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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(�
�) = 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 �
� = 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 4581 feet from the base of the tower. (c) The Sasquatch ran 200 cot(6◦) − 200 cot(6.5◦) ≈ 147 feet in those 10 seconds. This translates to ≈ 10 miles per hour. At the scene of the second sighting, the Sasquatch was ≈ 1755 feet from the tower, which means, if it keeps up this pace, it will reach the tower in about 2 minutes. 10.3 The Six Circular Functions and Fundamental Identities 769 79. The tree is about 41 feet tall. 80. The boat has traveled about 244 feet. 81. The tower is about 682 feet tall. The guy wire hits the ground about 731 feet away from the base of the tower. 770 Foundations of Trigonometry 10.4 Trigonometric Identities In Section 10.3, we saw the utility of the Pythagorean Identities in Theorem 10.8 along with the Quotient and Reciprocal Identities in Theorem 10.6. Not only did these identities help us compute the values of
the circular functions for angles, they were also useful in simplifying expressions involving the circular functions. In this section, we introduce several collections of identities which have uses in this course and beyond. Our ﬁrst set of identities is the ‘Even / Odd’ identities.1 Theorem 10.12. Even / Odd Identities: For all applicable angles θ, cos(−θ) = cos(θ) 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) + sin
2(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 simpliﬁes 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 Ident
ities 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 ≤ �
� ≤ 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 sin2(θ) results in the aptly-named ‘Power Reduction’ formulas below. Theorem 10.18. Power Reduction Formulas: For all angles θ, cos2(θ) = sin2(θ) = 1 + cos(2θ) 2 1 − cos(2θ) 2 Example 10.4.4. Rewrite sin2(θ) cos2(θ) as a sum and diﬀerence of cosines to the ﬁrst power. Solution. We begin with a straightforward application of Theorem 10.18 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 Found
ations 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 Theorem 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 trig
onometric 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 sinus
oid 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(2
x) − 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 sin2(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 √ √ − − 1 2 √ (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 sinusoid forms mentioned in Theorem 10.23. Since these quantities match those of the corresponding cosine and sine curves, we do not spell this out explicitly. Finally, since the ranges of secant and cosecant are unbounded, there is no amplitude associated with these curves. 10.5.3 Graphs of the Tangent and Cotangent Functions 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, mimicking the proof that the period of tan(x) is an option; for another approach, consider transforming tan(x) to cot(x) using identities. 806 Foundations of Trigonometry 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 − 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 c
ot 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 cos π 2 − 4x + 1 11. y = − cos 2x + 3 2 π 3 − 1 2 12. y = 4 sin(−2πx + π) In Exercises 13 - 24, graph one cycle of the given function. State the period of the function. 13. y = tan 16. y = sec x − x − π 3 π 2 14. y = 2 tan 17. y = − csc x − 3 1 4 x + π 3 15. y = 1 3 18. y = − 19. y = csc(2x − π) 20. y = sec(3x − 2π) + 4 21. y = csc −x − tan(−2x − π) + 1 1 3 sec 22. y = cot x + π 6 23. y = −11 cot x 1 5 24. y = cot 2x + 3π 2 + 1 1 3 In Exercises 25 - 34, use Example 10.5.3 as a guide to show
that the function is a sinusoid by rewriting it in the forms C(x) = A cos(ωx + φ) + B and S(x) = A sin(ωx + φ) + B for ω > 0 and 0 ≤ φ < 2π. 25. f (x) = √ 2 sin(x) + √ 2 cos(x) + 1 26. f (x) = 3 √ 3 sin(3x) − 3 cos(3x) 27. f (x) = − sin(x) + cos(x) − 2 28. f (x) = − 1 2 sin(2x) − √ 3 2 cos(2x) 29. f (x) = 2 √ 3 cos(x) − 2 sin(x) 31. f (x) = − 1 2 cos(5x) − √ 3 2 sin(5x) 30. f (x) = 3 2 cos(2x) − √ 3 3 2 sin(2x) + 6 32. f (x) = −6 √ 3 cos(3x) − 6 sin(3x) − 3 810 Foundations of Trigonometry 33. f (x) = √ 5 2 2 sin(x) − √ 5 2 2 cos(x) 34. f (x) = 3 sin x 6 √ − 3 3 cos x 6 35. In Exercises 25 - 34, you should have noticed a relationship between the phases φ for the S(x) and C(x). Show that if f (x) = A sin(ωx + α) + B, then f (x) = A cos(ωx + β) + B where β = α −. π 2 36. Let φ be an angle measured in radians and let P (a, b) be a point on the terminal side of φ when it is drawn in standard position. Use Theorem 10.3 and the sum identity for sine in Theorem 10.15 to show that f (x) = a sin(ωx) + b cos(ωx) + B (with ω > 0) can be rewritten as f (x) = a2 + b2 sin(ωx + φ) + B. √ 37. With the help of your classmates, express the domains of the functions in Examples
10.5.4 and 10.5.5 using extended interval notation. (We will revisit this in Section 10.7.) In Exercises 38 - 43, verify the identity by graphing the right and left hand sides on a calculator. 38. sin2(x) + cos2(x) = 1 39. sec2(x) − tan2(x) = 1 40. cos(x) = sin 41. tan(x + π) = tan(x) 42. sin(2x) = 2 sin(x) cos(x) 43. tan x 2 = − x π 2 sin(x) 1 + cos(x) In Exercises 44 - 50, graph the function with the help of your calculator and discuss the given questions with your classmates. 44. f (x) = cos(3x) + sin(x). Is this function periodic? If so, what is the period? 45. f (x) = sin(x) x. What appears to be the horizontal asymptote of the graph? 46. f (x) = x sin(x). Graph y = ±x on the same set of axes and describe the behavior of f. 47. f (x) = sin 1 x. What’s happening as x → 0? 48. f (x) = x − tan(x). Graph y = x on the same set of axes and describe the behavior of f. 49. f (x) = e−0.1x (cos(2x) + sin(2x)). Graph y = ±e−0.1x on the same set of axes and describe the behavior of f. 50. f (x) = e−0.1x (cos(2x) + 2 sin(x)). Graph y = ±e−0.1x on the same set of axes and describe the behavior of f. 51. Show that a constant function f is periodic by showing that f (x + 117) = f (x) for all real numbers x. Then show that f has no period by showing that you cannot ﬁnd a smallest number p such that f (x + p) = f (x) for all real numbers x. Said another way, show that f (x + p) = f (x) for all real numbers x for ALL values of p > 0, so no smallest value exists to satisfy the de�
�nition of ‘period’. 10.5 Graphs of the Trigonometric Functions 811 10.5.5 Answers 1. y = 3 sin(x) Period: 2π Amplitude: 3 Phase Shift: 0 Vertical Shift: 0 2. y = sin(3x) 2π 3 Period: Amplitude: 1 Phase Shift: 0 Vertical Shift: 0 3. y = −2 cos(x) Period: 2π Amplitude: 2 Phase Shift: 0 Vertical Shift: 0 4. y = cos x − π 2 Period: 2π Amplitude: 1 Phase Shift: Vertical Shift: 0 π 2 y 3 −3 y 1 −1 y 2 −2 y 1 −1 π 2 π 3π 2 x 2π π 6 π 3 π 2 2π 3 x π 2 π 3π 2 2π x π 2 π 3π 2 2π 5π 2 x 812 5. y = − sin x + Period: 2π Amplitude: 1 π Phase Shift: − 3 Vertical Shift: 0 π 3 Foundations of Trigonometry y 1 − π 3 π 6 2π 3 7π 6 5π 3 x 6. y = sin(2x − π) Period: π Amplitude: 1 Phase Shift: Vertical Shift: 0 π 2 7. y = − 1 3 cos 1 2 x + π 3 Period: 4π Amplitude: 1 3 2π Phase Shift: − 3 Vertical Shift: 0 8. y = cos(3x − 2π) + 4 2π 3 Period: Amplitude: 1 2π Phase Shift: 3 Vertical Shift: 4 −1 y 1 −1 y 1 3 π 2 3π 4 π 5π 4 3π 2 x − 2π 3 π 3 4π 3 7π 3 x 10π 3 − 1 3 y 5 4 3 2π 3 5π 6 π 7π 6 4π 3 x 10.5 Graphs of the Trigonometric Functions 813 9. y = sin −x − π 4 − 2 Period: 2π Amplitude: 1 Phase Shift: − π 4 π x + 4 Vertical Shift: −2 y = − sin (You need to use − 2 to ﬁnd this.)15 − 4x + 1 10. y = 2 3 Period: π 2 cos π 2 Amplitude: 2
3 π 8 4x − cos y = Vertical Shift: 1 Phase Shift: 2 3 (You need to use π 2 + 1 to ﬁnd this.)16 − 9π 4 − 7π 4 − 5π 4 − 3π 4 y π 4 3π 4 5π 4 x 7π 4 − π 4 −1 −2 − 3π 8 π 2 5π 8 x − 3π 8 − π 4 − π 8 11. y = − cos 2x + 3 2 π 3 − 1 2 y 1 Period: π Amplitude: 3 2 Phase Shift: − π 6 Vertical Shift: − 1 2 12. y = 4 sin(−2πx + π) Period: 1 Amplitude: 4 1 2 (You need to use Phase Shift: y = −4 sin(2πx − π) to ﬁnd this.)17 Vertical Shift: 0 − π 6 − 1 2 π 12 π 3 7π 12 x 5π 6 −4 15Two cycles of the graph are shown to illustrate the discrepancy discussed on page 796. 16Again, we graph two cycles to illustrate the discrepancy discussed on page 796. 17This will be the last time we graph two cycles to illustrate the discrepancy discussed on page 796. 814 Foundations of Trigonometry 13. y = tan x − Period: π π 3 14. y = 2 tan x − 3 1 4 Period: 4π y 1 − π 6 −1 π 12 π 3 7π 12 5π 6 x −2π −π y −1 −3 −5 π 2π x 15. y = tan(−2x − π) + 1 1 3 is equivalent to 1 3 tan(2x + π) + 1 y = − via the Even / Odd identity for tangent. Period − 3π 8 − π 4 x − 3π 4 − 5π 8 10.5 Graphs of the Trigonometric Functions 815 16. y = sec x − π 2 Start with y = cos x − π 2 Period: 2π 17. y = − csc x + π 3 Start with y = − sin Period: 2π x + π 3 y 1 −1 y 1 π 2 π 3π 2 2π x 5π 2 − π 3 −1 π 6 2π 3 7π 6
x 5π 3 18. y = − 1 3 sec 1 2 x + Start with y = − Period: 4π 1 3 π 3 cos 1 2 x + π 3 y 1 3 − 2π 3 − 1 3 π 3 4π 3 7π 3 x 10π 3 816 Foundations of Trigonometry 19. y = csc(2x − π) Start with y = sin(2x − π) Period: π 20. y = sec(3x − 2π) + 4 Start with y = cos(3x − 2π) + 4 Period: 2π 3 21. y = csc −x − π 4 Start with y = sin Period: 2π − 2 −x − π 4 − 2 y 1 −1 −2 −3 π 2 3π 4 π 5π 4 x 3π 2 2π 3 5π 6 π 7π 6 4π 3 x π 4 3π 4 5π 4 x 7π 4 10.5 Graphs of the Trigonometric Functions 817 22. y = cot x + Period: π π 6 23. y = −11 cot Period: 5π 1 5 x y 1 − π 6 π 12 −1 π 3 7π 12 x 5π 6 y 11 −11 5π 4 5π 2 15π 4 5π x 24. y = 1 3 Period: cot π 2 2x + 3π − 3π 8 − π 4 x − 3π 4 − 5π 8 818 Foundations of Trigonometry 25. f (x) = √ 2 sin(x) + √ 2 cos(x) + 1 = 2 sin x + π 4 + 1 = 2 cos x + 7π 4 + 1 26. f (x) = 3 √ 3 sin(3x) − 3 cos(3x) = 6 sin 3x + 11π 6 = 6 cos 3x + 4π 3 27. f (x) = − sin(x) + cos(x) − 2 = √ 2 sin x + 3π 4 − 2 = √ x + 2 cos π 4 − 2 28. f (x) = − 1 2 sin(2x) − √ 3 2 cos(2x) = sin 2x + 4π 3 = cos 2x + 5π 6 29. f (x)
= 2 √ 3 cos(x) − 2 sin(x) = 4 sin x + 30. f (x) = 3 2 cos(2x) − 31. f (x) = − 1 2 cos(5x) − 3 2 √ 3 3 2 √ 2π 3 = 4 cos x + π 6 5π 6 + 6 = 3 cos = cos 5x + 2π 3 sin(2x) + 6 = 3 sin 2x + sin(5x) = sin 5x + 7π 6 2x + π 3 + 6 32. f (x) = −6 √ 3 cos(3x) − 6 sin(3x) − 3 = 12 sin 3x + 4π 3 − 3 = 12 cos 3x + 5π 6 − 3 33. f (x) = √ 5 2 2 sin(x) − √ 5 2 2 cos(x) = 5 sin x + 34. f (x) = 3 sin x 6 √ − 3 3 cos x 6 = 6 sin x 6 + 7π 4 5π 3 = 5 cos x + 5π 4 = 6 cos x 6 + 7π 6 10.6 The Inverse Trigonometric Functions 819 10.6 The Inverse Trigonometric Functions As the title indicates, in this section we concern ourselves with ﬁnding inverses of the (circular) trigonometric functions. Our immediate problem is that, owing to their periodic nature, none of the six circular functions is one-to-one. To remedy this, we restrict the domains of the circular functions in the same way we restricted the domain of the quadratic function in Example 5.2.3 in Section 5.2 to obtain a one-to-one function. We ﬁrst consider f (x) = cos(x). Choosing the interval [0, π] allows us to keep the range as [−1, 1] as well as the properties of being smooth and continuous. y x Restricting the domain of f (x) = cos(x) to [0, π]. 1 Recall from Section 5.2 that the inverse of a function f is typically denoted f −1. For this reason, some textbooks use the notation f −1(x) = cos−1(x) for the inverse of f (x) = cos(x). The obvious pitfall here is our
convention of writing (cos(x))2 as cos2(x), (cos(x))3 as cos3(x) and so on. It is far too easy to confuse cos−1(x) with cos(x) = sec(x) so we will not use this notation in our text.1 Instead, we use the notation f −1(x) = arccos(x), read ‘arc-cosine of x’. To understand the ‘arc’ in ‘arccosine’, recall that an inverse function, by deﬁnition, reverses the process of the original function. The function f (t) = cos(t) takes a real number input t, associates it with the angle θ = t radians, and returns the value cos(θ). Digging deeper,2 we have that cos(θ) = cos(t) is the x-coordinate of the terminal point on the Unit Circle of an oriented arc of length \|t\| whose initial point is (1, 0). Hence, we may view the inputs to f (t) = cos(t) as oriented arcs and the outputs as x-coordinates on the Unit Circle. The function f −1, then, would take x-coordinates on the Unit Circle and return oriented arcs, hence the ‘arc’ in arccosine. Below are the graphs of f (x) = cos(x) and f −1(x) = arccos(x), where we obtain the latter from the former by reﬂecting it across the line y = x, in accordance with Theorem 5.3. y 1 −x) = cos(x), 0 ≤ x ≤ π reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x f −1(x) = arccos(x). 1But be aware that many books do! As always, be sure to check the context! 2See page 704 if you need a review of how we associate real numbers with angles in radian measure. 820 Foundations of Trigonometry We restrict g(x) = sin(x) in a similar manner, although the interval of choice is − π 2, π 2 y. x Restricting the domain of f (x) = sin(x
