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x)) 173. sin(arccos(2x)) 174. sin arccos x 5 175. cos arcsin x 2 176. cos (arctan (3x)) 177. sin(2 arcsin(7x)) 178. sin 2 arcsin √ x 3 3 179. cos(2 arcsin(4x)) 180. sec(arctan(2x)) tan(arctan(2x)) 181. sin (arcsin(x) + arccos(x)) 182. cos (arcsin(x) + arctan(x)) 183. tan (2 arcsin(x)) 184. sin arctan(x) 1 2 185. If sin(θ) = 186. If tan(θ) = 187. If sec(θ) = x 2 x 7 x 4 for − π 2 for − π 2 < θ < π 2, ﬁnd an expression for θ + sin(2θ) in terms of x. < θ < π 2, ﬁnd an expression for 1 2 θ − 1 2 sin(2θ) in terms of x. for 0 < θ < π 2, ﬁnd an expression for 4 tan(θ) − 4θ in terms of x. In Exercises 188 - 207, solve the equation using the techniques discussed in Example 10.6.7 then approximate the solutions which lie in the interval [0, 2π) to four decimal places. 188. sin(x) = 7 11 189. cos(x) = − 2 9 190. sin(x) = −0.569 191. cos(x) = 0.117 192. sin(x) = 0.008 193. cos(x) = 194. tan(x) = 117 195. cot(x) = −12 196. sec(x) = 197. csc(x) = − 200. cos(x) = − 90 17 7 16 198. tan(x) = − √ 10 199. sin(x) = 201. tan(x) = 0.03 202. sin(x) = 0.3502 359 360 3 2 3 8 846 Foundations of Trigonometry 203. sin(x) = −0.721 204. cos(x) = 0.9824 205. cos(x) =
−0.5637 206. cot(x) = 1 117 207. tan(x) = −0.6109 In Exercises 208 - 210, ﬁnd the two acute angles in the right triangle whose sides have the given lengths. Express your answers using degree measure rounded to two decimal places. 208. 3, 4 and 5 209. 5, 12 and 13 210. 336, 527 and 625 211. A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it touches level ground 360 feet from the base of the tower. What angle does the wire make with the ground? Express your answer using degree measure rounded to one decimal place. 212. At Cliﬀs of Insanity Point, The Great Sasquatch Canyon is 7117 feet deep. From that point, a ﬁre is seen at a location known to be 10 miles away from the base of the sheer canyon wall. What angle of depression is made by the line of sight from the canyon edge to the ﬁre? Express your answer using degree measure rounded to one decimal place. 213. Shelving is being built at the Utility Muﬃn Research Library which is to be 14 inches deep. An 18-inch rod will be attached to the wall and the underside of the shelf at its edge away from the wall, forming a right triangle under the shelf to support it. What angle, to the nearest degree, will the rod make with the wall? 214. A parasailor is being pulled by a boat on Lake Ippizuti. The cable is 300 feet long and the parasailor is 100 feet above the surface of the water. What is the angle of elevation from the boat to the parasailor? Express your answer using degree measure rounded to one decimal place. 