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3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 29. 0 33. − 7π 4 37. 180◦ 41. 60◦ 45. t = 5π 6 47. t = 6 30. 34. 4π 3 5π 6 38. −120◦ 42. 300◦ 1 x y 1 y 1 31. 35. 3π 4 π 4 39. 210◦ 43. −30◦ 46. t = −π 48. t = −2 32. − 36. − 3π 2 5π 4 40. 330◦ 44. 90 716 Foundations of Trigonometry 49. t = 12 (between 1 and 2 revolutions) y 1 1 x 50. About 30.12 miles per hour 51. About 6274.52 revolutions per minute 52. About 3.33 miles per hour 53. About 53.55 miles per hour 54. 70 miles per hour 55. About 4.32 miles per hour 57. 12π square units 59. 79.2825π ≈ 249.07 square units 61. 50π 3 square units 58. 6250π square units 60. π 2 square units 62. 38.025π ≈ 119.46 square units 10.2 The Unit Circle: Cosine and Sine 717 10.2 The Unit Circle: Cosine and Sine In Section 10.1.1, we introduced circular motion and derived a formula which describes the linear velocity of an object moving on a circular path at a constant angular velocity. One of the goals of this section is describe the position of such an object. To that end, consider an angle θ in standard position and let P denote the point where the terminal side of θ intersects the Unit Circle. By associating the point P with the angle θ, we are assigning a position on the Unit Circle to the angle θ. The x-coordinate of P is called the cosine of θ, written cos(θ), while the y-coordinate of P is called the sine of θ, written sin(θ).1 The reader is encouraged to verify that these rules used to match an angle with its cosine and sine do, in fact, satisfy the deﬁnition of a function. That is, for each angle θ, there is only one associated value of cos(θ) and only one associated value of sin(θ). y
1 θ 1 x y 1 P (cos(θ), sin(θ)) θ 1 x Example 10.2.1. Find the cosine and sine of the following angles. 1. θ = 270◦ 2. θ = −π 3. θ = 45◦ 4. θ = π 6 5. θ = 60◦ Solution. 1. To ﬁnd cos (270◦) and sin (270◦), we plot the angle θ = 270◦ in standard position and ﬁnd the point on the terminal side of θ which lies on the Unit Circle. Since 270◦ represents 3 4 of a counter-clockwise revolution, the terminal side of θ lies along the negative y-axis. Hence, the point we seek is (0, −1) so that cos (270◦) = 0 and sin (270◦) = −1. 2. The angle θ = −π represents one half of a clockwise revolution so its terminal side lies on the negative x-axis. The point on the Unit Circle that lies on the negative x-axis is (−1, 0) which means cos(−π) = −1 and sin(−π) = 0. 1The etymology of the name ‘sine’ is quite colorful, and the interested reader is invited to research it; the ‘co’ in ‘cosine’ is explained in Section 10.4. 718 Foundations of Trigonometry y 1 θ = 270◦ P (0, −1) y 1 P (−1, 0) 1 x 1 x θ = −π Finding cos (270◦) and sin (270◦) Finding cos (−π) and sin (−π) 3. When we sketch θ = 45◦ in standard position, we see that its terminal does not lie along any of the coordinate axes which makes our job of ﬁnding the cosine and sine values a bit more diﬃcult. Let P (x, y) denote the point on the terminal side of θ which lies on the Unit Circle. By deﬁnition, x = cos (45◦) and y = sin (45◦). If we drop a perpendicular line segment from P to the x-axis, we obtain a 45◦ − 45◦ − 90
◦ right triangle whose legs have lengths x and y units. From Geometry,2 we get y = x. Since P (x, y) lies on the Unit Circle, we have √ 2 x2 + y2 = 1. Substituting y = x into this equation yields 2x2 = 1, or x = ± 2. √ 2 Since P (x, y) lies in the ﬁrst quadrant, x > 0, so x = cos (45◦) = 2 and with y = x we have y = sin (45◦) = = 45◦ P (x, y) x 1 P (x, y) 45◦ y θ = 45◦ x 2Can you show this? 10.2 The Unit Circle: Cosine and Sine 719 4. As before, the terminal side of θ = π 6 does not lie on any of the coordinate axes, so we proceed using a triangle approach. Letting P (x, y) denote the point on the terminal side of θ which lies on the Unit Circle, we drop a perpendicular line segment from P to the x-axis to form = 1 a 30◦ − 60◦ − 90◦ right triangle. After a bit of Geometry3 we ﬁnd y = 1 2. Since P (x, y) lies on the Unit Circle, we substitute y = 1 4, or x = ± 2 into x2 + y2 = 1 to get x2 = 3 2. Here, x > 0 so x = cos π 2 so sin x, y) x 1 P (x, y) 60◦ y θ = π 6 = 30◦ x 5. Plotting θ = 60◦ in standard position, we ﬁnd it is not a quadrantal angle and set about using a triangle approach. Once again, we get a 30◦ − 60◦ − 90◦ right triangle and, after the usual computations, ﬁnd x = cos (60◦) = 1 2 and y = sin (60◦) = √ 3 2. y 1 P (x, y) θ = 60◦ x 1 P (x, y) 30◦ y θ = 60◦ x 3Again, can you show this? 720 Foundations of Trigonometry In Example 10.2.1, it
was quite easy to ﬁnd the cosine and sine of the quadrantal angles, but for In these latter cases, we made good non-quadrantal angles, the task was much more involved. use of the fact that the point P (x, y) = (cos(θ), sin(θ)) lies on the Unit Circle, x2 + y2 = 1. If we substitute x = cos(θ) and y = sin(θ) into x2 + y2 = 1, we get (cos(θ))2 + (sin(θ))2 = 1. An unfortunate4 convention, which the authors are compelled to perpetuate, is to write (cos(θ))2 as cos2(θ) and (sin(θ))2 as sin2(θ). Rewriting the identity using this convention results in the following theorem, which is without a doubt one of the most important results in Trigonometry. Theorem 10.1. The Pythagorean Identity: For any angle θ, cos2(θ) + sin2(θ) = 1. The moniker ‘Pythagorean’ brings to mind the Pythagorean Theorem, from which both the Distance Formula and the equation for a circle are ultimately derived.5 The word ‘Identity’ reminds us that, regardless of the angle θ, the equation in Theorem 10.1 is always true. If one of cos(θ) or sin(θ) is known, Theorem 10.1 can be used to determine the other, up to a (±) sign. If, in addition, we know where the terminal side of θ lies when in standard position, then we can remove the ambiguity of the (±) and completely determine the missing value as the next example illustrates. Example 10.2.2. Using the given information about θ, ﬁnd the indicated value. 1. If θ is a Quadrant II angle with sin(θ) = 3 5, ﬁnd cos(θ). 2. If π < θ < 3π 2 with cos(θ) = − √ 5 5, ﬁnd sin(θ). 3. If sin(θ) = 1, ﬁnd cos(θ). Solution. 1. When we substitute sin(θ) = 3 5
into The Pythagorean Identity, cos2(θ) + sin2(θ) = 1, we obtain cos2(θ) + 9 5. Since θ is a Quadrant II angle, its terminal side, when plotted in standard position, lies in Quadrant II. Since the x-coordinates are negative in Quadrant II, cos(θ) is too. Hence, cos(θ) = − 4 5. 25 = 1. Solving, we ﬁnd cos(θ) = ± 4 √ √ 5 5 2. Substituting cos(θ) = − 5. Since we 5 are given that π < θ < 3π 2, we know θ is a Quadrant III angle. Hence both its sine and cosine are negative and we conclude sin(θ) = − 2 into cos2(θ) + sin2(θ) = 1 gives sin(θ) = ± 2. When we substitute sin(θ) = 1 into cos2(θ) + sin2(θ) = 1, we ﬁnd cos(θ) = 0. Another tool which helps immensely in determining cosines and sines of angles is the symmetry inherent in the Unit Circle. Suppose, for instance, we wish to know the cosine and sine of θ = 5π 6. We plot θ in standard position below and, as usual, let P (x, y) denote the point on the terminal side of θ which lies on the Unit Circle. Note that the terminal side of θ lies π 6 radians short of one half revolution. In Example 10.2.1, we determined that cos π 2. This means 6 2 and sin π = 1 = √ 3 6 4This is unfortunate from a ‘function notation’ perspective. See Section 10.6. 5See Sections 1.1 and 7.2 for details. 10.2 The Unit Circle: Cosine and Sine 721 that the point on the terminal side of the angle π. From the ﬁgure below, it is clear that the point P (x, y) we seek can be obtained by reﬂecting that point about the y-axis. Hence, cos 5π 6 6, when plotted in standard position, is 2 and sin 5π = x, y) �
� = 5π = 5π is called the reference angle for the angle 5π In the above scenario, the angle π 6. In general, for a non-quadrantal angle θ, the reference angle for θ (usually denoted α) is the acute angle made between the terminal side of θ and the x-axis. If θ is a Quadrant I or IV angle, α is the angle between the terminal side of θ and the positive x-axis; if θ is a Quadrant II or III angle, α is the angle between the terminal side of θ and the negative x-axis. If we let P denote the point (cos(θ), sin(θ)), then P lies on the Unit Circle. Since the Unit Circle possesses symmetry with respect to the x-axis, y-axis and origin, regardless of where the terminal side of θ lies, there is a point Q symmetric with P which determines θ’s reference angle, α as seen below Reference angle α for a Quadrant I angle Reference angle α for a Quadrant II angle 722 Foundations of Trigonometry Reference angle α for a Quadrant III angle Reference angle α for a Quadrant IV angle We have just outlined the proof of the following theorem. Theorem 10.2. Reference Angle Theorem. Suppose α is the reference angle for θ. Then cos(θ) = ± cos(α) and sin(θ) = ± sin(α), where the choice of the (±) depends on the quadrant in which the terminal side of θ lies. In light of Theorem 10.2, it pays to know the cosine and sine values for certain common angles. In the table below, we summarize the values which we consider essential and must be memorized. Cosine and Sine Values of Common Angles θ(degrees) 0◦ 30◦ 45◦ 60◦ 90◦ θ(radians) cos(θ) sin(θ Example 10.2.3. Find the cosine and sine of the following angles. 1. θ = 225◦ 2. θ = 11π 6 3. θ = − 5π 4 4. θ = 7π 3 Solution. 1. We begin by plotting θ = 225◦ in standard position and ﬁnd its terminal side overshoots the negative x-axis to
land in Quadrant III. Hence, we obtain θ’s reference angle α by subtracting: α = θ − 180◦ = 225◦ − 180◦ = 45◦. Since θ is a Quadrant III angle, both cos(θ) < 0 and 10.2 The Unit Circle: Cosine and Sine sin(θ) < 0. The Reference Angle Theorem yields: cos (225◦) = − cos (45◦) = − sin (225◦) = − sin (45◦) = − √ 2 2. 723 √ 2 2 and 2. The terminal side of θ = 11π 6, when plotted in standard position, lies in Quadrant IV, just shy of the positive x-axis. To ﬁnd θ’s reference angle α, we subtract: α = 2π − θ = 2π − 11π 6 = π 6. Since θ is a Quadrant IV angle, cos(θ) > 0 and sin(θ) < 0, so the Reference Angle Theorem gives: cos 11π 6 2 and sin 11π = − sin π 6 = cos = 225◦ 45 = 11π 6 Finding cos (225◦) and sin (225◦) Finding cos 11π 6 and sin 11π 6 3. To plot θ = − 5π 4, we rotate clockwise an angle of 5π 4 from the positive x-axis. The terminal side of θ, therefore, lies in Quadrant II making an angle of α = 5π 4 − π = π 4 radians with respect to the negative x-axis. Since θ is a Quadrant II angle, the Reference Angle Theorem gives: cos − 5π 4 = − cos π 4 = sin. Since the angle θ = 7π one full revolution followed by an additional α = 7π coterminal, cos 7π 3 2 and sin 7π = cos π 3 = 1 = sin π 3 3 3, we ﬁnd the terminal side of θ by rotating 3 − 2π = π 3 radians. Since θ and α are √ = 3 2. = − 2 and sin − 5π 3 measures more than 2π = 6π 4 y 1 π 4 1 x θ = − 5π 4 y 1 θ = 7π
3 π 3 1 x Finding cos − 5π 4 and sin − 5π 4 Finding cos 7π 3 and sin 7π 3 724 Foundations of Trigonometry 6 as a reference angle, those with a denominator of 4 have π The reader may have noticed that when expressed in radian measure, the reference angle for a non-quadrantal angle is easy to spot. Reduced fraction multiples of π with a denominator of 6 have π 4 as their reference angle, and those with a denominator of 3 have π 3 as their reference angle.6 The Reference Angle Theorem in conjunction with the table of cosine and sine values on Page 722 can be used to generate the following ﬁgure, which the authors feel should be committed to memory. y (0, 1 5π 6 2π 3 3π (−1, 0) 0, 2π (1, 0) x 7π 6 11π 5π 4 4π 3 − 1 2, − √ 3 2 7π 4 5π 3π 2 (0, −1) Important Points on the Unit Circle 6For once, we have something convenient about using radian measure in contrast to the abstract theoretical nonsense about using them as a ‘natural’ way to match oriented angles with real numbers! 10.2 The Unit Circle: Cosine and Sine 725 The next example summarizes all of the important ideas discussed thus far in the section. Example 10.2.4. Suppose α is an acute angle with cos(α) = 5 13. 1. Find sin(α) and use this to plot α in standard position. 2. Find the sine and cosine of the following angles: (a) θ = π + α (b) θ = 2π − α (c) θ = 3π − α (d) θ = π 2 + α Solution. 1. Proceeding as in Example 10.2.2, we substitute cos(α) = 5 ﬁnd sin(α) = ± 12 Hence, sin(α) = 12 x-axis to the ray which contains the point (cos(α), sin(α)) = 5 13 into cos2(α) + sin2(α) = 1 and 13. Since α is an acute (and therefore Quadrant I) angle, sin(α) is positive. 13. To plot α in standard position, we begin our rotation
on the positive. 13, 12 13 y 1 5 13, 12 13 α 1 x Sketching α 2. (a) To ﬁnd the cosine and sine of θ = π + α, we ﬁrst plot θ in standard position. We can imagine the sum of the angles π+α as a sequence of two rotations: a rotation of π radians followed by a rotation of α radians.7 We see that α is the reference angle for θ, so by 13 and sin(θ) = ± sin(α) = ± 12 The Reference Angle Theorem, cos(θ) = ± cos(α) = ± 5 13. Since the terminal side of θ falls in Quadrant III, both cos(θ) and sin(θ) are negative, hence, cos(θ) = − 5 13 and sin(θ) = − 12 13. 7Since π + α = α + π, θ may be plotted by reversing the order of rotations given here. You should do this. 726 Foundations of Trigonometry Visualizing θ = π + α θ has reference angle α (b) Rewriting θ = 2π − α as θ = 2π + (−α), we can plot θ by visualizing one complete revolution counter-clockwise followed by a clockwise revolution, or ‘backing up,’ of α radians. We see that α is θ’s reference angle, and since θ is a Quadrant IV angle, the Reference Angle Theorem gives: cos(θ) = 5 13 and sin(θ) = − 12 13. y 1 y 1 θ θ 2π −α 1 x 1 x α Visualizing θ = 2π − α θ has reference angle α (c) Taking a cue from the previous problem, we rewrite θ = 3π − α as θ = 3π + (−α). The angle 3π represents one and a half revolutions counter-clockwise, so that when we ‘back up’ α radians, we end up in Quadrant II. Using the Reference Angle Theorem, we get cos(θ) = − 5 13 and sin(θ) = 12 13. 10.2 The Unit Circle: Cosine and Sine 727 y 1 −α 3π y 1
θ α 1 x 1 x Visualizing 3π − α θ has reference angle α (d) To plot θ = π 2 + α, we ﬁrst rotate π 2 radians and follow up with α radians. The reference angle here is not α, so The Reference Angle Theorem is not immediately applicable. (It’s important that you see why this is the case. Take a moment to think about this before reading on.) Let Q(x, y) be the point on the terminal side of θ which lies on the Unit Circle so that x = cos(θ) and y = sin(θ). Once we graph α in standard position, we use the fact that equal angles subtend equal chords to show that the dotted lines in the ﬁgure below are equal. Hence, x = cos(θ) = − 12 13. Similarly, we ﬁnd y = sin(θ) = 5 13. 13, 12 13 Q (x, y) α α 1 x 1 x Visualizing θ = π 2 + α Using symmetry to determine Q(x, y) 728 Foundations of Trigonometry Our next example asks us to solve some very basic trigonometric equations.8 Example 10.2.5. Find all of the angles which satisfy the given equation. 1. cos(θ) = 1 2 2. sin(θ) = − 1 2 3. cos(θ) = 0. Solution. Since there is no context in the problem to indicate whether to use degrees or radians, we will default to using radian measure in our answers to each of these problems. This choice will be justiﬁed later in the text when we study what is known as Analytic Trigonometry. In those sections to come, radian measure will be the only appropriate angle measure so it is worth the time to become “ﬂuent in radians” now. 1. If cos(θ) = 1 Unit Circle at x = 1 2, then the terminal side of θ, when plotted in standard position, intersects the 2. This means θ is a Quadrant I or IV angle with reference angle One solution in Quadrant I is θ = π coterminal with π IV case, we ﬁnd the solution to cos(θ) = 1 θ = 5π
3, we ﬁnd θ = π 3 + 2πk for integers k. 3, and since all other Quadrant I solutions must be 3 + 2πk for integers k.9 Proceeding similarly for the Quadrant 3, so our answer in this Quadrant is 2 here is 5π 2. If sin(θ) = − 1 2, then when θ is plotted in standard position, its terminal side intersects the Unit Circle at y = − 1 2. From this, we determine θ is a Quadrant III or Quadrant IV angle with reference angle π 6. 8We will study trigonometric equations more formally in Section 10.7. Enjoy these relatively straightforward exercises while they last! 9Recall in Section 10.1, two angles in radian measure are coterminal if and only if they diﬀer by an integer multiple of 2π. Hence to describe all angles coterminal with a given angle, we add 2πk for integers k = 0, ±1, ±2,.... 10.2 The Unit Circle: Cosine and Sine 729 In Quadrant III, one solution is 7π multiples of 2π: θ = 7π are of the form θ = 11π 6, so we capture all Quadrant III solutions by adding integer so all the solutions here 6 + 2πk. In Quadrant IV, one solution is 11π 6 + 2πk for integers k. 6 3. The angles with cos(θ) = 0 are quadrantal angles whose terminal sides, when plotted in standard position, lie along the y-axis While, technically speaking, π answers. If we follow the procedure set forth in the previous examples, we ﬁnd θ = π and θ = 3π θ = π 2 isn’t a reference angle we can nonetheless use it to ﬁnd our 2 + 2πk 2 + 2πk for integers, k. While this solution is correct, it can be shortened to 2 + πk for integers k. (Can you see why this works from the diagram?) One of the key items to take from Example 10.2.5 is that, in general, solutions to trigonometric equations consist of inﬁnitely many answers. To get a feel for these answers, the reader is encouraged to follow our mantra from Chapter 9 -