) to − π 2, π 2. It should be no surprise that we call g−1(x) = arcsin(x), which is read ‘arc-sine of x’. y 1 − π 2 x π 2 −1 g(x) = sin(x), − 1 1 x reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates − π 2 g−1(x) = arcsin(x). We list some important facts about the arccosine and arcsine functions in the following theorem. Theorem 10.26. Properties of the Arccosine and Arcsine Functions Properties of F (x) = arccos(x) – Domain: [−1, 1] – Range: [0, π] – arccos(x) = t if and only if 0 ≤ t ≤ π and cos(t) = x – cos(arccos(x)) = x provided −1 ≤ x ≤ 1 – arccos(cos(x)) = x provided 0 ≤ x ≤ π Properties of G(x) = arcsin(x) – Domain: [−1, 1] – Range – arcsin(x) = t if and only if − π – sin(arcsin(x)) = x provided − – arcsin(sin(x)) = x provided − π – additionally, arcsine is odd 2 2 and sin(t) = x 10.6 The Inverse Trigonometric Functions 821 Everything in Theorem 10.26 is a direct consequence of the facts that f (x) = cos(x) for 0 ≤ x ≤ π and F (x) = arccos(x) are inverses of each other as are g(x) = sin(x) for − π 2 and G(x) = arcsin(x). It’s about time for an example. 2 ≤ x ≤ π Example 10.6.1. 1. Find the exact values of the following. (a) arccos 1 2 (c) arccos − √ 2 2 (e) arccos cos π 6 (g) cos arccos − 3 5 (b) arcsin √ 2 2 (d) arcsin − 1 2 (f) arccos
cos 11π 6 (h) sin arccos − 3 5 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan (arccos (x)) (b) cos (2 arcsin(x)) Solution. 1. (a) To ﬁnd arccos 1 2, we need to ﬁnd the real number t (or, equivalently, an angle measuring 3 meets these 2. We know t = π 2 and π 2 with sin(t) = √ 2 2. The (b) The value of arcsin t radians) which lies between 0 and π with cos(t) = 1 = π criteria, so arccos 1 3. 2 √ 2 is a real number t between − π 2 √ 2 number we seek is t = π 4. Hence, arcsin 2 = π 4. √ (c) The number t = arccos √ − 2 2 lies in the interval [0, π] with cos(t) = − √ 2 2. Our answer 2 2 − is arccos (d) To ﬁnd arcsin − 1 2 answer is t = − π 6 so that arcsin − 1 = 3π 4., we seek the number t in the interval −, we could simply invoke Theorem 10.26 to get arccos cos π 6. However, in order to make sure we understand why this is the case, we choose to work the example through using the deﬁnition of arccosine. Working from the inside out, arccos cos π is the real number t with 0 ≤ t ≤ π 6 with sin(t) = − 1 (e) Since. The 2, π 2 6 2 = arccos √ 3 2. We ﬁnd t = π. Now, arccos 6, so that arccos cos π 6 = π 6. and cos(t) = 822 Foundations of Trigonometry (f) Since 11π 6 does not fall between 0 and π, Theorem 10.26 does not apply. We are forced to. From work through from the inside out starting with arccos cos 11π 6 = arccos 6. Hence, arccos cos 11π is
to use Theorem 10.26 directly. Since − the previous problem, we know arccos (g) One way to simplify cos arccos − 3 5 between −1 and 1, we have that cos arccos − 3 5 before, to really understand why this cancellation occurs, we let t = arccos − 3 5 by deﬁnition, cos(t) = − 3 in (nearly) the same amount of time. 5 is 5 and we are done. However, as. Then, 5, and we are ﬁnished 5. Hence, cos arccos − 3 = cos(t) = − 3 = − 3 5 2. (h) As in the previous example, we let t = arccos − 3 5 5 for some t where 0 ≤ t ≤ π. Since cos(t) < 0, we can narrow this down a bit and conclude that π 2 < t < π, so that t corresponds to an angle in Quadrant II. In terms of t, then, we need to ﬁnd = sin(t). Using the Pythagorean Identity cos2(t) + sin2(t) = 1, we get sin arccos − 3 5 − 3 5. Since t corresponds to a Quadrants II angle, we 5 choose sin(t) = 4 + sin2(t) = 1 or sin(t) = ± 4 5. Hence, sin arccos − 3 so that cos(ta) We begin this problem in the same manner we began the previous two problems. To help us see the forest for the trees, we let t = arccos(x), so our goal is to ﬁnd a way to express tan (arccos (x)) = tan(t) in terms of x. Since t = arccos(x), we know cos(t) = x where 0 ≤ t ≤ π, but since we are after an expression for tan(t), we know we need to throw out t = π 2 < t ≤ π so that, geometrically, t corresponds to an angle in Quadrant I or Quadrant II. One approach3 to ﬁnding tan(t) is to use the quotient identity tan(t) = sin(t) cos(t). Substituting cos(t) = x into the Pythagorean Identity cos2(
t) + sin2(t) = 1 gives x2 + sin2(t) = 1, from which we 1 − x2. Since t corresponds to angles in Quadrants I and II, sin(t) ≥ 0, get sin(t) = ± so we choose sin(t) = 2 from consideration. Hence, either 0 ≤ t < π 1 − x2. Thus, 2 or π √ √ tan(t) = sin(t) cos(t) = √ 1 − x2 x To determine the values of x for which this equivalence is valid, we consider our substitution t = arccos(x). Since the domain of arccos(x) is [−1, 1], we know we must restrict −1 ≤ x ≤ 1. Additionally, since we had to discard t = π 2, we need to discard x = cos π 2 = 0. Hence, tan (arccos (x)) = is valid for x in [−1, 0) ∪ (0, 1]. 1−x2 x √ 2, π (b) We proceed as in the previous problem by writing t = arcsin(x) so that t lies in the with sin(t) = x. We aim to express cos (2 arcsin(x)) = cos(2t) in terms interval − π of x. Since cos(2t) is deﬁned everywhere, we get no additional restrictions on t as we did in the previous problem. We have three choices for rewriting cos(2t): cos2(t) − sin2(t), 2 cos2(t) − 1 and 1 − 2 sin2(t). Since we know x = sin(t), it is easiest to use the last form: 2 cos (2 arcsin(x)) = cos(2t) = 1 − 2 sin2(t) = 1 − 2x2 3Alternatively, we could use the identity: 1 + tan2(t) = sec2(t). Since x = cos(t), sec(t) = 1 cos(t) = 1 x. The reader is invited to work through this approach to see what, if any, diﬃculties arise. 10.6 The Inverse Trigonometric Functions 823 To ﬁnd the restrictions on x, we once again appeal to our substitution t = arcsin(
x). Since arcsin(x) is deﬁned only for −1 ≤ x ≤ 1, the equivalence cos (2 arcsin(x)) = 1−2x2 is valid only on [−1, 1]. A few remarks about Example 10.6.1 are in order. Most of the common errors encountered in dealing with the inverse circular functions come from the need to restrict the domains of the original functions so that they are one-to-one. One instance of this phenomenon is the fact that arccos cos 11π 6 as opposed to 11π 6. This is the exact same phenomenon discussed in Section 6 (−2)2 = 2 as opposed to −2. Additionally, even though the expression we 5.2 when we saw arrived at in part 2b above, namely 1 − 2x2, is deﬁned for all real numbers, the equivalence cos (2 arcsin(x)) = 1 − 2x2 is valid for only −1 ≤ x ≤ 1. This is akin to the fact that while the x)2 = x is valid only for x ≥ 0. For expression x is deﬁned for all real numbers, the equivalence ( this reason, it pays to be careful when we determine the intervals where such equivalences are valid. = π √ The next pair of functions we wish to discuss are the inverses of tangent and cotangent, which are named arctangent and arccotangent, respectively. First, we restrict f (x) = tan(x) to its fundamental cycle on − π to obtain f −1(x) = arctan(x). Among other things, note that the 2, π 2 vertical asymptotes x = − π 2 and x = π 2 of the graph of f (x) = tan(x) become the horizontal 2 and y = π asymptotes y = − π 2 of the graph of f −1(x) = arctan(x). y 1 −x) = tan(x), − π 2 < x < π 2. reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates y π 2 π 4 −1(x) = arctan(x). Next, we restrict g(x) = cot(x) to