215. A tag-and-release program to study the Sasquatch population of the eponymous Sasquatch National Park is begun. From a 200 foot tall tower, a ranger spots a Sasquatch lumbering through the wilderness directly towards the tower. Let θ denote the angle of depression from the top of the tower to a point on the ground. If the range of the riﬂe with a tranquilizer dart is 300 feet, ﬁnd the smallest value of θ for which the corresponding point on the ground is in range of the riﬂe. Round your answer to the nearest hundreth of a
degree. In Exercises 216 - 221, rewrite the given function as a sinusoid of the form S(x) = A sin(ωx + φ) using Exercises 35 and 36 in Section 10.5 for reference. Approximate the value of φ (which is in radians, of course) to four decimal places. 216. f (x) = 5 sin(3x) + 12 cos(3x) 217. f (x) = 3 cos(2x) + 4 sin(2x) 218. f (x) = cos(x) − 3 sin(x) 219. f (x) = 7 sin(10x) − 24 cos(10x) 10.6 The Inverse Trigonometric Functions 847 220. f (x) = − cos(x) − 2 √ 2 sin(x) 221. f (x) = 2 sin(x) − cos(x) In Exercises 222 - 233, ﬁnd the domain of the given function. Write your answers in interval notation. 222. f (x) = arcsin(5x) 223. f (x) = arccos 3x − 1 2 224. f (x) = arcsin 2x2 225. f (x) = arccos 1 x2 − 4 226. f (x) = arctan(4x) 227. f (x) = arccot 2x x2 − 9 228. f (x) = arctan(ln(2x − 1)) 229. f (x) = arccot( √ 2x − 1) 230. f (x) = arcsec(12x) 231. f (x) = arccsc(x + 5) 232. f (x) = arcsec x3 8 233. f (x) = arccsc e2x 234. Show that arcsec(x) = arccos of f (x) = arcsec(x). 1 x for \|x\| ≥ 1 as long as we use 0, ∪ π 2, π π 2 as the range 235. Show that arccsc(x) = arcsin of f (x) = arccsc(x). 1 x for \|x\| ≥ 1 as long as we use − π 2, 0 0, ∪ π 2 as the range 236. Show
that arcsin(x) + arccos(x) = π 2 for −1 ≤ x ≤ 1. 237. Discuss with your classmates why arcsin 1 2 = 30◦. 238. Use the following picture and the series of exercises on the next page to show that arctan(1) + arctan(2) + arctan(3) = π y D(2, 3) A(0, 1) α β γ B(1, 0) x C(2, 0) O(0, 0) 848 Foundations of Trigonometry (a) Clearly AOB and BCD are right triangles because the line through O and A and the line through C and D are perpendicular to the x-axis. Use the distance formula to show that BAD is also a right triangle (with ∠BAD being the right angle) by showing that the sides of the triangle satisfy the Pythagorean Theorem. (b) Use AOB to show that α = arctan(1) (c) Use BAD to show that β = arctan(2) (d) Use BCD to show that γ = arctan(3) (e) Use the fact that O, B and C all lie on the x-axis to conclude that α + β + γ = π. Thus arctan(1) + arctan(2) + arctan(3) = π. 10.6 The Inverse Trigonometric Functions 849 10.6.6 Answers 1. arcsin (−1) = − π 2 2. arcsin − √ 3 2 = − π 3 3. arcsin − √ 2 2 = − π 4 4. arcsin 7. arcsin −. arcsin (0) = 0 8. arcsin √ 3 2 10. arccos (−1) = π 11. arccos − √ 3 2 = π 3 = 5π 6 13. arccos 16. arccos = 2π 14. arccos (0) = 17. arccos √ 3 2 π 2 = π 6 19. arctan − √ 3 = − π 3 20. arctan (−1) = − π 4 22. arctan (0) = 0 23. arctan √ 3 3 = π 6 25. ar