that is, ‘When in doubt, write it out!’ This is especially important when checking answers to the exercises. For example, another Quadrant IV solution to 2 is θ = − π sin(θ) = − 1 6. Hence, the family of Quadrant IV answers to number 2 above could just have easily been written θ = − π 6 + 2πk for integers k. While on the surface, this family may look 730 Foundations of Trigonometry diﬀerent than the stated solution of θ = 11π they represent the same list of angles. 6 + 2πk for integers k, we leave it to the reader to show 10.2.1 Beyond the Unit Circle We began the section with a quest to describe the position of a particle experiencing circular motion. In deﬁning the cosine and sine functions, we assigned to each angle a position on the Unit Circle. In this subsection, we broaden our scope to include circles of radius r centered at the origin. Consider for the moment the acute angle θ drawn below in standard position. Let Q(x, y) be the point on the terminal side of θ which lies on the circle x2 + y2 = r2, and let P (x, y) be the point on the terminal side of θ which lies on the Unit Circle. Now consider dropping perpendiculars from P and Q to create two right triangles, ∆OP A and ∆OQB. These triangles are similar,10 thus it follows that x 1 = r, so x = rx and, similarly, we ﬁnd y = ry. Since, by deﬁnition, x = cos(θ) and y = sin(θ), we get the coordinates of Q to be x = r cos(θ) and y = r sin(θ). By reﬂecting these points through the x-axis, y-axis and origin, we obtain the result for all non-quadrantal angles θ, and we leave it to the reader to verify these formulas hold for the quadrantal angles. x = r y r 1 Q (x, y) P (x, y) θ y 1 1 x r P (x, y) Q(x, y) = (r cos(θ), r sin(θ)) θ O A(x, 0)
B(x, 0) x Not only can we describe the coordinates of Q in terms of cos(θ) and sin(θ) but since the radius of x2 + y2, we can also express cos(θ) and sin(θ) in terms of the coordinates of Q. the circle is r = These results are summarized in the following theorem. Theorem 10.3. If Q(x, y) is the point on the terminal side of an angle θ, plotted in standard position, which lies on the circle x2 + y2 = r2 then x = r cos(θ) and y = r sin(θ). Moreover, cos(θ) = x r = x x2 + y2 and sin(θ) = y r = y x2 + y2 10Do you remember why? 10.2 The Unit Circle: Cosine and Sine 731 Note that in the case of the Unit Circle we have r = our deﬁnitions of cos(θ) and sin(θ). x2 + y2 = 1, so Theorem 10.3 reduces to Example 10.2.6. 1. Suppose that the terminal side of an angle θ, when plotted in standard position, contains the point Q(4, −2). Find sin(θ) and cos(θ). 2. In Example 10.1.5 in Section 10.1, we approximated the radius of the earth at 41.628◦ north latitude to be 2960 miles. Justify this approximation if the radius of the Earth at the Equator is approximately 3960 miles. Solution. 1. Using Theorem 10.3 with x = 4 and y = −2, we ﬁnd r = √ √ 5 5 5. 5 and sin(θ) = y that cos(θ) = x r = −4)2 + (−2)2 = √ 20 = 2 √ 5 so 2. Assuming the Earth is a sphere, a cross-section through the poles produces a circle of radius 3960 miles. Viewing the Equator as the x-axis, the value we seek is the x-coordinate of the point Q(x, y) indicated in the ﬁgure below. y 4 2 −4 −2 2 4 x Q(4, −2) −2 −4 y 3960 Q (x
, y) 41.628◦ 3960 x The terminal side of θ contains Q(4, −2) A point on the Earth at 41.628◦N Using Theorem 10.3, we get x = 3960 cos (41.628◦). Using a calculator in ‘degree’ mode, we ﬁnd 3960 cos (41.628◦) ≈ 2960. Hence, the radius of the Earth at North Latitude 41.628◦ is approximately 2960 miles. 732 Foundations of Trigonometry Theorem 10.3 gives us what we need to describe the position of an object traveling in a circular path of radius r with constant angular velocity ω. Suppose that at time t, the object has swept out an angle measuring θ radians. If we assume that the object is at the point (r, 0) when t = 0, the angle θ is in standard position. By deﬁnition, ω = θ t which we rewrite as θ = ωt. According to Theorem 10.3, the location of the object Q(x, y) on the circle is found using the equations x = r cos(θ) = r cos(ωt) and y = r sin(θ) = r sin(ωt). Hence, at time t, the object is at the point (r cos(ωt), r sin(ωt)). We have just argued the following. Equation 10.3. Suppose an object is traveling in a circular path of radius r centered at the origin with constant angular velocity ω. If t = 0 corresponds to the point (r, 0), then the x and y coordinates of the object are functions of t and are given by x = r cos(ωt) and y = r sin(ωt). Here, ω > 0 indicates a counter-clockwise direction and ω < 0 indicates a clockwise direction. y r 1 Q (x, y) = (r cos(ωt), r sin(ωt)) θ = ωt 1 x r Equations for Circular Motion Example 10.2.7. Suppose we are in the situation of Example 10.1.5. Find the equations of motion of Lakeland Community College as the earth rotates. Solution. From Example 10.1.5, we take r = 2960 miles and and ω =
π of motion are x = r cos(ωt) = 2960 cos π measured in miles and t is measured in hours. 12 hours. Hence, the equations 12 t, where x and y are 12 t and y = r sin(ωt) = 2960 sin π In addition to circular motion, Theorem 10.3 is also the key to developing what is usually called ‘right triangle’ trigonometry.11 As we shall see in the sections to come, many applications in trigonometry involve ﬁnding the measures of the angles in, and lengths of the sides of, right triangles. Indeed, we made good use of some properties of right triangles to ﬁnd the exact values of the cosine and sine of many of the angles in Example 10.2.1, so the following development shouldn’t be that much of a surprise. Consider the generic right triangle below with corresponding acute angle θ. The side with length a is called the side of the triangle adjacent to θ; the side with length b is called the side of the triangle opposite θ; and the remaining side of length c (the side opposite the 11You may have been exposed to this in High School. 10.2 The Unit Circle: Cosine and Sine 733 right angle) is called the hypotenuse. We now imagine drawing this triangle in Quadrant I so that the angle θ is in standard position with the adjacent side to θ lying along the positive x-axis. y c θ P (a, b) x c b c a θ According to the Pythagorean Theorem, a2 + b2 = c2, so that the point P (a, b) lies on a circle of radius c. Theorem 10.3 tells us that cos(θ) = a c, so we have determined the cosine and sine of θ in terms of the lengths of the sides of the right triangle. Thus we have the following theorem. c and sin(θ) = b Theorem 10.4. Suppose θ is an acute angle residing in a right triangle. If the length of the side adjacent to θ is a, the length of the side opposite θ is b, and the length of the hypotenuse is c, then cos(θ) = and sin(θ) = a c b c. Example 10.2.8. Find the measure
of the missing angle and the lengths of the missing sides of: 30◦ 7 Solution. The ﬁrst and easiest task is to ﬁnd the measure of the missing angle. Since the sum of angles of a triangle is 180◦, we know that the missing angle has measure 180◦ − 30◦ − 90◦ = 60◦. We now proceed to ﬁnd the lengths of the remaining two sides of the triangle. Let c denote the 7 length of the hypotenuse of the triangle. By Theorem 10.4, we have cos (30◦) = 7 cos(30◦). 2, we have, after the usual fraction gymnastics, c = 14 Since cos (30◦) =. At this point, we have two ways to proceed to ﬁnd the length of the side opposite the 30◦ angle, which we’ll denote b. We know the length of the adjacent side is 7 and the length of the hypotenuse is 14, so we 3 c, or c = √ √ √ 3 3 3 3 734 Foundations of Trigonometry could use the Pythagorean Theorem to ﬁnd the missing side and solve (7)2 + b2 = Alternatively, we could use Theorem 10.4, namely that sin (30◦) = b b = c sin (30◦) = 14 3 √ 3 3. The triangle with all of its data is recorded below. 2 = 7 · 1 √ 3 c. Choosing the latter, we ﬁnd 2 √ 14 3 3 for b. √ 3 c = 14 3 60◦ b = 7 √ 3 3 30◦ 7 We close this section by noting that we can easily extend the functions cosine and sine to real numbers by identifying a real number t with the angle θ = t radians. Using this identiﬁcation, we deﬁne cos(t) = cos(θ) and sin(t) = sin(θ). In practice this means expressions like cos(π) and sin(2) can be found by regarding the inputs as angles in radian measure or real numbers; the choice is the reader’s. If we trace the identiﬁcation of real numbers t with angles θ in radian measure to its roots on page 704,
we can spell out this correspondence more precisely. For each real number t, we associate an oriented arc t units in length with initial point (1, 0) and endpoint P (cos(t), sin(t)). cos(t), sin(t)) θ = t 1 x In the same way we studied polynomial, rational, exponential, and logarithmic functions, we will study the trigonometric functions f (t) = cos(t) and g(t) = sin(t). The ﬁrst order of business is to ﬁnd the domains and ranges of these functions. Whether we think of identifying the real number t with the angle θ = t radians, or think of wrapping an oriented arc around the Unit Circle to ﬁnd coordinates on the Unit Circle, it should be clear that both the cosine and sine functions are deﬁned for all real numbers t. In other words, the domain of f (t) = cos(t) and of g(t) = sin(t) is (−∞, ∞). Since cos(t) and sin(t) represent x- and y-coordinates, respectively, of points on the Unit Circle, they both take on all of the values between −1 an 1, inclusive. In other words, the range of f (t) = cos(t) and of g(t) = sin(t) is the interval [−1, 1]. To summarize: 10.2 The Unit Circle: Cosine and Sine 735 Theorem 10.5. Domain and Range of the Cosine and Sine Functions: • The function f (t) = cos(t) • The function g(t) = sin(t) – has domain (−∞, ∞) – has range [−1, 1] – has domain (−∞, ∞) – has range [−1, 1] Suppose, as in the Exercises, we are asked to solve an equation such as sin(t) = − 1 2. As we have already mentioned, the distinction between t as a real number and as an angle θ = t radians is often blurred. Indeed, we solve sin(t) = − 1 2 in the exact same manner12 as we did in Example 10.2.5 number 2. Our solution is only cosmetically diﬀerent in that the variable
used is t rather than θ: t = 7π 6 + 2πk for integers, k. We will study the cosine and sine functions in greater detail in Section 10.5. Until then, keep in mind that any properties of cosine and sine developed in the following sections which regard them as functions of angles in radian measure apply equally well if the inputs are regarded as real numbers. 6 + 2πk or t = 11π 12Well, to be pedantic, we would be technically using ‘reference numbers’ or ‘reference arcs’ instead of ‘reference angles’ – but the idea is the same. 736 Foundations of Trigonometry 10.2.2 Exercises In Exercises 1 - 20, ﬁnd the exact value of the cosine and sine of the given angle. 1. θ = 0 5. θ = 9. θ = 13. θ = 2π 3 5π 4 7π 4 17. θ = − 3π 4 2. θ = 6. θ = 10. θ = 14. θ = π 4 3π 4 4π 3 23π 6 18. θ = − π 6 3. θ = π 3 7. θ = π 11. θ = 3π 2 15. θ = − 13π 2 19. θ = 10π 3 4. θ = 8. θ = 12. θ = π 2 7π 6 5π 3 16. θ = − 43π 6 20. θ = 117π In Exercises 21 - 30, use the results developed throughout the section to ﬁnd the requested value. 21. If sin(θ) = − 7 25 with θ in Quadrant IV, what is cos(θ)? 22. If cos(θ) = 23. If sin(θ) = 4 9 5 13 with θ in Quadrant I, what is sin(θ)? with θ in Quadrant II, what is cos(θ)? 24. If cos(θ) = − 2 11 with θ in Quadrant III, what is sin(θ)? 25. If sin(θ) = − 2 3 with θ in Quadrant III, what is cos(θ)? 26. If cos(θ) = 27. If sin(�
�) = 28. If cos(θ) = 28 53 √ 2 5 √ 10 10 with θ in Quadrant IV, what is sin(θ)? 5 and π 2 < θ < π, what is cos(θ)? and 2π < θ < 5π 2 3π 2, what is sin(θ)?, what is cos(θ)? 29. If sin(θ) = −0.42 and π < θ < 30. If cos(θ) = −0.98 and π 2 < θ < π, what is sin(θ)? 10.2 The Unit Circle: Cosine and Sine 737 In Exercises 31 - 39, ﬁnd all of the angles which satisfy the given equation. 31. sin(θ) = 34. cos(θ) = 1 2 √ 2 2 32. cos(θ) = − √ 3 2 35. sin(θ) = √ 3 2 √ 3 2 33. sin(θ) = 0 36. cos(θ) = −1 39. cos(θ) = −1.001 37. sin(θ) = −1 38. cos(θ) = In Exercises 40 - 48, solve the equation for t. (See the comments following Theorem 10.5.) 40. cos(t) = 0 41. sin(t) = − 43. sin(t) = − 1 2 44. cos(t) = 1 2 √ 2 2 42. cos(t) = 3 45. sin(t) = −2 46. cos(t) = 1 47. sin(t) = 1 48. cos(t) = − √ 2 2 In Exercises 49 - 54, use your calculator to approximate the given value to three decimal places. Make sure your calculator is in the proper angle measurement mode! 49. sin(78.95◦) 50. cos(−2.01) 51. sin(392.994) 52. cos(207◦) 53. sin (π◦) 54. cos(e) In Exercises 55 - 58, ﬁnd the measurement of the missing angle and the lengths of the missing sides. (See Example 10.2.8) 55. Find θ, b, and c. 56. Find
θ, a, and c. c θ b 30◦ 1 45◦ c a θ 3 738 Foundations of Trigonometry 57. Find α, a, and b. 58. Find β, a, and c. b α 8 a 33◦ a 48◦ 6 c β In Exercises 59 - 64, assume that θ is an acute angle in a right triangle and use Theorem 10.4 to ﬁnd the requested side. 59. If θ = 12◦ and the side adjacent to θ has length 4, how long is the hypotenuse? 60. If θ = 78.123◦ and the hypotenuse has length 5280, how long is the side adjacent to θ? 61. If θ = 59◦ and the side opposite θ has length 117.42, how long is the hypotenuse? 62. If θ = 5◦ and the hypotenuse has length 10, how long is the side opposite θ? 63. If θ = 5◦ and the hypotenuse has length 10, how long is the side adjacent to θ? 64. If θ = 37.5◦ and the side opposite θ has length 306, how long is the side adjacent to θ? In Exercises 65 - 68, let θ be the angle in standard position whose terminal side contains the given point then compute cos(θ) and sin(θ). 65. P (−7, 24) 66. Q(3, 4) 67. R(5, −9) 68. T (−2, −11) In Exercises 69 - 72, ﬁnd the equations of motion for the given scenario. Assume that the center of the motion is the origin, the motion is counter-clockwise and that t = 0 corresponds to a position along the positive x-axis. (See Equation 10.3 and Example 10.1.5.) 69. A point on the edge of the spinning yo-yo in Exercise 50 from Section 10.1. Recall: The diameter of the yo-yo is 2.25 inches and it spins at 4500 revolutions per minute. 70. The yo-yo in exercise 52 from Section 10.1. Recall: The radius of the circle is 28 inches and it completes one revolution in 3 seconds. 71. A point on the edge of the hard drive in Exercise
53 from Section 10.1. Recall: The diameter of the hard disk is 2.5 inches and it spins at 7200 revolutions per minute. 10.2 The Unit Circle: Cosine and Sine 739 72. A passenger on the Big Wheel in Exercise 55 from Section 10.1. Recall: The diameter is 128 feet and completes 2 revolutions in 2 minutes, 7 seconds. 73. Consider the numbers: 0, 1, 2, 3, 4. Take the square root of each of these numbers, then divide each by 2. The resulting numbers should look hauntingly familiar. (See the values in the table on 722.) 74. Let α and β be the two acute angles of a right triangle. (Thus α and β are complementary angles.) Show that sin(α) = cos(β) and sin(β) = cos(α). The fact that co-functions of complementary angles are equal in this case is not an accident and a more general result will be given in Section 10.4. 75. In the scenario of Equation 10.3, we assumed that at t = 0, the object was at the point (r, 0). If this is not the case, we can adjust the equations of motion by introducing a ‘time delay.’ If t0 > 0 is the ﬁrst time the object passes through the point (r, 0), show, with the help of your classmates, the equations of motion are x = r cos(ω(t − t0)) and y = r sin(ω(t − t0)). 740 Foundations of Trigonometry 10.2.3 Answers 1. cos(0) = 1, sin(0) = 0 1 2 = 3. cos 5. cos π 3 2π 3, sin π 3 = = − 1 2, sin 2π 3 √ 3 2 √ 3 2 = 7. cos(π) = −1, sin(π) = 0 9. cos 11. cos 13. cos 5π 4 3π 2 7π 4 √ 2 2 = −, sin 5π 4 = − √ 2 2 = 0, sin 3π 2 = −1 √ 2 2 =, sin 7π 4 = − √ 2 2 2. cos 4. cos 6. cos 8. cos 10. cos 12. cos π 4 π 2 3π 4 7π 6 4π 3 5π 3 √
2 2 =, sin π 4 = √ 2 2 = 0, sin, sin, sin 3π 4 7π, sin 4π 3 = − √ 3 2 = 1 2, sin 5π 3 = − √ 3 2 14. cos 23π 6 √ 3 2, sin 23π 6 = − 1 2 = 15. cos − 13π 2 = 0, sin − 13π 2 = −1 16. cos − 43π 6 √ 3 2, sin − 43π 6 = 1 2 = − 17. cos 19. cos − 3π 4 10π 3 = − √ 2 2, sin − 3π 4 √ 2 2 = − 18. cos − π 6 = √ 3 2, sin −, sin 10π 3 = − √ 3 2 20. cos(117π) = −1, sin(117π) = 0 21. If sin(θ) = − 7 25 22. If cos(θ) = 23. If sin(θ) = 4 9 5 13 24. If cos(θ) = − 2 11 25. If sin(θ) = − 2 3 26. If cos(θ) = 28 53 with θ in Quadrant IV, then cos(θ) = with θ in Quadrant I, then sin(θ) = with θ in Quadrant II, then cos(θ) = − 12 13. with θ in Quadrant III, then sin(θ) = − 24 25.. √ 65 9 √ 117 11. √ 5 3. with θ in Quadrant III, then cos(θ) = − with θ in Quadrant IV, then sin(θ) = − 45 53. 10.2 The Unit Circle: Cosine and Sine 741 27. If sin(θ) = 28. If cos(θ) = √ 2 5 5 √ 10 10 and π 2 < θ < π, then cos(θ) = − and 2π < θ <, then sin(θ) = √. 5 5 √ 3 10. 10 √ 5π 2 3π 2 29. If sin(θ) = −0.42 and π < θ <, then cos(θ) = − 0.8236 ≈ −0.9075. 30. If
cos(θ) = −0.98 and < θ < π, then sin(θ) = √ 0.0396 ≈ 0.1990. π 2 31. sin(θ) = 1 2 32. cos(θ) = − √ 3 2 when θ = π 6 + 2πk or θ = 5π 6 + 2πk for any integer k. when θ = 5π 6 + 2πk or θ = 7π 6 + 2πk for any integer k. 33. sin(θ) = 0 when θ = πk for any integer k. 34. cos(θ) = 35. sin(θ) = √ 2 2 √ 3 2 when θ = when θ = π 4 π 3 + 2πk or θ = + 2πk or θ = 7π 4 2π 3 + 2πk for any integer k. + 2πk for any integer k. 36. cos(θ) = −1 when θ = (2k + 1)π for any integer k. 37. sin(θ) = −1 when θ = 38. cos(θ) = √ 3 2 when θ = 3π 2 π 6 + 2πk for any integer k. + 2πk or θ = 11π 6 + 2πk for any integer k. 39. cos(θ) = −1.001 never happens 40. cos(t) = 0 when t = π 2 + πk for any integer k. 41. sin(t) = − √ 2 2 when t = 5π 4 + 2πk or t = 7π 4 + 2πk for any integer k. 42. cos(t) = 3 never happens. 43. sin(t) = − 1 2 when t = 7π 6 + 2πk or t = 11π 6 + 2πk for any integer k. 44. cos(t) = 1 2 when t = π 3 + 2πk or t = 5π 3 + 2πk for any integer k. 45. sin(t) = −2 never happens 46. cos(t) = 1 when t = 2πk for any integer k. 742 Foundations of Trigonometry π 2 + 2πk for any integer k.