its fundamental cycle on (0, π) to obtain g−1(x) = arccot(x). Once again, the vertical asymptotes x = 0 and x = π of the graph of g(x) = cot(x) become the horizontal asymptotes y = 0 and y = π of the graph of g−1(x) = arccot(x). We show these graphs on the next page and list some of the basic properties of the arctangent and arccotangent functions. 824 Foundations of Trigonometry y 1 −1 π 4 π 2 3π 4 π x y π 3π 4 π 2 π 4 g(x) = cot(x), 0 < x < π. reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates g−1(x) = arccot(x). −1 1 x Theorem 10.27. Properties of the Arctangent and Arccotangent Functions Properties of F (x) = arctan(x) – Domain: (−∞, ∞) – Range: − π 2, π 2 – as x → −∞, arctan(x) → − π 2 – arctan(x) = t if and only if − π for x > 0 – arctan(x) = arccot 1 x – tan (arctan(x)) = x for all real numbers x 2 < x < π – arctan(tan(x)) = x provided − π – additionally, arctangent is odd 2 +; as x → ∞, arctan(x and tan(t) = x − Properties of G(x) = arccot(x) – Domain: (−∞, ∞) – Range: (0, π) – as x → −∞, arccot(x) → π−; as x → ∞, arccot(x) → 0+ – arccot(x) = t if and only if 0 < t < π and cot(t) = x – arccot(x) = arctan 1 x – cot (arccot(x)) = x for all real numbers x
for x > 0 – arccot(cot(x)) = x provided 0 < x < π 10.6 The Inverse Trigonometric Functions 825 Example 10.6.2. 1. Find the exact values of the following. √ (a) arctan( 3) (c) cot(arccot(−5)) √ 3) (b) arccot(− (d) sin arctan − 3 4 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan(2 arctan(x)) (b) cos(arccot(2x)) Solution. 1. 2. (a) We know arctan( 3, so arctan( t = π 3) is the real number t between − π 3) = π 3. √ √ √ (b) The real number t = arccot(− 3) = 5π 6. arccot(− √ 2 and π 2 with tan(t) = √ 3. We ﬁnd 3) lies in the interval (0, π) with cot(t) = − 3. We get √ (c) We can apply Theorem 10.27 directly and obtain cot(arccot(−5)) = −5. However, working it through provides us with yet another opportunity to understand why this is the case. Letting t = arccot(−5), we have that t belongs to the interval (0, π) and cot(t) = −5. Hence, cot(arccot(−5)) = cot(t) = −5. (a) If we let t = arctan(x), then − π (d) We start simplifying sin arctan − 3 4. Then tan(t) = − 3 by letting t = arctan −, we choose csc(t) = − 5 2 3, so sin(t) = − 3 3. Substituting, we get 1 + − 4 2. Since tan(t) < 0, we know, in fact, − π 4 for some − π 2 < t < 0. One way to proceed is to use The Pythagorean Identity, 1+cot2(t) = c
sc2(t), since this relates the reciprocals of tan(t) and sin(t) and is valid for all t under consideration.4 From tan(t) = − 3 4, we get cot(t) = − 4 3. Since = − 3 − π 5. = csc2(t) so that csc(t) = ± 5 5. Hence, sin arctan − 3 2 and tan(t) = x. We look for a way to express tan(2 arctan(x)) = tan(2t) in terms of x. Before we get started using identities, we note that tan(2t) is undeﬁned when 2t = π 2 + πk for integers k. Dividing both sides of this equation by 2 tells us we need to exclude values of t where t = π 4 + π 2 k, where k is are t = ± π an integer. The only members of this family which lie in − π 2, π 4, which 2. Returning ∪ − π means the values of t under consideration are − to arctan(2t), we note the double angle identity tan(2t) = 2 tan(t) 1−tan2(t), is valid for all the values of t under consideration, hence we get tan(2 arctan(x)) = tan(2t) = 2 tan(t) 1 − tan2(t) = 2x 1 − x2 4It’s always a good idea to make sure the identities used in these situations are valid for all values t under consideration. Check our work back in Example 10.6.1. Were the identities we used there valid for all t under consideration? A pedantic point, to be sure, but what else do you expect from this book? 826 Foundations of Trigonometry To ﬁnd where this equivalence is valid we check back with our substitution t = arctan(x). Since the domain of arctan(x) is all real numbers, the only exclusions come from the values of t we discarded earlier, t = ± π 4. Since x = tan(t), this means we exclude x = tan ± π 1−x2 holds for all x in 4 (−∞, −1) ∪ (−1, 1) ∪ (1, ∞).
= ±1. Hence, the equivalence tan(2 arctan(x)) = 2x (b) To get started, we let t = arccot(2x) so that cot(t) = 2x where 0 < t < π. In terms of t, cos(arccot(2x)) = cos(t), and our goal is to express the latter in terms of x. Since cos(t) is always deﬁned, there are no additional restrictions on t, so we can begin using identities to relate cot(t) to cos(t). The identity cot(t) = cos(t) sin(t) is valid for t in (0, π), so our strategy is to obtain sin(t) in terms of x, then write cos(t) = cot(t) sin(t). The identity 1 + cot2(t) = csc2(t) holds for all t in (0, π) and relates cot(t) and csc(t) = 1 sin(t). Substituting cot(t) = 2x, we get 1 + (2x)2 = csc2(t), or csc(t) = ± 4x2 + 1. Since t is. Hence, between 0 and π, csc(t) > 0, so csc(t) = 4x2 + 1 which gives sin(t) = 1√ √ √ 4x2+1 cos(arccot(2x)) = cos(t) = cot(t) sin(t) = √ 2x 4x2 + 1 Since arccot(2x) is deﬁned for all real numbers x and we encountered no additional restrictions on t, we have cos (arccot(2x)) = 2x√ for all real numbers x. 4x2+1 The last two functions to invert are secant and cosecant. A portion of each of their graphs, which were ﬁrst discussed in Subsection 10.5.2, are given below with the fundamental cycles highlighted. y y x x The graph of y = sec(x). The graph of y = csc(x). It is clear from the graph of secant that we cannot ﬁnd one single continuous piece of its
graph which covers its entire range of (−∞, −1] ∪ [1, ∞) and restricts the domain of the function so that it is one-to-one. The same is true for cosecant. Thus in order to deﬁne the arcsecant and arccosecant functions, we must settle for a piecewise approach wherein we choose one piece to cover the top of the range, namely [1, ∞), and another piece to cover the bottom, namely (−∞, −1]. There are two generally accepted ways make these choices which restrict the domains of these functions so that they are one-to-one. One approach simpliﬁes the Trigonometry associated with the inverse functions, but complicates the Calculus; the other makes the Calculus easier, but the Trigonometry less so. We present both points of view. 10.6 The Inverse Trigonometric Functions 827 10.6.1 Inverses of Secant and Cosecant: Trigonometry Friendly Approach In this subsection, we restrict the secant and cosecant functions to coincide with the restrictions on cosine and sine, respectively. For f (x) = sec(x), we restrict the domain to 0x) = sec(x) on 0, π 2 ∪ π 2, π reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x f −1(x) = arcsec(x) and we restrict g(x) = csc(x) to − π 2, 0 ∪ 0, π 2. y 1 −1 − π 2 x π 2 g(x) = csc(x) on − π 2, 0 ∪ 0, π 2 y π 2 −1 1 x reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates − π 2 g−1(x) = arccsc(x) Note that for both arcsecant and arccosecant, the domain is (−∞, −1] ∪ [1, ∞). Taking a page from Section 2.2, we can rewrite this as {x : \|x\| ≥ 1}. This is often done in Calculus textbooks,