ctan √ 3 = √ π 3 28. arccot − = 3 3 26. arccot − √ 3 = 5π 6 2π 3 29. arccot (0) = π 2 31. arccot (1) = 34. arccsc (2) = 37. arcsec √ 2 3 3 π 4 π 6 32. arccot √ 35. arcsec √ 38. arccsc. arcsin 1 2 = π 6 9. arcsin (1) = π 2 12. arccos − √ 2 2 = 3π 4 15. arccos 1 2 = π 3 18. arccos (1) = 0 21. arctan − √ 3 3 = − π 6 24. arctan (1) = π 4 27. arccot (−1) = 3π 4 30. arccot √ 3 3 = π 3 33. arcsec (2) = 36. arccsc √ π 3 2 = π 4 39. arcsec (1) = 0 40. arccsc (1) = π 2 41. arcsec (−2) = 4π 3 42. arcsec − √ 2 = 5π 4 850 Foundations of Trigonometry 44. arcsec (−1) = π 45. arccsc (−2) = 47. arccsc − √ 3 2 3 = 4π 3 48. arccsc (−1) = 7π 6 3π 2 50. arcsec − 2 = √ 51. arcsec 2 − √ 3 3 = 5π 6 54. arccsc − √ 2 = − π 4 3π 4 π 6 π 2 52. arcsec (−1) = π 53. arccsc (−2) = − 55. arccsc − √ 3 2 = − π 3 56. arccsc (−1) = − 43. arcsec − √ 3 2 3 = 7π 6 46. arccsc − √ 2 = 5π 4 49. arcsec (−2) = 2π 3 57. sin arcsin 59. sin arcsin 61. sin arcsin 58. sin arcsin − √ 2 2 √ 2 2 = − 60. sin (arcsin (−0.42)) = −0.42 √ 2 2 √
2 2 = is undeﬁned. 62. cos arccos 63. cos arccos − 1 2 = − 1 2 65. cos (arccos (−0.998)) = −0.998 67. tan (arctan (−1)) = −1 69. tan arctan 5 12 = 5 12 64. cos arccos 5 13 = 5 13 66. cos (arccos (π)) is undeﬁned. 68. tan arctan √ 3 = √ 3 70. tan (arctan (0.965)) = 0.965 71. tan (arctan (3π)) = 3π 72. cot (arccot (1)) = 1 73. cot arccot − √ √ 3 = − 3 75. cot (arccot (−0.001)) = −0.001 74. cot arccot 76. cot arccot − 7 24 17π 4 = − 7 24 = 17π 4 77. sec (arcsec (2)) = 2 78. sec (arcsec (−1)) = −1 10.6 The Inverse Trigonometric Functions 851 79. sec arcsec 1 2 is undeﬁned. 81. sec (arcsec (117π)) = 117π 83. csc arccsc 2 − √ 3 3 √ 3 2 3 = − 85. csc (arccsc (1.0001)) = 1.0001 87. arcsin sin 89. arcsin sin 91. arcsin sin 93. arccos cos π 6 3π 4 4π 3 2π 3 95. arccos − cos π 6 = π 4 = − π 3 = 2π 3 = π 6 = π 6 97. arctan tan π 3 = π 3 99. arctan (tan (π)) = 0 101. arctan tan 2π 3 = − π 3 80. sec (arcsec (0.75)) is undeﬁned. 82. csc arccsc √ √ 2 = √ 2 84. csc arccsc is undeﬁned. 2 2 86. csc arccsc π 4 is undeﬁned. 88. arcsin sin 90. arcsin sin − π 3 11π 6 = − π 3 = −
π 6 92. arccos cos 94. arccos cos 96. arccos cos π 4 3π 2 5π 4 = − tan 98. arctan 100. arctan tan π 4 π 4 = = π 2 3π 4 = − π 4 102. arccot cot π 2 π 3 is undeﬁned = π 3 103. arccot cot 105. arccot cot − π 4 3π 2 = 107. arcsec sec 109. arcsec sec 111. arcsec sec π 4 5π 6 5π 3 = 3π 4 π 2 7π 6 π 3 104. arccot (cot (π)) is undeﬁned 106. arccot cot 108. arcsec sec 2π 3 4π 3 = = 2π 3 4π 3 110. arcsec 112. arccsc sec − π 2 is undeﬁned. csc π 6 = π 6 = π 4 = = 852 Foundations of Trigonometry 113. arccsc csc 5π 4 = 5π 4 115. arccsc csc 117. arcsec sec 119. arcsec sec 121. arcsec sec 123. arcsec sec 125. arccsc csc 127. arccsc csc 129. arcsec sec − π 2 11π 12 = 3π 2 = 13π 12 π 4 = = 5π 5π 6 5π 3 5π 4 − π 2 11π 12 = 11π 12 131. sin arccos − = 1 2 133. sin (arctan (−2)) = − 135. sin (arccsc (−3)) = − 137. cos arctan √ 7 = 139. cos (arcsec (5)) = 1 5 114. arccsc csc 116. arccsc csc 118. arccsc csc 120. arcsec sec 2π 3 11π 6 9π 8 4π 3 = π 3 = 7π 6 = = 9π 8 2π 3 122. arcsec sec − π 2 is undeﬁned csc csc csc 128. arccsc 124. arccsc 126. arccsc π 6 2π 3 11π 6 9π 8 3 5 134. sin arccot √ 130. arccsc 132. sin ar
ccos csc 136. cos arcsin − 5 = = 4 5 √ 6 6 5 13 = 12 13 √ 3 10 138. cos (arccot (3)) = 140. tan arcsin − √ 2 5 = −2 10 5 = 4 3 141. tan arccos − √ 3 = − 1 2 143. tan (arccot (12)) = 1 12 142. tan arcsec 144. cot arcsin 5 3 12 13 = 5 12 10.6 The Inverse Trigonometric Functions 853 145. cot arccos √ 3 2 √ 3 = 147. cot (arctan (0.25)) = 4 = 13 5 149. sec arcsin − 12 13 √ 151. sec arccot − 10 10 153. csc arcsin 155. sin arcsin 3 5 = 5 3 5 13 + π 4 = 157. tan arctan(3) + arccos − 146. cot arccsc √ 5 = 2 148. sec arccos √ 3 2 √ 3 2 3 = 150. sec (arctan (10)) = √ 101 √ = − 11 152. csc (arccot (9)) = √ 82 154. csc arctan − 2 3 = − √ 13 2 156. cos (arcsec(3) + arctan(2)) = √ 10 √ 5 − 4 15 √ 17 2 26 3 5 = 1 3 158. sin 2 arcsin − 4 5 = − 24 25 159. sin 2arccsc 161. cos 2 arcsin 13 5 3 5 = 120 169 = 7 25 160. sin (2 arctan (2)) = 4 5 162. cos 2arcsec 25 7 = − 163. cos 2arccot − √ 5 = 2 3 165. sin (arccos (x)) = √ 1 − x2 for −1 ≤ x ≤ 1 164. sin arctan(2) 2 = 5 − 10 527 625 √ 5 166. cos (arctan (x)) = √ 167. tan (arcsin (x)) = √ 1 1 + x2 x 1 − x2 for all x for −1 < x < 1 168. sec (arctan (x)) = √ 1 + x2 for all x 169. csc (arccos (x))
= √ 1 1 − x2 for −1 < x < 1 170. sin (2 arctan (x)) = for all x 2x x2 + 1 √ 171. sin (2 arccos (x)) = 2x 1 − x2 for −1 ≤ x ≤ 1 854 Foundations of Trigonometry 172. cos (2 arctan (x)) = 173. sin(arccos(2x)) = 174. sin arccos 175. cos arcsin x 5 x 2 = = √ 1 − x2 1 + x2 for all x √ 1 − 4x2 for − 1 √ 25 − x2 5 2 ≤ x ≤ 1 2 for −5 ≤ x ≤ 5 4 − x2 2 for −2 ≤ x ≤ 2 176. cos (arctan (3x)) = √ for all x 1 1 + 9x2 √ 177. sin(2 arcsin(7x)) = 14x 1 − 49x2 for − 1 7 ≤ x ≤ 1 7 178. sin 2 arcsin √ x 3 3 √ 2x = 3 − x2 3 √ for − 3 ≤ x ≤ √ 3 179. cos(2 arcsin(4x)) = 1 − 32x2 for − 1 4 1 4 ≤ x ≤ √ 180. sec(arctan(2x)) tan(arctan(2x)) = 2x 1 + 4x2 for all x 181. sin (arcsin(x) + arccos(x)) = 1 for −1 ≤ x ≤ 1 182. cos (arcsin(x) + arctan(x)) = √ 1 − x2 − x2 √ 1 + x2 for −1 ≤ x ≤ 1 183. 10 tan (2 arcsin(x)) = √ 2x 1 − x2 1 − 2x2 for x in −1 184. sin arctan(x) = 1 2    √ x2 + 1 − 1 √ x2 + 1 2 √ x2 + 1 − 1 √ x2 + 1 2 − for x ≥ 0 for x < 0 185. If sin(θ) = 186. If tan(θ) = x 2 x 7
