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�csccoscot2sinsectancos2coscsccotsin1sinsectancos1coscsccotsin2cossec1sin2sincsc1cos2seccos1tan2cscsin1cot2tancotsec12cottancsc1224cos4sin4229cossin8�
�32tantansectan252sin1cossin41022sec1tansec23costantansincos4224secsectantancos11seccos11secsin11cscsin11csc1cottan11cottan11tancossin1tancossin�
��tancotseccsccscsincotcoscossectansincostancotcsc 94 The Trigonometric Functions 31. 33. 35. 37. 39. 41. 43. 45. 47. 32. 34. 36. 38. 40. 42. 44. 46. In Exercises 48 – 51, verify the identity. You may need to review the properties of absolute value and logarithms before proceeding. 48. 50. 49. 51. sintancotsec2112csc1cos1cos112csccotsec1sec1112sectancsc1csc1
112cotcsccotcsccotcossinsincos1tan1cot1sectansectan1sectansectan1csccotcsccot1csccotcsccot21secsectan1sin21secsectan1sin21csccsccot1cos21csccsccot1cos�
��cos1sin1sincossincsccot1cos21sinsectan1sinlnseclncoslncsclnsinlnsectanlnsectanlncsccotlncsccot 2.5 Beyond the Unit Circle 95 2.5 Beyond the Unit Circle Learning Objectives In this section you will:  Determine the values of the six circular functions from the coordinates of a point on a circle, centered at the origin, with any radius r.  Solve related application problems.  Describe the position of a particle experiencing circular motion. Recall that in defining the cosine and sine functions in Section 2.2, we assigned to each angle a position on the Unit Circle. Here we broaden our scope to include circles of radius r centered at the origin. Determining Cosine and Sine Consider for the moment the acute angle θ drawn below in standard position. Let be the point on the terminal side of θ which lies on the circle, and
let 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 triangles, ∆OPA and ∆OQB. These triangles are similar23. Thus, it follows that, from which. We similarly find. Since, by definition, and, we get the coordinates of Q to be and. 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. 23 Do you remember why? ,Qxy222xyr','Pxy'1xrrx'xrx'yry'cosx'sinycosxrsinyr 96 The Trigonometric Functions Not only can we describe the coordinates of Q in terms of and, but since the radius of the circle is, we can also express and in terms of the coordinates of Q. Throughout this textbook, by convention, the radius r of a circle is treated as positive as it relates to solving for trigonometric values. These results are summarized in the following theorem. Theorem 2.6. If is the point on the terminal side of an angle θ, plotted in standard position, which lies on the circle then and. Moreover, Note that in the case of the Unit Circle we have, so Theorem 2.6 reduces to our Unit Circle definitions of and. Example 2.5.1 1. Suppose that the terminal side of an angle θ, when plotted in standard position, contains the point. Find and. 2. In Example 1.3.4 in Section 1.3, we approximated the radius of the circle of revolution at 40.7608° North Latitude on Earth to be 2999 miles. Justify this approximation if the radius of the circle of revolution at the Equator is approximately 3960 miles. Solution. 1. Using Theorem 2.6 with and, we find so that cos�
�sin22rxycossin,Qxy222xyrcosxrsinyr2222cos= and sin.xxyyrrxyxy221rxycossin4,2Qcossin4x2y22422025r 2.5 Beyond the Unit Circle 97 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 indicated in the figure. Using Theorem 2.6, we get. Using a calculator in degree mode, we find. Hence, the radius of the circle of revolution at North Latitude 40.7608° is approximately 2999 miles. Position of a Particle in Circular Motion Theorem 2.6 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 when, the angle θ is in standard position. By definition,, which we rewrite as. According to Theorem 2.6, the location of the object on the circle is found using the equations and. With, we have and. Hence, at time t, the object is at the point. We have just argued the following. 
cos and sin42cos sin2525255cos sin.55xyrr,Qxy3960cos40.7608x3960cos40.76082999,0r0ttt,Qxycosxrsinyrt costxr sintyr cos,sinttrr 98 The Trigonometric Functions Equations for Circular Motion: Suppose an object is traveling on a circular path of radius r centered at the origin with constant angular velocity ω. If corresponds to the point, then the x and y coordinates of the object are functions of t and are given by and. Here, indicates a clockwise direction. Example 2.5.2. Suppose we are in the situation of Example 1.3.4. Find the equations of motion of Salt Lake Community College as the Earth rotates. Solution. From Example 1.3.4, we take miles and. Hence, using and, the equations of motion are Note that x and y are measured in miles and t is measured in hours. Determining the Other Four Circular Functions We have generalized the cosine and sine functions from coordinates on the Unit Circle to coordinates on circles of radius r. Using Theorem 2.6 in conjunction with Theorem 2.3, we generalize the remaining circular functions in kind. 0t,0r costxr sintyr0�
�2999r12 hours costxr sintyr2999cos and 2999sin.1212xtyt 2.5 Beyond the Unit Circle 99 Theorem 2.7. Suppose is the point on the terminal side of an angle θ, plotted in standard position, which lies on the circle. Then the circle has radius r and    , provided, provided.., provided, provided.. Example 2.5.3. 1. Suppose the terminal side of θ, when plotted in standard position, contains the point. Find the values of the six circular functions of θ. 2. Suppose θ is a Quadrant IV angle with. Find the values of the five remaining circular functions of θ. Solution. 1. The radius of the circle containing the point is With, and, we apply Theorems 2.6 and 2.7 to find the values of the six circular functions of θ.24 24 For convenience, the sketch shows 0 < θ < 2π. In reality, θ may be any angle, plotted in standard position, which contains the point Q (3,–4) on its terminal side. ,Qxy222xyr22cscxyryy0y22secxyrxx0xtanyx0xcotxy0y3,4Qcot43,4Q2222345.rxy3x4y
5r 100 The Trigonometric Functions 2. We look for a point which lies on the terminal side of θ 25, when θ is plotted in standard position. We are given that θ is a Quadrant IV angle, so we know and. Also,. Since, we may choose26 and, from which The five remaining circular function values follow. 25 Again, θ may be any angle, plotted in standard position, with Q on its terminal side. 26 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. 45sin csc5435cos sec5343tan cot.34yrryxrrxyxxy,Qxy0x0ycot4xy4414x1y22224117.rxy117sin17174417cos17171tan417csc17117sec4yrxryxryrx�
�� 2.5 Beyond the Unit Circle 101 We close this section by noting that we have not yet discussed the domains and ranges of the circular functions. In Chapter 3, we will graph the circular functions. This will provide a visual platform for the introduction of the domain and range of each circular function. 102 2.5 Exercises The Trigonometric Functions In Exercises 1 – 8, let θ be the angle in standard position whose terminal side contains the given point. Find the values of the six circular functions of θ. 1. 5. 2. 6. 3. 7. 4. 8. In Exercises 9 – 10, determine the radius of the circle of revolution at the given latitude on Earth. Assume that the radius of the circle of revolution at the Equator is approximately 3960 miles. 9. 55.39° North Latitude 10. 44.29° South Latitude In Exercises 11 – 12, determine the radius of the circle of revolution at the latitude corresponding to the given position on Earth. You may use the Internet for determining latitudes. Assume that the radius of the circle of revolution at the Equator is approximately 3960 miles. 11. Sydney, Australia 12. Nome, Alaska In Exercises 13 – 16, 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 corresponds to a position along the positive x-axis. 13. A point on the edge of a yo-yo which is 2.25 inches in diameter and spins at 4500 revolutions per minute. 14. A point on the edge of a yo-yo used in the trick ‘Around the World’ in which the performer throws the yo-yo so it sweeps out a vertical circle whose radius is the yo-yo string. Assume the yo-yo string is 28 inches long and the yo-yo takes 3 seconds to complete one revolution of the circle. 15. A point on the edge of the circular disk in a computer hard drive. The circular disk has diameter 2.5 inches and spins at a rate of 7200 RPM (revolutions per minute). 16. A passenger on the Giant Wheel at Cedar Point Amusement Part. The Giant Wheel is a circle with diameter 128 feet. It completes two revolutions in 2
minutes and 7 seconds. 1,5A3,1B6,2C10,12D7,24P3,4Q5,9R2,11T0t 2.5 Beyond the Unit Circle 103 In Exercises 17 – 30, use the given information to find the exact values of the remaining circular functions of θ. 17. 19. 21. 23. 25. 27. 29. with θ in Quadrant II. with θ in Quadrant I. 18. 20. with θ in Quadrant III. with θ in Quadrant IV. with θ in Quadrant III. 22. with θ in Quadrant II. with θ in Quadrant IV. 24. with θ in Quadrant II. with θ in Quadrant III. 26. with θ in Quadrant I. with. with. 28. 30. with. with. 31. In deriving the equations for circular motion, we assumed that at the object was at the point. If this is not the case, we can adjust the equations of motion by introducing a time delay. If is the first time the object passes through the point, show, with the help of your classmates, the equations of motion are and. 3sin512tan525csc24sec71091csc91cot23tan2sec4cot51cos3cot202�
