College Trigonometry Textbook PDF

College Trigonometry Version bπc Corrected Edition by Carl Stitz, Ph.D. Jeff Zeager, Ph.D. Lakeland Community College Lorain County Community College July 4, 2013 ii...

College Trigonometry Version bπc Corrected Edition by Carl Stitz, Ph.D. Jeff Zeager, Ph.D. Lakeland Community College Lorain County Community College July 4, 2013 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 book. 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 administrations 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 wish 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..................................... 874 10.7.2 Answers...................................... 877 11 Applications of Trigonometry 881 11.1 Applications of Sinusoids................................. 881 11.1.1 Harmonic Motion................................ 885 11.1.2 Exercises..................................... 891 11.1.3 Answers...................................... 894 11.2 The Law of Sines..................................... 896 11.2.1 Exercises..................................... 904 11.2.2 Answers...................................... 908 11.3 The Law of Cosines.................................... 910 viii Table of Contents 11.3.1 Exercises..................................... 916 11.3.2 Answers...................................... 918 11.4 Polar Coordinates..................................... 919 11.4.1 Exercises..................................... 930 11.4.2 Answers...................................... 932 11.5 Graphs of Polar Equations................................ 938 11.5.1 Exercises..................................... 958 11.5.2 Answers...................................... 963 11.6 Hooked on Conics Again................................. 973 11.6.1 Rotation of Axes................................. 973 11.6.2 The Polar Form of Conics............................ 981 11.6.3 Exercises..................................... 986 11.6.4 Answers...................................... 987 11.7 Polar Form of Complex Numbers............................ 991 11.7.1 Exercises..................................... 1004 11.7.2 Answers...................................... 1007 11.8 Vectors........................................... 1012 11.8.1 Exercises..................................... 1027 11.8.2 Answers...................................... 1031 11.9 The Dot Product and Projection............................ 1034 11.9.1 Exercises..................................... 1043 11.9.2 Answers...................................... 1045 11.10 Parametric Equations.................................. 1048 11.10.1 Exercises..................................... 1059 11.10.2 Answers...................................... 1063 Index 1069 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 [email protected] 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 Q P An angle with vertex P. 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. Q P A straight angle. 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 β α 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 ◦ ◦ 1 ◦ ◦ 2 (360 ) = 180 , a quarter of a revolution (a right angle) measures 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 32 of a revolution would measure 32 (360◦ ) = 240◦ , the measure of an angle which constitutes only 121 of a revolution measures 121 (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 1 constitutes 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 61 = 720 of a full rotation. This √ 360 ◦ 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◦ = 600 and 10 = 6000 , from which it follows that 1◦ = 360000. To convert a measure of 42.125◦ to the DMS system, we start by notingthat 42.125◦ = 42◦ + 0.125◦. Converting 0 the partial amount of degrees to minutes, we find 0.125◦ 60 1◦ = 7.50 = 70 + 0.50. Converting the  00  partial amount of minutes to seconds gives 0.50 6010 = 3000. Putting it all together yields 42.125◦ = 42◦ + 0.125◦ = 42◦ + 7.50 = 42◦ + 70 + 0.50 = 42◦ + 70 + 3000 = 42◦ 70 3000 1◦ 1◦ On the other hand, to convert 117◦ 150 4500 to decimal degrees, we first compute 150
600 = 4 and 1◦ 1 ◦ 4500 3600
00 = 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◦ 150 4500 = 117◦ + 150 + 4500 1◦ 1 ◦ = 117◦ + 4 + 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. β θ α γ Supplementary Angles 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◦ 280 1700. 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 ◦ ◦ ◦  0 α to the DMS system, we start with 111.371 = 111 + 0.371  00 . Next we convert 0.371◦ 1◦ = 22.260. Writing 22.260 = 220 + 0.260 , we convert 0.260 10 = 15.600. Hence, 60 60 111.371◦ = 111◦ + 0.371◦ = 111◦ + 22.260 = 111◦ + 220 + 0.260 = 111◦ + 220 + 15.600 = 111◦ 220 15.600 Rounding to seconds, we obtain α ≈ 111◦ 220 1600. 10.1 Angles and their Measure 697 1◦ 7 ◦ 1◦ 17 ◦ 2. To convert β to decimal degrees, we convert 280 and 1700

