 0. 22 and w = 1 . 0 q = 0. 7 16. 132 25 25 q= 18. 3 and w = 2. v = 3 1 2 .2 √ and w = − √ 6+ 2 4 √ √ 20.88◦ 1 3 10 . (200 pounds)(10 feet) cos (77. (13 pounds)(25 feet) cos (15◦ ) ≈ 313.11.35 foot-pounds 25.5◦ ) ≈ 432.92 foot-pounds 24.88 foot-pounds 1045 . (100 pounds)(42 feet) cos (13◦ ) ≈ 4092.9 The Dot Product and Projection 23. y P (x. However. As we shall see.1046 Applications of Trigonometry 11. neither represent y as a function of x nor x as a function of y. we imagine a bug crawling across a table top starting at the point O and tracing out a curve C in the plane. In this case. The angular frequency ω determines ‘how fast’ the 1 Note the use of the indefinite article ‘a’. then loops back around to cross its path2 at the point Q and finally heads off into the first quadrant. when plotted in the xy-plane. y) at any given time t. travels upwards to the left.1 The parametrization of C endows it with an orientation and the arrows on C indicate motion in the direction of increasing values of t. In this section. the bug reaches the point Q at two different times.) . our bug starts at the point O.) The independent variable t in this case is called a parameter and the system of equations x = f (t) y = g(t) is called a system of parametric equations or a parametrization of the curve C. if ω < 0. giving it a counter-clockwise orientation. g(t)) 5 4 3 2 1 1 2 3 4 5 O x Q The curve C does not represent y as a function of x because it fails the Vertical Line Test and it does not represent x as a function of y because it fails the Horizontal Line Test. If ω > 0. f (t) is used for x and g(t) is used for y. the equations of circular motion {x = r cos(ωt). 2 Here.56 in Section 1. y = sin(t) parametrizes the Unit Circle.5. y) = (f (t). To motivate the idea. the system of equations {x = cos(t). there are scores of interesting curves which. The parametrization determines the orientation and as we shall see. More generally. By definition. y = r sin(ωt) developed on page 732 in Section 10. but not necessarily) different function of t. Chapter 7 and most recently in Section 11. (Think about this before reading on. since the bug can be in only one place P (x. the orientation is counterclockwise. as shown below. the orientation is clockwise. it should. there are infinitely many different parametric representations for any given curve. It is important to note that the curve itself is a set of points and as such is devoid of any orientation.10 Parametric Equations As we have seen in Exercises 53 . it shows that neither f nor g can be one-to-one.1 are parametric equations which trace out a circle of radius r centered at the origin.2. If all of this seems hauntingly familiar.2. we can define the x-coordinate of P as a function of t and the y-coordinate of P as a (usually. different parametrizations can determine different orientations. we present a new concept which allows us to use functions to study these kinds of curves. (Traditionally. While this does not contradict our claim that f (t) and g(t) are functions of t. 7 to eliminate the parameter t and get an equation involving just x and y. we can use the technique of substitution as described in Section 8.10. we can never accurately describe the curve once we’ve eliminated the parameter. (y + = 4(x + 3). y = 2t − 1 Solution.1 certainly looks like a parabola. −1) −2 1 (−2. as required. and the presence of the t2 term in the equation x = t2 − 3 reinforces this hunch. we choose to solve the equation y = 2t − 1 for t to get t = y+1 2 .11. 3) 6 5 (6. In particular. Since the parametric equations x = t2 − 3.2. Sketch the curve described by y 5 t −2 −1 0 1 2 3 x(t) y(t) (x(t).1. y = 2960 sin 12 t that model the motion of Lakeland Community College as the earth rotates (see Example 10. while the parametric equations x = t2 − 3. then the usual techniques of substitution and elimination as learned in Section 8. In this case. can be a great help in graphing curves determined by parametric equations. Substituting this into the equation x = t2 − 3 yields x = 1)2 y+1 2 2 − 3 or.10 Parametric Equations 1047 π π object moves around the circle. we see that the graph of this equation is a parabola with vertex (−3. Technically speaking. If the system of parametric equations contains algebraic functions. The one piece of information we can never recover after eliminating the parameter is the orientation of the curve. 1) 1 3 (1. Thinking back to Section 7. Eliminating the parameter and obtaining an equation in terms of x and y.10.2) parameterize a circle of radius 2960 with a counter-clockwise rotation which completes one revolution as t runs through the interval [0. y(t)) 1 −5 (1.7 in Section 10. plot the corresponding points and connect the results in a pleasing fashion. x = t2 − 3 for t ≥ −2. we start there and as we plot successive points.1.3. Example 11. 3 . the equation (y + 1)2 = 4(x + 3) coupled with the restriction y ≥ −5 describes the same curve as the given parametric equations. y = 2t − 1 given to describe this curve are a system of equations. the equation (y + 1)2 = 4(x + 3) describes the entire parabola. as was the case in Example 11. 5) 4 3 2 1 −2−1 −1 −2 −3 −5 1 2 3 4 5 6 x The curve sketched out in Example 11. the equations x = 2960 cos 12 t . −5) −2 −3 (−2. whenever possible. To do so. after some rearrangement.3 we can remedy this situation by restricting the bounds on y. 24). −3) −3 −1 (−3. Since the portion of the parabola we want is exactly the part where y ≥ −5.7 can be applied to the We will have an example shortly where no matter how we restrict x and y. We follow the same procedure here as we have time and time again when asked to graph anything new – choose friendly values of t.10. −1) which opens to the right. we draw an arrow to indicate the direction of the path for increasing values of t. Since we are told t ≥ −2. It is time for another example. y = 2t − 1 for t ≥ −2 describe only a portion of the parabola. since that corresponds to a relative minimum4 on the graph of y = 2t2 . Sketch the curves described by the following parametric equations. 4. 1].6. Next. t = −1 and t = 1 are good values to pick since these are the extreme values of t. In this case. We note that as t takes on values in the interval [−1. 2) (where t = −1). ‘decreasing’ and ‘relative minimum’ mean. x = t3 ranges between −1 and 1. this means that in order to trace out the path described by the parametric equations.2. Our experience in Section 5. we √ eliminate the parameter. Substituting this into y = 2t2 √ 2 gives y = 2( 3 x) = 2x2/3 . 1]. y = g(t) to eliminate the parameter. We continue to move to the right (since x is still increasing) but now move upwards (since y is now increasing) until we reach (1. y = 2t2 we first sketch the graphs of x = t3 and y = 2t2 over the interval [−1. To get a feel for the curve described by the system x = t3 . y) | − 1 ≤ x ≤ 1. we see that x = t3 is increasing over the entire interval [−1. More generally. . 1. 1. Geometrically. Plugging in t = −1 gives the point (−1. 0) and t = 1 gives (1. the strategy changes slightly. We demonstrate these techniques in the following example. −1 ≤ t ≤ 1 x = t3 . 0) (where t = 0). 2). on the other hand. Solving x = t3 for t. This means that all of the action is happening on a portion of the plane. Example 11. to get a good sense of the shape of the curve. it is often best to solve for the trigonometric functions and relate them using an identity. we start at (−1. x = sin(t) for 0 < t < π y = csc(t) x = 1 + 3 cos(t) for 0 ≤ t ≤ y = 2 sin(t) 3π 2 Solution. We also choose t = 0.1 if you’ve forgotten what ‘increasing’. then move to the right (since x is increasing) and down (since y is decreasing) to (0. 