�csc52tan1032sec253220t,0r00t,0r 0()costtxr 0()sinttyr 104 CHAPTER 3 GRAPHS OF THE TRIGONOMETRIC FUNCTIONS Chapter Outline 3.1 Graphs of the Cosine and Sine Functions 3.2 Properties of the Graphs of Sinusoids 3.3 Graphs of the Tangent and Cotangent Functions 3.4 Graphs of the Secant and Cosecant Functions Introduction In Chapter 3, we graph the six trigonometric functions and learn about important properties of each function such as domain, range, period, and whether the function is even or odd. We begin with graphs of the cosine and sine functions in Section 3.1, which lead into further discussion of their designation as sinusoids in Section 3.2. We notice the similarities between sine and cosine graphs, along with horizontal shifts that will turn a sine graph into a cosine graph, or vice versa. Graphs of sinusoids and their applications, including harmonic motion, in Section 3.2 are followed by graphs and properties of tangent and cotangent functions in Section 3.3. We end with Section 3.4, graphing secant and cosecant functions and observing properties of each of these functions, as well as the relationship between the two. Throughout Chapter 3, applications are introduced. Particular attention is paid to the graphing techniques that allow us to graph transformations of each function. This is an essential chapter in our textbook as it provides many of the tools we will need in applying identities and formulas, as well as solving trigonometric equations, in future chapters. 3.1 Graphs of the Cosine and Sine Functions 105 3.1 Graph
s of the Cosine and Sine Functions Learning Objectives In this section you will:  Graph the cosine and sine functions and their transformations. Identify the period.  Learn the properties of the cosine and sine functions, including domain and range.  Determine whether a function is even or odd. We return to our discussion of the circular (trigonometric) functions as functions of real numbers and turn our attention to graphing the cosine and sine functions in the Cartesian Plane. Before proceeding with graphs of these trigonometric functions, we note that the graphs of both functions are continuous and smooth. Geometrically this means the graphs of the cosine and sine functions have no jumps, gaps, holes, vertical asymptotes, corners or cusps. As we shall see, the graphs of both functions meander nicely and don’t cause any trouble. Graph of the Cosine Function To graph the cosine function, we use x as the independent variable and y as the dependent variable.27 This allows us to turn our attention to graphing the cosine and sine functions in the Cartesian Plane. We graph by making a table using some of the common values of x in the interval. This generates a portion of the cosine graph, which we call the fundamental cycle of. 27 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 may help to avoid confusion. cosyx0,2cosyx 106 Graphs of the Trigonometric Functions x 0 1 0 π –1 x 2π 0 1 Noting that is defined for all real numbers x, we plot the points from the table to guide us in sketching the graph of on the interval. The fundamental cycle of. A few things about the graph above are worth mentioning: 1. This graph represents only part of the graph of. To get the entire graph, imagine copying and pasting this graph end to end infinitely in both directions (left and right) along the x-axis. 2. The vertical scale here has been greatly exaggerated for clarity and aesthetics. Below is an accurate to-
scale graph of showing several cycles with the fundamental cycle plotted thicker than the others. cosyx,cosxxcosyx,cosxx0,1542252,424222,42323,022,02742272,42342232,422,1,1cosyx,cosxxcosyx0,2cosyxcosyxcosyx 3.1 Graphs of the Cosine and Sine Functions 107 The graph of is usually described as ‘wavelike’ and, indeed, many of the applications involving the cosine and sine functions feature the modeling of wavelike phenomena. An accurately scaled graph of. Graph of the Sine Function We can plot the fundamental cycle of the graph of similarly. x 0 0 1 π 0 x –1 2π 0 The fundamental cycle of. cosyxcosyxsinyxsinyx�
��,sinxxsinyx,sinxx0,0542252,424222,42323,122,12742272,42342232,422,0,0sinyx 108 Graphs of the Trigonometric Functions As with the graph of, we can provide an accurately scaled graph of with the fundamental cycle highlighted. An accurately scaled graph of. It is no accident that the graphs of and are so similar. The graph of is a result of the graph of being shifted units to the left. Try it! Period of the Cosine and Sine Functions Not only can we obtain a graph of the sine function by shifting the graph of the cosine function units to the left, we can shift the graph of by 2π units to the left and obtain a graph that is equivalent to the original graph of. The same can be said for shifts of 4π, 6π, 8π, ··· units to the left. We say that the cosine function is periodic, as defined below. Periodic Functions: A function f is said to be periodic if there is a real number p so that for all real numbers x in the domain of f. The smallest positive number p, if it exists, is called the period of f. We see that by the definition of periodic functions, is periodic since for any integer k. To determine the period of f, we need to
find the smallest positive real number p so that for all real numbers x or, said differently, the smallest positive real number p such that for all real numbers x. We know that for all real numbers x but the question remains if any smaller real number will do the trick. Suppose and for all real numbers x. Then, in cosyxsinyxsinyxcosyxsinyxcosyxsinyx22cosyxcosyxfxpfxcosfxxcos2cosxkxfxpfxcoscosxpxcos2cosxx0pcoscosxpx 3.1 Graphs of the Cosine and Sine Functions 109 particular, so that. From this we know that 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 is periodic with 2π as its period. Having period 2π essentially means that we can completely understand everything about the functions and by studying one interval of length 2π, say. Even/Odd Properties of the Cosine and Sine Functions While we will explore the even and odd properties of the cosine and sine functions further in Section 4.1, for now it is worth revisiting the graphical test for symmetry as introduced in college algebra. Recall that we refer to a function as even if its graph is symmetric about the y-axis and a function as odd if
its graph is symmetric about the origin. Observe the symmetry of the ‘accurately scaled’ graphs of the cosine and sine functions from earlier in this section. The graph of is symmetric about the y-axis. As will be proved algebraically in Section 4.1, the cosine function is an even function. The graph of is symmetric about the origin. The sine function is an odd function. From a previous algebra class, you may recall that, for an even function f,. If f is odd then. This is true for all values of x, and applies to the trigonometric functions as well. For the cosine function, an even function,. The sine function is odd, and thus. Domain and Range of the Cosine and Sine Functions The two functions and, both trigonometric functions of x, are defined for all real values of x. This is evident in the preceding graphs of cosine and sine. Or, going back to our Unit Circle definitions for cosine and sine, whether we think of identifying the real number t with the angle 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 cosine and sine are defined for any real input number t. Thus, the domain of both and is. cos0cos0pcos1psingxxcosfxxsingxx0,2cosyxsinyxfxfxfxfxcoscosxxsinsinxxcosfxx�
��singxxtcosfttsingtt, 110 Graphs of the Trigonometric Functions Looking at the graphs of the cosine and sine functions, we see that the range includes all real numbers between –1 and 1, inclusive. Revisiting the Unit Circle, and represent x- and y-coordinates, respectively, of points on the Unit Circle. As points on the Unit Circle, they take on all of the values between –1 and 1, inclusive. In other words, the range of both and is the interval. Following is a summary of properties of both functions. Theorem 3.1. Properties of the Cosine and Sine Functions: The function  has domain  has range The function  has domain  has range   is continuous and smooth is even   is continuous and smooth is odd  has period 2π  has period 2π Graphs of Transformations of the Cosine and Sine Functions Now that we know the basic shapes of the graphs of and, we can use transformations to graph more complicated curves. The following theorem, borrowed from college algebra, will guide us in transformations of graphs of cosine and sine functions. cosfttsingttcosfttsingtt1,1cosftt,1,1singtt,1,1cosyx
sinyx 3.1 Graphs of the Cosine and Sine Functions 111 Theorem 3.2. Transformations of Periodic Functions: Suppose f is a periodic function. If and, then to graph 1. Divide the period of the function by ω to determine the period of the transformed function g. For sine and cosine, the period is 2π, so is the period of g. Note that ω is the number of times the cycle of g repeats on. 2. Find the vertical shift by determining the value of B. The graph will be shifted up by B if and down by if. The line will be the ‘baseline’. 3. Determine the vertical scaling,, also known as the amplitude. If, note that the graph will be reflected about the baseline. 4. The horizontal shift, also known as the phase shift, can be determined by rewriting as. This results in a horizontal shift to the left if or right if. In transforming one cycle of or, we need to keep track of the movement of some key points. We choose to track the points with x-values,,, and. 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. Example 3.1.1. Graph one cycle of the function. Solution. We start with the following graph of the fundamental cycle of. The locations of quarter marks are denoted with dashed blue lines, and intersect the graph at the key points,,, and. The baseline of is marked by a dashed pink line. 0A0gxAfxBfx20,20BB0ByBA0AyBxx0