so we include it here for completeness. Using these deﬁnitions, we get the following properties of the arcsecant and arccosecant functions. 828 Foundations of Trigonometry Theorem 10.28. Properties of the Arcsecant and Arccosecant Functionsa Properties of F (x) = arcsec(x) ∪ π – Domain: {x : \|x\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – Range: 0, π 2, π 2 – as x → −∞, arcsec(x) → π 2 – arcsec(x) = t if and only if 0 ≤ t < π – arcsec(x) = arccos 1 x – sec (arcsec(x)) = x provided \|x\| ≥ 1 – arcsec(sec(x)) = x provided 0 ≤ x < π provided \|x\| ≥ 1 2 or π 2 or π 2 < x ≤ π +; as x → ∞, arcsec(x and sec(t) = x Properties of G(x) = arccsc(x) 2 2, 0 ∪ 0, π – Domain: {x : \|x\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – Range: − π – as x → −∞, arccsc(x) → 0−; as x → ∞, arccsc(x) → 0+ – arccsc(x) = t if and only if − π – arccsc(x) = arcsin 1 x – csc (arccsc(x)) = x provided \|x\| ≥ 1 – arccsc(csc(x)) = x provided − π – additionally, arccosecant is odd 2 ≤ x < 0 or or 0 < t ≤ π provided \|x\| ≥ 1 2 2 and csc(t) = x a... assuming the “Trigonometry Friendly” ranges are used. Example 10.6.3. 1. Find the exact values of the following. (a) arcsec(2) (b) arccsc(−2) (c) arcsec sec 5π 4 (d) cot (arccsc (−3)) 2. Rewrite the following as
algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan(arcsec(x)) (b) cos(arccsc(4x)) 10.6 The Inverse Trigonometric Functions 829 Solution. 1. 2. (a) Using Theorem 10.28, we have arcsec(2) = arccos 1 2 (b) Once again, Theorem 10.28 comes to our aid giving arccsc(−2) = arcsin − 1 2 (c) Since 5π 4 doesn’t fall between 0 and π 2 and π, we cannot use the inverse property stated in Theorem 10.28. We can, nevertheless, begin by working ‘inside out’ which yields arcsec sec 5π 4 = arcsec(− = − π 6. 2) = arccos = π 3. 2 or π = 3π 4. 2 2 √ − √ (d) One way to begin to simplify cot (arccsc (−3)) is to let t = arccsc(−3). Then, csc(t) = −3 and, since this is negative, we have that t lies in the interval − π 2, 0. We are after cot (arccsc (−3)) = cot(t), so we use the Pythagorean Identity 1 + cot2(t) = csc2(t). Substituting, we have 1 + cot2(t) = (−3)2, or cot(t) = ± 2 ≤ t < 0, cot(t) < 0, so we get cot (arccsc (−3)) = −2 2. Since − π 8 = ±2 √ √ √ 2. ∪ π (a) We begin simplifying tan(arcsec(x)) by letting t = arcsec(x). Then, sec(t) = x for t in 0, π 2, π, and we seek a formula for tan(t). Since tan(t) is deﬁned for all t values 2 under consideration, we have no additional restrictions on t. To relate sec(t) to tan(t), we use the identity 1 + tan2(t) = sec2(t). This is valid for all values of t under consideration, √
and when we substitute sec(t) = x, we get 1 + tan2(t) = x2. Hence, tan(t) = ± x2 − 1. 2, π then then tan(t) ≥ 0; if, on the the other hand, t belongs to π If t belongs to 0, π 2 tan(t) ≤ 0. As a result, we get a piecewise deﬁned function for tan(t) tan(t) = √ √ − x2 − 1, x2 − 1, if 0 ≤ t < π 2 if π 2 < t ≤ π Now we need to determine what these conditions on t mean for x. Since x = sec(t), when, x ≤ −1. Since we encountered no further restrictions on t, the equivalence below holds for all x in (−∞, −1] ∪ [1, ∞). 2, x ≥ 1, and when π tan(arcsec(x)) = √ √ − x2 − 1, if x ≥ 1 x2 − 1, if x ≤ −1 2, 0∪0, π (b) To simplify cos(arccsc(4x)), we start by letting t = arccsc(4x). Then csc(t) = 4x for t in, and we now set about ﬁnding an expression for cos(arccsc(4x)) = cos(t). − π Since cos(t) is deﬁned for all t, we do not encounter any additional restrictions on t. From csc(t) = 4x, we get sin(t) = 1 4x, so to ﬁnd cos(t), we can make use if the identity cos2(t) + sin2(t) = 1. Substituting sin(t) = 1 = 1. Solving, we get 2 2 cos(t) = ± 4x gives cos2(t) + 1 √ 4x 16x2 − 1 16x2 = ± 16x2 − 1 4\|x\| Since t belongs to − π. (The absolute values here are necessary, since x could be negative.) To ﬁnd the values for, we know cos(t) ≥ 0, so we choose cos(t) = 2, 0 ∪ 0, π 16
−x2 4\|x\| 2 √ 830 Foundations of Trigonometry which this equivalence is valid, we look back at our original substution, t = arccsc(4x). Since the domain of arccsc(x) requires its argument x to satisfy \|x\| ≥ 1, the domain of arccsc(4x) requires \|4x\| ≥ 1. Using Theorem 2.4, we rewrite this inequality and solve to get x ≤ − 1 4. Since we had no additional restrictions on t, the equivalence ∪ 1 cos(arccsc(4x)) = holds for all x in −∞, − 1 4 4 or x ≥ 1 16x2−1 4\|x\| 4, ∞. √ Inverses of Secant and Cosecant: Calculus Friendly Approach 10.6.2 In this subsection, we restrict f (x) = sec(x) to 0, π 2 y 1 −1 π 2 π 3π 2 x ∪ π, 3π 2 y 3π 2 π π 2 f (x) = sec(x) on 0, π 2 ∪ π, 3π 2 reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x f −1(x) = arcsec(x) and we restrict g(x) = csc(x) to 0, π 2 ∪ π, 3π 2. y 1 −1 π 2 π 3π 2 x y 3π 2 π π 2 g(x) = csc(x) on 0, π 2 ∪ π, 3π 2 reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x g−1(x) = arccsc(x) Using these deﬁnitions, we get the following result. 10.6 The Inverse Trigonometric Functions 831 Theorem 10.29. Properties of the Arcsecant and Arccosecant Functionsa Properties of F (x) = arcsec(x) ∪ π, 3π 2 – Domain: {x : \|x\| ≥ 1} = (−∞, −1] ∪ [1, �
�) – Range: 0, π 2 – as x → −∞, arcsec(x) → 3π 2 – arcsec(x) = t if and only if 0 ≤ t < π – arcsec(x) = arccos 1 for x ≥ 1 onlyb x – sec (arcsec(x)) = x provided \|x\| ≥ 1 – arcsec(sec(x)) = x provided 0 ≤ x < π − 2 or π ≤ x < 3π 2 ; as x → ∞, arcsec(x) → π 2 2 or π ≤ t < 3π − 2 and sec(t) = x Properties of G(x) = arccsc(x) ∪ π, 3π 2 – Domain: {x : \|x\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – Range: 0, π 2 – as x → −∞, arccsc(x) → π+; as x → ∞, arccsc(x) → 0+ – arccsc(x) = t if and only if 0 < t ≤ π for x ≥ 1 onlyc – arccsc(x) = arcsin 1 x – csc (arccsc(x)) = x provided \|x\| ≥ 1 – arccsc(csc(x)) = x provided 0 < x ≤ π 2 or π < t ≤ 3π 2 or π < x ≤ 3π 2 2 and csc(t) = x a... assuming the “Calculus Friendly” ranges are used. bCompare this with the similar result in Theorem 10.28. cCompare this with the similar result in Theorem 10.28. Our next example is a duplicate of Example 10.6.3. The interested reader is invited to compare and contrast the solution to each. Example 10.6.4. 1. Find the exact values of the following. (a) arcsec(2) (b) arccsc(−2) (c) arcsec sec 5π 4 (d) cot (arccsc (−3)) 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan(arcsec(x)) (b) cos(arccsc(4x