47. sin(t) = 1 when t = 48. cos(t) = − √ 2 2 when t = 3π 4 + 2πk or t = 5π 4 + 2πk for any integer k. 49. sin(78.95◦) ≈ 0.981 50. cos(−2.01) ≈ −0.425 51. sin(392.994) ≈ −0.291 52. cos(207◦) ≈ −0.891 53. sin (π◦) ≈ 0.055 54. cos(e) ≈ −0.912 55. θ = 60◦, b = √ 3 3, c = 56. θ = 45◦, a = 3 57. α = 57◦, a = 8 cos(33◦) ≈ 6.709, b = 8 sin(33◦) ≈ 4.357 58. β = 42◦, c = 6 sin(48◦) ≈ 8.074, a = √ c2 − 62 ≈ 5.402 59. The hypotenuse has length 4 cos(12◦) ≈ 4.089. 60. The side adjacent to θ has length 5280 cos(78.123◦) ≈ 1086.68. 61. The hypotenuse has length 117.42 sin(59◦) ≈ 136.99. 62. The side opposite θ has length 10 sin(5◦) ≈ 0.872. 63. The side adjacent to θ has length 10 cos(5◦) ≈ 9.962. 64. The hypotenuse has length c = √ c2 − 3062 ≈ 398.797. 306 sin(37.5◦) 65. cos(θ) = − 7 25, sin(θ) = 24 25 ≈ 502.660, so the side adjacent to θ has length, sin(θ) = 4 5 66. cos(θ) = 67. cos(θ) = 3 5 √ 5, sin(θ) = − 68. cos(θ) = −, sin(θ) = − 106 106 √ 5 2 25 √ 106 9 106 √ 5 11 25 69. r = 1.125 inches, ω = 9000π radians
minute, x = 1.125 cos(9000π t), y = 1.125 sin(9000π t). Here x and y are measured in inches and t is measured in minutes. 10.2 The Unit Circle: Cosine and Sine 743 70. r = 28 inches, ω = 2π 3 radians second, x = 28 cos 2π 3 t, y = 28 sin 2π 3 t. Here x and y are measured in inches and t is measured in seconds. 71. r = 1.25 inches, ω = 14400π radians minute, x = 1.25 cos(14400π t), y = 1.25 sin(14400π t). Here x and y are measured in inches and t is measured in minutes. 127 t, y = 64 sin 4π second, x = 64 cos 4π 72. r = 64 feet, ω = 4π 127 in feet and t is measured in seconds radians 127 t. Here x and y are measured 744 Foundations of Trigonometry 10.3 The Six Circular Functions and Fundamental Identities In section 10.2, we deﬁned cos(θ) and sin(θ) for angles θ using the coordinate values of points on the Unit Circle. As such, these functions earn the moniker circular functions.1 It turns out that cosine and sine are just two of the six commonly used circular functions which we deﬁne below. Deﬁnition 10.2. The Circular Functions: Suppose θ is an angle plotted in standard position and P (x, y) is the point on the terminal side of θ which lies on the Unit Circle. The cosine of θ, denoted cos(θ), is deﬁned by cos(θ) = x. The sine of θ, denoted sin(θ), is deﬁned by sin(θ) = y. The secant of θ, denoted sec(θ), is deﬁned by sec(θ) = 1 x, provided x = 0. The cosecant of θ, denoted csc(θ), is deﬁned by csc(θ) = The tangent of θ, denoted tan(θ), is deﬁned by tan(θ) =
1 y y x The cotangent of θ, denoted cot(θ), is deﬁned by cot(θ) =, provided y = 0., provided x = 0. x y, provided y = 0. While we left the history of the name ‘sine’ as an interesting research project in Section 10.2, the names ‘tangent’ and ‘secant’ can be explained using the diagram below. Consider the acute angle θ below in standard position. Let P (x, y) denote, as usual, the point on the terminal side of θ which lies on the Unit Circle and let Q(1, y) denote the point on the terminal side of θ which lies on the vertical line x = 1. Q(1, y) = (1, tan(θ)) P (x, y) y 1 θ O A(x, 0) B(1, 0) x 1In Theorem 10.4 we also showed cosine and sine to be functions of an angle residing in a right triangle so we could just as easily call them trigonometric functions. In later sections, you will ﬁnd that we do indeed use the phrase ‘trigonometric function’ interchangeably with the term ‘circular function’. 10.3 The Six Circular Functions and Fundamental Identities 745 y = 1 x which gives y = y The word ‘tangent’ comes from the Latin meaning ‘to touch,’ and for this reason, the line x = 1 is called a tangent line to the Unit Circle since it intersects, or ‘touches’, the circle at only one point, namely (1, 0). Dropping perpendiculars from P and Q creates a pair of similar triangles ∆OP A and ∆OQB. Thus y x = tan(θ), where this last equality comes from applying Deﬁnition 10.2. We have just shown that for acute angles θ, tan(θ) is the y-coordinate of the point on the terminal side of θ which lies on the line x = 1 which is tangent to the Unit Circle. Now the word ‘secant’ means ‘to cut’, so a secant line is any line that ‘cuts through’ a circle at
two points.2 The line containing the terminal side of θ is a secant line since it intersects the Unit Circle in Quadrants I and III. With the point P lying on the Unit Circle, the length of the hypotenuse of ∆OP A is 1. If we let h denote the length of the hypotenuse of ∆OQB, we have from similar triangles that h x = sec(θ). Hence for an acute angle θ, sec(θ) is the length of the line segment which lies on the secant line determined by the terminal side of θ and ‘cuts oﬀ’ the tangent line x = 1. Not only do these observations help explain the names of these functions, they serve as the basis for a fundamental inequality needed for Calculus which we’ll explore in the Exercises. x, or h = 1 1 = 1 Of the six circular functions, only cosine and sine are deﬁned for all angles. Since cos(θ) = x and sin(θ) = y in Deﬁnition 10.2, it is customary to rephrase the remaining four circular functions in terms of cosine and sine. The following theorem is a result of simply replacing x with cos(θ) and y with sin(θ) in Deﬁnition 10.2. Theorem 10.6. Reciprocal and Quotient Identities: sec(θ) = 1 cos(θ), provided cos(θ) = 0; if cos(θ) = 0, sec(θ) is undeﬁned. csc(θ) = 1 sin(θ), provided sin(θ) = 0; if sin(θ) = 0, csc(θ) is undeﬁned. tan(θ) = sin(θ) cos(θ), provided cos(θ) = 0; if cos(θ) = 0, tan(θ) is undeﬁned. cot(θ) = cos(θ) sin(θ), provided sin(θ) = 0; if sin(θ) = 0, cot(θ) is undeﬁned. It is high time for an example. Example 10.3.1. Find the indicated
value, if it exists. 1. sec (60◦) 2. csc 7π 4 3. cot(3) 4. tan (θ), where θ is any angle coterminal with 3π 2. 5. cos (θ), where csc(θ) = − √ 5 and θ is a Quadrant IV angle. 6. sin (θ), where tan(θ) = 3 and π < θ < 3π 2. 2Compare this with the deﬁnition given in Section 2.1. 746 Solution. 1. According to Theorem 10.6, sec (60◦) = Foundations of Trigonometry 1 cos(60◦). Hence, sec (60◦) = 1 (1/2) = 2. 2. Since sin 7π 4 = − √ 2 2, csc 7π 4 = 1 sin( 7π 4 ) = 1 √ − 2/2 = − 2√ 2 √ = − 2. 3. Since θ = 3 radians is not one of the ‘common angles’ from Section 10.2, we resort to the calculator for a decimal approximation. Ensuring that the calculator is in radian mode, we ﬁnd cot(3) = cos(3) sin(3) ≈ −7.015. = 0 and sin(θ) = sin 3π 2 = −1. Attempting 4. If θ is coterminal with 3π to compute tan(θ) = sin(θ) 2, then cos(θ) = cos 3π cos(θ) results in −1 √ 2 0, so tan(θ) is undeﬁned. 5. We are given that csc(θ) = 1 √ 5 5 so sin(θ) = − 1√ 5. As we saw in Section 10.2, 5 we can use the Pythagorean Identity, cos2(θ) + sin2(θ) = 1, to ﬁnd cos(θ) by knowing sin(θ). Substituting, we get cos2(θ) + θ is a Quadrant IV angle, cos(θ) > 0, so cos(θ) = 2 = 1, which gives cos2(θ
) = 4 √ 5 5. 5, or cos(θ) = ± 2 √ 5 5. Since sin(θ. If tan(θ) = 3, then sin(θ) cos(θ) = 3. Be careful - this does NOT mean we can take sin(θ) = 3 and cos(θ) = 1. Instead, from sin(θ) cos(θ) = 3 we get: sin(θ) = 3 cos(θ). To relate cos(θ) and sin(θ), we once again employ the Pythagorean Identity, cos2(θ) + sin2(θ) = 1. Solving sin(θ) = 3 cos(θ) for cos(θ), we ﬁnd cos(θ) = 1 3 sin(θ). Substituting this into the Pythagorean Identity, we ﬁnd sin2(θ) + 1 3 sin(θ)2 10 so sin(θ) = ± 3 10. Since π < θ < 3π 2, θ is a Quadrant III angle. This means sin(θ) < 0, so our ﬁnal answer is sin(θ) = − 3 = 1. Solving, we get sin2(θ) = 9 10 √ √ 10 10. While the Reciprocal and Quotient Identities presented in Theorem 10.6 allow us to always reduce problems involving secant, cosecant, tangent and cotangent to problems involving cosine and sine, it is not always convenient to do so.3 It is worth taking the time to memorize the tangent and cotangent values of the common angles summarized below. 3As we shall see shortly, when solving equations involving secant and cosecant, we usually convert back to cosines and sines. However, when solving for tangent or cotangent, we usually stick with what we’re dealt. 10.3 The Six Circular Functions and Fundamental Identities 747 Tangent and Cotangent Values of Common Angles θ(degrees) 0◦ 30◦ 45◦ 60◦ 90◦ θ(radians) tan(θ) cot(θ undeﬁned undeﬁned √
3 1 √ 3 3 0 Coupling Theorem 10.6 with the Reference Angle Theorem, Theorem 10.2, we get the following. Theorem 10.7. Generalized Reference Angle Theorem. The values of the circular functions of an angle, if they exist, are the same, up to a sign, of the corresponding circular functions of its reference angle. More speciﬁcally, if α is the reference angle for θ, then: cos(θ) = ± cos(α), sin(θ) = ± sin(α), sec(θ) = ± sec(α), csc(θ) = ± csc(α), tan(θ) = ± tan(α) and cot(θ) = ± cot(α). The choice of the (±) depends on the quadrant in which the terminal side of θ lies. We put Theorem 10.7 to good use in the following example. Example 10.3.2. Find all angles which satisfy the given equation. 1. sec(θ) = 2 Solution. 2. tan(θ) = √ 3 3. cot(θ) = −1. 1. To solve sec(θ) = 2, we convert to cosines and get 1 cos(θ) = 2 or cos(θ) = 1 same equation we solved in Example 10.2.5, number 1, so we know the answer is: θ = π or θ = 5π 3 + 2πk for integers k. 2. This is the exact 3 + 2πk √ √ = 2. From the table of common values, we see tan π 3 3 must, therefore, have a reference angle of π 3. According to Theorem 10.7, we know 3. Our next task is the solutions to tan(θ) = to determine in which quadrants the solutions to this equation lie. Since tangent is deﬁned as the ratio y x of points (x, y) on the Unit Circle with x = 0, tangent is positive when x and y have the same sign (i.e., when they are both positive or both negative.) This happens in Quadrants I and III. In Quadrant I, we get the solutions: θ = π 3 + 2πk for integers k, and for Quadrant III
, we get θ = 4π 3 + 2πk for integers k. While these descriptions of the solutions are correct, they can be combined into one list as θ = π 3 + πk for integers k. The latter form of the solution is best understood looking at the geometry of the situation in the diagram below.4 4See Example 10.2.5 number 3 in Section 10.2 for another example of this kind of simpliﬁcation of the solution. 748 Foundations of Trigonometry. From the table of common values, we see that π 4 has a cotangent of 1, which means the solutions to cot(θ) = −1 have a reference angle of π 4. To ﬁnd the quadrants in which our solutions lie, we note that cot(θ) = x y for a point (x, y) on the Unit Circle where y = 0. If cot(θ) is negative, then x and y must have diﬀerent signs (i.e., one positive and one negative.) Hence, our solutions lie in Quadrants II and IV. Our Quadrant II solution is θ = 3π 4 + 2πk, and for Quadrant IV, we get θ = 7π 4 +2πk for integers k. Can these lists be combined? Indeed they can - one such way to capture all the solutions is: θ = 3π 4 + πk for integers k We have already seen the importance of identities in trigonometry. Our next task is to use use the Reciprocal and Quotient Identities found in Theorem 10.6 coupled with the Pythagorean Identity found in Theorem 10.1 to derive new Pythagorean-like identities for the remaining four circular functions. Assuming cos(θ) = 0, we may start with cos2(θ) + sin2(θ) = 1 and divide both sides by cos2(θ) to obtain 1 + sin2(θ) cos2(θ). Using properties of exponents along with the Reciprocal and Quotient Identities, this reduces to 1 + tan2(θ) = sec2(θ). If sin(θ) = 0, we can divide both sides of the identity cos2(θ) + sin2(θ) = 1 by sin2(
θ), apply Theorem 10.6 once again, and obtain cot2(θ) + 1 = csc2(θ). These three Pythagorean Identities are worth memorizing and they, along with some of their other common forms, are summarized in the following theorem. cos2(θ) = 1 10.3 The Six Circular Functions and Fundamental Identities 749 Theorem 10.8. The Pythagorean Identities: 1. cos2(θ) + sin2(θ) = 1. Common Alternate Forms: 1 − sin2(θ) = cos2(θ) 1 − cos2(θ) = sin2(θ) 2. 1 + tan2(θ) = sec2(θ), provided cos(θ) = 0. Common Alternate Forms: sec2(θ) − tan2(θ) = 1 sec2(θ) − 1 = tan2(θ) 3. 1 + cot2(θ) = csc2(θ), provided sin(θ) = 0. Common Alternate Forms: csc2(θ) − cot2(θ) = 1 csc2(θ) − 1 = cot2(θ) Trigonometric identities play an important role in not just Trigonometry, but in Calculus as well. We’ll use them in this book to ﬁnd the values of the circular functions of an angle and solve equations and inequalities. In Calculus, they are needed to simplify otherwise complicated expressions. In the next example, we make good use of the Theorems 10.6 and 10.8. Example 10.3.3. Verify the following identities. Assume that all quantities are deﬁned. 1. 1 csc(θ) = sin(θ) 3. (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = 1 5. 6 sec(θ) tan(θ) = 3 1 − sin(θ) − 3 1 + sin(θ) 2. tan(θ) = sin(θ) sec(θ) 4. 6. sec(θ) 1 − tan(θ) = 1 cos(θ) − sin(θ) sin(θ) 1
− cos(θ) = 1 + cos(θ) sin(θ) Solution. In verifying identities, we typically start with the more complicated side of the equation and use known identities to transform it into the other side of the equation. 1. To verify 1 csc(θ) = sin(θ), we start with the left side. Using csc(θ) = 1 sin(θ), we get: 1 csc(θ) = 1 1 sin(θ) = sin(θ), which is what we were trying to prove. 750 Foundations of Trigonometry 2. Starting with the right hand side of tan(θ) = sin(θ) sec(θ), we use sec(θ) = 1 cos(θ) and ﬁnd: sin(θ) sec(θ) = sin(θ) 1 cos(θ) = sin(θ) cos(θ) = tan(θ), where the last equality is courtesy of Theorem 10.6. 3. Expanding the left hand side of the equation gives: (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = sec2(θ) − tan2(θ). According to Theorem 10.8, sec2(θ) − tan2(θ) = 1. Putting it all together, (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = sec2(θ) − tan2(θ) = 1. 4. While both sides of our last identity contain fractions, the left side aﬀords us more opportu- nities to use our identities.5 Substituting sec(θ) = 1 cos(θ) and tan(θ) = sin(θ) cos(θ), we get: sec(θ) 1 − tan(θ) = 1 cos(θ) 1 − sin(θ) cos(θ) = 1 cos(θ) 1 − sin(θ) cos(θ) · cos(θ) cos(θ) = 1 cos(θ) 1 − sin(θ) cos(θ) (cos(θ)) = (cos(θ))
(1)(cos(θ)) − 1 sin(θ) cos(θ) (cos(θ)) = 1 cos(θ) − sin(θ), which is exactly what we had set out to show. 5. The right hand side of the equation seems to hold more promise. We get common denomina- tors and add: 3 1 − sin(θ) − 3 1 + sin(θ) = = = = 3(1 + sin(θ)) (1 − sin(θ))(1 + sin(θ)) − 3(1 − sin(θ)) (1 + sin(θ))(1 − sin(θ)) 3 + 3 sin(θ) 1 − sin2(θ) − 3 − 3 sin(θ) 1 − sin2(θ) (3 + 3 sin(θ)) − (3 − 3 sin(θ)) 1 − sin2(θ) 6 sin(θ) 1 − sin2(θ) 5Or, to put to another way, earn more partial credit if this were an exam question! 10.3 The Six Circular Functions and Fundamental Identities 751 At this point, it is worth pausing to remind ourselves of our goal. We wish to transform this expression into 6 sec(θ) tan(θ). Using a reciprocal and quotient identity, we ﬁnd 6 sec(θ) tan(θ) = 6. In other words, we need to get cosines in our denominator. Theorem 10.8 tells us 1 − sin2(θ) = cos2(θ) so we get: sin(θ) cos(θ) 1 cos(θ) 3 1 − sin(θ) − 3 1 + sin(θ) = = 6 sin(θ) cos2(θ) 6 sin(θ) 1 − sin2(θ) 1 cos(θ) = 6 sin(θ) cos(θ) = 6 sec(θ) tan(θ) 6. It is debatable which side of the identity is more complicated. One thing which stands out is that the denominator on the left hand side is 1 − cos(θ), while the numerator of the right hand side is 1 + cos(θ). This suggests the strategy of
starting with the left hand side and multiplying the numerator and denominator by the quantity 1 + cos(θ): sin(θ) 1 − cos(θ) = = = sin(θ) (1 − cos(θ)) · (1 + cos(θ)) (1 + cos(θ)) = sin(θ)(1 + cos(θ)) (1 − cos(θ))(1 + cos(θ)) sin(θ)(1 + cos(θ)) 1 − cos2(θ) sin(θ)(1 + cos(θ)) sin(θ) sin(θ) = = sin(θ)(1 + cos(θ)) sin2(θ) 1 + cos(θ) sin(θ) In Example 10.3.3 number 6 above, we see that multiplying 1 − cos(θ) by 1 + cos(θ) produces a diﬀerence of squares that can be simpliﬁed to one term using Theorem 10.8. This is exactly the √ same kind of phenomenon that occurs when we multiply expressions such as 1 − 2 or 3 − 4i by 3 + 4i. (Can you recall instances from Algebra where we did such things?) For this reason, the quantities (1 − cos(θ)) and (1 + cos(θ)) are called ‘Pythagorean Conjugates.’ Below is a list of other common Pythagorean Conjugates. 2 by 1 + √ Pythagorean Conjugates 1 − cos(θ) and 1 + cos(θ): (1 − cos(θ))(1 + cos(θ)) = 1 − cos2(θ) = sin2(θ) 1 − sin(θ) and 1 + sin(θ): (1 − sin(θ))(1 + sin(θ)) = 1 − sin2(θ) = cos2(θ) sec(θ) − 1 and sec(θ) + 1: (sec(θ) − 1)(sec(θ) + 1) = sec2(θ) − 1 = tan2(θ) sec(θ)−tan(θ) and sec(θ)+tan(θ): (sec(θ)−
tan(θ))(sec(θ)+tan(θ)) = sec2(θ)−tan2(θ) = 1 csc(θ) − 1 and csc(θ) + 1: (csc(θ) − 1)(csc(θ) + 1) = csc2(θ) − 1 = cot2(θ) csc(θ) − cot(θ) and csc(θ) + cot(θ): (csc(θ) − cot(θ))(csc(θ) + cot(θ)) = csc2(θ) − cot2(θ) = 1 752 Foundations of Trigonometry Verifying trigonometric identities requires a healthy mix of tenacity and inspiration. You will need to spend many hours struggling with them just to become proﬁcient in the basics. Like many things in life, there is no short-cut here – there is no complete algorithm for verifying identities. Nevertheless, a summary of some strategies which may be helpful (depending on the situation) is provided below and ample practice is provided for you in the Exercises. Try working on the more complicated side of the identity. Strategies for Verifying Identities Use the Reciprocal and Quotient Identities in Theorem 10.6 to write functions on one side of the identity in terms of the functions on the other side of the identity. Simplify the resulting complex fractions. Add rational expressions with unlike denominators by obtaining common denominators. Use the Pythagorean Identities in Theorem 10.8 to ‘exchange’ sines and cosines, secants and tangents, cosecants and cotangents, and simplify sums or diﬀerences of squares to one term. Multiply numerator and denominator by Pythagorean Conjugates in order to take advan- tage of the Pythagorean Identities in Theorem 10.8. If you ﬁnd yourself stuck working with one side of the identity, try starting with the other side of the identity and see if you can ﬁnd a way to bridge the two parts of your work. 10.3.1 Beyond the Unit Circle In Section 10.2, we generalized the cosine and sine functions from coordinates on the Unit Circle to coordinates on
circles of radius r. Using Theorem 10.3 in conjunction with Theorem 10.8, we generalize the remaining circular functions in kind. Theorem 10.9. Suppose Q(x, y) is the point on the terminal side of an angle θ (plotted in standard position) which lies on the circle of radius r, x2 + y2 = r2. Then: r x r y = = x2 + y2 x x2 + y2 y, provided x = 0., provided y = 0., provided x = 0., provided y = 0. sec(θ) = csc(θ) = tan(θ) = cot(θ) = y x x y 10.3 The Six Circular Functions and Fundamental Identities 753 Example 10.3.4. 1. Suppose the terminal side of θ, when plotted in standard position, contains the point Q(3, −4). Find the values of the six circular functions of θ. 2. Suppose θ is a Quadrant IV angle with cot(θ) = −4. Find the values of the ﬁve remaining circular functions of θ. Solution. 1. Since x = 3 and y = −4, r = x2 + y2 = (3)2 + (−4)2 = √ cos(θ) = 3 5, sin(θ) = − 4 5, sec(θ) = 5 3, csc(θ) = − 5 4, tan(θ) = − 4 25 = 5. Theorem 10.9 tells us 3 and cot(θ) = − 3 4. 2. In order to use Theorem 10.9, we need to ﬁnd a point Q(x, y) which lies on the terminal side y, and since θ is a −1, we may choose6 x = 4 17. Applying Theorem 10.9 once √ √ 17 17 4, csc(θ) = − 17 17, sec(θ) = of θ, when θ is plotted in standard position. We have that cot(θ) = −4 = x Quadrant IV angle, we also know x > 0 and y < 0. Viewing −4 = 4 x2 + y2 = and y = −1 so that r = √ = 4 more
, we ﬁnd cos(θ) = 4√ 17 and tan(θ) = − 1 4. (4)2 + (−1)2 = = − 17, sin(θ) = − 1√ √ √ 17 17 We may also specialize Theorem 10.9 to the case of acute angles θ which reside in a right triangle, as visualized below. b c a θ Theorem 10.10. Suppose θ is an acute angle residing in a right triangle. If the length of the side adjacent to θ is a, the length of the side opposite θ is b, and the length of the hypotenuse is c, then tan(θ) = sec(θ) = csc(θ) = cot(θ The following example uses Theorem 10.10 as well as the concept of an ‘angle of inclination.’ The angle of inclination (or angle of elevation) of an object refers to the angle whose initial side is some kind of base-line (say, the ground), and whose terminal side is the line-of-sight to an object above the base-line. This is represented schematically below. 6We may choose any values x and y so long as x > 0, y < 0 and x y = −4. For example, we could choose x = 8 and y = −2. The fact that all such points lie on the terminal side of θ is a consequence of the fact that the terminal side of θ is the portion of the line with slope − 1 4 which extends from the origin into Quadrant IV. 754 Foundations of Trigonometry object θ ‘base line’ The angle of inclination from the base line to the object is θ Example 10.3.5. 1. The angle of inclination from a point on the ground 30 feet away to the top of Lakeland’s Armington Clocktower7 is 60◦. Find the height of the Clocktower to the nearest foot. 2. In order to determine the height of a California Redwood tree, two sightings from the ground, one 200 feet directly behind the other, are made. If the angles of inclination were 45◦ and 30◦, respectively, how tall is the tree to the nearest foot? Solution. 1. We can represent the problem situation using a right triangle as shown below. If we let h 30. From this