0cosyxsinyx023222sin31fxxsinyx0,0,12,03,122,00y 112 Graphs of the Trigonometric Functions The steps in Theorem 3.2 help us determine transformations for graphing. Step 1: We first determine the period. The period for is 2π. For coefficient of x is 3, so the graph of the function will cycle three times as fast as other words, it will repeat itself three times on the interval, with one cycle being, the. In completed in a period of. Step 2: We next determine the vertical shift. The graph of has a vertical shift of 1. By shifting the baseline for the graph of up by 1, we have a baseline of for the graph of. Step 3: The vertical scaling, or amplitude, of is 1. For the function, the sine function is multiplied by 2 so the function values are twice as much. Thus the amplitude is 2. Step 4: The horizontal shift is 0. Finally, we are ready to graph one cycle of. Since the horizontal shift is 0, we can leave the first quarter mark at. We then divide the period of by 4 to determine a distance of between quarter marks. The positions for quarter marks and the baseline of are used as a guide in graphing. 2sin31fxxsinyx2sin31fxxsinyx0,223�
��2sin31fxxsinyx1y2sin31fxxsinyx2sin31fxx2sin31fxx0x2361y2sin31fxxxy0 sinyx 3.1 Graphs of the Cosine and Sine Functions 113 The graph of will maintain the shape of the sine function. With an amplitude of 2, the local maximums and minimums will occur 2 units above and 2 units below the baseline, respectively. We can plot 5 key points of this function using the quarter mark locations and baseline as a guide, and finish the graph by connecting the points with a continuous smooth curve. Example 3.1.2. Graph one cycle of the function. Solution. We begin with a graph of the cosine function, noting the key points of,,, and. 2sin31fxx3cos2fxx0,1,02,13,022,1xy0
xyxy 0,1,13 2,13,36,12 2sin31fxx 114 Graphs of the Trigonometric Functions Step 1: The period of is 2π. For, the coefficient of x is 2 so the period is Step 2: The vertical shift of is 0. Thus, the baseline will remain at. Step 3: The amplitude is. Since, the graph will be reflected about the baseline. Step 4: To determine the horizontal shift, the function can be rewritten as The horizontal shift is, indicating a shift to the right of units. Now we are ready to graph this transformation of. From the horizontal shift of, the first quarter mark will be at. The period is π, so the distance between quarter marks will be. We denote the positions of the quarter marks and the baseline. The amplitude is 3. While the basic shape is that of the cosine, the graph will be reflected about the baseline, resulting in reversed positions for the local maximum and minimum values. After determining the 5 key points for one cycle of of., we connect the points with a continuous smooth curve, replicating the shape cosyx3cos2fxx223cos2fxx0y3330
0y3cos23cos22fxxx22cosyx22x40y0y3cos2fxxcosyxxy01-1 cosyx 3.1 Graphs of the Cosine and Sine Functions 115 Above, we have graphed one complete period of the function. This graph could easily be extended in either direction. In the next example we sketch the graph of a sine function over the larger interval of two periods. Example 3.1.3. Graph two full periods of the function and state the period. Identify the maximum and minimum y-values and their corresponding x-values. Solution. To sketch one period of the function through transformations of the the fundamental cycle of, we note the following. 3cos2fxx4sin2gxx4sin2gxxsinyxxy
xyxy 3,04 ,3 5,04 3,323cos2fxx,32 116 Graphs of the Trigonometric Functions 1. The period is, so the distance between quarter marks will be. 2. There is a vertical shift of –2, resulting in a baseline of. 3. The amplitude is 4. 4. There is no horizontal shift, so the first quarter mark is at. To sketch two periods, we extend the graph for one period, as follows. The period of 2 appears twice in the preceding graph. A maximum y-value of 2 corresponds with x- values of and. The minimum y-value is and corresponds with x-values of and. The functions in this section are examples of sinusoids. Roughly speaking, a sinusoid is the result of performing transformations to the basic graph of or. Sinusoids will be looked at extensively in Section 3.2. 2221422y0x4sin2ygxx321261232cosfxxsingxxxy2-112-2-6-2�
��xy 3.1 Graphs of the Cosine and Sine Functions 117 3.1 Exercises 1. Why are the cosine and sine functions called periodic functions? 2. How does the graph of compare with the graph of? Explain how you could horizontally translate the graph of to obtain the graph of. 3. For the function, what constants affect the range and how do they affect the range? In Exercises 4 – 15, graph one cycle of the given function. State the period of the function. 4. 7. 10. 13. 5. 8. 11. 14. 6. 9. 12. 15. In Exercises 16 – 27, graph two full periods of each function and state the period. Identify the maximum and minimum y-values and their corresponding x-values. 16. 19. 22. 25. 17. 20. 23. 26. 18. 21. 24. 27. sinyxcosyxsinyxcosyxcosfxABxCD3sinyxsin3yx2cosyxcos2yxsin3yxsin2yx11cos323yxcos324yxsin24yx2cos4132yx31cos2232yx4sin2yx
2sinfxx2cos3fxx3sinfxx4sinfxx2cosfxxcos2fxx12sin2fxx4cosfxx63cos5fxx3sin845yx2sin3214yx5sin5202yx 118 Graphs of the Trigonometric Functions 28. Show that a constant function f is periodic by showing that for all real numbers x. Then show that f has no period by showing that you cannot find a smallest number p such that for all real numbers x. Said another way, show that for all real numbers x for ALL values of, so no smallest value exists to satisfy the definition of period. 117fxfxfxpfxfxpfx0p 3.2 Properties of the Graphs of Sinusoids 119 3.2 Properties of the Graphs of Sinusoids Learning Objectives In this section you will:  Learn the properties of graphs of sinusoidal functions, including period, phase shift, amplitude and vertical shift.  Use properties to graph sinusoidal functions.  Write an equation of the form or from the graph of a sinusoidal function.  Solve applications of sinusoids, including harmonic motion. Sinusoids can
be characterized by four properties: period, amplitude, phase shift, and vertical shift. 1. We have already discussed period, that is, how long it takes for the sinusoid to complete one cycle. The standard period of both and is 2π, but horizontal scalings will change the period of the resulting sinusoid. 2. 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. 3. The phase shift of the sinusoid is the horizontal shift experienced by the fundamental cycle. We have seen that a phase (horizontal) shift of to the right takes to since. As the reader can verify, a phase shift of to the left takes to. 4. The vertical shift of a sinusoid is assumed to be 0, but we will state the more general case. C=Acos++BωxxS=Asin++Bωxxcosfxxsingxx2cosfxxsingxxcossin2xx2singxxcosfxx 120 Graphs of the Trigonometric Functions The following theorem shows how to find these four fundamental quantities from the formula of a given sinusoid. Theorem 3.3. For, the functions  have period  have amplitude and  have phase shift  have vertical shift B We note that in some scientific and engineering circles, the quantity ϕ mentioned in Theorem 3.3 is called the phase angle of the sinusoid. Since our interest in this book is primarily with graphing sinusoids, we focus our attention on the horizontal shift induced by ϕ. Graphs of
Sinusoids The proof of Theorem 3.3 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 using the even property of the cosine function, or the odd property of the sine function, from which Theorem 3.3 using the functions following two examples. Example 3.2.1. Graph. and and. We now test out in the Solution. Using Theorem 3.3, we first write the function f in the form prescribed in the theorem. From Theorem 3.3,,, and, resulting in the following. 0cosCxAxBsinSxAxB2A0coscosxxsinsinxx3cos12xfx13sin222gxx3cos12xfx3cos123cos122xfxx3A22�
��1B 3.2 Properties of the Graphs of Sinusoids 121  The period of f is.  The amplitude is.  The phase shift is, indicating a shift to the right 1 unit.  The vertical shift is, indicating a shift up 1 unit, and a baseline of. The graph shows one cycle of. Using the period, amplitude, phase shift and vertical shift, in conjunction with techniques from Section 3.1, the graph of is sketched as a transformation of. Key points are emphasized on each graph. The baseline for is shown as a dashed line. Example 3.2.2. Graph. Solution. Before applying Theorem 3.3, we use the odd property of the sine function to write in the required form. 224233A2121B1y3cos12xfxyfxcosyxyfx13sin222gxxgxxy cosyx yfx 122 Graphs of the Trigonometric Functions We next determine that,, and. The properties follow from Theorem 3.3.  The period of g is  The amplitude is..  The phase shift is, indicating a shift right units.  The vertical shift is up units. We graph via transformations of the sine function, using the above properties as a guide. Before graphing, we note that.
Since, the graph of must be reflected about the baseline. 13sin22213sin22213sin2 2213sin222gxxxxxfrom odd property of sine12A232B2211222223213sin222gxx12A0Aygxxy ygx sinyx 3.2 Properties of the Graphs of Sinusoids 123 Remember that the cycle graphed through transformations of the sine function in Example 3.1.2 is only one portion of the graph of. Indeed, another complete cycle begins at, and a third at. Note that whatever cycle we choose is sufficient to completely determine the sinusoid. Determining an Equation from the Graph of a Sinusoid Example 3.2.3. Below is the graph of one complete cycle of a sinusoid. One cycle of. 1. Find a cosine function whose graph matches the graph of. 2. Find a
sine function whose graph matches the graph of. Solution. 1. We fit the data to a function of the form by determining A, ω, ϕ and B.  Since one cycle is graphed over the interval, its period is. According to Theorem 3.3,, so that. ygx2x32xyfxyfxyfxyfxcosCxAxB1,5516263xy 51,2 55,2 11,22 71,22 32,2 124 Graphs of the Trigonometric Functions  To find the amplitude, we note that the range of the sinusoid is. The midpoint of the range is, indicating a baseline of. After marking the graph with the baseline, we see that the amplitude is.  Next, we see that the phase shift is, so we have or.  Finally, we refer to the baseline to verify a vertical shift of. Our final answer is. 2. Most of the work to fit the data to a function of the form is done.  The period, amplitude and vertical shift are the same as part 1. Thus,, and.