0 ≤ y ≤ 2}. t = 0 gives (0. 2). Certainly. we plot a few points to get a sense of the position and orientation of the curve. Finally. y = 2t2 . namely {(x. 1] whereas y = 2t2 is decreasing over the interval [−1. we get t = 3 x. the parametrization involves the trigonometric functions.3 yields the graph of our final answer below. −1 ≤ t ≤ 1 You should review Section 1. and y = 2t2 ranges between 0 and 2. x = t3 for −1 ≤ t ≤ 1 y = 2t2 x = e−t for t ≥ 0 y = e−2t 3. 1]. 0] and then increasing over [0. 2) (where t = 1). 2. x 1 2 y y 2 1 −1 −1 −1 x = t3 .10. If. −1 ≤ t ≤ 1 4 1 1 t 1 t −1 1 x y = 2t2 .1048 Applications of Trigonometry system {x = f (t). but never reaches. (because y = csc(t) is undefined at these t-values). We find that the range of x in this case is (0. 1). Or. ∞). 3 9 we take the time to analyze the end behavior of both x and y. and since x = 2e−t means e−t = x . and we start to head to the left. we could 2 4 2 2 2 recognize that y = e−2t = e−t . Note the open circle at the origin. we see that t = π gives the point 1 . 1].1 and 6. Next. we get y = x = x 2 2 4 this way as well. Either way. the graph of x = 2e−t . 6 5 . 2 1 This means that we are moving to the right and downwards. 1) (where t = 0). through 2 . π). Since t = 0 and t = π aren’t included in the 6 domain for t. we get that for t near 0. Piecing all of this information together. 4 x 2 y y 1 1 1 1 x = 2e−t . we know that our final graph will start at (2. Substituting 2 2 2 2 this for t in y = e−2t gives y = e−2(− ln(x/2)) = e2 ln(x/2) = eln(x/2) = x = x . 1) and heads towards. ∞). Once t = 2 . the orientation reverses. We pass back 1 through 2 . Starting with t = ln(1) = 0 we get5 (2.1 for a review of this concept. we start by graphing x = sin(t) and y = csc(t) over the interval (0.1 in Section 4. Plotting a few friendly points. 0). we plug in some friendly values of t to get a sense of the orientation of the curve. t = π 6 2 2 1 gives (1. 2 when t = π to 6 π π (1. but very large positive y-values. we have points with very small positive x-values. we get x = sin(t) → 0+ and y = csc(t) → ∞. For the system {x = sin(t). We find that the range of x is (0. since y = csc(t) is now increasing. As t → ∞. y = e−2t approaches the point (0.2 as needed. We find that as t → 0+ as well as when t → π − . y = e−2t for t ≥ 0 is a portion of the 2 parabola y = x which starts at the point (2. 1 .1 on page 22 and Theorem 4. 1] while the range of y is [1. Since both x = 2e−t and y = e−2t are always decreasing for t ≥ 0. This means the graph of x = 2e−t . π . since x = sin(t) is now decreasing.11. y = csc(t) for 0 < t < π. we proceed as in the previous example and graph x = 2e−t and y = e−2t over the interval [0. 1) and t = 5π returns us to 2 . y = e−2t . ‘friendly’ values of t involve natural logarithms. For the system x = 2e−t . Since t is ranging over the unbounded interval [0. Since t lies in the exponent here. y = e−2t for t ≥ 0. See the solution to part 3 in Example 1. 2 . ∞). and move consistently to the left (since x is decreasing) and down (since y is decreasing) to approach the origin. 0). x = sin(t) is increasing and y = csc(t) is decreasing. 4 and for t = ln(3) we get 2 . 2 when t = 5π back to the points with small positive x-coordinates and large 6 The reader is encouraged to review Sections 6. and up.2. To eliminate the parameter. As t ranges through the interval 0. t ≥ 0 2 t 1 y = e−2t . one way to proceed is to solve x = 2e−t for t to get t = − ln x . 2 .6 (0. 2] and the range of y is (0. 1) when t = 2 . t ≥ 0 3. x = 2e−t → 0+ and y = e−2t → 0+ as well. t ≥ 0 2 t 1 2 x x = 2e−t . for t = ln(2) we get 1 1.10 Parametric Equations 1049 2. we analyze the behavior of the system as t approaches 0 and π. 0 < t < π 4. 0) to (1. From Section 3 2 9 7. namely cos(t) and sin(t). Our parametric equations here are tracing out three-quarters of this ellipse. 2 while y = 2 sin(t) is increasing. For π ≤ t ≤ π. 3π . 0 ≤ t ≤ 3π 2 y = 2 sin(t). Since x = sin(t). the curve traced out by this 1 parametrization is a portion of the graph of y = x . x is decreasing. we are hugging the vertical asymptote x = 0 of 1 the graph of y = x . 0) and (4. Hence. Hence. To eliminate the parameter here. respectively. y y x 1 π 2 3 2 1 π t π 2 1 π t x 1 {x = sin(t). while y continues to decrease. Proceeding as above. y = 2 sin(t) . or (x−1) + y4 = 1. We see that the parametrization given above traces out the portion of 1 y = x for 0 < x ≤ 1 twice as t runs through the interval (0. 3π . 0) with vertices at (−2. 2). as is y. 0) and continue moving to the left and upwards towards (1. 0) with a minor axis of length 4.4. As t ranges from 0 to π . π). y = csc(t) . −2). 2). 0) and (1. and we solve y = 2 sin(t) for sin(t) to get sin(t) = y . Using a 1 reciprocal identity. the motion becomes left 2 to right but continues downwards. 2 2 (1. 0 ≤ t ≤ {x = 1 + 3 cos(t). 0). π . we write y = csc(t) = sin(t) . On the interval π. we set about graphing {x = 1 + 3 cos(t). We now can explain the unusual behavior as t → 0+ and t → π − – for these values of t. connecting (−2. but now is downwards from (1. x = 1 + 3 cos(t) is decreasing. (−2. 0 < t < π y = csc(t). 2) to (−2. We see that x ranges 2 from −2 to 4 and y ranges from −2 to 2. This means that we start tracing out our answer at (4. −2). x 4 3 y 2 1 π 2 2 2 2 2 y 2 1 −1 −1 −2 3π 2 2 1 π 3π 2 t −1 −2 −1 −2 π 2 π 3π 2 t 1 2 3 4 x x = 1 + 3 cos(t). x begins to increase.1050 Applications of Trigonometry positive y-coordinates. π and 3π gives the points (4. 0). 0 ≤ t ≤ 3π 2 . y = 2 sin(t) for 0 ≤ t ≤ 3π by 2 first graphing x = 1 + 3 cos(t) and y = 2 sin(t) on the interval 0. are related by the Pythagorean Identity cos2 (t) + sin2 (t) = 1. Plugging in t = 0. we eliminate the parameter. in a counter-clockwise direction. we know that the graph of this equation is an ellipse centered at (1. we note that the trigonometric functions involved. 2 so the motion is still right to left. we solve x = 1 + 3 cos(t) for cos(t) to get cos(t) = x−1 . Substituting these 3 2 expressions into cos2 (t)+sin2 (t) = 1 gives x−1 + y = 1. To better explain this behavior. 0 < t < π x = sin(t). Find a parametrization for each of the following curves and check your answers. y = x2 from x = −3 to x = 2 2. b > 0. Doing so gives us the equation x = y 5 + 2y + 1. in order to trace out the entire graph of x = f (y) = y 5 + 2y + 1. Hence. we follow the procedure outlined on page 384 – we start with the equation y = f (x). we turn to the problem of finding parametric representations of curves. −3) and ends at (1. y = t for −∞ < t < ∞. Our final answer to this problem is x = t5 + 2t + 1. The check is almost trivial. The line segment which starts at (2. While we could attempt to solve this equation for y. we don’t need to. (Otherwise.10. 5) 4. Since y = x2 is written in the form y = f (x).) To find a formula y = f −1 (x). we need to let y run through all real numbers. 