we get denote the height of the tower, then Theorem 10.10 gives tan (60◦) = h h = 30 tan (60◦) = 30 3 ≈ 51.96. Hence, the Clocktower is approximately 52 feet tall. √ h ft. 60◦ 30 ft. Finding the height of the Clocktower 2. Sketching the problem situation below, we ﬁnd ourselves with two unknowns: the height h of the tree and the distance x from the base of the tree to the ﬁrst observation point. 7Named in honor of Raymond Q. Armington, Lakeland’s Clocktower has been a part of campus since 1972. 10.3 The Six Circular Functions and Fundamental Identities 755 h ft. 30◦ 200 ft. 45◦ x ft. Finding the height of a California Redwood Using Theorem 10.10, we get a pair of equations: tan (45◦) = h x+200. Since tan (45◦) = 1, the ﬁrst equation gives h x = 1, or x = h. Substituting this into the second equation gives 3. The result is a linear equation for h, so we proceed to expand the right hand side and gather all the terms involving h to one side. √ 3 3. Clearing fractions, we get 3h = (h + 200) x and tan (30◦) = h h+200 = tan (30◦) = √ h 3h = (h + 200) √ 3h = h √ 3 + 200 √ 3 = 200 3 √ 3 √ 3 3h − h √ (3 − 3)h = 200 √ 3 √ 3 √ 3 ≈ 273.20 Hence, the tree is approximately 273 feet tall. h = 200 3 − As we did in Section 10.2.1, we may consider all six circular functions as functions of real numbers. At this stage, there are three equivalent ways to deﬁne the functions sec(t), csc(t), tan(t) and cot(t) for real numbers t. First, we could go through the formality of the wrapping function on page 704 and deﬁne these functions as the appropriate ratios of x and y coordinates of points on the Unit Circle; second, we could deﬁne them by
associating the real number t with the angle θ = t radians so that the value of the trigonometric function of t coincides with that of θ; lastly, we could simply deﬁne them using the Reciprocal and Quotient Identities as combinations of the functions f (t) = cos(t) and g(t) = sin(t). Presently, we adopt the last approach. We now set about determining the domains and ranges of the remaining four circular functions. Consider the function F (t) = sec(t) deﬁned as F (t) = sec(t) = 1 cos(t). We know F is undeﬁned whenever cos(t) = 0. From Example 10.2.5 number 3, we know cos(t) = 0 whenever t = π 2 + πk for integers k. Hence, our domain for F (t) = sec(t), in set builder notation is {t : t = π 2 + πk, for integers k}. To get a better understanding what set of real numbers we’re dealing with, it pays to write out and graph this set. Running through a few values of k, we ﬁnd the domain to be {t : t = ± π 2,...}. Graphing this set on the number line we get 2, ± 5π 2, ± 3π 756 Foundations of Trigonometry − 5π 2 − 3π 2 − π 2 0 π 2 3π 2 5π 2 Using interval notation to describe this set, we get... ∪ − 5π 2, − 3π 2 ∪ − 3π, 3π 2 ∪ 3π 2, 5π 2 ∪... This is cumbersome, to say the least! In order to write this in a more compact way, we note that from the set-builder description of the domain, the kth point excluded from the domain, which we’ll call xk, can be found by the formula xk = π 2 +πk. (We are using sequence notation from Chapter 9.) Getting a common denominator and factoring out the π in the numerator, we get xk = (2k+1)π. The. domain consists of the intervals determined by successive points xk: (xk, xk + 1)
= In order to capture all of the intervals in the domain, k must run through all of the integers, that is, k = 0, ±1, ±2,.... The way we denote taking the union of inﬁnitely many intervals like this is to use what we call in this text extended interval notation. The domain of F (t) = sec(t) can now be written as 2, (2k+3)π 2 (2k+1)π 2 ∞ k=−∞ (2k + 1)π 2, (2k + 3)π 2 The reader should compare this notation with summation notation introduced in Section 9.2, in particular the notation used to describe geometric series in Theorem 9.2. In the same way the index k in the series ∞ k=1 ark−1 can never equal the upper limit ∞, but rather, ranges through all of the natural numbers, the index k in the union ∞ (2k + 1)π 2, (2k + 3)π 2 k=−∞ can never actually be ∞ or −∞, but rather, this conveys the idea that k ranges through all of the integers. Now that we have painstakingly determined the domain of F (t) = sec(t), it is time to discuss the range. Once again, we appeal to the deﬁnition F (t) = sec(t) = 1 cos(t). The range of f (t) = cos(t) is [−1, 1], and since F (t) = sec(t) is undeﬁned when cos(t) = 0, we split our discussion into two cases: when 0 < cos(t) ≤ 1 and when −1 ≤ cos(t) < 0. If 0 < cos(t) ≤ 1, then we can divide the inequality cos(t) ≤ 1 by cos(t) to obtain sec(t) = 1 cos(t) ≥ 1. Moreover, using the notation introduced in Section 4.2, we have that as cos(t) → 0+, sec(t) = 1 very small (+) ≈ very big (+). In other words, as cos(t) → 0+, sec(t) → ∞. If, on the other hand, if −1 ≤ cos(t) < 0, then dividing by cos
(t) causes a reversal of the inequality so that sec(t) = 1 sec(t) ≤ −1. In this case, as cos(t) → 0−, sec(t) = 1 very small (−) ≈ very big (−), so that as cos(t) → 0−, we get sec(t) → −∞. Since cos(t) ≈ cos(t) ≈ 1 1 10.3 The Six Circular Functions and Fundamental Identities 757 f (t) = cos(t) admits all of the values in [−1, 1], the function F (t) = sec(t) admits all of the values in (−∞, −1] ∪ [1, ∞). Using set-builder notation, the range of F (t) = sec(t) can be written as {u : u ≤ −1 or u ≥ 1}, or, more succinctly,8 as {u : \|u\| ≥ 1}.9 Similar arguments can be used to determine the domains and ranges of the remaining three circular functions: csc(t), tan(t) and cot(t). The reader is encouraged to do so. (See the Exercises.) For now, we gather these facts into the theorem below. Theorem 10.11. Domains and Ranges of the Circular Functions • The function f (t) = cos(t) • The function g(t) = sin(t) – has domain (−∞, ∞) – has range [−1, 1] – has domain (−∞, ∞) – has range [−1, 1] • The function F (t) = sec(t) = 1 cos(t) – has domain {t : t = π 2 + πk, for integers k} = – has range {u : \|u\| ≥ 1} = (−∞, −1] ∪ [1, ∞) ∞ k=−∞ (2k + 1)π 2, (2k + 3)π 2 • The function G(t) = csc(t) = 1 sin(t) – has domain {t : t = πk, for integers k} = ∞ k=−∞ (kπ, (k + 1)π) – has range {u : \|u\| ≥ 1} = (−∞, −
1] ∪ [1, ∞) • The function J(t) = tan(t) = sin(t) cos(t) – has domain {t : t = π 2 + πk, for integers k} = ∞ k=−∞ (2k + 1)π 2, (2k + 3)π 2 – has range (−∞, ∞) • The function K(t) = cot(t) = cos(t) sin(t) – has domain {t : t = πk, for integers k} = – has range (−∞, ∞) ∞ k=−∞ (kπ, (k + 1)π) 8Using Theorem 2.4 from Section 2.4. 9Notice we have used the variable ‘u’ as the ‘dummy variable’ to describe the range elements. While there is no mathematical reason to do this (we are describing a set of real numbers, and, as such, could use t again) we choose u to help solidify the idea that these real numbers are the outputs from the inputs, which we have been calling t. 758 Foundations of Trigonometry We close this section with a few notes about solving equations which involve the circular functions. First, the discussion on page 735 in Section 10.2.1 concerning solving equations applies to all six circular functions, not just f (t) = cos(t) and g(t) = sin(t). In particular, to solve the equation cot(t) = −1 for real numbers t, we can use the same thought process we used in Example 10.3.2, number 3 to solve cot(θ) = −1 for angles θ in radian measure – we just need to remember to write our answers using the variable t as opposed to θ. Next, it is critical that you know the domains and ranges of the six circular functions so that you know which equations have no solutions. For example, sec(t) = 1 2 is not in the range of secant. Finally, you will need to review the notions of reference angles and coterminal angles so that you can see why csc(t) = −42 has an inﬁnite set of solutions in Quadrant III and another inﬁnite set of solutions in Quadrant IV. 2 has no solution
because 1 10.3 The Six Circular Functions and Fundamental Identities 759 10.3.2 Exercises In Exercises 1 - 20, ﬁnd the exact value or state that it is undeﬁned. 1. tan π 4 5. tan − 11π 6 9. tan (117π) 13. tan 17. tan 31π 2 2π 3 2. sec π 6 6. sec − 10. sec − 3π 2 5π 3 14. sec π 4 18. sec (−7π) 3. csc 7. csc 5π 6 − π 3 4. cot 8. cot 4π 3 13π 2 11. csc (3π) 12. cot (−5π) 15. csc − 7π 4 19. csc π 2 16. cot 20. cot 7π 6 3π 4 In Exercises 21 - 34, use the given the information to ﬁnd the exact values of the remaining circular functions of θ. 21. sin(θ) = 3 5 with θ in Quadrant II 22. tan(θ) = 12 5 with θ in Quadrant III with θ in Quadrant I 24. sec(θ) = 7 with θ in Quadrant IV 25. csc(θ) = − with θ in Quadrant III 26. cot(θ) = −23 with θ in Quadrant II 27. tan(θ) = −2 with θ in Quadrant IV. 28. sec(θ) = −4 with θ in Quadrant II. 29. cot(θ) = √ 5 with θ in Quadrant III. 30. cos(θ) = 1 3 with θ in Quadrant I. 31. cot(θ) = 2 with 0 < θ < π 2. 33. tan(θ) = √ 10 with π < θ < 3π 2. 32. csc(θ) = 5 with π 2 < θ < π. 34. sec(θ) = 2 √ 5 with 3π 2 < θ < 2π. In Exercises 35 - 42, use your calculator to approximate the given value to three decimal places. Make sure your calculator is in the proper angle measurement mode! 35. csc(78.
95◦) 36. tan(−2.01) 37. cot(392.994) 38. sec(207◦) 39. csc(5.902) 40. tan(39.672◦) 41. cot(3◦) 42. sec(0.45) 23. csc(θ) = 25 24 10 91 √ 91 760 Foundations of Trigonometry In Exercises 43 - 57, ﬁnd all of the angles which satisfy the equation. 43. tan(θ) = √ 3 44. sec(θ) = 2 45. csc(θ) = −1 46. cot(θ) = √ 3 3 47. tan(θ) = 0 48. sec(θ) = 1 49. csc(θ) = 2 50. cot(θ) = 0 51. tan(θ) = −1 52. sec(θ) = 0 53. csc(θ) = − 55. tan(θ) = − √ 3 56. csc(θ) = −2 57. cot(θ) = −1 1 2 54. sec(θ) = −1 In Exercises 58 - 65, solve the equation for t. Give exact values. 58. cot(t) = 1 59. tan(t) = 62. cot(t) = − √ 3 63. tan(t 60. sec(t) = − 64. sec(t) = √ 2 3 61. csc(t) = 0 65. csc(t) = √ 2 3 3 In Exercises 66 - 69, use Theorem 10.10 to ﬁnd the requested quantities. 66. Find θ, a, and c. 67. Find α, b, and c. 60◦ a c θ 9 b α c 12 34◦ 68. Find θ, a, and c. 69. Find β, b, and c. 47◦ c a θ 6 b β c 2.5 50◦ 10.3 The Six Circular Functions and Fundamental Identities 761 In Exercises 70 - 75, use Theorem 10.10 to answer the question. Assume that θ is an angle in a right triangle. 70. If θ = 30�
� and the side opposite θ has length 4, how long is the side adjacent to θ? 71. If θ = 15◦ and the hypotenuse has length 10, how long is the side opposite θ? 72. If θ = 87◦ and the side adjacent to θ has length 2, how long is the side opposite θ? 73. If θ = 38.2◦ and the side opposite θ has lengh 14, how long is the hypoteneuse? 74. If θ = 2.05◦ and the hypotenuse has length 3.98, how long is the side adjacent to θ? 75. If θ = 42◦ and the side adjacent to θ has length 31, how long is the side opposite θ? 76. A tree standing vertically on level ground casts a 120 foot long shadow. The angle of elevation from the end of the shadow to the top of the tree is 21.4◦. Find the height of the tree to the nearest foot. With the help of your classmates, research the term umbra versa and see what it has to do with the shadow in this problem. 77. The broadcast tower for radio station WSAZ (Home of “Algebra in the Morning with Carl and Jeﬀ”) has two enormous ﬂashing red lights on it: one at the very top and one a few feet below the top. From a point 5000 feet away from the base of the tower on level ground the angle of elevation to the top light is 7.970◦ and to the second light is 7.125◦. Find the distance between the lights to the nearest foot. 78. On page 753 we deﬁned the angle of inclination (also known as the angle of elevation) and in this exercise we introduce a related angle - the angle of depression (also known as the angle of declination). The angle of depression of an object refers to the angle whose initial side is a horizontal line above the object and whose terminal side is the line-of-sight to the object below the horizontal. This is represented schematically below. horizontal θ observer object The angle of depression from the horizontal to the object is θ (a) Show that if the horizontal is above and parallel to level ground then the angle of depression (from observer to object) and the angle of inclination (from object to observer) will be congruent
because they are alternate interior angles. 762 Foundations of Trigonometry (b) From a ﬁretower 200 feet above level ground in the Sasquatch National Forest, a ranger spots a ﬁre oﬀ in the distance. The angle of depression to the ﬁre is 2.5◦. How far away from the base of the tower is the ﬁre? (c) The ranger in part 78b sees a Sasquatch running directly from the ﬁre towards the ﬁretower. The ranger takes two sightings. At the ﬁrst sighting, the angle of depression from the tower to the Sasquatch is 6◦. The second sighting, taken just 10 seconds later, gives the the angle of depression as 6.5◦. How far did the Saquatch travel in those 10 seconds? Round your answer to the nearest foot. How fast is it running in miles per hour? Round your answer to the nearest mile per hour. If the Sasquatch keeps up this pace, how long will it take for the Sasquatch to reach the ﬁretower from his location at the second sighting? Round your answer to the nearest minute. 79. When I stand 30 feet away from a tree at home, the angle of elevation to the top of the tree is 50◦ and the angle of depression to the base of the tree is 10◦. What is the height of the tree? Round your answer to the nearest foot. 80. From the observation deck of the lighthouse at Sasquatch Point 50 feet above the surface of Lake Ippizuti, a lifeguard spots a boat out on the lake sailing directly toward the lighthouse. The ﬁrst sighting had an angle of depression of 8.2◦ and the second sighting had an angle of depression of 25.9◦. How far had the boat traveled between the sightings? 81. A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it makes a 43◦ angle with the ground. How tall is the tower? How far away from the base of the tower does the wire hit the ground? In Exercises 82 - 128, verify the identity. Assume that all quantities are deﬁned. 82. cos(θ) sec(θ) = 1 84. sin(θ) csc(θ) = 1 86
. csc(θ) cos(θ) = cot(θ) 88. 90. 92. cos(θ) sin2(θ) = csc(θ) cot(θ) 1 − cos(θ) sin(θ) = csc(θ) − cot(θ) sin(θ) 1 − cos2(θ) = csc(θ) 83. tan(θ) cos(θ) = sin(θ) 85. tan(θ) cot(θ) = 1 87. 89. 91. 93. sin(θ) cos2(θ) = sec(θ) tan(θ) 1 + sin(θ) cos(θ) = sec(θ) + tan(θ) cos(θ) 1 − sin2(θ) sec(θ) 1 + tan2(θ) = sec(θ) = cos(θ) 10.3 The Six Circular Functions and Fundamental Identities 763 94. 96. csc(θ) 1 + cot2(θ) cot(θ) csc2(θ) − 1 = sin(θ) = tan(θ) 95. tan(θ) sec2(θ) − 1 = cot(θ) 97. 4 cos2(θ) + 4 sin2(θ) = 4 98. 9 − cos2(θ) − sin2(θ) = 8 99. tan3(θ) = tan(θ) sec2(θ) − tan(θ) 100. sin5(θ) = 1 − cos2(θ)2 sin(θ) 101. sec10(θ) = 1 + tan2(θ)4 sec2(θ) 102. cos2(θ) tan3(θ) = tan(θ) − sin(θ) cos(θ) 103. sec4(θ) − sec2(θ) = tan2(θ) + tan4(θ) 104. 106. cos(θ) + 1 cos(θ) − 1 1 − cot(θ) 1 + cot(θ) = = 1 +
sec(θ) 1 − sec(θ) tan(θ) − 1 tan(θ) + 1 105. sin(θ) + 1 sin(θ) − 1 = 1 + csc(θ) 1 − csc(θ) 107. 1 − tan(θ) 1 + tan(θ) = cos(θ) − sin(θ) cos(θ) + sin(θ) 108. tan(θ) + cot(θ) = sec(θ) csc(θ) 109. csc(θ) − sin(θ) = cot(θ) cos(θ) 110. cos(θ) − sec(θ) = − tan(θ) sin(θ) 111. cos(θ)(tan(θ) + cot(θ)) = csc(θ) 112. sin(θ)(tan(θ) + cot(θ)) = sec(θ) 113. 1 1 − cos(θ) + 1 1 + cos(θ) = 2 csc2(θ) 114. 1 sec(θ) + 1 + 1 sec(θ) − 1 = 2 csc(θ) cot(θ) 115. 1 csc(θ) + 1 + 1 csc(θ) − 1 = 2 sec(θ) tan(θ) 116. 1 csc(θ) − cot(θ) − 1 csc(θ) + cot(θ) = 2 cot(θ) 117. cos(θ) 1 − tan(θ) + sin(θ) 1 − cot(θ) = sin(θ) + cos(θ) 118. 120. 122. 124. 126. 128. 1 sec(θ) + tan(θ) 1 csc(θ) − cot(θ) = sec(θ) − tan(θ) = csc(θ) + cot(θ) 1 1 − sin(θ) 1 1 − cos(θ) = sec2(θ) + sec(θ) tan(θ) = csc2(θ) + csc(θ
) cot(θ) cos(θ) 1 + sin(θ) = 1 − sin(θ) cos(θ) 1 − sin(θ) 1 + sin(θ) = (sec(θ) − tan(θ))2 119. 121. 123. 125. 1 sec(θ) − tan(θ) 1 csc(θ) + cot(θ) = sec(θ) + tan(θ) = csc(θ) − cot(θ) 1 1 + sin(θ) 1 1 + cos(θ) = sec2(θ) − sec(θ) tan(θ) = csc2(θ) − csc(θ) cot(θ) 127. csc(θ) − cot(θ) = sin(θ) 1 + cos(θ) 764 Foundations of Trigonometry In Exercises 129 - 132, verify the identity. You may need to consult Sections 2.2 and 6.2 for a review of the properties of absolute value and logarithms before proceeding. 129. ln \| sec(θ)\| = − ln \| cos(θ)\| 130. − ln \| csc(θ)\| = ln \| sin(θ)\| 131. − ln \| sec(θ) − tan(θ)\| = ln \| sec(θ) + tan(θ)\| 132. − ln \| csc(θ) + cot(θ)\| = ln \| csc(θ) − cot(θ)\| 133. Verify the domains and ranges of the tangent, cosecant and cotangent functions as presented in Theorem 10.11. 134. As we did in Exercise 74 in Section 10.2, let α and β be the two acute angles of a right triangle. (Thus α and β are complementary angles.) Show that sec(α) = csc(β) and tan(α) = cot(β). The fact that co-functions of complementary angles are equal in this case is not an accident and a more general result will be given in Section 10.4. 135. We wish to establish the inequality cos(θ) < < 1 for 0 < θ <. Use the
diagram from the beginning of the section, partially reproduced below, to answer the following. sin(θ(1, 0) x (a) Show that triangle OP B has area 1 2 sin(θ). (b) Show that the circular sector OP B with central angle θ has area (c) Show that triangle OQB has area 1 2 tan(θ). (d) Comparing areas, show that sin(θ) < θ < tan(θ) for 0 < θ < π 2. (e) Use the inequality sin(θ) < θ to show that sin(θ) θ < 1 for 0 < θ < 1 2 θ. π 2. (f) Use the inequality θ < tan(θ) to show that cos(θ) < with the previous part to complete the proof. sin(θ) θ for 0 < θ < π 2. Combine this 10.3 The Six Circular Functions and Fundamental Identities 765 136. Show that cos(θ) < sin(θ) θ < 1 also holds for − π 2 < θ < 0. 137. Explain why the fact that tan(θ) = 3 = 3 solution to number 6 in Example 10.3.1.) 1 does not mean sin(θ) = 3 and cos(θ) = 1? (See the 766 Foundations of Trigonometry 10.3.3 Answers 1. tan 4. cot π 4 4π 3 = 1 = 7. csc − π 3 = − = 2 √ 3 = 10. sec 13. tan 16. cot 19. csc − 5π 3 31π 2 7π 6 π 2 = 1 2. sec 5. tan 8. cot 3 = π 6 − 11π 6 13π. csc 5π 6 = 2 6. sec − 3π 2 is undeﬁned 9. tan (117π) = 0 11. csc (3π) is undeﬁned 12. cot (−5π) is undeﬁned √ 3 3 √ 2 3 is undeﬁned 14. sec 15. csc − √ 2 = 7π 4 18. sec (−7π) = −1 = π 4 2π 3 3π 4 √ 2 √ = − 3
= −1 17. tan 20. cot 21. sin(θ) = 3 5, cos(θ) = − 4 5, tan(θ) = − 3 4, csc(θ) = 5 3, sec(θ) = − 5 4, cot(θ) = − 4 3 22. sin(θ) = − 12 13, cos(θ) = − 5 13, tan(θ) = 12 5, csc(θ) = − 13 12, sec(θ) = − 13 5, cot(θ) = 5 12 23. sin(θ) = 24 25, cos(θ) = 7 25, tan(θ) = 24 24, sec(θ) = 25 7, cot(θ) = 7 24 7, csc(θ) = 25 √ 3, cos(θ) = 1 7, tan(θ) = −4 3, csc(θ) = − 7 √ 12, sec(θ) = 7, cot(θ) = − 3 √ 3 12 √ 24. sin(θ) = −4 7 √ 25. sin(θ) = − √ 26. sin(θ) = 91 10, cos(θ) = − 3 530, cos(θ) = − 23 530 10, tan(θ) = √ 91 3, csc(θ) = − 10 91 √ 530, tan(θ) = − 1 530 23, csc(θ) = 3, cot(θ) = 3, sec(θ) = − 10 √ 530 23, cot(θ) = −23 91 91 √ √ 91 √ 27. sin(θ) = − 2 √ 28. sin(θ) = √ 5 5, cos(θ) = 4, cos(θ) = − 1 15 29. sin(θ) = − √ 30. sin(θ) = 2 √ 31. sin(θ) = 32. sin(θ) = 1 2 √ 6 6, cos(θ) = − 3, cos(θ) = 1 5, cos(θ) = 2 5, cos(θ) = − 2
5 √ 5 5, tan(θ) = −2, csc(θ) = − √ 4, tan(θ) = − √ 30 6, tan(θ) = √ 15, csc(θ) = 4 √ 5 5, csc(θ) = − 530, sec(θ) = − √ 5, cot(θ) = − 1 2 √ 5 2, sec(θ) = √ 15 15, sec(θ) = −4, cot(θ) = − √ √ 30 5, cot(θ) = 6, sec(θ) = − √ 15 15 √ 5 3, tan(θ) = 2 √ 5, tan(θ) = 1 5 2, csc(θ) = 3 √ 2, csc(θ) = √ 2 4, sec(θ) = 3, cot(θ) = √ 5 2, cot(θ) = 2 5, sec(θ) = √ 2 4 √ 6 5, tan(θ) = − √ 6 12, csc(θ) = 5, sec(θ) = − 5 √ 12, cot(θ) = −2 6 √ 6 33. sin(θ) = − 34. sin(θ) = − √ 110 11, cos(θ) = − √ √ 95 5 10, tan(θ) = − 10, cos(θ) = √ 11 11, tan(θ) = √ √ 10, csc(θ) = − √ 95 19, sec(θ) = 2 √ 110 10, sec(θ) = − √ 19, csc(θ) = − 2 √ 11, cot(θ) = √ 5, cot(θ) = − 19 19 √ 10 10 10.3 The Six Circular Functions and Fundamental Identities 767 35. csc(78.95◦) ≈ 1.019 37. cot(392.994) ≈ 3.292 39. csc(5.902) ≈ −2.688 41. cot(3◦) ≈ 19.