 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. 35,221212y51222A11312B12cos332CxxsinSxAxB32A12B71,22xybaseline 71,22 55,2 131,22 38,2 191,22xybaseline 51,2 55,2 11,22�
�� 71,22 32,2 3.2 Properties of the Graphs of Sinusoids 125 Taking the phase shift to be, we get, or Hence, our answer is. Note that each of the answers given in Example 3.2.3 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 taking. In this case, the phase shift is so for an answer of. Alternately, we could have extended the graph of to the left and considered a sine function starting at, and so on. Each of these formulas determine the same sinusoidal curve and the formulas are equivalent using identities. Applications of Sinusoids Sinusoids are used to model a fair number of behaviors that possess a wavelike motion, such as sound, voltage, and spring action. The following examples look at circular motion that can be expressed as a sinusoidal function. Example 3.2.4. A circle with radius 3 feet is mounted with its center 4 feet off the ground. The point closest to the ground is labeled P, as shown below. Sketch a graph of the height above the ground of the point P as the circle is rotated. Find a function h that gives the height in terms of the angle x of rotation. 7272727237.6712sin362Sxx11,222A12612sin362Sxxyfx51,22�
� 126 Graphs of the Trigonometric Functions Solution. Sketching the height, we note that it will start 1 foot above the ground, then increase up to 7 feet above the ground, and continue to oscillate 3 feet above and below the center value of 4 feet. Although we could use a transformation of either the cosine or sine function, we start by looking for characteristics that would make one function easier to use than the other. Since this graph starts at its lowest value, when, using the cosine function would not require a horizontal shift. Thus, we choose to model the graph with a cosine function. We note that a standard cosine graph starts at the highest value so we do need to incorporate a vertical reflection. We see that the graph oscillates 3 feet above and below the horizontal center of the graph. The basic cosine graph has an amplitude of 1, so this graph has been vertically stretched by 3. Finally, to move the center of the circle up to a height of 4 feet, the graph has been vertically shifted up by 4. Putting these transformations together, we find that. 0x3cos4hxxxh(x)baseline 3.2 Properties of the Graphs of Sinusoids 127 Example 3.2.5. The London Eye is a huge Ferris wheel with a diameter of 135 meters (443 feet). It completes one rotation every 30 minutes. Riders board from a platform 2 meters above the ground. Express a rider’s height above ground as a function of time in minutes. Solution. The wheel takes 30 minutes to complete 1 revolution, so the height will oscillate with a period of 30 minutes. Because the rider boards at the lowest point, the height will start at the smallest value and increase, following the shape of a vertically reflected cosine curve:, for time t and height H. A phase shift is not required. With a diameter of 135 meters, the
wheel has a radius of 67.5 meters. The height will oscillate with amplitude 67.5 meters above and below the horizontal center of the wheel. Passengers board 2 meters above the ground level, so the center of the wheel must be located meters above ground level. The horizontal midline of the oscillation will be at 69.5 meters. Putting this all together, we have  Period:  Amplitude: ; (due to the vertical reflection of the cosine curve)  Phase Shift:  Vertical Shift: An equation for the rider’s height, with t in minutes and H in meters, is cosHtAtB67.5269.52301567.5A67.5A0069.5B67.5cos69.515Htt 128 Graphs of the Trigonometric Functions Harmonic Motion One of the major applictions 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. Here, 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,28 while its weight could change. 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 formula29. Here, w is the weight of the object, m is the mass and g is the acceleration due to gravity. In the English system, and in the SI system,. Hence, on Earth a mass of 1 slug weighs 32 lbs. and a mass of 1 kg weighs 9.8 N.30 Suppose we attach an object with mass m to a spring as detected below. The weight of the object will stretch the spring. The system is said to be in ‘equilibrim’ 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. 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 28 Well, assuming the object isn’t subjected to relativistic speeds... 29 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. 30 Note that 1 pound = 1 slug foot / second2 and 1 Newton = 1 kg meter / second2. wmg232feetsecondg29.8 meterssecondgFkd 3.2 Properties of the Graphs of Sinusoids 129 external force stops it. If we let denote the object’s displacement from the equilibrium position at time t, then    means the object is at the equilibrium position, means the object is above the equilibrium position, means the object is below the equilibrium position. The function is called the ‘equation of motion’ of the object.31 at the above the 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). In the following theorem, which comes from Differential Equations,   is a function of the mass m of the object, the spring constant k, the initial displacement of the object and initial velocity of the object. means the object is released from the equilibrium position, means the object is released above the equilibrium position and means the object is released below the equilibrium position. 31 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). xt0xt0xt0xtxt0xt0xt0xtxt0x0v00x00x00x 130 Graphs of the Trigonometric Functions  means the object is released from rest, means the object is heading upwards and means the object is heading downwards.32 Theorem 3.4. 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 and the initial velocity of the object is, then the displacement x from the equilibrium position at time t is given by where and   and. It is a great exercise in ‘dimensional analysis’ to verify that the formulas in Theorem 3.4 work out so that ω has units and A has units ft. or m, depending on which system we choose. Example 3.2.6. Suppose an object weighing 64 pounds stretches a spring 8 feet. 1. If an object is attached to the spring and released 3 feet below the equilibrium position from rest, find the equation of motion of
the object,. 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,. What is the longest distance the object travels above the equilibrium position? When does this first happen? Solution. In order to use the formulas in Theorem 3.4, we first need to determine the spring constant k and the mass m of the object. We know the object weighs 64 lbs. and stretches the spring 8 ft. Using Hooke’s Law with and, we get 32 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 oppostition to gravity, any movement upwards is considered negative. 00v00v00v0x0vsinxtAtkm2200vAx0sinAx0cosAv1sxtxt64F8d 3.2 Properties of the Graphs of Sinusoids 131 To find m, we use with lbs. and. We get slugs. We can now proceed to apply Theorem 3..4. 1. To find the equation of motion,, we must determine values for A, ω and ϕ. We begin with ω. Now, since the object is released 3 feet below the equilibrium position, from rest, and. We use these values, along with, to find A. To determine the phase ϕ, we use and, along with a formula from Theorem 3.4. Hence, the equation of motion is. 648lbs.8ft.Fkdkk Hooke's Lawwmg64w2.32ftgs�
��2msinxtAt822km from Theorem 3.4 since =8 lbs./ft. and =2 slugs km03x00v22200220323vAx from Theorem 3.403x3A0sin3sin3sin12Ax from Theorem 3.43sin22xtt 132 Graphs of the Trigonometric Functions Next, to find when the object passes through the equilibrium position, we solve. The object first passes through the equilibrium position at the smallest positive t value, which in this case is 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. Since our applied domain in this situation is, and the period of, we graph over the interval is. Remembering that means the object is below the equilibrium position and means the object is above the equilibrium position, the fact our graph is crossing through the t-axis at 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 and, but now we have, the negative indicating the velocity is directed upwards. Here, we get 0xt3sin202sin202 for integers 42tttkk�
� after the usual analysis0.784txt0txt222xt0,0xt0xt4t203x08vtx 3sin22xtt 3.2 Properties of the Graphs of Sinusoids 133 We use and to determine ϕ. We will need to identify ϕ using the arcsine since is not a common angle. With the sine being positive, ϕ is in Quadrant I or Quadrant II. Knowing whether the cosine is positive or negative will determine which of those quadrants ϕ resides in. Since we know, and, the formula for from Theorem 3.4. will help us find the cosine. This tells us that ϕ is in Quadrant II, so we have. Hence,. Now, since the amplitude is 5, the object will travel at most 5 feet above the equilibrium position. This happens when, the negative sign once again signifying that the object is above the equilibrium position. 2200228325.vAx5A03x0sin5sin33sin5Ax from Theorem 3.4355A208v0
v0cos(5)(2)cos84cos5Av3arcsin535sin2arcsin5xtt5xt35sin2arcsin553sin2arcsin15tt 134 Graphs of the Trigonometric Functions Going through the usual machinations, we get for integers k. The smallest of these values is when, that is, seconds after the start of the motion. 13arcsin254tk0k13arcsin1.107254t 3.2 Properties of the Graphs of Sinusoids 135 3.2 Exercises In Exercises 1 – 12, state the period, phase shift, amplitude and vertical shift of the given function. Graph one cycle of the function. 1. 4
. 7. 10. 2. 5. 8. 11. 3. 6. 9. 12. 13. Write an equation of the form for the sine function whose graph is shown below. 14. Write an equation of the form for the cosine function whose graph is shown below. 3sinyxsin3yx2cosyxcos2yxsin3yxsin2yx11cos323yxcos324yxsin24yx2cos4132yx31cos2232yx4sin2yxsinSxAxBcosCxAxB 136 Graphs of the Trigonometric Functions 15. Write an equation of the form for the cosine function whose graph is shown below. 16. Write an equation of the form for the sine function whose graph is shown below. 17. Write an equation of the form for the cosine function whose graph is shown below. cosCxAxBsinSxAxBcosCxAxB�
� 3.2 Properties of the Graphs of Sinusoids 137 18. Write an equation of the form for the sine function whose graph is shown below. 19. Write an equation of the form for the cosine function whose graph is shown below. 20. Write an equation of the form for the sine function whose graph is shown below. sinSxAxBcosCxAxBsinSxAxB 138 Graphs of the Trigonometric Functions In Exercises 21 – 22, verify the identity by using technology to graph the right and left hand sides. 21. 22. In Exercises 23 – 26, graph the function with the help of technology and discuss the given questions with your classmates. 23. 24. 25. 26.. Is this function periodic? If so, what is the period?. What appears to be the horizontal asymptote of the graph?. Graph on the same set of axes and describe the behavior of f.. What’s happening as? 27. A Ferris wheel is 25 meters in diameter and boarded from a platform that is 1 meter above the ground. The six o’clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. The function gives a person’s height in meters above the ground t minutes after the wheel begins to turn. a. Find the period, amplitude and vertical shift of. b. Find a formula for the height function. c. How high off the ground is a person after 5 minutes? 28. 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 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? �
��22cossin1xxcossin2xxcos3sinfxxxsinxfxxsinfxxxyx1sinfxx0xhththtlbs.ft. 3.2 Properties of the Graphs of Sinusoids 139 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. 140 3.3 Graphs of the Tangent and Cotangent Functions Learning Objectives In this section you will:  Graph the tangent and cotangent functions and their transformations. Identify the period and vertical asymptotes.  Learn the properties of the tangent and cotangent functions, including domain and range; determine whether a function is even or odd. We now turn our attention to the graphs of the tangent and cotangent functions. Graph of the Tangent Function When constructing a table of values for the tangent function, we recall that. We can use our common values for the fundamental cycles of and in determining values of the tangent. It follows that the tangent function,, is undefined at and, both points at which. sintancosxxxcosyxsinyxtanyx2x3/2xcos0x 141 x undefined undefined cosxsinx�
��sintancosxxx,tanxx01000,0422221,1420134222213,14100,054222215,14320174222217,1421002,0 142 To determine the behavior of the graph of when x is close to or, we will look at some values for on both sides of and on both sides of.  The following chart shows some values for when x is less than, but close to. We note that and include approximate values of the tangent for the indicated radian measures of x. We note that values of are positive, getting larger and larger, as x approaches from the left. The result is a vertical asymptote at. x 1.5 1.55 1.56 1.57 14 48 93 1256 undefined Mathematical notation:  When x is greater than, but close to, as x approaches from the right, the values of get smaller and smaller, approaching negative infinity:. x 1.58 1.59 1.6 1.7 undefined –109 –52 –34 –8 Noting that, we look at values for when x is close to. x 4.6 4.65 4.7 4.71 9 16 81 419 undefined tanyx232tanx2x32xtanx221.571�
��tanx22x1.5712tanxAs, tan.2xx22tanxAs, tan2xx1.5712tanx324.712tanx3234.7122tanx3As, tan.2xx 143 x 4.72 4.73 4.75 4.8 undefined –131 –57 –27 –11 Thus we have vertical asymptotes at and at. We also know something about the behavior of the graph as it approaches these vertical asymptotes from each side. Plotting this information followed by the usual ‘copy and paste’ produces the following two graphs. The graph of over. The graph of, with fundamental cycle highlighted. From the graph, it appears the tangent function is periodic with period π. This is, in fact, the case as we will prove in Section 4.2, following the introduction of the sum identity for tangent. We take as our fundamental cycle for the interval. 34.7122tanx3As, tan2xx2x32xtanyx0,2tanyxtanyx2,2 144 From the graph, it appears that the domain of the tangent function,, includes all real numbers x except for. These are the x-values where and are the only numbers for
which is undefined. Thus, the domain of is all real numbers x, excluding for any integer k. The range of, as observed from the graph, includes all real numbers. Graph of the Cotangent Function It should be no surprise that the graph of the cotangent function behaves similarly to the graph of the tangent function. Plotting over the interval results in the graph below. x x undefined undefined undefined The graph of over. tanyx2,32,52,xcos0xtanyxtanyx2xktanyxcotyx0,2cotx,cotxxcotx,cotxx05415,1441,143203,0220,027417,143413,142cotyx0,2 145 It clearly appears that the period of is π, which is indeed the case and will be revisited in Section 4.2. The vertical asymptotes in the
interval are, and. We take as one fundamental cycle the interval. A more complete graph of is below, with the fundamental cycle highlighted. The graph of. We see, from the graph, the apparent domain of the cotangent function is all real numbers x except for. These are the x-values where and are the only numbers for which is undefined. Thus, the domain of is all real numbers x, excluding, for any integer k. The range of includes all real numbers. Note that on the intervals between their vertical asymptotes, both and are continuous and smooth. In other words, they are continuous and smooth on their domains. Both functions are odd, as you can see by the symmetry of the graphs about the origin, and as we will verify algebraically in Section 4.1. The following theorem summarizes the properties of the tangent and cotangent functions. cotx0,20xx2x0,cotyxcotyx0,,2,xsin0xcotyxcotyxxkcotyxtanyxcotyx 146 Theorem 3.4. Properties of the Tangent and Cotangent Functions:  The function – has domain – has range is continuous and smooth on its domain is odd – – – has period π  The function – has domain – has range – is continuous and smooth on its domain – is odd – has period π Graphs of Transformations of the Tangent and Cotangent Functions Graphing transformations of the tangent and cotangent functions is similar to graphing transformations of the sine and cosine. Theorem 3.2, in which, can be used for both, but there are a few differences to be aware of.  The period of both tangent and cotangent is π, and so
the period of transformations will be.  The vertical scaling is, but tangent and cotangent do not have amplitude since they do not possess the wavelike characteristics of the sine and cosine functions.  The vertical shift is B, but a baseline is not defined for transformations of the tangent and cotangent functions.  Since we do not have a baseline, vertical scaling should be completed before the vertical shift.  The horizontal shift,, is unchanged and will be our last step in graphing transformations of the tangent and cotangent. tanJxx:, is any integer2xxkk,cotKxx:, is any integerxxkk,gxAfxBA Example 3.3.1. Graph one cycle of the following functions. Find the period. 147 1. Solution. 2. 1. We will graph the function through a series of transformations to the fundamental cycle of the graph of. Before proceeding, we rewrite the function in a format that will simplify this process: The fundamental cycle of is restricted to the interval. In tracking transformations of the graph of, we will start with the quarter marks,,, and. The first and last of these quarter marks are place markers for vertical asymptotes, but it will be important to track vertical asymptotes as well as points in the transformation process. Hence, we start with the two vertical asymptotes and, and the three points, and.  We begin by determining the period of. Since the coefficient of x is, the period is. Stretching the period π of the tangent function to the period 2π of the transformed function will result in quarter marks of,, 0, and. �
��3tan2xfx3cot142gxx3tan2xfxtanyxfx1tan032fxxtanyx2,2tanyx2x40422x2x4,10,04,11tan032fxx1221222xy tanyx 148  The vertical scaling is, so the graph will not be stretched in the vertical direction. However, tells us the graph will be reflected about the x-axis, as follows.  The vertical shift is 3, indicating a shift up by 3 units.  Finally, the horizontal shift is 0, so the graph will not be shifted left or right. One cycle of. 11103tan2xfx�
�xyxy 149 2. We graph the function using transformations of. The fundamental cycle of the cotangent function is on the domain. We use the quarter marks,,, and. Before proceeding, the function can be written as follows.  The period of is, resulting in quarter marks 0, 1, 2, 3 and 4, shown below.  The vertical scaling is. This will stretch the graph vertically by a factor of 3. We multiply the y-coordinates of the points at the second and fourth quarter marks, i.e. and, by 3 to obtain the transformed points and. The middle point,, does not move. Note these transformations in the graph to the right.  Next, the graph is shifted up by 1 unit.  Lastly, a horizontal shift of –2 moves the graph two units to the left. 3cot142gxxcotyx0,04234gx3cot214gxxgx44331,13,11,33,32,0�
��xy 1,1 3,1xy cotyxxy 1,3 3,3 2,0 150 One cycle of. Since the number of classical applications involving sinusoids far outnumber those involving tangent and cotangent functions, we omit the more extensive coverage here that was given to sinusoidal functions. The ambitious reader is invited to explore further results from this section. We next move on to graphs of secant and cosecant functions. 3cot142gxxxy 0,1 1,4 1,2 151 3.3 Exercises In Exercises 1 – 6, graph one cycle of the given function. State the period of the function. 1. 4. 2. 5. 3. 6. In Exercises 7 – 15, graph two full periods of each function. State the period and asymptotes. 7. 10. 13. 8. 11. 14. 9. 12. 15. In Exercises 16 – 17, find an equation for the graph of each function. 16. tan3yx12tan34yx
1tan213yxcot6yx111cot5yx13cot2132yxtanfxxcotfxx2tan432fxxtan2fxxtan2fxx4tanfxxtan4fxxtanfxx3cot2fxx 17. 152 18. Verify the identity by using technology to graph the right and left hand sides. 19. Graph the function with the help of technology. Graph on the same set of axes and describe the behavior of f. 20. The function marks the distance in the movement of a light beam from a police car across a wall for time x, in seconds, and distance, in feet. a. Graph the function on the interval. b. Find and interpret the vertical stretching factor, the period and any asymptotes. c. Evaluate and and discuss the function’s values at those inputs. tantanxxtanfxxxyx20tan10fxxfxfx0,5