600 = 15 36000 = 3600. Putting it all together, we have 37◦ 280 1700 = 37◦ + 280 + 1700 7 ◦ 17 ◦ = 37◦ + 15 + 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◦ 280 1700. While we could reach for the calculator to obtain an approximate answer, we choose instead to do a bit of sexagesimal7 arithmetic. We first rewrite 90◦ = 90◦ 00 000 = 89◦ 600 000 = 89◦ 590 6000. In essence, we are ‘borrowing’ 1◦ = 600 from the degree place, and then borrowing 10 = 6000 from the minutes place.8 This yields, γ = 90◦ − 37◦ 280 1700 = 89◦ 590 6000 − 37◦ 280 1700 = 52◦ 310 4300. 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-clockwise9 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 de Si al in T er m m er in T al Si d e Initial Side 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 y 4 3 2 α = 120◦ 1 −4 −3 −2 −1 1 2 3 4 x −1 β = −240◦ −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 60◦ 1 counter-clockwise 360 ◦ = 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 5 360◦ = 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◦. y y 4 4 3 3 2 2 1 α = 60◦ 1 −4 −3 −2 −1 1 2 3 4 x −4 −3 −2 −1 1 2 3 4 x −1 −1 −2 −2 β = −225◦ −3 −3 −4 −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 12 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 12 1 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◦. y y 4 4 3 3 γ = 540◦ 2 2 1 1 −4 −3 −2 −1 1 2 3 4 x −4 −3 −2 −1 1 2 3 4 x −1 −1 −2 −2 φ = −750◦ −3 −3 −4 −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, C π= 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 C equation in Definition 10.1 to get a formula more useful for our purposes, namely: 2π = 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 s the arc. Using proportionality arguments, it stands to reason that the ratio should also be a r constant among all circles, and it is this ratio which defines the radian measure of an angle. s θ r r s The radian measure of θ is. 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. r r β r r r α 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 = 2π. From this we can find the radian measure of other central angles using proportions, r 702 Foundations of Trigonometry just like we did with degrees. For instance, half of a revolution has radian measure 12 (2π) = π, a quarter revolution has radian measure 14 (2π) = π2 , 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 = π2 ’, we mean θ is an angle which measures π2 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 = π2. 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. π 4π 9π 5π 1. α = 2. β = − 3. γ = 4. φ = − 6 3 4 2 Solution. 1. The angle α = π6 is positive, so we draw an angle with its initial side on the positive x-axis and rotate counter-clockwise (π/6) 1 2π = 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 π 12π 13π 6 , thus when k = 1, we get θ = 6 + 6 = 6. Substituting k = −1 gives θ = 6 − 6 = − 6 and when we let k = 2, we get θ = 6 + 6 = 25π π 12π 11π π 24π 6. (4π/3) 2. Since β = − 4π3 is negative, we start at the positive x-axis and rotate clockwise 2π = 23 of a revolution. We find β to be a Quadrant II angle. To find coterminal angles, we proceed as before using 2π = 6π 4π 6π 3 , and compute θ = − 3 + 3 · k for integer values of k. We obtain 3 , 2π − 10π 8π 3 and 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 y y 4 4 3 3 2 2 1 π 1 α= 6 −4 −3 −2 −1 1 2 3 4 x −4 −3 −2 −1 1 2 3 4 x −1 −1 −2 −2 β = − 4π 3 −3 −3 −4 −4 π α= 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 π 1 4 of the radian measure with 4 or 8 of a revolution remaining. We have γ as a Quadrant I angle. All angles coterminal with γ are of the form θ = 9π 8π 4 + 4 · k, where k is an integer. Working through the arithmetic, we find: π4 , − 7π 17π 4 and 4. 4. To graph φ = − 5π 4π 2 , we begin our rotation clockwise from the positive x-axis. As 2π = 2 , after one full revolution clockwise, we have π2 or 41 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 4π π 3π 7π 2 + 2 · k for a few integers k and obtain − 2 , 2 and 2. y y 4 4 3 3 φ = − 5π 2 2 2 1 1 −4 −3 −2 −1 1 2 3 4 x −4 −3 −2 −1 1 2 3 4 x −1 −1 −2 9π −2 γ= 4 −3 −3 −4 −4 9π γ= 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 360 radians ◦ , or its reduced equivalent, π radians 180◦ , as the conversion factor between ◦ ◦ π radians = π3 radians, or