2.11. y = f −1 (x) where f (x) = x5 + 2x + 1 3. The circle x2 + 2x + y 2 − 4y = 4 5. the range of y = f (x) = x5 + 2x + 1 is (−∞. with x = t we have y = t2 = x2 as t = x runs from −3 to 2.10 Parametric Equations 1051 Now that we have had some good practice sketching the graphs of parametric equations. checking the orientation. let x = g(t) and y = t and let t run through I. 1. The left half of the ellipse Solution. y1 ). our solution is trivial to check. of course.7 7 2 2 x2 4 + y2 9 =1 Provided you followed the inverse function theory. ∞). let x = x0 + (x1 − x0 )t and y = y0 + (y1 − y0 )t for 0 ≤ t ≤ 1. the bounds on t match precisely the bounds on x so we get x = t. y0 ) and terminal point (x1 . let x = t and y = f (t) and let t run through I. We start with the following. .1 that since f (x) = x5 + 2x + 1 is an odd-degree polynomial. We put these formulas to good use in the following example. • To parametrize (x−h) + (y−k) = 1 where a. f −1 would not exist. let x = h + a cos(t) and y = k + b sin(t) a2 b2 for 0 ≤ t < 2π. • To parametrize a directed line segment with initial point (x0 . We know from our work in Section 3. As in the previous problem. when indicated. (This will impart a counter-clockwise orientation. interchange x and y and solve for y. y = t2 for −3 ≤ t ≤ 2. • To parametrize x = g(y) as y runs through some interval I. 1.3. We are told to parametrize y = f −1 (x) for f (x) = x5 + 2x + 1 so it is safe to assume that f is one-to-one. Since x = t. Example 11. Parametrizations of Common Curves • To parametrize y = f (x) as x runs through some interval I. We can parametrize x = f (y) = y 5 + 2y + 1 by setting y = t so that x = t5 + 2t + 1.) The reader is encouraged to verify the above formulas by eliminating the parameter and. we let x = t and y = f (t) = t2 . Substituting this into y = −3 + 8t gives y = −3 + 8t = −3 + 8(2 − x). While these equations at first glance are quite a handful. y = −3 + 8t for 0 ≤ t ≤ 1. 5). Once again. we opt for a more direct approach. To find the equation for x. 2π). Rearranging 9 9 3 3 these last two equations. as required. for an initial point of (2.8 they can be summarized as ‘starting point + (displacement)t’. and when t = 1. . Plugging in x = 2 gives y = −8(2) + 13 = −3. the displacement in the y-direction is (5 − (−3)) = 8. we let t range through the interval [0.2 to show that the entire circle is covered as t ranges through the interval [0. x = 2 − t = 1. so we get y = −3 + 8t. x2 + 2x + y 2 − 4y = 4 (−1 + 3 cos(t))2 + 2(−1 + 3 cos(t)) + (2 + 3 sin(t))2 − 4(2 + 3 sin(t)) = 4 1 − 6 cos(t) + 9 cos2 (t) − 2 + 6 cos(t) + 4 + 12 sin(t) + 9 sin2 (t) − 8 − 12 sin(t) = 4 9 cos2 (t) + 9 sin2 (t) − 5 = 4 9 cos2 (t) + sin2 (t) − 5 = 4 9 (1) − 5 = 4 4 = 4 Now that we know the parametric equations give us points on the circle. To parametrize line segment which starts at (2. x = 2 − t = 2. we identify cos(t) = x+1 and sin(t) = y−2 . We substitute x = −1 + 3 cos(t) and y = 2 + 3 sin(t) into the equation x2 + 2x + y 2 − 4y = 4 and show that the latter is satisfied for all t such that 0 ≤ t < 2π. we have that the line segment starts at x = 2 and ends at x = 1. the equation for x is x = 2 + (−1)t = 2 − t. and then manipulating the resulting equation in x and y into the original equation x2 + 2x + y 2 − 4y = 4. 4. but they come from the Pythagorean Identity cos2 (t) + sin2 (t) = 1. invoking a Pythagorean Identity. or y = −8x + 13.1052 Applications of Trigonometry 3. so all we need to check is that it starts and stops at the correct points. Hence. we get (x + 1)2 + (y − 2)2 = 9.8. Plugging in x = 1 gives y = −8(1) + 13 = 5 for an ending point of (1. In order to use the formulas above to parametrize the circle x2 +2x+y 2 −4y = 4. 2π). we can solve x = 2 − t for t to get t = 2 − x. To check. the formulas x = h + a cos(t) and y = k + b sin(t) can be 9 9 a challenge to memorize. we make use of the formulas x = x0 +(x1 −x0 )t and y = y0 +(y1 −y0 )t for 0 ≤ t ≤ 1. We get as our final answer {x = −1 + 3 cos(t). After completing the squares. we could eliminate the parameter by solving x = −1 + 3 cos(t) for cos(t) and y = 2 + 3 sin(t) for sin(t). 5). Hence. we get x = −1 + 3 cos(t) and y = 2 + 3 sin(t). For y. y = 2 + 3 sin(t) for 0 ≤ t < 2π. When t = 0. Our final answer is {x = 2 − t. To check our answer. we first need to put it into the correct form. −3) and ends at (1. In order to complete one revolution around the circle. Instead. 8 ? ? ? ? ? Compare and contrast this with Exercise 65 in Section 11. 2 2 or (x+1) + (y−2) = 1.10. we note that the line segment starts at y = −3 and ends at y = 5. we can go through the usual analysis as demonstrated in Example 11. −3). 2 2 In the equation (x+1) + (y−2) = 1. This means the displacement in the x-direction is (1 − 2) = −1. We know this is the graph of a line. Two easy ways to alter parametrizations are given below. We note that the formulas given on page 1051 offer only one of literally infinitely many ways to parametrize the common curves listed there. 2. so we need to introduce a ‘time delay’ of 2. This parametrization. we restrict t to the values which correspond to Quadrant II and Quadrant III angles. with t = 0 corresponding to (0.10 Parametric Equations 2 2 1053 5. 4). • Shift of Parameter: Replacing every occurrence of t with (t − c) in a parametric description for a curve (including any inequalities which describe the bounds on t) shifts the start of the parameter t ahead by c units. We would like t to begin at t = 0 instead of t = −2. y = 3 sin(t) lie on the ellipse x + y9 = 1. We can parametrize y = x2 from x = −1 to x = 2 using the formula given on Page 1051 as x = t. 1) and ends at (2. namely 3π π 3π π 2 ≤ t ≤ 2 . 0). y = t2 − 4t + 4 for 0 ≤ t ≤ 3. The problem here is that the parametrization we have starts 2 units ‘too soon’. The curve which starts at (2. y = t2 for −2 ≤ t ≤ 1. Our final answer is {x = 2 cos(t). After simplifying. y = (t − 2)2 for −2 ≤ t − 2 ≤ 1. oriented clockwise. we replace every occurrence of t with −t to get x = −t. 1). y = t2 for −1 ≤ t ≤ 2. then travels along a line to (5. however. as required. we find that the restriction π ≤ t ≤ 3π 2 2 generates the left half of the ellipse. y = 3 sin(t) for 2 ≤ t ≤ 2 . but since we are interested in only the left half of the ellipse. we can either use the formulas above or think back to the 4 Pythagorean Identity to get x = 2 cos(t) and y = 3 sin(t). Shift the parameter so that the path starts at t = 0.2. y = (−t)2 for −1 ≤ −t ≤ 2. Replacing every occurrence of t with (t − 2) gives x = −(t − 2). 1.11. Find a parametrization for the following curves. which reduces 4 4 9 to the Pythagorean Identity cos2 (t) + sin2 (t) = 1. The Unit Circle. the formulas offered there need to be altered to suit the situation. Solution. 3. Employing 4 the techniques demonstrated in Example 11. We demonstrate these techniques in the following example. This proves that the points generated by 2 2 the parametric equations {x = 2 cos(t). Example 11. we need to reverse the orientation. Hence. 2 2 2 2 . 0). In the equation x + y9 = 1. starts at (−1.10. 4) and follows the parabola y = x2 to end at (−1. The normal range on the parameter in this case is 0 ≤ t < 2π. we get x = −t.4. Adjusting Parametric Equations • Reversing Orientation: Replacing every occurrence of t with −t in a parametric description for a curve (including any inequalities which describe the bounds on t) reverses the orientation of the curve.10. 1. Substituting x = 2 cos(t) and y = 3 sin(t) into x + y9 = 1 gives 4 cos (t) + 9 sin (t) = 1. The two part path which starts at (0. To do so. −1). At times. 4). Simplifying yields x = 2 − t. travels along a line to (3. Our current description of the second part starts at t = 0. For the first part of the path. We put our answer to Example 11. we get {x = 3t. for 0 ≤ t ≤ 1 8 − 4t. This parametrization gives a clockwise orientation.4 number 3 to good use to derive the equation of a cycloid. we think: ‘starting point + (displacement)t’. the point (0. for 1 ≤ t ≤ 2 and g(t) = 4t. y = − sin(t) 2 2 for − 3π ≤ t < π traces out the Unit Circle clockwise starting at the point (0. as required. Simplifying yields {x = 1 + 2t. Let θ be the angle in radians which measures the amount of clockwise rotation experienced by the radius highlighted in the figure. We now 2 2 shift the parameter by introducing a ‘time delay’ of 3π units by replacing every occurrence 2 of t with t − 3π . y) r θ x 9 10 courtesy of the Even/Odd Identities courtesy of the Sum/Difference Formulas . so the first order of business is to reverse the orientation. This 2 2 2 2 2 2 simplifies10 to {x = − sin(t). 