1f2.5f 153 3.4 Graphs of the Secant and Cosecant Functions In this section you will: Learning Objectives  Graph the secant and cosecant functions and their transformations. Identify the period and vertical asymptotes.  Learn the properties of the secant and cosecant functions, including domain and range; determine whether a function is even or odd. Finally, we turn our attention to graphing the secant and cosecant functions. Graph of the Secant Function To get started, we graph the secant function using our table of values for the fundamental cycle of. We can take reciprocals of these cosine values since. x undefined undefined cosyx1seccosxxcosx1seccosxx,secxx0110,14222,2420342223,2411,1542225,24320742227,242112,1 154 The domain of the secant function excludes all odd multiples of since these are the values of x for which at. In our table based on the fundamental cycle of, the secant is undefined and. These are both x-values at which vertical asymptotes occur. To determine the behavior of the graph of when x is close to or, we will look at some values for on both sides
of and on both sides of.  The following chart shows some values for when x is less than, but close to. We note that and include approximate values of the secant for the indicated radian measures of x. x 1.5 1.55 1.56 1.57 14 48 93 1256 undefined We note that values of are positive, getting larger and larger, as x approaches from the left:  We next look at values of when x is greater than, but close to. x 1.58 1.59 1.6 1.7 undefined –109 –52 –34 –8 As x approaches from the right, the values of get smaller and smaller, approaching negative infinity:  Using a similar analysis, which we leave to the reader,. 2cos0xcosyx2x32xsecyx232secx2x32xsecx221.5711.57121seccosxxsecx2As, sec.2xxsecx21.57121seccosxx2secxAs, sec2xx3as, sec, and23as, sec.2xxxx 155 Plotting points and asymptotes, with graph behavior echoing the above results, we have the following two graphs. The second graph is an extension
of the first. The graph of over. The graph of, with fundamental cycle highlighted. In the above illustration, the dotted graph of is included for reference. It is helpful to graph the secant function by starting with a graph of the cosine function and sketching vertical asymptotes at each x-value for which. The points where are also points where. After drawing the asymptotes and marking the points where, a rough graph of can quickly be completed by sketching the ‘U’ shapes of the secant function. Since is periodic with period 2π, it follows that is also periodic with period 2π.1 Due to the close relationship between the cosine and secant, the fundamental cycle of the secant function is the same as that of the cosine function. We previously noted that the domain of the secant function excludes all odd multiples of. The range of, as observed graphically, includes all real numbers y 1 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). secyx0,2secyxcosyxcos0xcos1xsec1xsec1xsecyxcosxsecx2secyx 156 such that or, or equivalently. By thinking of the secant function as being the reciprocal of the cosine function, a similar result can be obtained algebraically. Graph of the Cosecant Function As one would expect, to graph, we begin with and take reciprocals of the corresponding y-values. Here, we encounter issues at, and. These are locations of vertical asymptotes. Proceeding with an analysis of graph behavior near these asymptotes, we graph the fundamental cycle of followed by an extended graph of. A dotted graph of
is included for reference. x undefined undefined undefined 1y1y1ycscyxsinyx0xx2xcscyxcscyxsinyxsinx1cscsinxx,cscxx004222,24211,12342223,240542225,2432113,12742227,2420 157 The fundamental cycle of. The graph of. Since and are merely phase shifts of each other, so too are and. As with the tangent and cotangent functions, both and are continuous and smooth on their domains. The following theorem summarizes the properties of the secant and cosecant functions. Note that all of these properties are direct results of them being reciprocals of the cosine and sine functions, respectively. cscyxcscyxsinyxcosyxcscyx�
��secyxsecyxcscyx 158 Theorem 3.5. Properties of the Secant and Cosecant Functions:  The function – has domain – has range is continuous and smooth on its domain is even – – – has period 2π  The function – has domain – has range is continuous and smooth on its domain is odd – – – has period 2π Graphs of Transformations of the Secant and Cosecant Functions In the next example, we discuss graphing more general secant and cosecant functions Example 3.4.1. Graph one cycle of the following functions. State the period of each. 1. Solution. 2. 1. Before graphing we will graph to use as a guide. Using the technique from Section 3.1, we will start with a graph of the cosine function and apply appropriate transformations. First, however, it helps to rewrite in a format suggested by Theorem 3.2.  The period is.  The vertical shift is 1, for a baseline of. secFxx:, is any integer2xxkk:1,11,yycscGxx:, is any integerxxkk:1,11,yy12sec2fxxcsc53xgx12sec2fxx12cos
2yx12cos2yx2cos201yx221y 159  The amplitude is and, since, the graph will be reflected about the baseline.  There is no horizontal shift. One cycle of. Next, to graph, we observe the following.  Points where the graph of the cosine function crosses the baseline are vertical asymptotes of the secant function.  Maximum values of the cosine function occur at the lowest points on the secant curve.  Minimum values of the cosine function occur at points that are maximum values for the secant.  Transformations of the secant function retain the familiar “U” shape between vertical asymptotes. Below, we use as a guide in graphing one cycle of. 222012cos2yx12sec2fxx12cos2yxfxxybaseline 12cos2yx 160 One cycle of. Since one cycle is graphed on the interval, the period is. 2. We graph the function by first graphing the corresponding sine function,, and then using the sine function as a guide. The first step is to rewrite the sine function in the form suggested by Theorem 3.2. We proceed to graph as a transformation of, referring to Theorem 3.2 as necessary.  The period is.  The vertical shift is for a baseline of. 12sec2fxx0,0�
��csc53xgxsin53xysin5315sin3315sin3315sin33xyxxx from odd property of sine15sin33yx sinyx225353yxybaseline 161  The amplitude is and, since, the graph will be reflected about the baseline.  Since, the graph will be shifted to the right by 1 unit. Locating quarter marks and corresponding points results in the following graph of. One cycle of. We use the transformed sine graph as a guide in sketching one cycle of. One cycle of. 11331031xxsin53xysin53xycsc53xgx�
�csc53xgxxybaseline 12cos2yx 51,3 3,22 52,3 54,23 53,3 53yxybaseline 12cos2yx 162 We find the period to be. While real world applications of secants and cosecants are limited, at least in comparison to the large number of available sinusoidal applications, a couple of examples are included in the Exercises. We conclude Chapter 3 with the expectation of putting to good use the properties and graphs of the trigonometric functions that have been introduced in this chapter. 312 163 3.4 Exercises In Exercises 1 – 6, graph one cycle of the given function. State the period of the function. 1. 4. 2. 5. 3. 6. In Exercises 7 – 18, graph two full periods of each function. State the period and asymptotes. 7. 10. 13. 16. 8. 11. 14. 17. 9. 12. 15. 18. In Exercises 19 – 22, find an equation for the graph of each function. 19. sec2yxcsc3yx11sec323yx
csc2yxsec324yxcsc24yxsecfxxcscfxx2sec14fxx6csc3fxx2cscfxx1csc4fxx4sec3fxx7sec5fxx9csc10fxx2csc14fxxsec23fxx7csc54fxx 164 20. 21. 22. 165 23. Standing on the shore of a lake, a fisherman sights a boat far in the distance to his left. Let x, measured in radians, be the angle formed by the line of sight to the ship and a line due north from his position. Assume due north is 0 and x is measured negative to the left and positive to the right. The boat travels from due west to due east and, ignoring the curvature of the Earth, the distance, in kilometers, from the fisherman to the boat is given by the function. a. What is a reasonable domain for? b. Graph on the domain. c. Find and discuss the meaning of any vertical asym
ptotes on the graph of. d. Calculate and interpret. Round to the nearest hundredth. e. Calculate and interpret. Round to the nearest hundredth. f. What is the minimum distance between the fisherman and the boat? When does this occur? 24. A laser rangefinder is locked on a comet approaching Earth. The distance, in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by. a. Graph on the interval. b. Evaluate and interpret the information. c. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? d. Find and discuss the meaning of any vertical asymptotes. dx1.5secdxxdxdxdx3d6dgx250,000csc30gxxgx0,305g 166 Trigonometric Identities and Formulas CHAPTER 4 TRIGONOMETRIC IDENTITIES AND FORMULAS Chapter Outline 4.1 The Even/Odd Identities 4.2 Sum and Difference Identities 4.3 Double Angle Identities 4.4 Power Reduction and Half Angle Formulas 4.5 Product to Sum and Sum to Product Formulas 4.6 Using Sum Identities in Determining Sinusoidal Formulas Introduction In Chapter 4, we begin exploring new trigonometric identities and formulas which provide us with varying ways to represent the same trigonometric expression. Section 4.1 advances our study of even/odd identities and includes algebraic proofs of these identities. In Section 4.2, we discover the essential sum and difference identities upon which many of the remaining identities and formulas depend. Section 4.3 introduces the double angles identities, from which the power reduction and half angle formulas are derived in Section 4.4. Along with the sum and difference
identities, the half angle formulas will help us obtain exact values for trigonometric functions of some ‘non-common’ angles. The product to sum and sum to product formulas introduced in Section 4.5 will allow further manipulation of trigonometric expressions. Finally, Section 4.6 revisits the formulas of sinusoids first introduced in Section 3.2. With the assistance of the sum and difference identities, we may now write sinusoidal formulas in a standard format that will simplify graphing. Throughout Chapter 4, attention will be paid to finding exact values of trigonometric functions, writing trigonometric expressions in varying formats, and verifying trigonometric identities. These skills will be important in the future study of calculus. 167 4.1 The Even/Odd Identities Learning Objectives In this section you will:  Learn the even/odd identities.  Use the even/odd identities in simplifying trigonometric expressions.  Use the even/odd identities in verifying trigonometric identities. In Section 2.4, we saw the utility of the Pythagorean identities, Theorem 2.5, along with the reciprocal and quotient identities from Theorem 2.3. 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 formally introduce the even/odd identities1, while recalling their graphical significance from Chapter 3. The Even/Odd Identities Theorem 4.1. The Even/Odd Identities: For all applicable angles θ,       In light of the reciprocal and quotient identities, Theorem 2.3, it suffices to show and. The remaining four circular functions can be expressed in terms of and so the proofs of their even/odd identities are left as exercises. Consider an angle θ plotted in standard position. Let be the angle coterminal with θ with. (We can construct the angle by rotating counter-clockwise from the positive x-axis to the terminal side of θ as pictured to the right.) Since θ and are coterminal, and. 1 As mentioned at the end of Section 2.3, 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. sinsincsccsccoscossecsectantancotcotcoscossinsincossin0002000coscos0sinsin 168 We now consider the angles and. Since θ is coterminal with, there is some integer k so that. Therefore, Since k is an integer, so is –k, which means –θ is coterminal with. Hence, and. Let P and Q denote the points on the terminal sides of and, respectively, which lie on the Unit Circle. By definition, the coordinates of P are and the coordinates of Q are. Since and sweep out congruent central sectors of the Unit Circle, it follows that the points P and Q are symmetric about the x-axis. Thus, and. Since the cosines and sines of and are the same as those for θ and –θ, respectively, we get and, as