the two systems. For example, to convert 60 to radians we find 60 180◦ ◦ simply π3. To convert from radian measure back to degrees, we multiply by the ratio π 180 radian. For 180◦ ◦.15 Of particular interest is the example, − 5π 5π

6 radians is equal to − 6 radians π radians = −150 ◦ 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: π radians ˆ To convert degree measure to radian measure, multiply by 180◦ 180◦ ˆ To convert radian measure to degree measure, multiply by 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, x2 +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 s s radian measure of θ is = = s so that, once again blurring the distinction between an angle r 1 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 y y y 1 1 1 s t θ t 1 x 1 x 1 x 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. 3π 2. t = −2π 3. t = −2 4. t = 117 1. t = 4 Solution. 1. The arc associated with t = 3π 3π 4 is the arc on the Unit Circle which subtends the angle 4 in 3π 3 radian measure. Since 4 is 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. y y 1 1 1 x x 1 3π t= 4 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 π2 ≈ 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 5π8 ≈ 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 1172π as 18.62 which tells us we complete 18 revolutions counter-clockwise with 0.62, or just shy of 58 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. y y 1 1 1 x 1 x t = −2 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. 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. Note that with this convention s the formula we used to define radian measure, namely θ = , 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. 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 s have v = displacement time =. The quantity v has units of length time and conveys two ideas: the direction t 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 s is moving - it is the speed of the object. Measuring θ in radians we have θ = thus s = rθ and r s rθ θ v= = =r· t t t θ The quantity is called the average angular velocity of the object. It is denoted by ω and is t 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 shown17 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 time. Thus the left hand side of the equation v = rω has units length radians time , whereas radians length·radians the right hand side has units length · time = 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 shown19 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 = 2π24radians π hours = 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 π miles v = 2960 miles · ≈ 775 12 hours 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, 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. 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◦ 500 6. −32◦ 100 1200 7. 502◦ 350 8. 237◦ 580 4300 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◦ 5π 11π 5π 13. −270◦ 14. 15. − 16. 6 3 4 3π π 7π π 17. 18. − 19. 20. 4 3 2 4 π 7π 5π 21. − 22. 23. − 24. 3π 2 6 3 π 15π 13π 25. −2π 26. − 27. 28. − 4 4 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. 2π 7π 11π 37. π 38. − 39. 40. 3 6 6 π 5π π π 41. 42. 43. − 44. 3 3 6 2 710 Foundations of Trigonometry In Exercises 45 - 49, sketch the oriented arc on the Unit Circle which corresponds to the given real number. 5π 45. t = 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 = 21 r2 θ. A s (Hint: Use the proportion area of the circle = circumference of the circle.) r s θ 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. π 5π 57. θ = , r = 12 58. θ = , r = 100 59. θ = 330◦ , r = 9.3 6 4 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◦ 450 2. 200◦ 190 3000 3. −317◦ 30 3600 4. 179◦ 590 5600 5. 125.833◦ 6. −32.17◦ 7. 502.583◦ 8. 237.979◦ 9. 330◦ is a Quadrant IV angle 10. −135◦ is a Quadrant III angle coterminal with 690◦ and −30◦ coterminal with 225◦ and −495◦ y y 4 4 3 3 2 2 1 1 x x −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 −1 −1 −2 −2 −3 −3 −4 −4 11. 120◦ is a Quadrant II angle 12. 