0). but t = 0 still corresponds to the point (1. we may parametrize the path as {x = f (t). The reader can verify that {x = cos(t).10. We get x = cos t − 3π . and for the second part we get {x = 3 + 2t. We know that any interval of length 2π will parametrize the entire circle. y = − cos(t) for 0 ≤ t < 2π. we shift the parameter in the second part so it starts at t = 1. y = sin(t) for 0 ≤ t < 2π gives a counter-clockwise parametrization of the Unit Circle with t = 0 corresponding to (1.1054 Applications of Trigonometry 2. Our strategy is to first get the parametrization to ‘start’ at the 2 point (0. y = g(t) for 0 ≤ t ≤ 2 where f (t) = 3t. 0). −1) and then shift the parameter accordingly so the ‘start’ coincides with t = 0. y = − sin(t) . y = 4 − 4(t − 1) for 0 ≤ t − 1 ≤ 1. y = − sin t − 3π for − 3π ≤ t − 3π < π . Replacing t with (t − 1) in the second set of parametric equations gives {x = 3 + 2(t − 1). y P (x. y = 8 − 4t for 1 ≤ t ≤ 2. for 1 ≤ t ≤ 2 3. so we keep the equations {x = cos(t). Suppose a circle of radius r rolls along the positive x-axis at a constant velocity v as pictured below. y = − sin(t) for −2π < t ≤ 0. so we introduce a ‘time delay’ of 1 unit to the second set of parametric equations. Hence. −1). −1) is reached when t = − 3π . and find the upper 2 bound by adding 2π so − 3π ≤ t < π . y = sin(−t) for 0 ≤ −t < 2π. y = 4t for 0 ≤ t ≤ 1. Replacing t with −t gives {x = cos(−t). y = 4 − 4t for 0 ≤ t ≤ 1. which simplifies9 to {x = cos(t). When parameterizing line segments. but start the parameter t at − 3π . Since the first parametrization leaves off at t = 1. for 0 ≤ t ≤ 1 1 + 2t. We know that {x = cos(t). we set θ ≥ 0.10 Parametric Equations 1055 Our goal is to find parametric equations for the coordinates of the point P (x. we obtain x + y = 1 r r r r which reduces to x2 + y 2 = r2 . (We have renamed the parameter ‘θ’ to match the context of this problem. the center of the circle is at the point (vt.and y-coordinates of points on the graph by r. Hence. y = − cos(θ) for 0 ≤ θ < 2π. Since the velocity v is constant. In the language of Section 1. From our work in Example 11. in terms of the parameter θ.5. the center of the circle.1. all we need to do11 is multiply both x and y by the factor r which yields {x = −r sin(θ). we are stretching the graph by a factor of r in both the x. we use a graphing utility. As always. which can be written as {x = r(θ − sin(θ)). 0). 12 Does this seem familiar? See Example 11. t is (rθ. As a result. y = r(1 − cos(θ)) . r). 11 2 2 . Since the motion starts at θ = 0 and proceeds indefinitely. Using a calculator to graph parametric equations is very similar to graphing polar equations on a calculator. we have that at time t. Graph your answer using a calculator.and y-directions. From Section 10.11. −1) can be modeled by the equations {x = − sin(θ). but for the purposes of this book. we know that clockwise motion along the Unit Circle starting at the point (0. we know v = rθ .7. we multiply both the x. y = −r cos(θ) . We know that one full revolution of the circle occurs over the interval 0 ≤ t < 2π. We have r = 3 which gives the equations {x = 3(t − sin(t)). since the radius of the circle is r and the circle isn’t moving vertically. We end the section with a demonstration of the graphing calculator. Solution. or vt = rθ.) To model this motion on a circle of radius r.13 Ensuring that the calculator is in ‘Parametric Mode’ and ‘radian mode’ we enter the equations and advance to the ‘Window’ screen. so If we replace x with x and y with y in the equation for the Unit Circle x2 + y 2 = 1.1. 13 See page 957 in Section 11. we know that the center of the circle is always r units above the x-axis. but rather. y = −r cos(θ) + r .1.1.1 in Section 11.10. Furthermore. (Here we have returned to the convention of using t as the parameter. We now need to adjust for the fact that the circle isn’t stationary with center (0. t.) Sketching the cycloid by hand is a wonderful exercise in Calculus. Find the parametric equations of a cycloid which results from a circle of radius 3 rolling down the positive x-axis as described above. Example 11. is rolling along the positive x-axis. the challenge is to determine appropriate bounds on the parameter. We get {x = −r sin(θ) + rθ.4 number 3. y = −r cos(θ) by shifting the x-coordinate to the right rθ units (by adding rθ to the expression for x) and the y-coordinate up r units12 (by adding r to the expression for y). the center of the circle has traveled a distance vt down the positive x-axis.10.5. y) in terms of θ. r). we need to modify the equations {x = −r sin(θ). Hence. y = 3(1 − cos(t)) for t ≥ 0. we know that at time t. Putting these two facts together. as well as for x and y. r). see page 957 in Section 11. so y ranges between 0 and 6. Below we graph the cycloid with these settings. The reader is invited to use the Zoom Square feature on the graphing calculator to see what window gives a true geometric perspective of the three arches.14 We know from our derivation of the equations of the cycloid that the center of the generating circle has coordinates (rθ. The values of y go from the bottom of the circle to the top. (It is instructive to note that keeping the y settings between 0 and 6 messes up the geometry of the cycloid. and then extend t to range from 0 to 6π which forces x to range from 0 to 18π yielding three arches of the cycloid.) 14 Again. . Since t ranges between 0 and 2π. we set x to range between 0 and 6π. 3).5. or in this case.1056 Applications of Trigonometry it seems reasonable to keep these as our bounds on t. (3t. The ‘Tstep’ seems reasonably small – too large a value here can lead to incorrect graphs. x = 4t − 3 for 0 ≤ t ≤ 1 y = 6t − 2 x = 2t for − 1 ≤ t ≤ 2 y = t2 x = t2 + 2t + 1 y =t+1 for t ≤ 1 2. 18 − t2 for t ≥ −3 7. plot the set of parametric equations with the help of a graphing utility.10. x = t3 − 3t for − 2 ≤ t ≤ 2 y = t2 − 4 x = et + e−t for − 2 ≤ t ≤ 2 y = et − e−t 22. 6.20. 18. 10. 5. 20. 16. 11. 24. 14. Be sure to indicate the orientation imparted on the curve by the parametrization. In Exercises 21 . x = 4 cos3 (t) for 0 ≤ t ≤ 2π y = 4 sin3 (t) x = cos(3t) for 0 ≤ t ≤ 2π y = sin(4t) 23.11. x=t for − ∞ < t < ∞ y = t3 π π x = cos(t) for − ≤ t ≤ y = sin(t) 2 2 x = −1 + 3 cos(t) for 0 ≤ t ≤ 2π y = 4 sin(t) π x = 2 cos(t) for 0 < t < y = sec(t) 2 π π x = sec(t) for − < t < y = tan(t) 2 2 π π x = tan(t) for − < t < y = 2 sec(t) 2 2 x = cos(t) for 0 ≤ t ≤ π y=t 8. 13. 1. 19. 12. Be sure to indicate the orientation imparted on the curve by the parametrization.1 Exercises In Exercises 1 . 17. 4. plot the set of parametric equations by hand. x = 4t − 1 for 0 ≤ t ≤ 1 y = 3 − 4t x=t−1 for 0 ≤ t ≤ 3 y = 3 + 2t − t2 x= y= 1 9 1 3t 3. 15. x = t3 for − ∞ < t < ∞ y=t x = 3 cos(t) for 0 ≤ t ≤ π y = 3 sin(t) π x = 3 cos(t) for ≤ t ≤ 2π y = 2 sin(t) + 1 2 π x = 2 tan(t) for 0 < t < y = cot(t) 2 π 3π x = sec(t) for < t < y = tan(t) 2 2 π 3π x = tan(t) for < t < y = 2 sec(t) 2 2 π π x = sin(t) for − ≤ t ≤ y=t 2 2 9. . 21.10 Parametric Equations 1057 11.24. ) 29. . 41. y = r sin(θ) = f (θ) sin(θ) . (3. 0). oriented counter-clockwise 32.39. oriented counter-clockwise 38. oriented counter-clockwise 36. −5) to (−2.1058 Applications of Trigonometry In Exercises 25 . Every polar curve r = f (θ) can be translated to a system of parametric equations with parameter θ by {x = r cos(θ) = f (θ) cos(θ). the ellipse (x − 1)2 + 9y 2 = 9. find a parametric description for the given oriented curve. 25.1 to find the parametric equations which model a passenger’s position as they ride the London Eye. 30. the directed line segment from (3. the circle (x − 1)2 + y 2 = 4. 