)) 832 Solution. Foundations of Trigonometry 1. 2. (a) Since 2 ≥ 1, we may invoke Theorem 10.29 to get arcsec(2) = arccos 1 2 (b) Unfortunately, −2 is not greater to or equal to 1, so we cannot apply Theorem 10.29 to arccsc(−2) and convert this into an arcsine problem. Instead, we appeal to the deﬁnition. and satisﬁes csc(t) = −2. The t The real number t = arccsc(−2) lies in 0, π 2 we’re after is t = 7π ∪ π, 3π 2 6, so arccsc(−2) = 7π 6. = π 3. lies between π and 3π (c) Since 5π 4 arcsec sec 5π 4 tion as we have done in the previous examples to see how it goes. 2, we may apply Theorem 10.29 directly to simplify 4. We encourage the reader to work this through using the deﬁni- = 5π (d) To simplify cot (arccsc (−3)) we let t = arccsc (−3) so that cot (arccsc (−3)) = cot(t).. Using the identity 2. Since √ 2. We know csc(t) = −3, and since this is negative, t lies in π, 3π 2 1 + cot2(t) = csc2(t), we ﬁnd 1 + cot2(t) = (−3)2 so that cot(t) = ± t is in the interval π, 3π 2, we know cot(t) > 0. Our answer is cot (arccsc (−3)) = 2 8 = ±2 √ √ ∪ π, 3π 2 (a) We begin simplifying tan(arcsec(x)) by letting t = arcsec(x). Then, sec(t) = x for t in 0, π, and we seek a formula for tan(t). Since tan(t) is deﬁned for all t values 2 under consideration, we have no additional restrictions on t. To relate sec(t) to tan(t), we use the identity 1 + tan2(
t) = sec2(t). This is valid for all values of t under consideration, and when we substitute sec(t) = x, we get 1 + tan2(t) = x2. Hence, tan(t) = ± x2 − 1. Since t lies in 0, π, tan(t) ≥ 0, so we choose tan(t) = x2 − 1. Since we found √ 2 x2 − 1 holds for all x no additional restrictions on t, the equivalence tan(arcsec(x)) = in the domain of t = arcsec(x), namely (−∞, −1] ∪ [1, ∞). ∪ π, 3π 2 √ √ ∪ π, 3π 2 (b) To simplify cos(arccsc(4x)), we start by letting t = arccsc(4x). Then csc(t) = 4x for t in, and we now set about ﬁnding an expression for cos(arccsc(4x)) = cos(t). 0, π 2 Since cos(t) is deﬁned for all t, we do not encounter any additional restrictions on t. From csc(t) = 4x, we get sin(t) = 1 4x, so to ﬁnd cos(t), we can make use if the identity cos2(t) + sin2(t) = 1. Substituting sin(t) = 1 = 1. Solving, we get 2 cos(t) = ± 4x gives cos2(t) + 1 √ 4x 16x2 − 1 16x2 = ± 16x2 − 1 4\|x\|, then cos(t) ≥ 0, and we choose cos(t) = in which case cos(t) ≤ 0, so, we choose cos(t) = − If t lies in 0, π 2 to π, 3π 2 (momentarily) piecewise deﬁned function for cos(t) √ 16x2−1 4\|x\| √. Otherwise, t belongs 16x2−1 4\|x\| This leads us to a cos(t) = √ √ −    16x2 − 1 4\|x
\| 16x2 − 1 4\|x\|,, if 0 ≤ t ≤ π 2 if π < t ≤ 3π 2 10.6 The Inverse Trigonometric Functions 833 We now see what these restrictions mean in terms of x. Since 4x = csc(t), we get that for 0 ≤ t ≤ π 2, 4x ≥ 1, or x ≥ 1 4. In this case, we can simplify \|x\| = x so √ √ cos(t) = 16x2 − 1 4\|x\| = 16x2 − 1 4x Similarly, for π < t ≤ 3π get 2, we get 4x ≤ −1, or x ≤ − 1 4. In this case, \|x\| = −x, so we also cos(t) = − √ 16x2 − 1 4\|x\| = − √ 16x2 − 1 4(−x) = √ 16x2 − 1 4x 16x2−1 Hence, in all cases, cos(arccsc(4x)) = 4x the domain of t = arccsc(4x), namely −∞, − 1 4, and this equivalence is valid for all x in ∪ 1 4, ∞ √ 10.6.3 Calculators and the Inverse Circular Functions. In the sections to come, we will have need to approximate the values of the inverse circular functions. On most calculators, only the arcsine, arccosine and arctangent functions are available and they are usually labeled as sin−1, cos−1 and tan−1, respectively. If we are asked to approximate these values, it is a simple matter to punch up the appropriate decimal on the calculator. If we are asked for an arccotangent, arcsecant or arccosecant, however, we often need to employ some ingenuity, as our next example illustrates. Example 10.6.5. 1. Use a calculator to approximate the following values to four decimal places. (a) arccot(2) (b) arcsec(5) (c) arccot(−2) (d) arccsc − 3 2 2. Find the domain and range of the following functions. Check your answers using a calculator. (a) f (x) = π 2 − arccos x 5 (b) f (x) =
3 arctan (4x). (c) f (x) = arccot x 2 + π Solution. 1. (a) Since 2 > 0, we can use the property listed in Theorem 10.27 to rewrite arccot(2) as arccot(2) = arctan 1 2. In ‘radian’ mode, we ﬁnd arccot(2) = arctan 1 2 ≈ 0.4636. (b) Since 5 ≥ 1, we can use the property from either Theorem 10.28 or Theorem 10.29 to write arcsec(5) = arccos 1 5 ≈ 1.3694. 834 Foundations of Trigonometry (c) Since the argument −2 is negative, we cannot directly apply Theorem 10.27 to help us ﬁnd arccot(−2). Let t = arccot(−2). Then t is a real number such that 0 < t < π and cot(t) = −2. Moreover, since cot(t) < 0, we know π 2 < t < π. Geometrically, this means t corresponds to a Quadrant II angle θ = t radians. This allows us to proceed using a ‘reference angle’ approach. Consider α, the reference angle for θ, as pictured below. By deﬁnition, α is an acute angle so 0 < α < π 2, and the Reference Angle Theorem, Theorem 10.2, tells us that cot(α) = 2. This means α = arccot(2) radians. Since the argument of arccotangent is now a positive 2, we can use Theorem 10.27 to get α = arccot(2) = arctan 1 ≈ 2.6779 radians, 2 we get arccot(−2) ≈ 2.6779. radians. Since θ = π − α = π − arctan 1 2 y 1 θ = arccot(−2) radians α 1 x. By deﬁnition, the real number Another way to attack the problem is to use arctan − 1 2 t = arctan − 1 satisﬁes tan(t) = − 1 2. Since tan(t
) < 0, we know 2 more speciﬁcally that − π 2 < t < 0, so t corresponds to an angle β in Quadrant IV. To ﬁnd the value of arccot(−2), we once again visualize the angle θ = arccot(−2) radians and note that it is a Quadrant II angle with tan(θ) = − 1 2. This means it is exactly π units away from β, and we get θ = π + β = π + arctan − 1 ≈ 2.6779 radians. Hence, 2 as before, arccot(−2) ≈ 2.6779. 2 < t < π 2 with − π 10.6 The Inverse Trigonometric Functions 835 y 1 θ = arccot(−2) radians, π (d) If the range of arccosecant is taken to be − π = arcsin − 2 3 ∪ π, 3π 2, we can use Theorem 10.28 to ≈ −0.7297. If, on the other hand, the range of arccosecant get arccsc − 3 2 is taken to be 0, π, then we proceed as in the previous problem by letting 2. Then t is a real number with csc(t) = − 3 t = arccsc − 3 2. Since csc(t) < 0, we have 2 that π < θ ≤ 3π 2, so t corresponds to a Quadrant III angle, θ. As above, we let α be 2, which means α = arccsc 3 the reference angle for θ. Then 0 < α < π 2 radians. Since the argument of arccosecant is now positive, we may use Theorem 10.29 to get α = arccsc 3 ≈ 3.8713 = arcsin 2 2 3 ≈ 3.8713. radians, arccsc − 3 2 radians. Since θ = π + α = π + arcsin 2 3 2 and csc(α) = 3 y 1 θ = arccsc − 3 2 radians α 1 x 836 2. Foundations of Trigonometry 5 by setting the argument of the arccosine, in this case x (a) Since
the domain of F (x) = arccos(x) is −1 ≤ x ≤ 1, we can ﬁnd the domain of 2 − arccos x f (x) = π 5, between −1 and 1. Solving −1 ≤ x 5 ≤ 1 gives −5 ≤ x ≤ 5, so the domain is [−5, 5]. To determine the range of f, we take a cue from Section 1.7. Three ‘key’ points on the graph of and (1, 0). Following the procedure outlined in F (x) = arccos(x) are (−1, π), 0, π 2 Theorem 1.7, we track these points to −5, − π. Plotting these values 2 tells us that the range5 of f is − π, (0, 0) and 5, π 2. Our graph conﬁrms our results. 2, π 2 (b) To ﬁnd the domain and range of f (x) = 3 arctan (4x), we note that since the domain of F (x) = arctan(x) is all real numbers, the only restrictions, if any, on the domain of f (x) = 3 arctan (4x) come from the argument of the arctangent, in this case, 4x. Since 4x is deﬁned for all real numbers, we have established that the domain of f is all real numbers. To determine the range of f, we can, once again, appeal to Theorem 1.7. Choosing our ‘key’ point to be (0, 0) and tracking the horizontal asymptotes y = − π 2 and y = π 2, we ﬁnd that the graph of y = f (x) = 3 arctan (4x) diﬀers from the graph of y = F (x) = arctan(x) by a horizontal compression by a factor of 4 and a vertical stretch by a factor of 3. It is the latter which aﬀects the range, producing a range of − 3π. We conﬁrm our ﬁndings on the calculator below. 2, 3π 2 y = f (x) = − arccos y = f (x) = 3 ar