081 43. tan(θ) = √ 3 when θ = π 3 36. tan(−2.01) ≈ 2.129 38. sec(207◦) ≈ −1.122 40. tan(39.672◦) ≈ 0.829 42. sec(0.45) ≈ 1.111 + πk for any integer k 44. sec(θ) = 2 when θ = π 3 + 2πk or θ = 5π 3 + 2πk for any integer k 45. csc(θ) = −1 when θ = 46. cot(θ) = √ 3 3 when θ = 3π 2 π 3 + 2πk for any integer k. + πk for any integer k 47. tan(θ) = 0 when θ = πk for any integer k 48. sec(θ) = 1 when θ = 2πk for any integer k 49. csc(θ) = 2 when θ = 50. cot(θ) = 0 when θ = π 6 π 2 51. tan(θ) = −1 when θ = + 2πk or θ = 5π 6 + 2πk for any integer k. + πk for any integer k 3π 4 + πk for any integer k 52. sec(θ) = 0 never happens 53. csc(θ) = − 1 2 never happens 54. sec(θ) = −1 when θ = π + 2πk = (2k + 1)π for any integer k 55. tan(θ) = − √ 3 when θ = 2π 3 + πk for any integer k 56. csc(θ) = −2 when θ = 57. cot(θ) = −1 when θ = 7π 6 3π 4 + 2πk or θ = 11π 6 + 2πk for any integer k + πk for any integer k 58. cot(t) = 1 when t = π 4 + πk for any integer k 59. tan(t) = √ 3 3 when t = π 6 + πk for any integer k 768 Foundations of Trigonometry 60. sec(t)
= − √ 2 3 3 when t = 5π 6 + 2πk or t = 7π 6 61. csc(t) = 0 never happens + 2πk for any integer k 62. cot(t) = − √ 3 when t = 5π 6 5π 6 + πk for any integer k + πk for any integer k when t = √ 3 3 63. tan(t) = − 64. sec(t) = 65. csc(t when t = π 6 + 2πk or t = 11π 6 + 2πk for any integer k when t = + 2πk or t = 2π 3 + 2πk for any integer k π 3 √ 66. θ = 30◦, a = 3 √ 3, c = √ 108 = 6 3 67. α = 56◦, b = 12 tan(34◦) = 8.094, c = 12 sec(34◦) = 12 cos(34◦) ≈ 14.475 68. θ = 43◦, a = 6 cot(47◦) = 6 tan(47◦) ≈ 5.595, c = 6 csc(47◦) = 6 sin(47◦) ≈ 8.204 69. β = 40◦, b = 2.5 tan(50◦) ≈ 2.979, c = 2.5 sec(50◦) = 2.5 cos(50◦) ≈ 3.889 70. The side adjacent to θ has length 4 √ 3 ≈ 6.928 71. The side opposite θ has length 10 sin(15◦) ≈ 2.588 72. The side opposite θ is 2 tan(87◦) ≈ 38.162 73. The hypoteneuse has length 14 csc(38.2◦) = 14 sin(38.2◦) ≈ 22.639 74. The side adjacent to θ has length 3.98 cos(2.05◦) ≈ 3.977 75. The side opposite θ has length 31 tan(42◦) ≈ 27.912 76. The tree is about 47 feet tall. 77. The lights are about 75 feet apart. 78. (b) The ﬁre is about 4
581 feet from the base of the tower. (c) The Sasquatch ran 200 cot(6◦) − 200 cot(6.5◦) ≈ 147 feet in those 10 seconds. This translates to ≈ 10 miles per hour. At the scene of the second sighting, the Sasquatch was ≈ 1755 feet from the tower, which means, if it keeps up this pace, it will reach the tower in about 2 minutes. 10.3 The Six Circular Functions and Fundamental Identities 769 79. The tree is about 41 feet tall. 80. The boat has traveled about 244 feet. 81. The tower is about 682 feet tall. The guy wire hits the ground about 731 feet away from the base of the tower. 770 Foundations of Trigonometry 10.4 Trigonometric Identities In Section 10.3, we saw the utility of the Pythagorean Identities in Theorem 10.8 along with the Quotient and Reciprocal Identities in Theorem 10.6. Not only did these identities help us compute the values of the circular functions for angles, they were also useful in simplifying expressions involving the circular functions. In this section, we introduce several collections of identities which have uses in this course and beyond. Our ﬁrst set of identities is the ‘Even / Odd’ identities.1 Theorem 10.12. Even / Odd Identities: For all applicable angles θ, cos(−θ) = cos(θ) sin(−θ) = − sin(θ) tan(−θ) = − tan(θ) sec(−θ) = sec(θ) csc(−θ) = − csc(θ) cot(−θ) = − cot(θ) In light of the Quotient and Reciprocal Identities, Theorem 10.6, it suﬃces to show cos(−θ) = cos(θ) and sin(−θ) = − sin(θ). The remaining four circular functions can be expressed in terms of cos(θ) and sin(θ) so the proofs of their Even / Odd Identities are left as exercises. Consider an angle θ plotted in standard position. Let θ0 be the angle coterminal with θ with 0 ≤ θ0 < 2π.
(We can construct the angle θ0 by rotating counter-clockwise from the positive x-axis to the terminal side of θ as pictured below.) Since θ and θ0 are coterminal, cos(θ) = cos(θ0) and sin(θ) = sin(θ0). y 1 θ0 y 1 θ0 P (cos(θ0), sin(θ0)) 1 x θ Q(cos(−θ0), sin(−θ0)) 1 x −θ0 We now consider the angles −θ and −θ0. Since θ is coterminal with θ0, there is some integer k so that θ = θ0 + 2π · k. Therefore, −θ = −θ0 − 2π · k = −θ0 + 2π · (−k). Since k is an integer, so is (−k), which means −θ is coterminal with −θ0. Hence, cos(−θ) = cos(−θ0) and sin(−θ) = sin(−θ0). Let P and Q denote the points on the terminal sides of θ0 and −θ0, respectively, which lie on the Unit Circle. By deﬁnition, the coordinates of P are (cos(θ0), sin(θ0)) and the coordinates of Q are (cos(−θ0), sin(−θ0)). Since θ0 and −θ0 sweep out congruent central sectors of the Unit Circle, it 1As mentioned at the end of Section 10.2, properties of the circular functions when thought of as functions of angles in radian measure hold equally well if we view these functions as functions of real numbers. Not surprisingly, the Even / Odd properties of the circular functions are so named because they identify cosine and secant as even functions, while the remaining four circular functions are odd. (See Section 1.6.) 10.4 Trigonometric Identities 771 follows that the points P and Q are symmetric about the x-axis. Thus, cos(−θ0) = cos(θ0) and sin(−θ0) = − sin(θ0). Since the cosines and sines of θ0 and −θ0 are the
same as those for θ and −θ, respectively, we get cos(−θ) = cos(θ) and sin(−θ) = − sin(θ), as required. The Even / Odd Identities are readily demonstrated using any of the ‘common angles’ noted in Section 10.2. Their true utility, however, lies not in computation, but in simplifying expressions involving the circular functions. In fact, our next batch of identities makes heavy use of the Even / Odd Identities. Theorem 10.13. Sum and Diﬀerence Identities for Cosine: For all angles α and β, cos(α + β) = cos(α) cos(β) − sin(α) sin(β) cos(α − β) = cos(α) cos(β) + sin(α) sin(β) We ﬁrst prove the result for diﬀerences. As in the proof of the Even / Odd Identities, we can reduce the proof for general angles α and β to angles α0 and β0, coterminal with α and β, respectively, each of which measure between 0 and 2π radians. Since α and α0 are coterminal, as are β and β0, it follows that α − β is coterminal with α0 − β0. Consider the case below where α0 ≥ β0. y P (cos(α0), sin(α0)) α0 − β0 Q(cos(β0), sin(β0)) y 1 A(cos(α0 − β0), sin(α0 − β0)) α0 β0 α0 − β0 O 1 x O B(1, 0) x Since the angles P OQ and AOB are congruent, the distance between P and Q is equal to the distance between A and B.2 The distance formula, Equation 1.1, yields (cos(α0) − cos(β0))2 + (sin(α0) − sin(β0))2 = (cos(α0 − β0) − 1)2 + (sin(α0 − β0) − 0)2 Squaring both sides, we expand the left hand side of this equation as (cos(α0) − cos(β0))2 + (sin(α0) − sin(β0))2 =
cos2(α0) − 2 cos(α0) cos(β0) + cos2(β0) + sin2(α0) − 2 sin(α0) sin(β0) + sin2(β0) = cos2(α0) + sin2(α0) + cos2(β0) + sin2(β0) −2 cos(α0) cos(β0) − 2 sin(α0) sin(β0) 2In the picture we’ve drawn, the triangles P OQ and AOB are congruent, which is even better. However, α0 − β0 could be 0 or it could be π, neither of which makes a triangle. It could also be larger than π, which makes a triangle, just not the one we’ve drawn. You should think about those three cases. 772 Foundations of Trigonometry From the Pythagorean Identities, cos2(α0) + sin2(α0) = 1 and cos2(β0) + sin2(β0) = 1, so (cos(α0) − cos(β0))2 + (sin(α0) − sin(β0))2 = 2 − 2 cos(α0) cos(β0) − 2 sin(α0) sin(β0) Turning our attention to the right hand side of our equation, we ﬁnd (cos(α0 − β0) − 1)2 + (sin(α0 − β0) − 0)2 = cos2(α0 − β0) − 2 cos(α0 − β0) + 1 + sin2(α0 − β0) = 1 + cos2(α0 − β0) + sin2(α0 − β0) − 2 cos(α0 − β0) Once again, we simplify cos2(α0 − β0) + sin2(α0 − β0) = 1, so that (cos(α0 − β0) − 1)2 + (sin(α0 − β0) − 0)2 = 2 − 2 cos(α0 − β0) Putting it all together, we get 2 − 2 cos(α0) cos(β0) − 2 sin(α0) sin(β0) = 2 − 2 cos(α0 − β0), which simpl
iﬁes to: cos(α0 − β0) = cos(α0) cos(β0) + sin(α0) sin(β0). Since α and α0, β and β0 and α − β and α0 − β0 are all coterminal pairs of angles, we have cos(α − β) = cos(α) cos(β) + sin(α) sin(β). For the case where α0 ≤ β0, we can apply the above argument to the angle β0 − α0 to obtain the identity cos(β0 − α0) = cos(β0) cos(α0) + sin(β0) sin(α0). Applying the Even Identity of cosine, we get cos(β0 − α0) = cos(−(α0 − β0)) = cos(α0 − β0), and we get the identity in this case, too. To get the sum identity for cosine, we use the diﬀerence formula along with the Even/Odd Identities cos(α + β) = cos(α − (−β)) = cos(α) cos(−β) + sin(α) sin(−β) = cos(α) cos(β) − sin(α) sin(β) We put these newfound identities to good use in the following example. Example 10.4.1. 1. Find the exact value of cos (15◦). 2. Verify the identity: cos π 2 − θ = sin(θ). Solution. 1. In order to use Theorem 10.13 to ﬁnd cos (15◦), we need to write 15◦ as a sum or diﬀerence of angles whose cosines and sines we know. One way to do so is to write 15◦ = 45◦ − 30◦. cos (15◦) = cos (45◦ − 30◦) = cos (45◦) cos (30◦) + sin (45◦) sin (30◦) √ √ √ 10.4 Trigonometric Identities 773 2. In a straightforward application of Theorem 10.13, we ﬁnd cos − θ π 2 = cos π 2 π 2 = (0) (cos(θ)) + (1) (sin
(θ)) cos (θ) + sin sin (θ) = sin(θ) The identity veriﬁed in Example 10.4.1, namely, cos π 2 − θ = sin(θ), is the ﬁrst of the celebrated ‘cofunction’ identities. These identities were ﬁrst hinted at in Exercise 74 in Section 10.2. From sin(θ) = cos π 2 − θ, we get: π 2 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. = cos(θ), π 2 π 2 = cos − θ − θ sin − Theorem 10.14. Cofunction Identities: For all applicable angles θ, cos sin − θ − θ π 2 π 2 = sin(θ) = cos(θ) sec csc − θ − θ π 2 π 2 = csc(θ) = sec(θ) tan cot − θ − θ π 2 π 2 = cot(θ) = tan(θ) With the Cofunction Identities in place, we are now in the position to derive the sum and diﬀerence formulas for sine. To derive the sum formula for sine, we convert to cosines using a cofunction identity, then expand using the diﬀerence formula for cosine sin(α + β) = cos = cos − (α + β = cos π 2 = sin(α) cos(β) + cos(α) sin(β) cos(β) + sin − α − α sin(β) We can derive the diﬀerence formula for sine by rewriting sin(α − β) as sin(α + (−β)) and using the sum formula and the Even / Odd Identities. Again, we leave the details to the reader. Theorem 10.15. Sum and Diﬀerence Identities for Sine: For all angles α and β, sin(α + β) = sin(α) cos(β) + cos(α) sin
(β) sin(α − β) = sin(α) cos(β) − cos(α) sin(β) 774 Example 10.4.2. 1. Find the exact value of sin 19π 12 Foundations of Trigonometry 2. If α is a Quadrant II angle with sin(α) = 5 13, and β is a Quadrant III angle with tan(β) = 2, ﬁnd sin(α − β). 3. Derive a formula for tan(α + β) in terms of tan(α) and tan(β). Solution. 1. As in Example 10.4.1, we need to write the angle 19π 12 as a sum or diﬀerence of common angles. The denominator of 12 suggests a combination of angles with denominators 3 and 4. One such combination is 19π 4. Applying Theorem 10.15, we get 12 = 4π 3 + π 19π 12 sin = sin 4π 3 4π 3 √ 3 2 π 4 + cos √ π 4 2 2 √ 2 6 − 4 = sin − √ − = = π 4 + cos + − 1 2 4π 3 √ 2 2 sin 13 2 = 1, or cos(α) = ± 12 2. In order to ﬁnd sin(α − β) using Theorem 10.15, we need to ﬁnd cos(α) and both cos(β) and sin(β). To ﬁnd cos(α), we use the Pythagorean Identity cos2(α) + sin2(α) = 1. Since 13, we have cos2(α) + 5 sin(α) = 5 13. Since α is a Quadrant II angle, cos(α) = − 12 13. We now set about ﬁnding cos(β) and sin(β). We have several ways to proceed, but the Pythagorean Identity 1 + tan2(β) = sec2(β) is a quick way to get sec(β), and hence, √ cos(β). With tan(β) = 2, we get 1 + 22 = sec2(β) so that sec(β) = ± 5. Since β is a √ 5 = − Quadrant III angle, we choose sec(β) = − 5. We now need to determine sin(β). We
could use The Pythagorean Identity cos2(β) + sin2(β) = 1, but we opt instead to use a quotient identity. From tan(β) = sin(β) cos(β), we have sin(β) = tan(β) cos(β) √ 5 5. We now have all the pieces needed to ﬁnd sin(α − β): so we get sin(β) = (2) 5 so cos(β) = 1 sec(β sin(α − β) = sin(α) cos(β) − cos(α) sin(β) √ 5 5 − − 12 13 √ 2 5 5 − = 5 13 29 − 5 √ = − 65 10.4 Trigonometric Identities 775 3. We can start expanding tan(α + β) using a quotient identity and our sum formulas tan(α + β) = = sin(α + β) cos(α + β) sin(α) cos(β) + cos(α) sin(β) cos(α) cos(β) − sin(α) sin(β) Since tan(α) = sin(α) cos(α) and tan(β) = sin(β) denominator by cos(α) cos(β) we will have what we want cos(β), it looks as though if we divide both numerator and tan(α + β) = sin(α) cos(β) + cos(α) sin(β) cos(α) cos(β) − sin(α) sin(β) · 1 cos(α) cos(β) 1 cos(α) cos(β) sin(α) cos(β) cos(α) cos(β) cos(α) cos(β) cos(α) cos(β) sin(α) cos(β) cos(α) cos(β) cos(α) cos(β) cos(α) cos(β) + − + − cos(α) sin(β) cos(α) cos(β) sin(α) sin(β) cos(α) cos(β) cos(α) sin(β) cos(α) cos(β) sin(α) sin(β) cos(α) cos(β) tan(α) + tan(β) 1 − tan(α) tan(β) = =
= Naturally, this formula is limited to those cases where all of the tangents are deﬁned. The formula developed in Exercise 10.4.2 for tan(α +β) can be used to ﬁnd a formula for tan(α −β) by rewriting the diﬀerence as a sum, tan(α+(−β)), and the reader is encouraged to ﬁll in the details. Below we summarize all of the sum and diﬀerence formulas for cosine, sine and tangent. Theorem 10.16. Sum and Diﬀerence Identities: For all applicable angles α and β, cos(α ± β) = cos(α) cos(β) ∓ sin(α) sin(β) sin(α ± β) = sin(α) cos(β) ± cos(α) sin(β) tan(α ± β) = tan(α) ± tan(β) 1 ∓ tan(α) tan(β) In the statement of Theorem 10.16, we have combined the cases for the sum ‘+’ and diﬀerence ‘−’ of angles into one formula. The convention here is that if you want the formula for the sum ‘+’ of 776 Foundations of Trigonometry two angles, you use the top sign in the formula; for the diﬀerence, ‘−’, use the bottom sign. For example, tan(α − β) = tan(α) − tan(β) 1 + tan(α) tan(β) If we specialize the sum formulas in Theorem 10.16 to the case when α = β, we obtain the following ‘Double Angle’ Identities. Theorem 10.17. Double Angle Identities: For all applicable angles θ, cos(2θ) =    cos2(θ) − sin2(θ) 2 cos2(θ) − 1 1 − 2 sin2(θ) sin(2θ) = 2 sin(θ) cos(θ) tan(2θ) = 2 tan(θ) 1 − tan2(θ) The three diﬀerent forms for cos(2θ) can be explained by our ability to ‘exchange�
� squares of cosine and sine via the Pythagorean Identity cos2(θ) + sin2(θ) = 1 and we leave the details to the reader. It is interesting to note that to determine the value of cos(2θ), only one piece of information is required: either cos(θ) or sin(θ). To determine sin(2θ), however, it appears that we must know both sin(θ) and cos(θ). In the next example, we show how we can ﬁnd sin(2θ) knowing just one piece of information, namely tan(θ). Example 10.4.3. 1. Suppose P (−3, 4) lies on the terminal side of θ when θ is plotted in standard position. Find cos(2θ) and sin(2θ) and determine the quadrant in which the terminal side of the angle 2θ lies when it is plotted in standard position. 2. If sin(θ) = x for − π 2 ≤ θ ≤ π 2, ﬁnd an expression for sin(2θ) in terms of x. 3. Verify the identity: sin(2θ) = 2 tan(θ) 1 + tan2(θ). 4. Express cos(3θ) as a polynomial in terms of cos(θ). Solution. 1. Using Theorem 10.3 from Section 10.2 with x = −3 and y = 4, we ﬁnd r = x2 + y2 = 5. Hence, cos(θ) = − 3 5. Applying Theorem 10.17, we get cos(2θ) = cos2(θ) − sin2(θ) = − 3 25, and sin(2θ) = 2 sin(θ) cos(θ) = 2 4 2 − 4 25. Since both 5 5 cosine and sine of 2θ are negative, the terminal side of 2θ, when plotted in standard position, lies in Quadrant III. 5 and sin(θ) = 4 = − 7 = − 24 − 3 5 2 5 10.4 Trigonometric Identities 777 2. If your ﬁrst reaction to ‘sin(θ) = x’ is ‘No it’s not, cos