3). 0) (Shift the parameter so t = 0 corresponds to (−2. the ellipse 9x2 + 4y 2 + 24y = 0. the triangle with vertices (0. the curve y = 4 − x2 from (−2. 2) 26. the ellipse 9x2 + 4y 2 + 24y = 0. oriented clockwise (Shift the parameter so t = 0 corresponds to (0. −2). (0. 0). the curve x = y 2 − 9 from (−5.) 35. oriented counter-clockwise 34. the circle x2 + y 2 = 25. the circle x2 + y 2 − 6y = 0. 0). (Shift the parameter so t = 0 corresponds to (0.) 40. 0). 28.) 39. Use parametric equations and a graphing utility to graph the inverse of f (x) = x3 + 3x − 4. oriented clockwise (Shift the parameter so t begins at 0.5 and then graphing the parametric equations you found using a graphing utility. −4) 27. the curve x = y 2 − 9 from (0. the circle x2 + y 2 − 6y = 0. 4). oriented counter-clockwise 37. Check your answer by graphing r = 6 cos(2θ) by hand using the techniques presented in Section 11. Use your results from Exercises 3 and 4 in Section 11. 0). oriented counter-clockwise 33. −2) to (0. 42. 3) to (−5. 0) to (2. 0). 0) to (2. the curve y = 4 − x2 from (−2. −1) to (3. the circle (x − 3)2 + (y + 1)2 = 117.) 31. oriented counter-clockwise (Shift the parameter so t = 0 corresponds to (0. 3). Convert r = 6 cos(2θ) to a system of parametric equations. the directed line segment from (−2. (Here. the ‘hammer throw’. . 44. is launched into the air. the path of the projectile is given by15   x = v0 cos(θ) t 1  y = − gt2 + v0 sin(θ) t + s0 2 for 0 ≤ t ≤ T where v0 is the initial speed of the object. 0) x 43. We’ve seen this before. In one event. Eliminate the parameter in the equations for projectile motion to show that the path of the projectile follows the curve y=− g sec2 (θ) 2 x + tan(θ)x + s0 2 2v0 Use the vertex formula (Equation 2. θ is the angle from the horizontal at which the projectile is launched. Ignoring everything except the force gravity. s0 is the initial height of the projectile above the ground and T is the time when the object returns to the ground.16 g is the acceleration due to gravity. find the parametric equations for the flight of the hammer. Carl’s friend Jason competes in Highland Games Competitions across the country.) y θ s0 (x(T ). called a projectile.4) to show the maximum height of the projectile is y= 15 16 2 v0 sin2 (θ) + s0 2g when x = 2 v0 sin(2θ) 2g A nice mix of vectors and Calculus are needed to derive this.) When will the hammer hit the ground? How far away will it hit the ground? s2 Check your answer using a graphing utility. (See the figure below. . If the weight is released 6 feet above the ground at an angle of 42◦ with respect to the horizontal with an initial speed of 33 feet per second.10 Parametric Equations 1059 Suppose an object. It’s the angle of elevation which was defined on page 753. use g = 32 ft.11. he throws a 56 pound weight for distance. ∞).8 s2 and compare that to the formula for s(t) given in Exercise 25 in Section 2. (The projectile was launched vertically. In another event. 51.7. (Hence the use of the adjective ‘hyperbolic. Using a graphing utility as needed. we explore the hyperbolic cosine function. Suppose θ = π . Four other hyperbolic functions are waiting to be defined: the hyperbolic secant sech(t). defined below: cosh(t) = et + e−t 2 and sinh(t) = et − e−t 2 47. Show that {x(t) = cosh(t). Enjoy some nostalgia and revisit Exercise 35 in Section 6. denoted sinh(t). they should.) Simplify the general parametric 2 m formula given for y(t) above using g = 9.52. use your results from part 44 to determine how fast the sheaf was launched into the air. . Exercise 47 in Section 6. then convert them to formulas involving et and e−t . the hyperbolic cosecant csch(t). the hyperbolic tangent tanh(t) and the hyperbolic cotangent coth(t). . Define these functions in terms of cosh(t) and sinh(t).3 and the answer to Exercise 38 in Section 6. Using a graphing utility as needed. 52. use g = 32 ft. 49. 48. If the weight is released 5 feet above the ground at an angle of 85◦ with respect to the horizontal and the sheaf reaches a maximum height of 31. ∞) and the range of cosh(t) is [1. the ‘sheaf toss’.3. and the hyperbolic sine function.5. What is x(t) in this case? In Exercises 47 . If these functions look familiar.1060 Applications of Trigonometry 45. Compare the definitions of cosh(t) and sinh(t) to the formulas for cos(t) and sin(t) given in Exercise 83f in Section 11. verify that the domain and range of sinh(t) are both (−∞. Consult a suitable reference (a Calculus book. ∞). y(t) = sinh(t) parametrize the right half of the ‘unit’ hyperbola x2 − y 2 = 1. verify that the domain of cosh(t) is (−∞.5 feet. or this entry on the hyperbolic functions) and spend some time reliving the thrills of trigonometry with these ‘hyperbolic’ functions. Jason throws a 20 pound weight for height. denoted cosh(t).) s2 46.’) 50. (Once again.4. 2 Answers 1. x = 4t − 3 for 0 ≤ t ≤ 1 y = 6t − 2 y 4 3 2 1 −3 −2 −1 −1 −2 1 x 2. 1 x = 9 18 − t2 1 y = 3t y 2 1 −3 −2 −1 −1 1 2 for t ≥ −3 x . x = 2t for − 1 ≤ t ≤ 2 y = t2 y 4 3 2 1 −3 −2 −1 1 2 3 4 x 4.10 Parametric Equations 1061 11. x=t−1 for 0 ≤ t ≤ 3 y = 3 + 2t − t2 y 4 3 2 1 −1 1 2 x 5. x = t2 + 2t + 1 for t ≤ 1 y =t+1 y 2 1 1 −1 −2 2 3 4 5 x 6.11.10. x = 4t − 1 for 0 ≤ t ≤ 1 y = 3 − 4t y 3 2 1 −1 −1 1 2 3 x 3. π π x = cos(t) for − ≤ t ≤ y = sin(t) 2 2 y 1 10. π x = 3 cos(t) for ≤ t ≤ 2π y = 2 sin(t) + 1 2 y 3 2 1 −3 −1 −1 1 3 x .1062 x=t for − ∞ < t < ∞ y = t3 y 4 3 2 1 −1 −1 −2 −3 −4 1 x Applications of Trigonometry x = t3 for − ∞ < t < ∞ y=t y 1 −4 −3 −2 −1 −1 1 2 3 4 x 7. x = 3 cos(t) for 0 ≤ t ≤ π y = 3 sin(t) y 3 2 1 −1 1 x −3 −2 −1 −1 1 2 3 x 11. 8. x = −1 + 3 cos(t) for 0 ≤ t ≤ 2π y = 4 sin(t) y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 x 12. 9. π 3π x = sec(t) for < t < y = tan(t) 2 2 y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 x . 1 2 3 4 x 1 2 3 4 x 15.10 Parametric Equations π x = 2 cos(t) for 0 < t < y = sec(t) 2 y 4 3 2 1 4 3 2 1 1063 π x = 2 tan(t) for 0 < t < y = cot(t) 2 y 13.11. π π x = sec(t) for − < t < y = tan(t) 2 2 y 4 3 2 1 1 −1 −2 −3 −4 2 3 4 x 16. 14. x = t3 − 3t for − 2 ≤ t ≤ 2 y = t2 − 4 22. x = cos(t) for 0 < t < π y=t y π 20. x = 4 cos3 (t) for 0 ≤ t ≤ 2π y = 4 sin3 (t) y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x y −2 −1 −1 −2 −3 −4 1 2 x . π π x = sin(t) for − < t < y=t 2 2 y π 2 π 2 −1 1 x −π 2 −1 1 x 21. −2 −1 −1 −2 −3 −4 1 2 x −2 −1 1 2 x 19. 18.1064 π π x = tan(t) for − < t < y = 2 sec(t) 2 2 y 4 3 2 1 Applications of Trigonometry 3π π x = tan(t) for < t < y = 2 sec(t) 2 2 y 17. 11. 36. 2 ≤ t ≤ 3   0. 32. 34. 28. 27. 29. 1 ≤ t ≤ 2 y(t) =  12 − 4t. 25. y(t) where: 3t. 1 ≤ t ≤ 2 x(t) =  0. x = 5t − 2 for 0 ≤ t ≤ 1 y = −1 − 3t x=t−2 for 0 ≤ t ≤ 4 y = 4t − t2 x = t2 − 6t for 0 ≤ t ≤ 5 y =3−t x = 1 + 2 cos(t) for 0 ≤ t < 2π y = 2 sin(t) x = 3 cos(t) for 0 ≤ t < 2π y = 3 − 3 sin(t) x = 1 + 3 cos(t) for 0 ≤ t < 2π y = sin(t)   x = 2 cos t − π = 2 sin(t) 2 38. for 0 ≤ t < 2π  y = −3 − 3 sin t − π = −3 + 3 cos(t) 2 39. 24. {x(t). 37. 35. 33. 0 ≤ t ≤ 1 6 − 3t.10 Parametric Equations x = et + e−t for − 2 ≤ t ≤ 2 y = et − e−t y 7 5 3 1 −1 −3 −5 −7 −1 1 2 3 4 5 6 7 x −1 1 x 1065 x = cos(3t) for 0 ≤ t ≤ 2π y = sin(4t) y 1 23. 30. 0 ≤ t ≤ 1 4t − 4. 31. x = 3 − 5t for 0 ≤ t ≤ 1 y = −5 + 7t x=t for − 2 ≤ t ≤ 2 y = 4 − t2 x = t2 − 9 for − 2 ≤ t ≤ 3 y=t x = 5 cos(t) for 0 ≤ t < 2π y = 5 sin(t) x = 3 cos(t) for 0 ≤ t < 2π y = 3 + 3 sin(t) √ x = 3 + √ cos(t) 117 for 0 ≤ t < 2π y = −1 + 117 sin(t) x = 2 cos(t) for 0 ≤ t < 2π y = 3 sin(t) − 3 26. 2 ≤ t ≤ 3   . y = 6 cos(2θ) sin(θ) 42. .34 feet per second. To find when the hammer hits the ground. The parametric equations for the inverse are 41.61 seconds after it was launched into the air.5 − 67. 