ctan (4x) π 2 x 5 (c) To ﬁnd the domain of g(x) = arccot x + π, we proceed as above. Since the domain of 2 G(x) = arccot(x) is (−∞, ∞), and x 2 is deﬁned for all x, we get that the domain of g is (−∞, ∞) as well. As for the range, we note that the range of G(x) = arccot(x), like that of F (x) = arctan(x), is limited by a pair of horizontal asymptotes, in this case y = 0 and y = π. Following Theorem 1.7, we graph y = g(x) = arccot x + π starting with 2 y = G(x) = arccot(x) and ﬁrst performing a horizontal expansion by a factor of 2 and following that with a vertical shift upwards by π. This latter transformation is the one which aﬀects the range, making it now (π, 2π). To check this graphically, we encounter a bit of a problem, since on many calculators, there is no shortcut button corresponding to the arccotangent function. Taking a cue from number 1c, we attempt to rewrite +π in terms of the arctangent function. Using Theorem 10.27, we have g(x) = arccot x 2 when x = arctan 2 that arccot x 2 > 0, or, in this case, when x > 0. Hence, for x > 0, x 2 + π. When x we have g(x) = arctan 2 2 < 0, we can use the same argument in number x. 1c that gave us arccot(−2) = π + arctan − 1 2 to give us arccot x 2 = π + arctan 2 x 5It also conﬁrms our domain! 10.6 The Inverse Trigonometric Functions 837 Hence, for x < 0, g(x) = π + arctan 2 + 2π. What about x = 0? We x know g(0) = arccot(0) + π = π, and neither of the formulas for g
involving arctangent will produce this result.6 Hence, in order to graph y = g(x) on our calculators, we need to write it as a piecewise deﬁned function: + π = arctan 2 x g(x) = arccot x 2 + π =    We show the input and the result below. arctan 2 x arctan 2 x + 2π, π, + π, if x < 0 if x = 0 if x > 0 y = g(x) in terms of arctangent y = g(x) = arccot x 2 + π The inverse trigonometric functions are typically found in applications whenever the measure of an angle is required. One such scenario is presented in the following example. Example 10.6.6. 7 The roof on the house below has a ‘6/12 pitch’. This means that when viewed from the side, the roof line has a rise of 6 feet over a run of 12 feet. Find the angle of inclination from the bottom of the roof to the top of the roof. Express your answer in decimal degrees, rounded to the nearest hundredth of a degree. Front View Side View Solution. If we divide the side view of the house down the middle, we ﬁnd that the roof line forms the hypotenuse of a right triangle with legs of length 6 feet and 12 feet. Using Theorem 10.10, we 6Without Calculus, of course... 7The authors would like to thank Dan Stitz for this problem and associated graphics. 838 Foundations of Trigonometry ﬁnd the angle of inclination, labeled θ below, satisﬁes tan(θ) = 6 we can use the arctangent function and we ﬁnd θ = arctan 1 2 12 = 1 radians ≈ 26.56◦. 2. Since θ is an acute angle, 6 feet θ 12 feet 10.6.4 Solving Equations Using the Inverse Trigonometric Functions. In Sections 10.2 and 10.3, we learned how to solve equations like sin(θ) = 1 2 for angles θ and tan(t) = −1 for real numbers t. In each case, we ultimately appealed to the Unit Circle and relied on
the fact that the answers corresponded to a set of ‘common angles’ listed on page 724. If, on the other hand, we had been asked to ﬁnd all angles with sin(θ) = 1 3 or solve tan(t) = −2 for real numbers t, we would have been hard-pressed to do so. With the introduction of the inverse trigonometric functions, however, we are now in a position to solve these equations. A good parallel to keep in mind is how the square root function can be used to solve certain quadratic equations. The equation x2 = 4 is a lot like sin(θ) = 1 2 in that it has friendly, ‘common value’ answers x = ±2. The equation x2 = 7, on the other hand, is a lot like sin(θ) = 1 3. We know8 there are answers, but we can’t express them using ‘friendly’ numbers.9 To solve x2 = 7, we make use of the square root function and write x = ± 7. We can certainly approximate these answers using a calculator, but as far as exact answers go, we leave them as x = ± 7. In the same way, we will use the arcsine function to solve sin(θ) = 1 3, as seen in the following example. √ √ Example 10.6.7. Solve the following equations. 1. Find all angles θ for which sin(θ) = 1 3. 2. Find all real numbers t for which tan(t) = −2 3. Solve sec(x) = − 5 3 for x. Solution. 1. If sin(θ) = 1 3, then the terminal side of θ, when plotted in standard position, intersects the Unit Circle at y = 1 3. Geometrically, we see that this happens at two places: in Quadrant I and Quadrant II. If we let α denote the acute solution to the equation, then all the solutions 8How do we know this again? 9This is all, of course, a matter of opinion. For the record, the authors ﬁnd ± √ 7 just as ‘nice’ as ±2. 10.6 The Inverse Trigonometric Functions 839 to this equation in Quadrant I are coterminal with α, and α serves as the reference angle
for all of the solutions to this equation in Quadrant II = arcsin 1 3 radians α 1 x Since 1 3 isn’t the sine of any of the ‘common angles’ discussed earlier, we use the arcsine functions to express our answers. The real number t = arcsin 1 is deﬁned so it satisﬁes 3 radians. Since the solutions in Quadrant I 0 < t < π + 2πk are all coterminal with α, we get part of our solution to be θ = α + 2πk = arcsin 1 3 for integers k. Turning our attention to Quadrant II, we get one solution to be π − α. Hence, the Quadrant II solutions are θ = π − α + 2πk = π − arcsin 1 3 3. Hence, α = arcsin 1 + 2πk, for integers k. 2 with sin(t) = 1 3 2. We may visualize the solutions to tan(t) = −2 as angles θ with tan(θ) = −2. Since tangent is negative only in Quadrants II and IV, we focus our eﬀorts there = arctan(−2) radians π β Since −2 isn’t the tangent of any of the ‘common angles’, we need to use the arctangent function to express our answers. The real number t = arctan(−2) satisﬁes tan(t) = −2 and − π 2 < t < 0. If we let β = arctan(−2) radians, we see that all of the Quadrant IV solutions 840 Foundations of Trigonometry to tan(θ) = −2 are coterminal with β. Moreover, the solutions from Quadrant II diﬀer by exactly π units from the solutions in Quadrant IV, so all the solutions to tan(θ) = −2 are of the form θ = β + πk = arctan(−2) + πk for some integer k. Switching back to the variable t, we record our ﬁnal answer to tan(t) = −2 as t = arctan(−2) + πk for integers k. 3. The last equation we are asked to solve