(θ) = x!’ then you have indeed learned something, and we take comfort in that. However, context is everything. Here, ‘x’ is just a variable - it does not necessarily represent the x-coordinate of the point on The Unit Circle which lies on the terminal side of θ, assuming θ is drawn in standard position. Here, x represents the quantity sin(θ), and what we wish to know is how to express sin(2θ) in terms of x. We will see more of this kind of thing in Section 10.6, and, as usual, this is something we need for Calculus. Since sin(2θ) = 2 sin(θ) cos(θ), we need to write cos(θ) in terms of x to ﬁnish the problem. We substitute x = sin(θ) into the Pythagorean Identity, cos2(θ) + sin2(θ) = 1, 2 ≤ θ ≤ π to get cos2(θ) + x2 = 1, or cos(θ) = ± 2, cos(θ) ≥ 0, and thus √ 1 − x2. cos(θ) = 1 − x2. Our ﬁnal answer is sin(2θ) = 2 sin(θ) cos(θ) = 2x 1 − x2. Since − π √ √ 3. We start with the right hand side of the identity and note that 1 + tan2(θ) = sec2(θ). From this point, we use the Reciprocal and Quotient Identities to rewrite tan(θ) and sec(θ) in terms of cos(θ) and sin(θ): 2 tan(θ) 1 + tan2(θ) = 2 tan(θ) sec2(θ) = = 2 sin(θ) cos(θ) 2 sin(θ) cos(θ) 1 cos2(θ) = 2 sin(θ) cos(θ) cos2(θ) cos(θ) cos(θ) = 2 sin(θ) cos(θ) = sin(2θ) 4. In Theorem 10.17, one of the formulas for cos(2θ), namely
cos(2θ) = 2 cos2(θ) − 1, expresses cos(2θ) as a polynomial in terms of cos(θ). We are now asked to ﬁnd such an identity for cos(3θ). Using the sum formula for cosine, we begin with cos(3θ) = cos(2θ + θ) = cos(2θ) cos(θ) − sin(2θ) sin(θ) Our ultimate goal is to express the right hand side in terms of cos(θ) only. We substitute cos(2θ) = 2 cos2(θ) − 1 and sin(2θ) = 2 sin(θ) cos(θ) which yields cos(3θ) = cos(2θ) cos(θ) − sin(2θ) sin(θ) = 2 cos2(θ) − 1 cos(θ) − (2 sin(θ) cos(θ)) sin(θ) = 2 cos3(θ) − cos(θ) − 2 sin2(θ) cos(θ) Finally, we exchange sin2(θ) for 1 − cos2(θ) courtesy of the Pythagorean Identity, and get cos(3θ) = 2 cos3(θ) − cos(θ) − 2 sin2(θ) cos(θ) = 2 cos3(θ) − cos(θ) − 2 1 − cos2(θ) cos(θ) = 2 cos3(θ) − cos(θ) − 2 cos(θ) + 2 cos3(θ) = 4 cos3(θ) − 3 cos(θ) and we are done. 778 Foundations of Trigonometry In the last problem in Example 10.4.3, we saw how we could rewrite cos(3θ) as sums of powers of cos(θ). In Calculus, we have occasion to do the reverse; that is, reduce the power of cosine and sine. Solving the identity cos(2θ) = 2 cos2(θ) − 1 for cos2(θ) and the identity cos(2θ) = 1 − 2 sin2(θ) for sin
2(θ) results in the aptly-named ‘Power Reduction’ formulas below. Theorem 10.18. Power Reduction Formulas: For all angles θ, cos2(θ) = sin2(θ) = 1 + cos(2θ) 2 1 − cos(2θ) 2 Example 10.4.4. Rewrite sin2(θ) cos2(θ) as a sum and diﬀerence of cosines to the ﬁrst power. Solution. We begin with a straightforward application of Theorem 10.18 1 − cos(2θ) 2 1 − cos2(2θ) 1 + cos(2θ) 2 sin2(θ) cos2(θ cos2(2θ) Next, we apply the power reduction formula to cos2(2θ) to ﬁnish the reduction sin2(θ) cos2(θ cos2(2θ) 1 + cos(2(2θ)) 2 − 1 8 cos(4θ) cos(4θ) Another application of the Power Reduction Formulas is the Half Angle Formulas. To start, we apply the Power Reduction Formula to cos2 θ 2 cos2 θ 2 We can obtain a formula for cos θ by extracting square roots. In a similar fashion, we may obtain 2 a half angle formula for sine, and by using a quotient formula, obtain a half angle formula for tangent. We summarize these formulas below. 1 + cos(θ) 2 1 + cos 2 θ 2 2 = =. 10.4 Trigonometric Identities 779 Theorem 10.19. Half Angle Formulas: For all applicable angles θ, cos sin θ 2 θ 2 = ± 1 + cos(θ) 2 = ± 1 − cos(θ) 2 tan = ± θ 2 1 − cos(θ) 1 + cos(θ) where the choice of ± depends on the quadrant in which the terminal side of θ 2 lies. Example 10.4.5. 1. Use a half angle formula to ﬁnd the exact value of cos (15◦). 2. Suppose −π ≤ θ ≤ 0 with cos(θ) = − 3 5. Find sin θ 2. 3. Use the identity given in number 3 of Example 10
.4.3 to derive the identity tan θ 2 = sin(θ) 1 + cos(θ) Solution. 1. To use the half angle formula, we note that 15◦ = 30◦ 2 and since 15◦ is a Quadrant I angle, its cosine is positive. Thus we have cos (15◦) = + 1 + cos (30◦) = = Back in Example 10.4.1, we found cos (15◦) by using the diﬀerence formula for cosine. In that case, we determined cos (15◦) =. The reader is encouraged to prove that these two expressions are equal. 6+ 4 √ √ 2 2. If −π ≤ θ ≤ 0, then − π < 0. Theorem 10.19 gives 2 sin 2 ≤ 0, which means sin θ 2 ≤ θ θ 1 − cos (θ 10 √ 2 5 5 780 Foundations of Trigonometry 3. Instead of our usual approach to verifying identities, namely starting with one side of the equation and trying to transform it into the other, we will start with the identity we proved in number 3 of Example 10.4.3 and manipulate it into the identity we are asked to prove. The identity we are asked to start with is sin(2θ) = 2 tan(θ) 1+tan2(θ). If we are to use this to derive an identity for tan θ, it seems reasonable to proceed by replacing each occurrence of θ with θ 2 2 2 tan θ 2 1 + tan2 θ 2 2 tan θ 2 1 + tan2 θ 2 sin 2 θ 2 sin(θ) = = We now have the sin(θ) we need, but we somehow need to get a factor of 1 + cos(θ) involved.. We continue to manipulate our To get cosines involved, recall that 1 + tan2 θ 2 given identity by converting secants to cosines and using a power reduction formula = sec2 θ 2 sin(θ) = 2 tan θ 2 1 + tan2 θ 2 2 tan θ 2 sec2 θ 2 sin(θ) = 2 tan θ 2 sin(θ) = sin(θ) = 2 tan θ 2 sin(θ) = tan θ 2 θ sin(
θ) 1 + cos(θ) 2 = tan cos2 θ 2 1 + cos 2 θ 2 2 (1 + cos(θ)) Our next batch of identities, the Product to Sum Formulas,3 are easily veriﬁed by expanding each of the right hand sides in accordance with Theorem 10.16 and as you should expect by now we leave the details as exercises. They are of particular use in Calculus, and we list them here for reference. Theorem 10.20. Product to Sum Formulas: For all angles α and β, cos(α) cos(β) = 1 2 [cos(α − β) + cos(α + β)] sin(α) sin(β) = 1 2 [cos(α − β) − cos(α + β)] sin(α) cos(β) = 1 2 [sin(α − β) + sin(α + β)] 3These are also known as the Prosthaphaeresis Formulas and have a rich history. The authors recommend that you conduct some research on them as your schedule allows. 10.4 Trigonometric Identities 781 Related to the Product to Sum Formulas are the Sum to Product Formulas, which we will have need of in Section 10.7. These are easily veriﬁed using the Product to Sum Formulas, and as such, their proofs are left as exercises. Theorem 10.21. Sum to Product Formulas: For all angles α and β, cos(α) + cos(β) = 2 cos cos(α) − cos(β) = −2 sin sin(α) ± sin(β) = 2 sin Example 10.4.6 cos sin cos 1. Write cos(2θ) cos(6θ) as a sum. 2. Write sin(θ) − sin(3θ) as a product. Solution. 1. Identifying α = 2θ and β = 6θ, we ﬁnd cos(2θ) cos(6θ) = 1 = 1 = 1 2 [cos(2θ − 6θ) + cos(2θ + 6θ)] 2 cos(−4θ) + 1 2 cos(4θ) + 1 2 cos(8θ) 2 cos(8θ), where the last equality is courtesy of the even identity for
cosine, cos(−4θ) = cos(4θ). 2. Identifying α = θ and β = 3θ yields sin(θ) − sin(3θ) = 2 sin θ − 3θ 2 cos θ + 3θ 2 = 2 sin (−θ) cos (2θ) = −2 sin (θ) cos (2θ), where the last equality is courtesy of the odd identity for sine, sin(−θ) = − sin(θ). The reader is reminded that all of the identities presented in this section which regard the circular functions as functions of angles (in radian measure) apply equally well to the circular (trigonometric) functions regarded as functions of real numbers. In Exercises 38 - 43 in Section 10.5, we see how some of these identities manifest themselves geometrically as we study the graphs of the these functions. In the upcoming Exercises, however, you need to do all of your work analytically without graphs. 782 Foundations of Trigonometry 10.4.1 Exercises In Exercises 1 - 6, use the Even / Odd Identities to verify the identity. Assume all quantities are deﬁned. 1. sin(3π − 2θ) = − sin(2θ − 3π) 2. cos − π 4 − 5t = cos 5t + π 4 3. tan(−t2 + 1) = − tan(t2 − 1) 4. csc(−θ − 5) = − csc(θ + 5) 5. sec(−6t) = sec(6t) 6. cot(9 − 7θ) = − cot(7θ − 9) In Exercises 7 - 21, use the Sum and Diﬀerence Identities to ﬁnd the exact value. You may have need of the Quotient, Reciprocal or Even / Odd Identities as well. 7. cos(75◦) 10. csc(195◦) 8. sec(165◦) 11. cot(255◦) 13. cos 16. cos 19. cot 13π 12 7π 12 11π 12 14. sin 17. tan 20. csc 11π 12 17π 12 5π 12 9. sin
(105◦) 12. tan(375◦) 15. tan 13π 12 18. sin π 12 21. sec − π 12 22. If α is a Quadrant IV angle with cos(α) = √ 5 5, and sin(β) = √ 10 10, where π 2 < β < π, ﬁnd (a) cos(α + β) (b) sin(α + β) (c) tan(α + β) (d) cos(α − β) (e) sin(α − β) (f) tan(α − β) 23. If csc(α) = 3, where 0 < α < π 2, and β is a Quadrant II angle with tan(β) = −7, ﬁnd (a) cos(α + β) (b) sin(α + β) (c) tan(α + β) (d) cos(α − β) (e) sin(α − β) (f) tan(α − β) 24. If sin(α) = 3 5, where 0 < α < π 2, and cos(β) = 12 13 where 3π 2 < β < 2π, ﬁnd (a) sin(α + β) (b) cos(α − β) (c) tan(α − β) 10.4 Trigonometric Identities 783 25. If sec(α) = − 5 3, where π 2 < α < π, and tan(β) = 24 7, where π < β < 3π 2, ﬁnd (a) csc(α − β) (b) sec(α + β) (c) cot(α + β) In Exercises 26 - 38, verify the identity. 26. cos(θ − π) = − cos(θ) 27. sin(π − θ) = sin(θ) 28. tan θ + π 2 = − cot(θ) 29. sin(α + β) + sin(α − β) = 2 sin(α) cos(β) 30. sin(α + β) − sin(α − β) = 2 cos(α) sin(β) 31. cos(α + β) + cos(α − β) = 2 cos(α
) cos(β) 32. cos(α + β) − cos(α − β) = −2 sin(α) sin(β) 33. sin(α + β) sin(α − β) = 1 + cot(α) tan(β) 1 − cot(α) tan(β) 34. 36. 37. cos(α + β) cos(α − β) = 1 − tan(α) tan(β) 1 + tan(α) tan(β) 35. tan(α + β) tan(α − β) = sin(α) cos(α) + sin(β) cos(β) sin(α) cos(α) − sin(β) cos(β) sin(t + h) − sin(t) h = cos(t) sin(h) h + sin(t) cos(h) − 1 h cos(t + h) − cos(t) h = cos(t) cos(h) − 1 h − sin(t) sin(h) h 38. tan(t + h) − tan(t) h = tan(h) h sec2(t) 1 − tan(t) tan(h) In Exercises 39 - 48, use the Half Angle Formulas to ﬁnd the exact value. You may have need of the Quotient, Reciprocal or Even / Odd Identities as well. 39. cos(75◦) (compare with Exercise 7) 40. sin(105◦) (compare with Exercise 9) 41. cos(67.5◦) 43. tan(112.5◦) 45. sin 47. sin π 12 5π 8 42. sin(157.5◦) 44. cos (compare with Exercise 16) 7π 12 π 8 7π 8 (compare with Exercise 18) 46. cos 48. tan 784 Foundations of Trigonometry In Exercises 49 - 58, use the given information about θ to ﬁnd the exact values of sin(2θ) θ 2 sin cos(2θ) θ 2 cos tan(2θ) θ 2 tan 49. sin(θ) = − 7 25 where 3π 2 < θ < 2π 50. cos(θ) = 28 53 where 0 < θ < π 2 51
. tan(θ) = 53. cos(θ) = 55. cos(θ) = 57. sec(θ) = 12 5 3 5 12 13 √ where π < θ < 3π 2 where 0 < θ < π 2 52. csc(θ) = 4 where π 2 < θ < π 54. sin(θ) = − 4 5 where π < θ < 3π 2 where 5 where 3π 2 3π 2 < θ < 2π 56. sin(θ) = 5 13 where < θ < 2π 58. tan(θ) = −2 where In Exercises 59 - 73, verify the identity. Assume all quantities are deﬁned. 59. (cos(θ) + sin(θ))2 = 1 + sin(2θ) 60. (cos(θ) − sin(θ))2 = 1 − sin(2θ) 61. tan(2θ) = 1 1 − tan(θ) − 1 1 + tan(θ) 62. csc(2θ) = cot(θ) + tan(θ) 2 63. 8 sin4(θ) = cos(4θ) − 4 cos(2θ) + 3 64. 8 cos4(θ) = cos(4θ) + 4 cos(2θ) + 3 65. sin(3θ) = 3 sin(θ) − 4 sin3(θ) 66. sin(4θ) = 4 sin(θ) cos3(θ) − 4 sin3(θ) cos(θ) 67. 32 sin2(θ) cos4(θ) = 2 + cos(2θ) − 2 cos(4θ) − cos(6θ) 68. 32 sin4(θ) cos2(θ) = 2 − cos(2θ) − 2 cos(4θ) + cos(6θ) 69. cos(4θ) = 8 cos4(θ) − 8 cos2(θ) + 1 70. cos(8θ) = 128 cos8(θ) − 256 cos6(θ) + 160 cos4(θ) − 32 cos2
(θ) + 1 (HINT: Use the result to 69.) 71. sec(2θ) = cos(θ) cos(θ) + sin(θ) + sin(θ) cos(θ) − sin(θ) 72. 73. 1 cos(θ) − sin(θ) 1 cos(θ) − sin(θ) + − 1 cos(θ) + sin(θ) 1 cos(θ) + sin(θ) = = 2 cos(θ) cos(2θ) 2 sin(θ) cos(2θ) 10.4 Trigonometric Identities 785 In Exercises 74 - 79, write the given product as a sum. You may need to use an Even/Odd Identity. 74. cos(3θ) cos(5θ) 75. sin(2θ) sin(7θ) 76. sin(9θ) cos(θ) 77. cos(2θ) cos(6θ) 78. sin(3θ) sin(2θ) 79. cos(θ) sin(3θ) In Exercises 80 - 85, write the given sum as a product. You may need to use an Even/Odd or Cofunction Identity. 80. cos(3θ) + cos(5θ) 81. sin(2θ) − sin(7θ) 82. cos(5θ) − cos(6θ) 83. sin(9θ) − sin(−θ) 84. sin(θ) + cos(θ) 85. cos(θ) − sin(θ) 86. Suppose θ is a Quadrant I angle with sin(θ) = x. Verify the following formulas (a) cos(θ) = √ 1 − x2 (b) sin(2θ) = 2x √ 1 − x2 (c) cos(2θ) = 1 − 2x2 87. Discuss with your classmates how each of the formulas, if any, in Exercise 86 change if we change assume θ is a Quadrant II, III, or IV angle. 88. Suppose θ is a Quadrant I angle with tan(θ) = x. Verify the following formulas (a
) cos(θ) = √ 1 x2 + 1 (c) sin(2θ) = 2x x2 + 1 (b) sin(θ) = √ x x2 + 1 (d) cos(2θ) = 1 − x2 x2 + 1 89. Discuss with your classmates how each of the formulas, if any, in Exercise 88 change if we change assume θ is a Quadrant II, III, or IV angle. for − for − 90. If sin(θ) = 91. If tan(θ) = 92. If sec(θ) = x 2 x 7 x 4, ﬁnd an expression for cos(2θ) in terms of x., ﬁnd an expression for sin(2θ) in terms of x. for 0 < θ < π 2, ﬁnd an expression for ln \| sec(θ) + tan(θ)\| in terms of x. 93. Show that cos2(θ) − sin2(θ) = 2 cos2(θ) − 1 = 1 − 2 sin2(θ) for all θ. 94. Let θ be a Quadrant III angle with cos(θ) = −. Show that this is not enough information to 1 5 by ﬁrst assuming 3π < θ < 7π 2 and then assuming π < θ < 3π 2 determine the sign of sin θ 2 and computing sin in both cases. θ 2 786 Foundations of Trigonometry 95. Without using your calculator, show that 96. In part 4 of Example 10.4.3, we wrote cos(3θ) as a polynomial in terms of cos(θ). In Exercise 69, we had you verify an identity which expresses cos(4θ) as a polynomial in terms of cos(θ). Can you ﬁnd a polynomial in terms of cos(θ) for cos(5θ)? cos(6θ)? Can you ﬁnd a pattern so that cos(nθ) could be written as a polynomial in cosine for any natural number n? 97. In Exercise 65, we has you verify an identity which expresses sin(3θ) as a polynomial in terms of sin(θ). Can you
do the same for sin(5θ)? What about for sin(4θ)? If not, what goes wrong? 98. Verify the Even / Odd Identities for tangent, secant, cosecant and cotangent. 99. Verify the Cofunction Identities for tangent, secant, cosecant and cotangent. 100. Verify the Diﬀerence Identities for sine and tangent. 101. Verify the Product to Sum Identities. 102. Verify the Sum to Product Identities. 10.4 Trigonometric Identities 787 = 2 − √ 3 12. tan(375◦) = 10.4.2 Answers 7. cos(75◦) = √ 2 √ 6 − 4 9. sin(105◦) = √ 2 √ 6 + 4 11. cot(255◦) = √ √ 13. cos 15. tan 17. tan 19. cot 21. sec 13π 12 13π 12 17π 12 11π 12 − π 12 = 2 + √ 3 = −(2 + √ 3) √ √ 2 6 − = 18. sin 20. csc π 12 5π 12 8. sec(165◦) = − √ 4 2 + √ 6 √ √ 6 2 − = 10. csc(195◦) = √ √ √ = −( √ 6 + √ 14. sin 11π 12 = 16. cos √ 7π 12 = √ − = 22. (a) cos(α + β) = − √ 2 10 (b) sin(α + β) = √ 2 7 10 √ 2 2 (c) tan(α + β) = −7 (d) cos(α − β) = − (e) sin(α − β) = √ 2 2 23. (a) cos(α + β) = − √ 4 + 7 30 2 (c) tan(α + β) = (e) sin(α − β28 + 4 + 7 28 + 30 (f) tan(α − β) = −1 63 − 100 √ 2 41 = (b) sin(α + β) = (d) cos(α − β) = (f) tan(α − β) = √ 2 √ 28 −
30 2 −4 + 7 30 √ √ 28 + 4 − 7 2 2 = − 24. (a) sin(α + β) = 16 65 (b) cos(α − β) = 33 65 (c) tan(α − β) = 63 + 100 √ 2 41 56 33 788 Foundations of Trigonometry 25. (a) csc(α − β) = − 5 4 (b) sec(α + β) = 125 117 (c) cot(α + β) = 117 44 39. cos(75◦) = 41. cos(67.5◦) = √ 40. sin(105◦) = 42. sin(157.5◦) = √ 1 − √ 2 44. cos 7π 12 = − √ 3 2 − 2 46. cos 48. tan 2 + 2 = − π 8 = 7π 43. tan(112.5◦) = − 45. sin 47. sin π 12 5π 49. sin(2θ) = − √ sin θ 2 = 50. sin(2θ) = sin θ 2 = 51. sin(2θ) = sin θ 2 = 52. sin(2θ) = − 336 625 2 10 2520 2809 √ 5 106 106 120 169 √ 3 13 13 √ 15 8 cos(2θ) = cos θ 2 = − cos(2θ) = − cos θ 2 = 9 527 625 √ 2 7 10 1241 2809 √ 106 106 119 169 √ 2 13 13 cos(2θ) = − cos θ 2 = − cos(2θ) = 7 8 sin θ 2 = √ 15 8 + 2 4 cos θ 2 = √ 15 8 − 2 4 tan(2θ) = − tan θ 2 = − tan(2θ) = − tan θ 2 = 5 9 tan(2θ) = − tan θ 2 = − 336 527 1 7 2520 1241 120 119 3 2 √ 15 7 tan(2θ) = − tan θ 2 tan √ 15 √ √ 15 15 53. sin(2θ) = sin θ 2 = 24 25 √ 5 5 cos(2θ) = − cos θ 2 = 2 7 25
√ 5 5 24 7 tan(2θ) = − tan θ 2 = 1 2 10.4 Trigonometric Identities 789 54. sin(2θ) = sin θ 2 = 24 25 √ 2 5 sin(2θ) = − √ sin θ 2 = sin(2θ) = − sin θ 2 = 5 5 120 169 26 26 120 169 √ 26 26 4 5 55. 56. 57. cos(2θ) = − cos θ 2 = − cos(2θ) = cos θ 2 = − 26 26 cos(2θ) = cos θ 2 = 7 25 √ 5 5 119 169 5 √ 119 169 √ 26 26 3 5 tan(2θ) = − 24 7 = −2 tan θ 2 120 119 1 5 120 119 tan(2θ) = − tan θ 2 = − tan(2θ) = − tan θ 2 = 5 tan(2θ) = 4 3 tan θ 2 = − sin(2θ) = − cos(2θ) = − sin θ 2 = 50 − 10 10 √ 5 cos θ 2 = − √ 5 50 + 10 10 58. sin(2θ) = − 4 5 cos(2θ) = − 3 5 sin θ 2 = √ 5 50 + 10 10 cos θ 2 = 50 − 10 10 √ tan θ 2 = tan(2θ) = 5 tan θ 2 = tan 10 10 74. 77. cos(2θ) + cos(8θ) 2 cos(4θ) + cos(8θ) 2 80. 2 cos(4θ) cos(θ) 83. 2 cos(4θ) sin(5θ) 90. 1 − x2 2 75. 78. cos(5θ) − cos(9θ) 2 cos(θ) − cos(5θ) 2 9 2 sin θ 81. −2 cos θ 5 2 76. 79. sin(8θ) + sin(10θ) 2 sin(2θ) + sin(4θ) 2 11 2 sin θ 1 2 θ 82. 2 sin √ θ − 2 cos 84. π 4 91. 14x x2 + 49 �
� 85. − θ − 2 sin π 4 √ 92. ln \|x + x2 + 16\| − ln(4) 790 Foundations of Trigonometry 10.5 Graphs of the Trigonometric Functions In this section, we return to our discussion of the circular (trigonometric) functions as functions of real numbers and pick up where we left oﬀ in Sections 10.2.1 and 10.3.1. As usual, we begin our study with the functions f (t) = cos(t) and g(t) = sin(t). 10.5.1 Graphs of the Cosine and Sine Functions From Theorem 10.5 in Section 10.2.1, we know that the domain of f (t) = cos(t) and of g(t) = sin(t) is all real numbers, (−∞, ∞), and the range of both functions is [−1, 1]. The Even / Odd Identities in Theorem 10.12 tell us cos(−t) = cos(t) for all real numbers t and sin(−t) = − sin(t) for all real numbers t. This means f (t) = cos(t) is an even function, while g(t) = sin(t) is an odd function.1 Another important property of these functions is that for coterminal angles α and β, cos(α) = cos(β) and sin(α) = sin(β). Said diﬀerently, cos(t+2πk) = cos(t) and sin(t+2πk) = sin(t) for all real numbers t and any integer k. This last property is given a special name. Deﬁnition 10.3. Periodic Functions: A function f is said to be periodic if there is a real number c so that f (t + c) = f (t) for all real numbers t in the domain of f. The smallest positive number p for which f (t + p) = f (t) for all real numbers t in the domain of f, if it exists, is called the period of f. We have already seen a family of periodic functions in Section 2.1: the constant functions. However, despite being periodic a constant function has no period. (We’ll leave that odd gem as
an exercise for you.) Returning to the circular functions, we see that by Deﬁnition 10.3, f (t) = cos(t) is periodic, since cos(t + 2πk) = cos(t) for any integer k. To determine the period of f, we need to ﬁnd the smallest real number p so that f (t + p) = f (t) for all real numbers t or, said diﬀerently, the smallest positive real number p such that cos(t + p) = cos(t) for all real numbers t. We know that cos(t + 2π) = cos(t) for all real numbers t but the question remains if any smaller real number will do the trick. Suppose p > 0 and cos(t + p) = cos(t) for all real numbers t. Then, in particular, cos(0 + p) = cos(0) so that cos(p) = 1. From this we know p is a multiple of 2π and, since the smallest positive multiple of 2π is 2π itself, we have the result. Similarly, we can show g(t) = sin(t) is also periodic with 2π as its period.2 Having period 2π essentially means that we can completely understand everything about the functions f (t) = cos(t) and g(t) = sin(t) by studying one interval of length 2π, say [0, 2π].3 One last property of the functions f (t) = cos(t) and g(t) = sin(t) is worth pointing out: both of these functions are continuous and smooth. Recall from Section 3.1 that geometrically this means the graphs of the cosine and sine functions have no jumps, gaps, holes in the graph, asymptotes, 1See section 1.6 for a review of these concepts. 2Alternatively, we can use the Cofunction Identities in Theorem 10.14 to show that g(t) = sin(t) is periodic with period 2π since g(t) = sin(t) = cos. 3Technically, we should study the interval [0, 2π),4since whatever happens at t = 2π is the same as what happens at t = 0. As we will see shortly, t = 2π gives us an extra ‘check’ when
we go to graph these functions. 4In some advanced texts, the interval of choice is [−π, π). 10.5 Graphs of the Trigonometric Functions 791 corners or cusps. As we shall see, the graphs of both f (t) = cos(t) and g(t) = sin(t) meander nicely and don’t cause any trouble. We summarize these facts in the following theorem. Theorem 10.22. Properties of the Cosine and Sine Functions • The function f (x) = cos(x) • The function g(x) = sin(x) – has domain (−∞, ∞) – has range [−1, 1] – has domain (−∞, ∞) – has range [−1, 1] – is continuous and smooth – is continuous and smooth – is even – has period 2π – is odd – has period 2π In the chart above, we followed the convention established in Section 1.6 and used x as the independent variable and y as the dependent variable.5 This allows us to turn our attention to graphing the cosine and sine functions in the Cartesian Plane. To graph y = cos(x), we make a table as we did in Section 1.6 using some of the ‘common values’ of x in the interval [0, 2π]. This generates a portion of the cosine graph, which we call the ‘fundamental cycle’ of y = cos(x). x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π cos(xx, cos(x)) (0, 1 √ 3π 4, − 2 2 (π, −1 3π √ 7π 4, (2π, 1) 2 2 5π y 1 −1 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x The ‘fundamental cycle’ of y = cos(x). A few things about the graph above are worth mentioning. First, this graph represents only part of the graph of y = cos(x). To get the entire graph, we imagine ‘copying and pasting’ this graph end to end inﬁnitely in both directions (left and right) on the x-axis. Secondly, the vertical scale here has been greatly exaggerated for clarity and
aesthetics. Below is an accurate-to-scale graph of y = cos(x) showing several cycles with the ‘fundamental cycle’ plotted thicker than the others. The 5The use of x and y in this context is not to be confused with the x- and y-coordinates of points on the Unit Circle which deﬁne cosine and sine. Using the term ‘trigonometric function’ as opposed to ‘circular function’ can help with that, but one could then ask, “Hey, where’s the triangle?” 792 Foundations of Trigonometry graph of y = cos(x) is usually described as ‘wavelike’ – indeed, many of the applications involving the cosine and sine functions feature modeling wavelike phenomena. y x An accurately scaled graph of y = cos(x). We can plot the fundamental cycle of the graph of y = sin(x) similarly, with similar results. x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π sin(xx, sin(x)) (0, 0 √ 3π 4, 2 2 0 √ 2 2 −1 √ 2 2 − − 5π 7π (π, 01 3π √ 4, − 2 2 y 1 −1 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x 0 (2π, 0) The ‘fundamental cycle’ of y = sin(x). As with the graph of y = cos(x), we provide an accurately scaled graph of y = sin(x) below with the fundamental cycle highlighted. y x An accurately scaled graph of y = sin(x). It is no accident that the graphs of y = cos(x) and y = sin(x) are so similar. Using a cofunction identity along with the even property of cosine, we have sin(x) = cos π 2 − x = cos − x − = cos x − π 2 π 2 Recalling Section 1.7, we see from this formula that the graph of y = sin(x) is the result of shifting the graph of y = cos(x) to the right π Now that we know the basic shapes of the graphs of y = cos(x) and y = sin(x), we can use The
orem 1.7 in Section 1.7 to graph more complicated curves. To do so, we need to keep track of 2 units. A visual inspection conﬁrms this. 10.5 Graphs of the Trigonometric Functions 793 the movement of some key points on the original graphs. We choose to track the values x = 0, π 2, π, 3π 2 and 2π. These ‘quarter marks’ correspond to quadrantal angles, and as such, mark the location of the zeros and the local extrema of these functions over exactly one period. Before we begin our next example, we need to review the concept of the ‘argument’ of a function as ﬁrst introduced in Section 1.4. For the function f (x) = 1 − 5 cos(2x − π), the argument of f is x. We shall have occasion, however, to refer to the argument of the cosine, which in this case is 2x − π. Loosely stated, the argument of a trigonometric function is the expression ‘inside’ the function. Example 10.5.1. Graph one cycle of the following functions. State the period of each. 1. f (x) = 3 cos πx−π 2 + 1 Solution. 1. We set the argument of the cosine, πx−π 2 2. g(x) = 1 2 sin(π − 2x) + 3 2, equal to each of the values: 0, π 2, π, 3π 2, 2π and solve for x. We summarize the results below. πx−π a πx−π πx− = 3π 4 2 2 = 2π 5 0 π 2 π 3π 2 2π πx−π πx−π πx−π 2 Next, we substitute each of these x values into f (x) = 3 cos πx−π corresponding y-values and connect the dots in a pleasing wavelike fashion. 2 + 1 to determine the x 1 2 3 4 5 f (x) (x, f (x)) 4 1 (1, 4) (2, 1) −2 (3, −2) 1 4 (4, 1) (5, 4) y 4 3 2 1 −1 −2 1 2 3 4 5 x One
cycle is graphed on [1, 5] so the period is the length of that interval which is 4. One cycle of y = f (x). 2. Proceeding as above, we set the argument of the sine, π − 2x, equal to each of our quarter marks and solve for x. 794 Foundations of Trigonometry a π − 2x = a x π π − 2x = 0 0 2 π π π − 2x = π 2 4 2 π − 2x = π π 0 3π π − 2x = 3π 2 − π 2 4 2π π − 2x = 2π − π 2 We now ﬁnd the corresponding y-values on the graph by substituting each of these x-values into g(x) = 1 2. Once again, we connect the dots in a wavelike fashion. 2 sin(π − 2x(xx, g(x)) π 2, 3 π 4, 2 0 One cycle of y = g(x). One cycle was graphed on the interval − π 2, π 2 so the period is π 2 − − π 2 = π. The functions in Example 10.5.1 are examples of sinusoids. Roughly speaking, a sinusoid is the result of taking the basic graph of f (x) = cos(x) or g(x) = sin(x) and performing any of the transformations6 mentioned in Section 1.7. Sinusoids can be characterized by four properties: period, amplitude, phase shift and vertical shift. We have already discussed period, that is, how long it takes for the sinusoid to complete one cycle. The standard period of both f (x) = cos(x) and g(x) = sin(x) is 2π, but horizontal scalings will change the period of the resulting sinusoid. The amplitude of the sinusoid is a measure of how ‘tall’ the wave is, as indicated in the ﬁgure below. The amplitude of the standard cosine and sine functions is 1, but vertical scalings can alter this. 6We have already seen how the Even/Odd and Cofunction Identities can be used to rewrite g(x) = sin(x) as a transformed version of f (x) = cos(
x), so of course, the reverse is true: f (x) = cos(x) can be written as a transformed version of g(x) = sin(x). The authors have seen some instances where sinusoids are always converted to cosine functions while in other disciplines, the sinusoids are always written in terms of sine functions. We will discuss the applications of sinusoids in greater detail in Chapter 11. Until then, we will keep our options open. 10.5 Graphs of the Trigonometric Functions 795 amplitude baseline period = sin(x). As the reader can verify, a phase shift of π The phase shift of the sinusoid is the horizontal shift experienced by the fundamental cycle. We have seen that a phase (horizontal) shift of π 2 to the right takes f (x) = cos(x) to g(x) = sin(x) since cos x − π 2 to the left takes g(x) = sin(x) to 2 f (x) = cos(x). The vertical shift of a sinusoid is exactly the same as the vertical shifts in Section 1.7. In most contexts, the vertical shift of a sinusoid is assumed to be 0, but we state the more general case below. The following theorem, which is reminiscent of Theorem 1.7 in Section 1.7, shows how to ﬁnd these four fundamental quantities from the formula of the given sinusoid. Theorem 10.23. For ω > 0, the functions C(x) = A cos(ωx + φ) + B and S(x) = A sin(ωx + φ) + B have period 2π ω have amplitude \|A\| have phase shift − φ ω have vertical shift B We note that in some scientiﬁc and engineering circles, the quantity φ mentioned in Theorem 10.23 is called the phase of the sinusoid. Since our interest in this book is primarily with graphing sinusoids, we focus our attention on the horizontal shift − φ The proof of Theorem 10.23 is a direct application of Theorem 1.7 in Section 1.7 and is left to the reader. The parameter ω, which is stipulated to be positive, is called the (angular) frequency of the sinusoid and is the number of cycles the sinusoid completes over a 2π interval
. We can always ensure ω > 0 using the Even/Odd Identities.7 We now test out Theorem 10.23 using the functions f and g featured in Example 10.5.1. First, we write f (x) in the form prescribed in Theorem 10.23, ω induced by φ. f (x) = 3 cos πx − π 2 + 1 = 3 cos π 2 x + − π 2 + 1, 7Try using the formulas in Theorem 10.23 applied to C(x) = cos(−x + π) to see why we need ω > 0. 796 Foundations of Trigonometry 2, φ = − π π/2 = 4, the amplitude is \|A\| = \|3\| = 3, the phase shift is − φ so that A = 3, ω = π 2 and B = 1. According to Theorem 10.23, the period of f is ω = 2π 2π π/2 = 1 (indicating a shift to the right 1 unit) and the vertical shift is B = 1 (indicating a shift up 1 unit.) All of these match with our graph of y = f (x). Moreover, if we start with the basic shape of the cosine graph, shift it 1 unit to the right, 1 unit up, stretch the amplitude to 3 and shrink the period to 4, we will have reconstructed one period of the graph of y = f (x). In other words, instead of tracking the ﬁve ‘quarter marks’ through the transformations to plot y = f (x), we can use ﬁve other pieces of information: the phase shift, vertical shift, amplitude, period and basic shape of the cosine curve. Turning our attention now to the function g in Example 10.5.1, we ﬁrst need to use the odd property of the sine function to write it in the form required by Theorem 10.23 ω = − −π/2 g(x) = 1 2 sin(π − 2x) + 3 2 = 1 2 sin(−(2x − π)) + 3 2 = − 1 2 sin(2x − π) + 3 2 = − 1 2 sin(2x + (−π)) + 3 2 = 1 2, the phase shift is − −π 2. The period
is then 2π Instead of the graph starting at x = π 2, ω = 2, φ = −π and B = 3 2 = π We ﬁnd A = − 1 2 = π, the amplitude is − 1 2 (indicating a shift right π 2 units) and the vertical shift is up 2 3 2. Note that, in this case, all of the data match our graph of y = g(x) with the exception of the phase shift. 2, it ends there. Remember, however, that the graph presented in Example 10.5.1 is only one portion of the graph of y = g(x). Indeed, another complete cycle begins at x = π 2, and this is the cycle Theorem 10.23 is detecting. The reason for the discrepancy is that, in order to apply Theorem 10.23, we had to rewrite the formula for g(x) using the odd property of the sine function. Note that whether we graph y = g(x) using the ‘quarter marks’ approach or using the Theorem 10.23, we get one complete cycle of the graph, which means we have completely determined the sinusoid. Example 10.5.2. Below is the graph of one complete cycle of a sinusoid y = f (x). −1, 5 2 y 3 2 1 51 1 2 3 4 5 x −1 −2 2, − 3 2 One cycle of y = f (x). 1. Find a cosine function whose graph matches the graph of y = f (x). 10.5 Graphs of the Trigonometric Functions 797 2. Find a sine function whose graph matches the graph of y = f (x). Solution. ω, so that ω = π 1. We ﬁt the data to a function of the form C(x) = A cos(ωx + φ) + B. Since one cycle is graphed over the interval [−1, 5], its period is 5 − (−1) = 6. According to Theorem 10.23, 3. Next, we see that the phase shift is −1, so we have − φ 6 = 2π ω = −1, or. As a result the amplitude A = 1 2 (4) = 2. Finally, to determine the vertical shift, we 2 average the endpoints of the range to ﬁnd
B = 1 2. Our ﬁnal answer is 2 + 1 C(x) = 2 cos π 2. 3. To ﬁnd the amplitude, note that the range of the sinusoid is − 3 2 + − 3 5 2 (1. Most of the work to ﬁt the data to a function of the form S(x) = A sin(ωx + φ) + B is done. The period, amplitude and vertical shift are the same as before with ω = π 3, A = 2 and B = 1 2. The trickier part is ﬁnding the phase shift. To that end, we imagine extending the graph of the given sinusoid as in the ﬁgure below so that we can identify a cycle beginning at 7 = − 7π 2, we get − φ 6. + 1 Hence, our answer is S(x) = 2 sin π 2.. Taking the phase shift to be 7 3 x − 7π 2, or, 5 2 − 10 x 7 2, 1 2 13 2, 1 2 19 2, 5 2 −1 −2 8, − 3 2 Extending the graph of y = f (x). Note that each of the answers given in Example 10.5.2 is one choice out of many possible answers. taking For example, when ﬁtting a sine function to the data, we could have chosen to start at 1 + 1 6 for an answer of S(x) = −2 sin π A = −2. In this case, the phase shift is 1 2. Alternatively, we could have extended the graph of y = f (x) to the left and considered a sine, and so on. Each of these formulas determine the same sinusoid curve function starting at − 5 2 and their formulas are all equivalent using identities. Speaking of identities, if we use the sum identity for cosine, we can expand the formula to yield 2, 1 3 x − π 2 so (x) = A cos(ωx + φ) + B = A cos(ωx) cos(φ) − A sin(ωx) sin(φ) + B. 798 Foundations of Trigonometry Similarly, using the sum identity for sine, we get S(x) = A sin(ωx + φ) + B = A sin(ωx)