for − π 2 for − π 2 < θ < π 2, then θ + sin(2θ) = arcsin x 2 x + √ 4 − x2 2 < θ < π 2, then 1 2 θ − 1 2 sin(2θ) = 1 2 arctan x 7 − 7x x2 + 49 10The equivalence for x = ±1 can be veriﬁed independently of the derivation of the formula, but Calculus is required to fully understand what is happening at those x values. You’ll see what we mean when you work through the details of the identity for tan(2t). For now, we exclude x = ±1 from our answer. 10.6 The Inverse Trigonometric Functions 855 187. If sec(θ) = 188. x = arcsin x 4 7 11 for 0 < θ < π 2, then 4 tan(θ) − 4θ = √ x2 − 16 − 4arcsec x 4 + 2πk or x = π − arcsin 7 11 + 2πk, in [0, 2π), x ≈ 0.6898, 2.4518 189. x = arccos − 2 9 + 2πk or x = − arccos − 2 9 + 2πk, in [0, 2π), x ≈ 1.7949, 4.4883 190. x = π + arcsin(0.569) + 2πk or x = 2π − arcsin(0.569) + 2πk, in [0, 2π), x ≈ 3.7469, 5.6779 191. x = arccos(0.117) + 2πk or x = 2π − arccos(0.117) + 2πk, in [0, 2π), x ≈ 1.4535, 4.8297 192. x = arcsin(0.008) + 2πk or x = π − arcsin(0.008) + 2πk, in [0, 2π), x ≈ 0.0080, 3.1336 193. x = arccos 359 360 + 2πk or x = 2π − arccos 359 360 + 2πk, in [0, 2π), x ≈ 0
.0746, 6.2086 194. x = arctan(117) + πk, in [0, 2π), x ≈ 1.56225, 4.70384 195. x = arctan − 1 12 + πk, in [0, 2π), x ≈ 3.0585, 6.2000 196. x = arccos 2 3 + 2πk or x = 2π − arccos 2 3 + 2πk, in [0, 2π), x ≈ 0.8411, 5.4422 197. x = π + arcsin 198. x = arctan − 3 8 199. x = arcsin 200. x = arccos − 7 16 + 2πk or x = 2π − arcsin 17 90 10 + πk, in [0, 2π), x ≈ 1.8771, 5.0187 17 90 √ + 2πk or x = π − arcsin 3 8 + 2πk, in [0, 2π), x ≈ 0.3844, 2.7572 7 16 + 2πk, in [0, 2π), x ≈ 2.0236, 4.2596 + 2πk or x = − arccos − + 2πk, in [0, 2π), x ≈ 3.3316, 6.0932 201. x = arctan(0.03) + πk, in [0, 2π), x ≈ 0.0300, 3.1716 202. x = arcsin(0.3502) + 2πk or x = π − arcsin(0.3502) + 2πk, in [0, 2π), x ≈ 0.3578, 2.784 203. x = π + arcsin(0.721) + 2πk or x = 2π − arcsin(0.721) + 2πk, in [0, 2π), x ≈ 3.9468, 5.4780 204. x = arccos(0.9824) + 2πk or x = 2π − arccos(0.9824) + 2πk, in [0, 2π), x ≈ 0.1879, 6.0953 205. x = arccos(
−0.5637) + 2πk or x = − arccos(−0.5637) + 2πk, in [0, 2π), x ≈ 2.1697, 4.1135 206. x = arctan(117) + πk, in [0, 2π), x ≈ 1.5622, 4.7038 207. x = arctan(−0.6109) + πk, in [0, 2π), x ≈ 2.5932, 5.7348 856 Foundations of Trigonometry 208. 36.87◦ and 53.13◦ 209. 22.62◦ and 67.38◦ 210. 32.52◦ and 57.48◦ 211. 68.9◦ 212. 7.7◦ 213. 51◦ 216. f (x) = 5 sin(3x) + 12 cos(3x) = 13 sin 3x + arcsin 214. 19.5◦ 12 13 ≈ 13 sin(3x + 1.1760) 215. 41.81◦ 217. f (x) = 3 cos(2x) + 4 sin(2x) = 5 sin 2x + arcsin 3 5 ≈ 5 sin(2x + 0.6435) 218. f (x) = cos(x) − 3 sin(x) = √ 10 sin x + arccos − 10 219. f (x) = 7 sin(10x) − 24 cos(10x) = 25 sin 10x + arcsin − √ 3 10 √ ≈ 10 sin(x + 2.8198) 24 25 ≈ 25 sin(10x − 1.2870) 220. f (x) = − cos(x) − 2 √ 2 sin(x) = 3 sin x + π + arcsin 221. f (x) = 2 sin(x) − cos(x) = √ 5 sin x + arcsin − 1 3 ≈ 3 sin(x + 3.4814) √ ≈ 5 sin(x − 0.4636) √ 5 5 223. −, 1 1 3 1 5 222. − 224. −, 1 5 √ √ 2 2, 225. (−∞, − √ √ 5] ∪ [− √