45.5 cos 15 t 2 43.48 feet from where it was thrown into the air.61. r = 6 cos(2θ) translates to x = 6 cos(2θ) cos(θ) for 0 ≤ θ < 2π. Since t ≥ 0.5 to get v0 = ±41.23 or 1. The parametric equations which describe the locations of passengers on the London Eye are π π x = 67. we find x(1.5 = 67.61) ≈ 39.1066 Applications of Trigonometry x = t3 + 3t − 4 for − ∞ < t < ∞ y=t 40. The parametric equations for the hammer throw are x = 33 cos(42◦ )t for y = −16t2 + 33 sin(42◦ )t + 6 t ≥ 0. we solve y(t) = 0 and get t ≈ −0.5 sin 15 t − π + 67.5 sin 15 t 2 for − ∞ < t < ∞ π π y = 67.34. The initial speed 2g 2(32) of the sheaf was approximately 41. We solve y = 2 v 2 sin2 (85◦ ) v0 sin2 (θ) + s0 = 0 + 5 = 31.5 cos 15 t − π = 67. To find how far away the hammer hits the ground. the hammer hits the ground after approximately t = 1. 667 applied domain of a function. 698 decimal degrees. 693 supplementary. 273 x-axis. 831 graph of. 831 1067 . 701 complementary. 698 vertex. 753 oriented. 696 coterminal. 6 absolute value definition of. 879 angle acute. 708 annuity annuity-due. 698 obtuse. 694 DMS. 6 y-intercept. 693 negative. 397 nth Roots of Unity. 894 angular frequency. 999 principal. 432 amplitude. 666 future value. 432 acute angle. 698 straight. 698 radian measure. 25 abscissa. 698 quadrantal.Index nth root of a complex number. 1033. 6 x-coordinate. 693 degree. 211 properties of. 830 properties of. 761 of elevation. 173 acidity of a solution pH. 693 angle side opposite pairs. 622 alkalinity of a solution pH. 697 positive. 696 terminal side. 761 of depression. 694 adjoint of a matrix. 25 y-axis. 60 arccosecant calculus friendly definition of. 695 initial side. 794. 6 y-coordinate. 1034 central angle. 6 x-intercept. 998. 694 between two vectors. 694 standard position. 698 measurement. 695 definition. 173 inequality. 694 of declination. 701 reference. 753 of inclination. 1004 u-substitution. 667 ordinary definition of. 721 right. 55 of a logarithm. body mass index. 498 of a hyperbola. 568 average angular velocity. 793 arithmetic sequence. 903 binomial coefficient. 585 scalar multiplication. 819 properties of. 827 properties of. 355 Boyle’s Law. 425 of a trigonometric function. 827 properties of. 304 location of. 830 properties of. 824 graph of. 824 properties of. 304 intuitive definition of. 683 Binomial Theorem. 820 arctangent definition of. 581 vector addition. 831 graph of. 824 argument of a complex number definition of. 820 properties of. 277 BMI.1068 trigonometry friendly definition of. 304 intuitive definition of. 820 graph of. 828 graph of. 1013 scalar multiplication. 6 Cartesian coordinates. 308 of a hyperbola. 823 properties of. 824 graph of. 350 buffer solution. 949 Cartesian coordinate plane. 820 arccotangent definition of. 269 center of a circle. 828 arccosine definition of. 828 arcsine definition of. 684 Bisection Method. 516 Index . 993 of a function. 346 average cost function. 366 matrix addition. 304 location of. 989 properties of. 824 arcsecant calculus friendly definition of. 312 formal definition of. 311 vertical formal definition of. 191 back substitution. 82 average rate of change. 560 bearings. 1016 asymptote horizontal formal definition of. 820 graph of. 6 Cauchy’s Bound. 531 slant determination of. 531 of an ellipse. 831 trigonometry friendly definition of. 306 augmented matrix. 160 average velocity. 707 average cost. 311 slant (oblique). 828 graph of. 579 matrix multiplication. 706 axis of symmetry. 478 cardioid. 654 associative property for function composition. 288 definition of. 752 properties of. 1011 composite function definition of. 6 polar. 773 common base. 917 rectangular. 701 change of base formulas. 717. alternate. 366 matrix addition. 289 complex number nth root. 989 real part. 989 imaginary part. 616 Cofunction Identities. 744. 989 rectangular form. 519 circular function. 498 from slicing a cone. 286. 989 imaginary unit. 744 properties of. 226 cofactor. 553 constant function as a horizontal line. 350 constant term of a polynomial. 744 cis(θ). 989 properties of. 696 Complex Factorization Theorem. 989 set of. 993 principal argument. 917 correlation coefficient. 791 of an angle. 470 conic sections definition. 532 conjugate of a complex number definition of. 1004 argument definition of. 241 continuously compounded interest. 286 modulus definition of. 999 nth Roots of Unity. 287 properties of. 1013 dot product. 290 consistent system. 993 coefficient of determination. 730. 991 polar form cis-notation. 993 conjugate definition of. 626 Charles’s Law. 226 cosecant graph of. 101 intuitive definition of. 579 vector addition. 442 characteristic polynomial. 1032 complementary angles. i. 156 formal definition of. 2. 100 constant of proportionality. 498 standard equation. 791 1069 . 2 complex plane. 802 cosine graph of. 495 conjugate axis of a hyperbola. 367 compound interest. 288 Conjugate Pairs Theorem. 495 radius of. 998. 472 contradiction. 989 properties of.Index central angle. 549 coordinates Cartesian. 498 definition of. 498 standard equation. 355 circle center of. 360 properties of. 420 common logarithm. 236 continuous. 287 properties of. 422 commutative property function composition does not have. 989 component form of a vector. 801 of an angle. 159 cost function. 567 direct variation. 82 cotangent graph of. 58 dot product commutative property of. 775 difference quotient. 585 scalar multiplication.1070 cost average. 776 earthquake Richter Scale. 11 distributive property matrix matrix multiplication. 626 eigenvector. 516 from slicing a cone. 519 major axis. 236 DeMoivre’s Theorem. 626 ellipse center. 775 for tangent. 1032 scalar multiplication. 581 vector dot product. 995 dependent system. 1016 DMS. start-up. 695 decreasing function formal definition of. 45 implied. 431 eccentricity. 55 depreciation. 346 fixed. 1032 work. 10 distance formula. 1040 Double Angle Identities. 505 discriminant of a conic. 979 of a parabola. 771. 694 degree of a polynomial. 554 dependent variable. 806 coterminal angle. 773. 1035 relation to vector magnitude. 1032 definition of. 516 definition of. 195 distance definition. 698 Coulomb’s Law. 100 degree measure. 1033 properties of. 619 curve orientated. 1032 geometric interpretation. 350 directrix of a conic section in polar form. 775 for sine. 101 intuitive definition of. 60 definition of. 752 properties of. 805 of an angle. 516 minor axis. 614 properties of. 82. 1032 distributive property of. 744. 977 of a quadratic equation. 1054 decibel. 1046 cycloid. 979 eigenvalue. 431 decimal degrees. 420 Descartes’ Rule of Signs. 273 determinant of a matrix definition of. 79 dimension of a matrix. 82 variable. 516 Index . 355 Cramer’s Rule. 496 guide rectangle. 695 domain applied. 195 trichotomy. 516 eccentricity. 522. 1032 relation to orthogonality. 522 foci. 616 Difference Identity for cosine. 55. 770 exponential function algebraic properties of. 654. 241 cost. 756 Factor Theorem. 531 of a parabola. n even. 437 natural base. 156 local (relative) maximum. 442 common base. 168 increasing. 379 properties of. 378 average cost. 239 polynomial. 43 dependent variable of. 76 difference quotient. 422 1071 . 2 end behavior of f (x) = axn . 506 focus of a conic section in polar form. 505 of an ellipse. 384 uniqueness of. 708. 156 continuous. 744 composite definition of. global) minimum. 420 one-to-one properties of. 100 independent variable of. 549 linear of n variables. 979 focus (foci) of a hyperbola. 437 solving equations with. 82 circular. 879 function (absolute) maximum. 418 Fundamental Graphing Principle. 100 definition as a relation. 507 focal length of a parabola.Index reflective property. 