, sec(x) = − 5 3, poses two immediate problems. First, we are not told whether or not x represents an angle or a real number. We assume the latter, but note that we will use angles and the Unit Circle to solve the equation regardless. Second, as we have mentioned, there is no universally accepted range of the arcsecant function. For that reason, we adopt the advice given in Section 10.3 and convert this to the cosine problem cos(x) = − 3 5. Adopting an angle approach, we consider the equation cos(θ) = − 3 5 and note that our solutions lie in Quadrants II and III. Since − 3 5 isn’t the cosine of any of the ‘common angles’, we’ll need to express our solutions in terms of the arccosine function. The real number 5. If we let β = arccos − 3 t = arccos − 3 5 radians, we see that β is a Quadrant II angle. To obtain a Quadrant III angle solution, we may simply use −β = − arccos − 3. Since all angle solutions are coterminal with β 5 + 2πk or 5 to be θ = β + 2πk = arccos − 3 or −β, we get our solutions to cos(θ) = − 3 + 2πk for integers k. Switching back to the variable x, we θ = −β + 2πk = − arccos − 3 5 + 2πk record our ﬁnal answer to sec(x) = − 5 for integers k. + 2πk or x = − arccos − 3 5 2 < t < π with cos(t) = − 3 3 as x = arccos − 3 is deﬁned so that π 5 5 5 y 1 y 1 β = arccos − 3 5 radians β = arccos − 3 5 radians 1 x 1 −β = − arccos − 3 5 x radians The reader is encouraged to check the answers found in Example 10.6.7 - both analytically and with the calculator (see Section 10.6.3). With practice, the inverse trigonometric functions will become as familiar to you as the square root function. Speaking of practice... 10.6 The Inverse Trigonometric Functions 8
41 10.6.5 Exercises In Exercises 1 - 40, ﬁnd the exact value. 1. arcsin (−1) 2. arcsin − √ 3 2 5. arcsin (0) 6. arcsin 1 2 3. arcsin − √ 2 2 7. arcsin √ 2 2 4. arcsin − 1 2 8. arcsin √ 3 2 9. arcsin (1) 10. arccos (−1) 11. arccos − √ 3 2 12. arccos − √ 2 2 13. arccos − 1 2 17. arccos √ 3 2 21. arctan − √ 3 3 14. arccos (0) 15. arccos 1 2 16. arccos √ 2 2 18. arccos (1) 19. arctan − √ 3 20. arctan (−1) 22. arctan (0) 23. arctan 25. arctan √ 3 26. arccot − √ 3 29. arccot (0) 30. arccot √ 3 3 33. arcsec (2) 34. arccsc (2) 37. arcsec √ 2 3 3 38. arccsc √ 2 3 3 √ 3 3 24. arctan (1) 27. arccot (−1) 28. arccot − √ 3 3 31. arccot (1) 32. arccot √ 3 35. arcsec √ 2 36. arccsc √ 2 39. arcsec (1) 40. arccsc (1) In Exercises 41 - 48, assume that the range of arcsecant is 0, π 2 arccosecant is 0, π 2 when ﬁnding the exact value. ∪ π, 3π 2 ∪ π, 3π 2 and that the range of 41. arcsec (−2) 45. arccsc (−2) 42. arcsec − √ 46. arccsc − √ 2 2 43. arcsec − 47. arccsc − √ 2 3 3 √ 2 3 3 44. arcsec (−1) 48. arccsc (−1) 842 Foundations of Trigonometry In Exercises 49 - 56, assume that the range of arcsecant is
0, π 2 arccosecant is − π when ﬁnding the exact value. 2, 0 ∪ 0, π 2 ∪ π 2, π and that the range of 49. arcsec (−2) 53. arccsc (−2) 50. arcsec − √ 54. arccsc − √ 2 2 51. arcsec − 55. arccsc − √ 2 3 3 √ 2 3 3 52. arcsec (−1) 56. arccsc (−1) In Exercises 57 - 86, ﬁnd the exact value or state that it is undeﬁned. 57. sin arcsin 1 2 58. sin arcsin − √ 2 2 60. sin (arcsin (−0.42)) 61. sin arcsin 5 4 63. cos arccos − 1 2 64. cos arccos 5 13 66. cos (arccos (π)) 67. tan (arctan (−1)) 59. sin arcsin 3 5 √ 62. cos arccos 2 2 65. cos (arccos (−0.998)) 68. tan arctan √ 3 70. tan (arctan (0.965)) 71. tan (arctan (3π)) 69. tan arctan 5 12 72. cot (arccot (1)) 73. cot arccot − √ 3 75. cot (arccot (−0.001)) 76. cot arccot 78. sec (arcsec (−1)) 79. sec arcsec 17π 4 1 2 82. csc arccsc √ 2 81. sec (arcsec (117π)) 84. csc arccsc √ 2 2 74. cot arccot − 7 24 77. sec (arcsec (2)) 80. sec (arcsec (0.75)) 83. csc arccsc − √ 2 3 3 85. csc (arccsc (1.0001)) 86. csc arccsc π 4 In Exercises 87 - 106, ﬁnd the exact value or state that it is undeﬁned. 87. arcsin sin π 6 88. arcsin − sin π 3 89. arcsin sin 3π 4 10.6 The Inverse Trigonometric Functions 843 90. arcsin sin 93.
arccos 96. arccos cos cos 11π 6 2π 3 5π 4 91. arcsin sin 94. arccos 97. arctan 4π 3 3π 2 cos 92. arccos 95. arccos cos π 4 cos − π 6 tan tan π 3 π 2 98. arctan tan 101. arctan tan − π 4 2π 3 99. arctan (tan (π)) 100. arctan 102. arccot 105. arccot cot cot π 3 π 2 103. arccot cot 106. arccot cot − π 4 2π 3 104. arccot (cot (π)) In Exercises 107 - 118, assume that the range of arcsecant is 0, π 2 arccosecant is 0, π 2 when ﬁnding the exact value. ∪ π, 3π 2 117. arcsec sec 118. arccsc csc In Exercises 119 - 130, assume that the range of arcsecant is 0, π 2 arccosecant is − π when ﬁnding the exact value. 2, 0 ∪ 0, π 2 sec π 4 107. arcsec 110. arcsec sec 113. arccsc csc 116. arccsc csc − π 2 5π 4 11π 6 sec π 4 119. arcsec 122. arcsec sec 125. arccsc csc 128. arccsc csc − π 2 5π 4 11π 6 108. arcsec sec 111. arcsec sec 114. arccsc csc 120. arcsec sec 123. arcsec sec 126. arccsc csc 4π 3 5π 3 2π 3 11π 12 4π 3 5π 3 2π 3 11π 12 ∪ π, 3π 2 and that the range of 109. arcsec sec 5π 6 csc π 6 112. arccsc 115. arccsc csc − π 2 9π 8 ∪ π 2, π and that the range of 121. arcsec sec 5π 6 csc π 6 124. arccsc 127. arccsc csc − π 2 9π 8 129. arcsec sec 130. arccsc csc 844 Foundations of Trigonometry In Ex
ercises 131 - 154, ﬁnd the exact value or state that it is undeﬁned. 131. sin arccos − 1 2 132. sin arccos 3 5 134. sin arccot √ 5 137. cos arctan √ 7 135. sin (arccsc (−3)) 140. tan arcsin − √ 2 5 5 141. tan arccos − 1 2 143. tan (arccot (12)) 144. cot arcsin 12 13 138. cos (arccot (3)) 139. cos (arcsec (5)) 133. sin (arctan (−2)) 136. cos arcsin − 5 13 142. tan arcsec 145. cot arccos 5 3 √ 3 2 √ 3 2 146. cot arccsc √ 5 147. cot (arctan (0.25)) 148. sec arccos 149. sec arcsin − 12 13 150. sec (arctan (10)) 151. sec arccot − √ 10 10 152. csc (arccot (9)) 153. csc arcsin 3 5 154. csc arctan − 2 3 In Exercises 155 - 164, ﬁnd the exact value or state that it is undeﬁned. 155. sin arcsin 5 13 + π 4 156. cos (arcsec(3) + arctan(2)) 157. tan arctan(3) + arccos − 3 5 158. sin 2 arcsin − 4 5 159. sin 2arccsc 13 5 161. cos 2 arcsin 3 5 163. cos 2arccot − √ 5 160. sin (2 arctan (2)) 162. cos 2arcsec 25 7 164. sin arctan(2) 2 10.6 The Inverse Trigonometric Functions 845 In Exercises 165 - 184, rewrite the quantity as algebraic expressions of x and state the domain on which the equivalence is valid. 165. sin (arccos (x)) 166. cos (arctan (x)) 167. tan (arcsin (x)) 168. sec (arctan (x)) 169. csc (arccos (x)) 170. sin (2 arctan (x)) 171. sin (2 arccos (x)) 172. cos (2 arctan (

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