required. Simplifying Expressions The even/odd identities are readily demonstrated using any of the common angles noted in Section 2.2. Their true utility, however, lies not in computation, but in simplifying expressions involving the circular functions. Example 4.1.1. Use identities to fully simplify the expression:. 0002k0022.kk00coscos0sinsin0000cos,sin00cos,sin0000coscos00sinsin00coscossinsin1sin1sinxx Solution. We begin with the odd identity of the sine function. 169 Verifying Identities This section ends with the proof of a trigonometric identity. Looking back at Section 2.4, where we began verifying identities, we can now
add the even/odd identities to the Pythagorean, reciprocal and quotient identities as tools in proving that identities are true. Example 4.1.2. Verify the identity:. Solution. We begin with the left, more complicated, side. 2222 sinsin 1s cosin1sin1sin1sin1s1sinincosxxxxxxxxxxsince difference of squares22sincoscossinsincos22222222sincossincossincossincossincossincossincossincossincossinc
os sincossincos1coseven/odd identitiesdifference of squares sin 170 4.1 Exercises 1. We know is an even function, and and are odd functions. What about, and? Are they even, odd or neither? Why? 2. Examine the graph of on the interval. How can we tell whether the function is even or odd by only observing the graph of? In Exercises 3 – 8, use identities to fully simplify the expression. 3. 5. 7. 4. 6. 8. In Exercises 9 – 14, use the even/odd identities to verify the identity. Assume all quantities are defined. 9. 11. 13. In Exercises 15 – 18, prove or disprove the identity. 15. 17. 10. 12. 14. 16. 18. 19. Verify the even/odd identities for tangent, cosecant, secant and cotangent. cosgxxsinfxxtanhxx2cosGxx2sinFxx2tanHxxsecfxx�
�,secfxxsincoscscxxxcsccoscotxxxcottansecttt323sincsccos2coscosttttttancotxxsincosseccsctancotxxxxxxsin32sin23cos5cos544tt22tan1tan1tt55csccscsec6sec6tt9779cotcot112cotcsc1cos1cosxxxx
2tansincossecxxxxsecsintancotxxxx1sincoscos1sinxxxx 171 4.2 The Sum and Difference Identities In this section you will: Learning Objectives  Learn the sum and difference identities for cosine, sine and tangent.  Use the sum and difference identities to find values of trigonometric functions.  Use the sum and difference identities in verifying trigonometric identities.  Learn and apply the cofunction identities. We begin with a theorem introducing the sum and difference identities for the cosine function, followed by a proof of the theorem. The Sum and Difference Identities for the Cosine Function Theorem 4.2: Sum and Difference Identities for Cosine.   We first prove the result for differences. As in the proof of the even/odd identities, we can reduce the proof for general angles and to angles and, coterminal with and, respectively, each of which measure between 0 and 2π radians. Since and are coterminal, as are and, it follows that is coterminal with. Consider the following case where. coscoscossinsincoscoscossinsin0000
0000 172 Since the angles POQ and AOB are congruent, the distance between P and Q is equal to the distance between A and B.1 Using the distance formula to determine distances QP and BA, we have or, after squaring both sides: Expanding the left hand side, then using the Pythagorean identities, we get the following:. and Turning our attention to the right hand side, we will use : Putting in all together, we get which simplifies as follows:. Since and, and, and, are all coterminal pairs of angles, we have 1 In the picture, the triangles POQ and AOB are congruent. However, α0 – β0 could be 0 or it could be π, neither of which makes a triangle. Or, α0 – β0 could be larger than π, which makes a triangle, just not the one we’ve drawn. You should think about these three cases. 222200000000coscossinsincos1sin0220002000002cos1sincoscossinsin0
2200cossin12200cossin1220000coscossinsin2222000000002222000000000000cos2coscoscossin2sinsinsincossincossin2coscos2sinsin22coscos2sinin.s220000cossin1222200000000002200000000cos1sin0cos2cos1sin
1cossin2co.s22cos00000022coscos2sinsi2cn2os00000000000000000022cos22coscos2sinsin2cos2coscos2sinsincoscoscossinsin after swapping sides after subtracting 2 from each side after dividing through by -20000 173 For the case where, we can apply the above argument to the angle to obtain the identity Applying the even identity
of cosine, we get from which it follows that. To verify the sum identity for cosine, we use the difference identity along with the even/odd identities: We put these newfound identities to good use in the following example. Example 4.2.1. 1. Find the exact value of. 2. Verify the identity. Solution. 1. In order to use Theorem 4.2 to find, 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. coscoscossinsin.0000000000coscoscossinsin.000000coscoscos,coscoscossinsin�
�coscoscoscossinsincoscossinsincoscossinsincoscossinsin. difference identity for cosine even identity of cosine odd identity of sinecos15cossin2cos15154530 174 2. This is a straightforward application of Theorem 4.2. The identity verified in Example 4.2.1, namely, is the first of the celebrated cofunction identities. These identities were first hinted at in the 2.1 Exercises, problem 41. The Cofunction Identities From, we get 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. cos15cos4530cos45cos30sin45sin3023212222624 difference identity for cosine�
�coscoscossinsin2220cos1sinsincossin2sincos2sincos222cos 175 Theorem 4.3. Cofunction Identities: For all applicable angles θ,       With the cofunction identities in place, we are now in the position to derive the sum and difference identities for sine. The Sum and Difference Identities for the Sine Function We begin with the sum identity. We can derive the difference identity for sine by rewriting as and using the sum identity and the even/odd identities. Again, we leave the details to the reader. Theorem 4.4. Sum and Difference Identities for Sine: For all angles α and β,   Example 4.2.2. 1. Find the exact value of. 2. If α is a Quadrant II angle with and β is a Quadrant III angle with, find. 3. Derive a formula for in terms of and. sincos2cossin2�
��seccsc2cscsec2tancot2cottan2 ssincos2cos2coscossinsin22incos2from difference identity for cosine sinsinsinsincoscossinsinsincoscossin
19sin125sin13tan2sintantantan Solution. 176 1. As in Example 4.2.1, we need to write the angle 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 from which we have 2. Using the difference identity for sine,. We know that, but need to find, and. To find, we use along with a Pythagorean identity. We next use a different Pythagorean identity, along with, to find. 191219163121212434194sinsin123444sincoscossin343432122222624 sum identity for sinesinsincoscossin5
sin13coscossincos5sin13222225cos113144cos16912cos1312cos1cossin13  from since is a Quadrant II angle tan2cos 177 We need to determine, knowing that and. We now have all the pieces needed to find. 3. We can start expanding. 22222 112secsec5sec515cos1cotans5csefrom since is a Quadrant III angle reciprocal identity for secantsin1cos5tan2sintancos1sin252sinsintan5cos
 from the quotient identity sinsinsincoscossin511221313552913529565tan 178 The last step is to replace with and with. Naturally, this result is limited to those cases where all of the tangents are defined. The Sum and Difference Identities for the Tangent Function The formula developed in Example 4.2.2 for can be used to find a formula for by rewriting the difference as a sum,. The reader is encouraged to fill in the details. Below we summarize all of the sum and difference formulas for sine, cosine and tangent. sintancossincoscossincoscossinsin quotient identity for tangent sum identitie
sinsincosc1sincoscossincoscos1coscossinsincoscossincoscossincoscoscooscoscoscossinsincoscoscoscossinsincoscossins1coss for cosine and sine goal: & sincossincostansincostan
tantantan1tantantantantan 179 Theorem 4.5. Sum and Difference Identities: For all applicable angles α and β,    In the statement of Theorem 4.5, we have combined the cases for the sum ‘+’ and difference ‘–‘ of angles into one formula. The convention is that if you want the formula for the sum ‘+’ of two angles, use the top sign in the formula; for the difference ‘–‘ use the bottom sign. For example,. We finish this section by, as promised in Section 3.3, proving algebraically that the period of the tangent function is π. Recall that a function f is periodic if there is a real number p so that for all real numbers t in the domain of f. The smallest positive number p, if it exists, is called the period of f. To prove that the period of is π, we appeal to the sum identity for tangents. This tells us that the tangent is a periodic function and that the period of is at most π. To show that it is exactly π, suppose p is a positive real number so that for all real numbers x. For, we have which means p is a multiple of π. The smallest positive multiple of π is π itself, so we have established that the period of the tangent function is π. sinsincoscossincoscoscossinsin