cos(φ) + A cos(ωx) sin(φ) + B. Making these observations allows us to recognize (and graph) functions as sinusoids which, at ﬁrst glance, don’t appear to ﬁt the forms of either C(x) or S(x). Example 10.5.3. Consider the function f (x) = cos(2x) − √ 3 sin(2x). Find a formula for f (x): 1. in the form C(x) = A cos(ωx + φ) + B for ω > 0 2. in the form S(x) = A sin(ωx + φ) + B for ω > 0 Check your answers analytically using identities and graphically using a calculator. Solution. 1. The key to this problem is to use the expanded forms of the sinusoid formulas and match up 3 sin(2x) with the expanded form of corresponding coeﬃcients. Equating f (x) = cos(2x) − C(x) = A cos(ωx + φ) + B, we get √ √ cos(2x) − 3 sin(2x) = A cos(ωx) cos(φ) − A sin(ωx) sin(φ) + B It should be clear that we can take ω = 2 and B = 0 to get √ cos(2x) − 3 sin(2x) = A cos(2x) cos(φ) − A sin(2x) sin(φ) To determine A and φ, a bit more work is involved. We get started by equating the coeﬃcients of the trigonometric functions on either side of the equation. On the left hand side, the coeﬃcient of cos(2x) is 1, while on the right hand side, it is A cos(φ). Since this equation is to hold for all real numbers, we must have8 that A cos(φ) = 1. Similarly, we ﬁnd by equating the coeﬃcients of sin(2x) that A sin(φ) = 3. What we have here is a system of nonlinear equations! We can temporarily eliminate the dependence on φ by using the Pythagorean Identity. We
know cos2(φ) + sin2(φ) = 1, so multiplying this by A2 gives A2 cos2(φ)+A2 sin2(φ) = A2. Since A cos(φ) = 1 and A sin(φ) = 3)2 = 4 or A = ±2. Choosing A = 2, we have 2 cos(φ) = 1 and 2 sin(φ) = 3 or, after some √ 3 rearrangement, cos(φ) = 1 2. One such angle φ which satisﬁes this criteria is. We can easily φ = π check our answer using the sum formula for cosine 3. Hence, one way to write f (x) as a sinusoid is f (x) = 2 cos 2x + π 3, we get A2 = 12+( √ 2 and sin(φ) = √ √ √ 3 f (x) = 2 cos 2x + π 3 = 2 cos(2x) cos π 3 cos(2x) 1 − sin(2x) 2 √ 3 sin(2x) = 2 = cos(2x) − − sin(2x) sin π 3 √ 3 2 8This should remind you of equation coeﬃcients of like powers of x in Section 8.6. 10.5 Graphs of the Trigonometric Functions 799 2. Proceeding as before, we equate f (x) = cos(2x) − S(x) = A sin(ωx + φ) + B to get √ 3 sin(2x) with the expanded form of √ cos(2x) − 3 sin(2x) = A sin(ωx) cos(φ) + A cos(ωx) sin(φ) + B Once again, we may take ω = 2 and B = 0 so that √ cos(2x) − 3 sin(2x) = A sin(2x) cos(φ) + A cos(2x) sin(φ) We equate9 the coeﬃcients of cos(2x) on either side and get A sin(φ) = 1 and A cos(φ) = − 3. Using A2 cos2(φ) + A2 sin
2(φ) = A2 as before, we get A = ±2, and again we choose A = 2. √ 3 This means 2 sin(φ) = 1, or sin(φ) = 1 2.. One such angle which meets these criteria is φ = 5π 6 Checking our work analytically, we have 3, which means cos(φ) = − 6. Hence, we have f (x) = 2 sin 2x + 5π 2, and 2 cos(φ) = − √ √ f (x) = 2 sin 2x + 5π 6 + cos(2x) sin 5π = 2 sin(2x) cos 5π 6 6 √ + cos(2x) 1 3 2 2 − = 2 sin(2x) = cos(2x) − √ 3 sin(2x) Graphing the three formulas for f (x) result in the identical curve, verifying our analytic work. √ It is important to note that in order for the technique presented in Example 10.5.3 to ﬁt a function into one of the forms in Theorem 10.23, the arguments of the cosine and sine function much match. 3 sin(3x) is not.10 It That is, while f (x) = cos(2x) − is also worth mentioning that, had we chosen A = −2 instead of A = 2 as we worked through Example 10.5.3, our ﬁnal answers would have looked diﬀerent. The reader is encouraged to rework Example 10.5.3 using A = −2 to see what these diﬀerences are, and then for a challenging exercise, use identities to show that the formulas are all equivalent. The general equations to ﬁt a function of the form f (x) = a cos(ωx) + b sin(ωx) + B into one of the forms in Theorem 10.23 are explored in Exercise 35. 3 sin(2x) is a sinusoid, g(x) = cos(2x) − √ 9Be careful here! 10This graph does, however, exhibit sinusoid-like characteristics! Check it out! 800 Foundations of Trigonometry 10.5.2 Graphs of the Secant and Cosecant Functions We now turn our attention to graphing
y = sec(x). Since sec(x) = 1 cos(x), we can use our table of values for the graph of y = cos(x) and take reciprocals. We know from Section 10.3.1 that the domain of F (x) = sec(x) excludes all odd multiples of π 2, and sure enough, we run into trouble at 2 and x = 3π x = π 2 since cos(x) = 0 at these values. Using the notation introduced in Section 4.2, −, cos(x) → 0+, so sec(x) → ∞. (See Section 10.3.1 for a more detailed we have that as x → π 2 analysis.) Similarly, we ﬁnd that as x → π, sec(x) → −∞; and as 2 x → 3π, sec(x) → ∞. This means we have a pair of vertical asymptotes to the graph of y = sec(x), 2 x = π 2. Since cos(x) is periodic with period 2π, it follows that sec(x) is also.11 Below we graph a fundamental cycle of y = sec(x) along with a more complete graph obtained by the usual ‘copying and pasting.’12 +, sec(x) → −∞; as x → 3π 2 2 and x = 3π − + y x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π √ 2 − − √ √ sec(x) 1 cos(x) 1 √ 2 2 0 undeﬁned 2 2 −1 √ 2 2 0 undeﬁned 2 2 1 −1 √ √ − − 2 1 √ (x, sec(x)) (0, 1) √ 2 π 4, 2 3π 2 5π √ 2 4, − (π, −1) √ 2 4, − √ 2 7π 4, (2π, 1) 3 2 1 −1 −2 −3 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x The ‘fundamental cycle’ of y = sec(x). y x The graph of y = sec(x). 11Provided 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). 12In Section 10.3.1, we argued the range of F (x) = sec(x) is (−∞, −1] ∪ [1, ∞). We can now see this graphically. 10.5 Graphs of the Trigonometric Functions 801 As one would expect, to graph y = csc(x) we begin with y = sin(x) and take reciprocals of the corresponding y-values. Here, we encounter issues at x = 0, x = π and x = 2π. Proceeding with the usual analysis, we graph the fundamental cycle of y = csc(x) below along with the dotted graph of y = sin(x) for reference. Since y = sin(x) and y = cos(x) are merely phase shifts of each other, so too are y = csc(x) and y = sec(x). y x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π sin(x) √ 1 2 √ √ csc(x) 0 undeﬁned 2 2 1 √ 2 2 0 undeﬁned 2 2 −1 √ 2 2 0 undeﬁned −x, csc(x)) √ 3π 4, 2 5π 2 7π √ 2 4, − 2, −1 3π √ 2 4, − 3 2 1 −1 −2 −3 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x Once again, our domain and range work in Section 10.3.1 is veriﬁed geometrically in the graph of y = G(x) = csc(x). The ‘fundamental cycle’ of y = csc(x). y x The graph of y = csc(x). Note that, on the intervals between the vertical asymptotes, both F (x) = sec(x) and G(x) = csc(x) are continuous and smooth. In other words, they are continuous and smooth on their domains.13 The following theorem summarizes the properties of
the secant and cosecant functions. Note that 13Just like the rational functions in Chapter 4 are continuous and smooth on their domains because polynomials are continuous and smooth everywhere, the secant and cosecant functions are continuous and smooth on their domains since the cosine and sine functions are continuous and smooth everywhere. 802 Foundations of Trigonometry all of these properties are direct results of them being reciprocals of the cosine and sine functions, respectively. Theorem 10.24. Properties of the Secant and Cosecant Functions The function F (x) = sec(x) – has domain x : x = π 2 + πk, k is an integer = ∞ k=−∞ (2k + 1)π 2, (2k + 3)π 2 – has range {y : \|y\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – is continuous and smooth on its domain – is even – has period 2π The function G(x) = csc(x) ∞ k=−∞ (kπ, (k + 1)π) – has domain {x : x = πk, k is an integer} = – has range {y : \|y\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – is continuous and smooth on its domain – is odd – has period 2π In the next example, we discuss graphing more general secant and cosecant curves. Example 10.5.4. Graph one cycle of the following functions. State the period of each. 1. f (x) = 1 − 2 sec(2x) Solution. 2. g(x) = csc(π − πx) − 5 3 1. To graph y = 1 − 2 sec(2x), we follow the same procedure as in Example 10.5.1. First, we set 2 and 2π and solve for x. the argument of secant, 2x, equal to the ‘quarter marks’ 0, π 2, π, 3π a 2x = a 0 2x = 0 π 2x = π 2 2 π 2x = π 3π 2x = 3π 2 2 2π 2x = 2π x 0 π 4 π 2 3π 4 π 10.
5 Graphs of the Trigonometric Functions 803 Next, we substitute these x values into f (x). If f (x) exists, we have a point on the graph; otherwise, we have found a vertical asymptote. In addition to these points and asymptotes, we have graphed the associated cosine curve – in this case y = 1 − 2 cos(2x) – dotted in the picture below. Since one cycle is graphed over the interval [0, π], the period is π − 0 = π. y 3 2 1 −1 π 4 π 2 3π 4 π x x 0 π 4 π 2 3π 4 π f (x) −1 (x, f (x)) (0, −1) undeﬁned 3 π 2, 3 undeﬁned −1 (π, −1) 2. Proceeding as before, we set the argument of cosecant in g(x) = csc(π−πx)−5 3 equal to the quarter marks and solve for x. One cycle of y = 1 − 2 sec(2x). a π − πx = a x π − πx = 0 1 0 π − πx = π 1 π 2 2 2 π − πx = π 0 π 2 − 1 π − πx = 3π 3π 2 2 2π π − πx = 2π −1 Substituting these x-values into g(x), we generate the graph below and ﬁnd the period to be 1 − (−1) = 2. The associated sine curve, y = sin(π−πx)−5, is dotted in as a reference1 g(x) undeﬁned − 4 3 undeﬁned (x, g(x)) 1 2, − 4 3 −2 − 1 2, −2 undeﬁned y −1 − 1 2 1 1 2 x −1 −2 One cycle of y = csc(π−πx)−5 3. 804 Foundations of Trigonometry Before moving on, we note that it is possible to speak of the period, phase shift and vertical shift of secant and cosecant graphs and use even/odd identities to put them in a form similar to the sinus
oid forms mentioned in Theorem 10.23. Since these quantities match those of the corresponding cosine and sine curves, we do not spell this out explicitly. Finally, since the ranges of secant and cosecant are unbounded, there is no amplitude associated with these curves. 10.5.3 Graphs of the Tangent and Cotangent Functions 2 and x = 3π Finally, we turn our attention to the graphs of the tangent and cotangent functions. When constructing a table of values for the tangent function, we see that J(x) = tan(x) is undeﬁned at −, sin(x) → 1− x = π 2, in accordance with our ﬁndings in Section 10.3.1. As x → π and cos(x) → 0+, so that tan(x) = sin(x) 2. Using a + similar analysis, we get that as x → π, 2 tan(x) → −∞. Plotting this information and performing the usual ‘copy and paste’ produces: cos(x) → ∞ producing a vertical asymptote at x = π +, tan(x) → −∞; as x → 3π 2, tan(x) → ∞; and as x → 3π 3π 4 π 5π 4 3π 2 7π 4 2π tan(x) 0 1 (x, tan(x)) (0, 0) 4, 1 π undeﬁned −1 0 1 undeﬁned −1 0 4, −1 3π (π, 0) 4, 1 5π 4, −1 7π (2π, 0) 1 −1 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x The graph of y = tan(x) over [0, 2π]. y x The graph of y = tan(x). 10.5 Graphs of the Trigonometric Functions 805 From the graph, it appears as if the tangent function is periodic with period π. To prove that this is the case, we appeal to the sum formula for tangents. We have: tan(x + π) = tan(x) + tan(π) 1 − tan(x) tan(π) = tan(x) + 0 1 − (tan
(x))(0) = tan(x), which tells us the period of tan(x) is at most π. To show that it is exactly π, suppose p is a positive real number so that tan(x + p) = tan(x) for all real numbers x. For x = 0, we have tan(p) = tan(0 + p) = tan(0) = 0, which means p is a multiple of π. The smallest positive multiple of π is π itself, so we have established the result. We take as our fundamental cycle for y = tan(x) the interval − π 2, − π 2. From the graph, we see conﬁrmation of our domain and range work in Section 10.3.1., and use as our ‘quarter marks’ x = − π 4 and π 4, 0, π 2, π 2 It should be no surprise that K(x) = cot(x) behaves similarly to J(x) = tan(x). Plotting cot(x) over the interval [0, 2π] results in the graph below. y 1 −1 x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π cot(x) undeﬁned (x, cot(x)) 1 0 −1 undeﬁned 1 0 −1 undeﬁned 1 3π 4, 1 5π 3π 2, 0 4, −1 7π π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x The graph of y = cot(x) over [0, 2π]. From these data, it clearly appears as if the period of cot(x) is π, and we leave it to the reader to prove this.14 We take as one fundamental cycle the interval (0, π) with quarter marks: x = 0, π 4, π 4 and π. A more complete graph of y = cot(x) is below, along with the fundamental cycle highlighted as usual. Once again, we see the domain and range of K(x) = cot(x) as read from the graph matches with what we found analytically in Section 10.3.1. 2, 3π 14Certainly, mim
icking the proof that the period of tan(x) is an option; for another approach, consider transforming tan(x) to cot(x) using identities. 806 Foundations of Trigonometry y x The graph of y = cot(x). The properties of the tangent and cotangent functions are summarized below. As with Theorem 10.24, each of the results below can be traced back to properties of the cosine and sine functions and the deﬁnition of the tangent and cotangent functions as quotients thereof. Theorem 10.25. Properties of the Tangent and Cotangent Functions The function J(x) = tan(x) – has domain x : x = π 2 + πk, k is an integer = ∞ k=−∞ (2k + 1)π 2, (2k + 3)π 2 – has range (−∞, ∞) – is continuous and smooth on its domain – is odd – has period π The function K(x) = cot(x) – has domain {x : x = πk, k is an integer} = – has range (−∞, ∞) – is continuous and smooth on its domain – is odd – has period π ∞ k=−∞ (kπ, (k + 1)π) 10.5 Graphs of the Trigonometric Functions 807 Example 10.5.5. Graph one cycle of the following functions. Find the period. 1. f (x) = 1 − tan x 2. 2. g(x) = 2 cot π 2 x + π + 1. Solution. 1. We proceed as we have in all of the previous graphing examples by setting the argument of 4, 0, π 2, equal to each of the ‘quarter marks’ − π, namely x 2, − π 4 tangent in f (x) = 1 − tan x 2 and π 2, and solving for x Substituting these x-values into f (x), we ﬁnd points on the graph and the vertical asymptotes. x −x) undeﬁned (x, f (x)) 2 1 0 − π 2, 2 (0, 1) 2, 0 π undeﬁned y 2 1 −1 −2 −π − �
� 2 π 2 π x We see that the period is π − (−π) = 2π. One cycle of y = 1 − tan x 2. 2. The ‘quarter marks’ for the fundamental cycle of the cotangent curve are 0, π To graph g(x) = 2 cot π and solving for x. 2 x + π + 1, we begin by setting π 4 and π. 2 x + π equal to each quarter mark 4, π 2, 3π 808 Foundations of Trigonometry a 0 π 4 π 2 3π 1 4 − 1 2 x + π = 3π We now use these x-values to generate our graph. x −2 − 3 2 −1 − 1 2 0 g(x) undeﬁned (x, g(x)) 3 1 2, 3 − 3 (−1, 1) 2, −1 −1 − 1 undeﬁned y 3 2 1 −2 −1 x −1 One cycle of y = 2 cot π 2 x + π + 1. We ﬁnd the period to be 0 − (−2) = 2. As with the secant and cosecant functions, it is possible to extend the notion of period, phase shift and vertical shift to the tangent and cotangent functions as we did for the cosine and sine functions in Theorem 10.23. Since the number of classical applications involving sinusoids far outnumber those involving tangent and cotangent functions, we omit this. The ambitious reader is invited to formulate such a theorem, however. 10.5 Graphs of the Trigonometric Functions 809 10.5.4 Exercises In Exercises 1 - 12, graph one cycle of the given function. State the period, amplitude, phase shift and vertical shift of the function. 1. y = 3 sin(x) 2. y = sin(3x) 3. y = −2 cos(x) 4. y = cos x − 7. y = − 1 3 cos π 2 1 2 x + π 3 5. y = − sin x + π 3 6. y = sin(2x − π) 8. y = cos(3x − 2π) + 4 9. y = sin −x − π 4 − 2 10. y = 2 3

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