93 identity. 380 linear. 243 entry in a matrix. 82 focal diameter of a parabola. 31. 55 arithmetic. 708. 418 graphical properties of. 879 of a sinusoid. 516 free variable. 101 (absolute. 55 difference. 23 identity. 82 decreasing. 419 inverse properties of. 95 exponential. 101 logarithmic. 552 frequency angular. 795 ordinary. 79 domain. 448 extended interval notation. 101 local (relative) minimum. 240 of f (x) = axn . n odd. 516 ellipsis (. 379 solving for. 523 standard equation. ). 95 Even/Odd Identities. . 100. 76 as a process. 549 graph of. 437 change of base formula. 45 even. 549 even function. 240 of a function graph. 567 equation contradiction. 554 linear of two variables. 519 vertices. . 399 argument. 173 algebraic. 55 inverse definition of. 101 absolute value. 651 empty set. 681 fixed cost. 367 constant. 420 definition of. 258 factorial. 360 properties of. 519 Half-Angle Formulas. additive. 531 branch. 93 for polar equations. 579 Index . 306 Hooke’s Law. 95 fundamental cycle of y = cos(x). 76 profit. 531 vertices. 82 quadratic. 67 growth model limited. 557 geometric sequence. 121 greatest integer function. 132 horizontal shift. 93 of a relation. 23 rational function.1072 notation. 779 harmonic motion. 241 sum. 531 hyperbolic cosine. 367 matrix. 235 price-demand. 1060 hyperbolic sine. 289 Gauss-Jordan Elimination. 62 polynomial. 120. 55 odd. 534 vertical. 304 intuitive definition of. 82 product. 446 Heron’s Formula. 123 of a function. 475 uninhibited. 381 hyperbola asymptotes. 20 of an equation. 130 vertical shift. 45 rational. 571 Gaussian Elimination. 305 horizontal scaling. 936 Fundamental Theorem of Algebra. 912 hole in a graph. 531 foci. 532 standard equation horizontal. 304 location of. 883 Henderson-Hasselbalch Equation. 188 quotient. 23 Horizontal Line Test (HLT). 542 identity function. 475 logistic. 76 range. 381 periodic. 532 for an ellipse. 791 Fundamental Graphing Principle for equations. 531 conjugate axis. 305 location of. 534 transverse axis. 654 geometric series. 531 center. 532 definition of. 472 guide rectangle for a hyperbola. 82 smooth. 669 graph hole in. 135 vertical scaling. 1060 hyperboloid. 76 transformation of graphs. 308 horizontal line. 301 revenue. 135 zero. 23 for functions. 126 transformations. 790 piecewise-defined. 350 horizontal asymptote formal definition of. 321 reflection about an axis. 531 from slicing a cone. 496 guide rectangle. 95 one-to-one. 315 least squares regression line. 549 imaginary axis. 379 solving for. 472. 156 of best fit. 100 independent system. 225 parallel. 948 c line horizontal. 643 quadratic. 698 instantaneous rate of change. 382 invertible function. 477 information entropy. 2 intercept definition of. 989 imaginary part of a complex number. 554 independent variable. 605 latus rectum of a parabola. 553 increasing function formal definition of. 350 1073 Kepler’s Third Law of Planetary Motion. 209 non-linear. multiplicative. 673 inductive step. 581 matrix. 236 leading term of a polynomial. 241 interrobang. 25 interest compound. 379 properties of. 286 implied domain of a function. 2 irreducible quadratic. 211 graphical interpretation. 477 initial side of an angle. 895 leading coefficient of a polynomial. 384 uniqueness of. 579. 673 induction hypothesis. 602 of a function definition of. 3 notation. 321 intersection of two sets. 379 matrix. i. 948 lima¸on. 2 greatest integer function. 4 interval definition of. 161. 214 inflection point. 585 statement which is always true. 908 Law of Sines.Index matrix. 673 inequality absolute value. 23 least squares regression. 55 index of a root. 2 set of. 469 Intermediate Value Theorem polynomial zero version. 507 Law of Cosines. 707 integer definition of. 470 compounded continuously. multiplicative. 472 simple. 236 Learning Curve Equation. 756 inverse matrix. 350 invertibility function. 3 notation for. 380 inverse variation. 166 . 602 irrational number definition of. 225 linear function. 397 induction base step. 25 location of. extended. 290 joint variation. 989 imaginary unit. 67 set of. 355 Kirchhoff’s Voltage Law. 101 intuitive definition of. additive. 58 inconsistent system. 225 lemniscate. 215 sign diagram. 151 slope-intercept form. 593 main diagonal. 984 row echelon form. 431 logistic growth. “base b”. 570 rotation. 614 properties of. 567 square matrix. 626 cofactor. 585 definition of. 585 identity for. 593 maximum formal definition of. 422 one-to-one properties of. 602 leading entry. 459 logarithmic scales. 578 upper triangular. 568 characteristic polynomial. 585 matrix multiplication associative property of. 23 linear equation n variables. 101 local minimum formal definition of. 437 solving equations with. 549 linear function. 516 Markov Chain. 579 additive identity. 423 inverse properties of. 585 major axis of an ellipse. 592 mathematical model. 569 lower triangular. 581 properties of. 616 dimension. 156 local maximum formal definition of. 442 common. 554 two variables.1074 perpendicular. 422 graphical properties of. 155 slope of. 422 general. 101 measure of an angle. 581 zero product property. 616 multiplicative inverse. 622 augmented. 567 equality. 475 LORAN. 567 entry. 102 intuitive definition of. 593 main diagonal. 579 additive inverse. 584 reduced row echelon form. 581 size. 585 minor. 155 vertical. 567 determinant definition of. 102 intuitive definition of. 578 invertible. 584 distributive property. 579 definition of. 586 sum. 569 row operations. 101 logarithm algebraic properties of. 568 scalar multiplication associative property of. 579 adjoint. 437 natural. 102 intuitive definition of. 60 matrix addition associative property. 580 distributive properties. 693 Index . 579 commutative property. 581 identity for. 581 definition of. 578 properties of. 585 properties of. 538 lower triangular matrix. 438 change of base formula. 167 point-slope form. 616 definition. 602 product of row and column. 510 parallel vectors. 697 point-slope form of a line. 422 parametric equations. 191 definition of. 708 pi. 1036 definition of. 879 ordered pair. 605 pH. 101 focus. 989 horizontal. 1035 equivalent representations of. 505 effect on the graph of a polynomial. 688 Newton’s Law of Cooling. 917 orthogonal vectors. 420 parameter. 244 paraboloid. 1046 set of. 60 reflective property. 477 Newton’s Law of Universal Gravitation. 6 piecewise-defined function. 506 multiplicity vertex. 1046 natural number parametric solution. 12 definition of. 351 period circular motion. 700 ordinate. 249 vertex formulas. 879 ordinary frequency. 245. 496 minor axis of an ellipse. 510 modulus of a complex number standard equation definition of. 194 of a zero. 505 midpoint formula. 95 periodic function.Index midpoint axis of symmetry. 1028 natural base. 616 from slicing a cone. 188. 505 minimum focal diameter. 432 one-to-one function. 552 definition of. 7 conversion into rectangular. 879 odd function. 102 focal length. 350. 1046 natural logarithm. 921 overdetermined system. 477 oriented angle. 507 formal definition of. 421. 188 model latus rectum. 991 vertical. 507 mathematical. π. 508 properties of. 13 directrix. 381 phase. 554 polar axis. 795. 628 negative angle. 922 orthogonal projection. 2 partial fractions. 516 graph of a quadratic function. 917 1075 . 505 minor. 795. 917 parabola pole. 790 Ohm’s Law. 790 obtuse angle. 506 intuitive definition of. 1046 point of diminishing returns. 311 of a function. 2 parametrization. 704 polar coordinates origin. 694 of a sinusoid. 62 orientation. 6 phase shift. 698 Pascal’s Triangle. 708 oblique asymptote. 155 oriented arc. 474 password strength. 188 general form. 8 quadratic formula. 258 divisor. 438 for radicals. 995 Index for exponential functions. ˆ. 778 power rule for absolute value. 