405◦ is a Quadrant I angle coterminal with 480◦ and −240◦ coterminal with 45◦ and −315◦ y y 4 4 3 3 2 2 1 1 x x −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 −1 −1 −2 −2 −3 −3 −4 −4 5π 13. −270◦ lies on the positive y-axis 14. is a Quadrant II angle 6 17π 7π coterminal with 90◦ and −630◦ coterminal with and − 6 6 y y 4 4 3 3 2 2 1 1 x x −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 −1 −1 −2 −2 −3 −3 −4 −4 10.1 Angles and their Measure 713 11π 5π 15. − is a Quadrant I angle 16. is a Quadrant III angle 3 4 π 5π 13π 3π coterminal with and − coterminal with and − 3 3 4 4 y y 4 4 3 3 2 2 1 1 x x −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 −1 −1 −2 −2 −3 −3 −4 −4 3π π 17. is a Quadrant II angle 18. − is a Quadrant IV angle 4 3 11π 5π 5π 7π coterminal with and − coterminal with and − 4 4 3 3 y y 4 4 3 3 2 2 1 1 x x −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 −1 −1 −2 −2 −3 −3 −4 −4 7π π 19. lies on the negative y-axis 20. is a Quadrant I angle 2 4 3π π 9π 7π coterminal with and − coterminal with and − 2 2 4 4 y y 4 4 3 3 2 2 1 1 x x −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 −1 −1 −2 −2 −3 −3 −4 −4 714 Foundations of Trigonometry π 7π 21. − lies on the negative y-axis 22. is a Quadrant III angle 2 6 3π 5π 19π 5π coterminal with and − coterminal with and − 2 2 6 6 y y 4 4 3 3 2 2 1 1 x x −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 −1 −1 −2 −2 −3 −3 −4 −4 5π 23. − is a Quadrant I angle 24. 3π lies on the negative x-axis 3 π 11π coterminal with and − coterminal with π and −π 3 3 y y 4 4 3 3 2 2 1 1 x x −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 −1 −1 −2 −2 −3 −3 −4 −4 π 25. −2π lies on the positive x-axis 26. − is a Quadrant IV angle 4 7π 9π coterminal with 2π and −4π coterminal with and − 4 4 y y 4 4 3 3 2 2 1 1 x x −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 −1 −1 −2 −2 −3 −3 −4 −4 10.1 Angles and their Measure 715 15π 13π 27. is a Quadrant IV angle 28. − is a Quadrant IV angle 4 6 7π π 11π π coterminal with and − coterminal with and − 4 4 6 6 y y 4 4 3 3 2 2 1 1 x x −4 −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 −1 −1 −2 −2 −3 −3 −4 −4 4π 3π 3π 29. 0 30. 31. 32. − 3 4 2 7π 5π π 5π 33. − 34. 35. 36. − 4 6 4 4 37. 180◦ 38. −120◦ 39. 210◦ 40. 330◦ 41. 60◦ 42. 300◦ 43. −30◦ 44. 90◦ 5π 45. t = 46. t = −π 6 y y 1 1 1 x 1 x 47. t = 6 48. t = −2 y y 1 1 1 x 1 x 716 Foundations of Trigonometry 49. t = 12 (between 1 and 2 revolutions) y 1 1 x 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 π 59. 79.2825π ≈ 249.07 square units 60. square units 2 50π 61. square units 62. 38.025π ≈ 119.46 square units 3 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). y y 1 1 P (cos(θ), sin(θ)) θ θ x x 1 1 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 34 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 (270◦ ) = 0 and sin (270◦ ) = −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 y y 1 1 θ = 270◦ P (−1, 0) x x 1 1 θ = −π P (0, −1) Finding cos (270◦ ) and sin (270◦ ) 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, q we have √ 1 2 x2 + y 2 = 1. Substituting y = x into this equation yields 2x2 = 1, or x = ± 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. y 1 P (x, y) P (x, y) θ = 45◦ 45◦ x y 1 θ = 45◦ x 2 Can you show this? 10.2 The Unit Circle: Cosine and Sine 719 4. As before, the terminal side of θ = π6 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 we find y = 2 so sin 6 = 12. 3 1 π Since P (x, y) lies on the Unit Circle, we substitute y = 12 into x2 + y 2 = 1 to get x2 = 34 , or √
√ x = ± 23. Here, x > 0 so x = cos π6 = 23. y 1 P (x, y) P (x, y) θ= π 60◦ 6 x y 1 θ= π 6 = 30◦ x 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◦ ) = 12 and y = sin (60◦ ) = 23. y 1 P (x, y) P (x, y) 30◦ θ= 60◦ y x 1 θ = 60◦ x 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, x2 + y 2 = 1. If we substitute x = cos(θ) and y = sin(θ) into x2 + y 2 = 1, we get (cos(θ))2 + (sin(θ))2 = 1. An unfortunate4 convention, which the authors are compelled to perpetuate, is to write (cos(θ))2 as cos2 (θ) and (sin(θ))2 as sin2 (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, cos2 (θ) + sin2 (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) = 53 , find cos(θ). √ 3π 5 2. If π < θ < 2 with cos(θ) = − 5 , find sin(θ). 3. If sin(θ) = 1, find cos(θ). Solution. 1. When we substitute sin(θ) = 35 into The Pythagorean Identity, cos2 (θ) + sin2 (θ) = 1, we 9 obtain cos2 (θ) + 25 = 1. Solving, we find cos(θ) = ± 54. 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) = − 45. √ √ 5 2. Substituting cos(θ) = − 2 2 5 into cos (θ) + sin (θ) = 1 gives sin(θ) = ± √25 = ± 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 2 5 are negative and we conclude sin(θ) = − 5. 3. When we substitute sin(θ) = 1 into cos2 (θ) + sin2 (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 π6 radians short of one
√ half revolution. In Example 10.2.1, we determined that cos π6 = 23 and sin π6 = 12. 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 π6 , when plotted in standard position, is 23 , 12. From the figure below, it is clear that the point √ P (x, y) we seek can be obtained by reflecting that 3 point about the y-axis. Hence, cos 6 = − 2 and sin 5π 5π

1 6 = 2. y y 1 1