tantantan1tantantantantan1tantanftpfttanJxxtantantan1tantantan01tan0tanJxxxxxxxJxtanxtantanxpx0xtantan0tan0tantan0xpxpp from We leave it to the reader to prove algebraically that the period of the cotangent function is also π.2 180 2 Certainly, mimicking the proof for the period of tan(x) is an option. For another approach, consider transforming tan(x) to cot(x) using identities. 181 4.2 Exercises In Exercises 1 – 15, 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. 1. 4. 7. 10. 13. 2. 5. 8. 11. 14. 3. 6. 9. 12. 15. 16. If α is a Quadrant IV angle with, and, where, find (a) (d) (b) (e) (c) (f) 17. If, where, and β is a Quadrant II angle with, find (a) (d) (b) (e) (c) (f) 18. If, where, and, where, find (a) (b) (c) 19. If, where, and, where, find (a) (b) (c) cos75sec165sin105csc195cot255tan37513cos1211sin1213tan127cos1217tan12sin1211cot125csc12sec125cos510sin102cossintancossintancsc302tan7cos�
�sintancossintan3sin50212cos13322sincostan5sec3224tan732cscseccot 182 21. 23. 25. 27. In Exercises 20 – 32, verify the identity. 20. 22. 24. 26. 28. 29. 30. 31. 32. 33. Verify the cofunction identities for tangent, secant, cosecant and cotangent. 34. Verify the difference identities for sine and tangent. coscossinsintancot2sinsin2sincos
tansincossincostansincossincossinsinsincos1cossinththhtthhhcoscoscos1sincossinththhtthhh2tantantansec1tantanthththhth 183 4.3 Double Angle Identities Learning Objectives In this section you will:  Learn the double angle identities for sine, cosine and tangent.  Find trigonometric values of double angles.  Verify identities involving double angles. In Section 4.2, the sum identities for the trigonometric functions were introduced: Double Angle Identities Using the sum identities, in the case where, we let to attain the double angle identities in the following theorem. Theorem 4.6. Double Angle Identities: For all applicable angles θ,    The three different forms for can be explained by our ability to
exchange squares of cosine and sine via a Pythagorean identity. We verify that : sinsincoscossincoscoscossinsintantantan.1tantansin22sincos2222cossincos22cos112sin22tantan21tancos22cos212sin 184 Trigonometric Values of Double Angles Now that we have established the double angle identities, we put them to good use in determining trigonometric values of double angles. Example 4.3.1. 1. Suppose lies on the terminal side of θ when θ is plotted in standard position. Find and. Determine the quadrant in which the terminal side of the angle 2
θ lies when it is plotted in standard position. 2. If for, find an expression for in terms of x. Solution. 1. Using, from Theorem 2.6 in Section 2.5, with and, we find. Hence, and. It follows that and 222222222cos2coscoscossinsincosscos2cossincoin1sinsin12sin.ssin1 sum identity for cosine verifies first form: from 3,4Pcos2sin2sinx22sin2222xyr3x4y225rxy3cos5xr4sin5yr2222cos2cossin3455725
 from double angle identitysin22sincos4325524.25 from double angle identity 185 Since both the cosine and sine of 2θ are negative, the terminal side of 2θ, when plotted in standard position, lies in Quadrant III. 2. If your first reaction to is that x should be the cosine of θ, then you have indeed learned something. However, context is everything. Here, x is just a variable. It does not necessarily represent the x-coordinate of a point on the Unit Circle. Here, x represents the quantity, and what we wish to know is how to express in terms of x. We will see more of this kind of thing in Chapter 5 and, as usual, this is something we need for calculus. We start with the double angle identity for sine: We need to write in terms of x to finish the problem. There are two different methods that come readily to mind, both of which are good to know. The first is purely algebraic, using the Pythagorean identity: The second method, preferred by many, provides a visual approach for determining triangle with acute angle θ and, noting that. We sketch a right we label the hypotenuse with length 1 and the side opposite θ with length x. We then use the Pythagorean Theorem to determine the length of the side adjacent to θ. sinxsinsin2sin22sincos2cssinoxxfrom the problem statement that �
�cos222222cossin1cos1cos1cos1cos022xxx Pythagorean identity from the problem statement since cosoppositehypotenusesin,1xx222222adjacent lengthadjacent lengthadjacent length111xxx1x  186 This results in the following triangle. From the triangle we see that. Then, back to solving for, we have a final answer is. Verifying Identities that Include Double Angles Establishing trigonometric identities using the double angle identities is our next task. As before, starting with the more complicated side of an equation is usually a good strategy. Example 4.3.2. 1. Verify the identity. 2. Verify the identity. 221cos11xxsin22sin221xx22tansin21tan2tan2cottan1x 21x 187 Solution. 1. We start with the right hand side of the identity. 2. In this case, we begin with the left side of the equation. Part 2 of the previous example is a case where the more complicated side of the initial equation appeared on the right, but we chose
to start with the left side. Beginning with the right side would have required some thinking ahead, possibly working backwards. Try it! When using identities to simplify a trigonometric expression, solve a trigonometric equation or verify a trigonometric identity, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated, although generally one of the paths will result in a simpler solution. In verifying identities, the strategies established in Section 2.4 will help, but there is no substitute for practice. 22222tan2tan1tansecsin2cos1cossin2coscos2sincossin2from Pythagorean identityfrom reciprocal & quotient identitiesfrom double angle identity for sine222tantan21tan12tantan11tantan21tantan2cottan�
�double angle identitygoal: numerator of 2reciprocal identity for cotangent 188 One last note before we move on to Section 4.4. While double angle identities could be established for secant, cosecant and cotangent, the identities already established in this section may be used in their place. Recall that secant, cosecant and cotangent are reciprocal identities of cosine, sine and tangent, respectively. Thus, for example, and so any of the three double angle identities for cosine may be used in determining. 1sec2cos2sec2 189 4.3 Exercises 1. Use the double angle identity to verify the double angle identity. 2. Use the double angle identities for and to verify the double angle identity for. In Exercises 3 – 12, use the given information about θ to find the exact values of (a) 3. 5. 7. 9. where where where where 11. where (b) (c) 4. 6. 8. 10. 12. where where where where where In Exercises 13 – 22, verify the identity. Assume all quantities are defined. 13. 15. 17. 19. 14. 16. 18. 20. 22cos2cossin2cos22cos1cos2sin2tan2sin2cos2tan27sin2532228cos530212tan5�
��32csc423cos5024sin53212cos133225sin132sec5322tan2222tansin21tan221tancos21tan22sincostan22cos12cossin1sin22cossin1sin211tan21tan1tan�
��cottancsc22cossinsec2cossincossin 190 21. 22. 23. Suppose θ is a Quadrant I angle with. Verify the following formulas. (a) (b) (c) 24. Discuss with your classmates how each of the formulas, if any, in Exercise 23 change if we assume θ is a Quadrant II, III or IV angle. 25. Suppose θ is a Quadrant I angle with. Verify the following formulas. (a) (c) (b) (d) 26. Discuss with your classmates how each of the formulas, if any, in Exercise 25 change if we assume θ is a Quadrant II, III or IV angle. 27. If for, find an expression for in terms of x. 28. If for, find an expression for in terms of x. 112coscossincossincos2112sincossincossincos2sinx2cos1x2sin221xx2cos212x
tanx21cos1x2sin1xx22sin21xx221cos21xxsin2x22cos2tan7x22sin2 191 4.4 Power Reduction and Half Angle Formulas Learning Objectives In this section you will:  Learn and apply the power reduction formulas for sine and cosine.  Learn and apply the half angle formulas for sine, cosine and tangent. In Section 4.3, the double angle identities allowed us to write as powers of sine and/or cosine. In calculus, we have occasion to do the reverse; that is, reduce the power of sine and cosine. Power Reduction Formulas Solving the identity for and the identity for result in the aptly-named ‘power reduction’ formulas below. Theorem 4.7. Power Reduction Formulas: For all angles θ,   Example 4.4.1. Rewrite as a sum and/or difference of cosines to the first power. Solution. We begin with a straightforward application of Theorem 4.7. cos22cos212sin2sin2cos22cos12cos21cos2sin221cos2cos2�
�22sincos 192 Another application of the power reduction formulas is the half angle formulas. Half Angle Formulas To start, we apply the power reduction formula to : We can obtain a formula for by extracting square roots. In a similar fashion, we may obtain a half angle formula for sine. By using a quotient identity, we obtain a half angle formula for tangent. These formulas are summarized below. 2222 21cos21cos2sincos2211cos2411cos2441cos2211442111cos448811cos488from power reduction formulasreplacing with in power reduction formula2cos221cos22cos221cos.2cos2 193 Theorem 4.8. Half Angle Formulas. For all applicable angles θ,    where the choice of ± depends on the quadrant in which the terminal side of lies. Example 4.4.2. 1. Use
a half angle formula to find the exact value of. 2. Suppose with. Find. 3. Use the identity, verified in Example 4.3.2, to derive the identity. Solution. 1. To use the half angle formula, we note that. 1cossin221coscos221costan21cos2cos1503cos5sin222tansin21tansintan21cos30152 194 Back in Example 4.2.1, we found by using the difference identity for cosine. In that case, we determined. The reader is encouraged to prove these two expressions are equal. 2. If, then, which means. 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, 30cos15cos21cos3023122312222232 from half angle formula for cosine positive since 15 is in Quadrant Icos1562cos1540�
��022sin021cossin223152315525810255 half angle formula for sine22tansin21tan 195 and we will manipulate it into the identity. If we are to use to derive an identity for, it seems reasonable to proceed by replacing each occurrence of θ with. sintan21cos22tansin21tantan2222222tansin21tan2tan2sin221tan22tan2sinsec2sin2tanc22os2 replacing with from a Pythagorean identity �
��1cos22sin2tan22sintan1cos2sintan21cos from reciprocal identity for secant from power reduction formula for cosine 196 4.4 Exercises In Exercises 1 – 15, use half angle formulas to find the exact value. You may have need of the quotient, reciprocal or even/odd identities as well. 1. 4. 7. 10. 13. 2. 5. 8. 11. 14. 3. 6. 9. 12. 15. In Exercises 16 – 29, use the given information about θ to find the exact values of (a) 16. 18. 20. 22. 24. 26. 28. (b) (c) 17. 19. 21. 23. 25. where where where where where where where where where where, where θ is in Quadrant IV 27., where θ is in Quadrant III, where θ is in Quadrant II 29., where θ is in Quadrant II cos75sin105cos67.5sin157.5tan112.57cos12sin12cos85sin87tan811cos1211sin125tan123tan123tan8�

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