290 over the real numbers. 473 radius of a circle. 995 for exponential functions. 173 for complex numbers. 273 zero lower bound. 498 range definition of. 215 irreducible quadratic. . 82 projection x−axis. 698 quadrants. 46 orthogonal. 1036 Pythagorean Conjugates. 995 for exponential functions. 397 principal argument of a complex number. 751 Pythagorean Identities. 290 constant term. 398 for the modulus of a complex number. 437 for logarithms. 173 for complex numbers. 989 principal unit vectors. 190 inequality. 991 radian measure. 438 for radicals. 438 for radicals. 274 positive angle. 991 Product to Sum Formulas. 993 polar rose. 398 for the modulus of a complex number. 228 Quotient Identities. 1022 ı ˆ Principle of Mathematical Induction. 698 Power Reduction Formulas. 160 . 244 upper bound. 991 price-demand function. 236 end behavior. 236 variations in sign. 194 quadratic function definition of. 749 quadrantal angle. 190 quadratic regression. 437 for logarithms. 469 principal nth root. 948 polynomial division dividend. 780 profit function. 45 rate of change average. 258 factor. 274 multiplicity. 235 degree. 258 synthetic division. 397 radioactive decay. 45 y−axis.1076 polar form of a complex number. 258 quotient. 701 radical properties of. 437 for logarithms. 239 leading coefficient. 260 polynomial function completely factored over the complex numbers. 236 definition of. 398 for the modulus of a complex number. 258 remainder. 173 for complex numbers. 290 standard form. 673 product rule for absolute value. 745 quotient rule for absolute value. 398 radicand. 82 principal. 236 leading term. 269 ray definition of. 654 reduced row echelon form. 2 set of. 745 rectangular coordinates also known as Cartesian coordinates. 580 distributive properties of. 23 definition. 1016 scalar projection. 228 total squared error. 722 for the circular functions. 226 least squares line. 10 regression coefficient of determination. 752 properties of. 569 row operations for a matrix. 654 formula for nth term. 693 real axis. 800 of an angle. 154 rational exponent. 693 initial point. 581 vector associative property of. 1015 distributive properties of. 431 right angle. 1004 rotation matrix. 570 reference angle. 922 rectangular form of a complex number. 291 real number definition of. 666 definition of. 721 Reference Angle Theorem for cosine and sine. 1016 properties of. 82 Richter Scale. 656 sum of first n terms. 654 definition of. 398 rational functions. 652 geometric common ratio. 472 slope of a line. 581 definition of. 654 definition of. 654 formula for nth term. 2 Rational Zeros Theorem. 656 sum of first n terms. 258 revenue function. 20 Fundamental Graphing Principle. 802 secant line. 581 properties of. 23 Remainder Theorem. 397 radicand. 226 correlation coefficient. 989 Reciprocal Identities. 2 real part of a complex number. 694 root index. 2 set of. 652 alternating. 652 arithmetic common difference. 160 sequence nth term. 984 rotation of axes. 161. 568 scalar multiplication matrix associative property of. 1037 secant graph of. 301 rational number definition of. 666 1077 . 744. 972 row echelon form. 225 quadratic. 126 of a point. 989 recursion equation. 397 Roots of Unity. 989 Real Factorization Theorem. 747 reflection of a function graph. 225 relation algebraic description. 1016 definition of.Index instantaneous. 917 conversion into polar. 311 slope definition. 260 system of equations back-substitution. 9 testing a function graph for. 554 free variable. 560 coefficient matrix. 95 testing an equation for. 661 index of summation. 4 verbal description. 879 slant asymptote. 795. 241 sound intensity level decibel. 908 Side-Side-Side triangle. 312 formal definition of. 661 properties of. 879 graph of. 2 intersection. 9 about the y-axis. 792 of an angle. 399 for quadratic inequality.1078 recursive. 586 standard position of a vector. 908 sign diagram algebraic function. 1 sets of numbers. 879 phase shift. 214 polynomial function. 1017 standard position of an angle. 654 series. 698 start-up cost. 549 dependent. 155 smooth. 693 Sum Identity for cosine. 590 definition. 151 of a line. 791 sinusoid amplitude. 9 about the origin. 668 set definition of. 879 phase. 744 properties of. 773. 571 Gaussian Elimination. 311 slant asymptote determination of. 557 Index . 661 supplementary angles. 151 rate of change. 794. 730. 781 summation notation definition of. 321 simple interest. 661 lower limit of summation. 431 square matrix. 2 union. 879 frequency angular. 592 stochastic process. 879 baseline. 879 properties of. 552 Gauss-Jordan Elimination. 4 roster method. 154 slope-intercept form of a line. 879 vertical shift. 1 Side-Angle-Side triangle. 26 synthetic division tableau. 242 rational function. 469 sine graph of. 1 set-builder notation. 1 set-builder notation. 879 ordinary. 1 empty. 664 upper limit of summation. 880 period. 717. 696 symmetry about the x-axis. 82 steady state. 775 for sine. 590 consistent. 775 for tangent. 592 straight angle. 795. 775 Sum to Product Formulas. 553 constant matrix. 771. 1022 definition of. 350 joint. 590 tangent graph of. 531 Triangle Inequality. 554 unknowns matrix. 1032 distributive property of. 1036 1079 . 752 properties of. 1021 Upper and Lower Bounds Theorem. 554 leading variable. 744. 1032 relation to magnitude. 552 triangular form. 1034 component form. 4 Unit Circle definition of. 350 variations in sign. 1040 head. 1016 angle between two. 274 upper triangular matrix. Louis Leon. 120. 1013. 556 underdetermined system. 1012 properties of. 1018 dot product commutative property of. 315 total squared error. 554 two variables. 724 unit vector. 556 linear n variables. 646 non-linear. 554 uninhibited growth. 554 parametric solution. 1013 commutative property. 1032 definition of. 1035 work. 1013 definition of. 350 inverse. 273 vector x-component. 1032 normalization. 1033. 1010 addition associative property. 806 terminal side of an angle. 129 rigid. 1010 Decomposition Theorem Generalized. 637 overdetermined. 501 important points. 556 underdetermined. 1010 y-component. 804 of an angle. 1010 magnitude definition of. 1033 properties of. 1032 relation to orthogonality. 55 independent. 55 variable cost. 698 Thurstone. 472 union of two sets. 1018 properties of. 1010 direction definition of. 129 transformations of function graphs. 553 independent. 1038 Principal. 225 transformation non-rigid. 1032 geometric interpretation. 550 linear in form. 1018 relation to dot product. 350 direct. 593 variable dependent.Index inconsistent. 1013 additive identity. 159 variation constant of proportionality. 183 triangular form. 1022 orthogonal projection. 1010 initial point. 1013 additive inverse. 135 transverse axis of a hyperbola. 1018 properties of. . 2 set of. 244 of a function. 95 upper and lower bounds. 1016 identity for. 693 of an ellipse. 707 vertex of a hyperbola. 1039 wrapping function. 1037 standard position. 505 of an angle. 1015 distributive properties. 1010 terminal point. 516 vertical asymptote formal definition of.1080 orthogonal vectors. 2 work. 1010 triangle inequality. 1021 velocity average angular. 274 Index . 1035 parallel. 1016 scalar product definition of. 1016 zero product property. 707 instantaneous angular. 306 vertical line. ˆ. 704 zero multiplicity of. 23 Vertical Line Test (VLT). 1016 definition of. 1042 unit vector. 304 location of. 43 whole number definition of. 1028 principal unit vectors. 1032 properties of. 531 of a parabola. 1016 properties of. 1032 scalar projection. 1017 tail. 188. 1022 ı ˆ resultant. 304 intuitive definition of. 1011 scalar multiplication associative property of. 707 instantaneous. 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