3, √ 3] ∪ [ 5, ∞) 2 2 226. (−∞, ∞) 227. (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) 228., ∞ 1 2 230. −∞, − 1 12 ∪ 1 12, ∞ 229., ∞ 1 2 231. (−∞, −6] ∪ [−4, ∞) 232. (−∞, −2] ∪ [2, ∞) 233. [0, ∞) 10.7 Trigonometric Equations and Inequalities 857 10.7 Trigonometric Equations and Inequalities In Sections 10.2, 10.3 and most recently 10.6, we solved some basic equations involving the trigonometric functions. Below we summarize the techniques we’ve employed thus far. Note that we use the neutral letter ‘u’ as the argument1 of each circular function for generality. Strategies for Solving Basic Equations Involving Trigonometric Functions To solve cos(u) = c or sin(u) = c for −1 ≤ c ≤ 1, ﬁrst solve for u in the interval [0, 2π) and add integer multiples of the period 2π. If c < −1 or of c > 1, there are no real solutions. To solve sec(u) = c or csc(u) = c for c ≤ −1 or c ≥ 1, convert to cosine or sine, respectively, and solve as above. If −1 < c < 1, there are no real solutions. To solve tan(u) = c for any real number c, ﬁrst solve for u in the interval − π 2, π 2 and add integer multiples of the period π. To solve cot(u) = c for c = 0, convert to tangent and solve as above. If c = 0, the solution to cot(u) = 0 is u = π 2 + πk for integers k. 2 and ﬁnd the solution x = π 2, we know the solutions take the form u = π 6 + 2πk for integers k. How do we solve something like sin(3x) = 1 Using the above guidelines, we can comfortably solve sin(x) = 1 or x = 5
π has the form sin(u) = 1 integers k. Since the argument of sine here is 3x, we have 3x = π integers k. To solve for x, we divide both sides2 of these equations by 3, and obtain x = π or x = 5π 6 + 2πk 2? Since this equation 6 + 2πk for 6 + 2πk for 18 + 2π 3 k 3 k for integers k. This is the technique employed in the example below. Example 10.7.1. Solve the following equations and check your answers analytically. List the solutions which lie in the interval [0, 2π) and verify them using a graphing utility. 2. csc 1 6 + 2πk or u = 5π 6 + 2πk or 3x = 5π 3. cot (3x) = 0 18 + 2π 1. cos(2x3 5. tan x 2 6. sin(2x) = 0.87 4. sec2(x) = 4 Solution. 1. The solutions to cos(u) = − √ the argument of cosine here is 2x, this means 2x = 5π k. Solving for x gives x = 5π analytically, we substitute them into the original equation. For any integer k we have 6 + 2πk for integers k. Since 6 + 2πk for integers 12 + πk for integers k. To check these answers 6 + 2πk or 2x = 7π 12 + πk or x = 7π 6 + 2πk or u = 7π 2 are u = 5π 3 cos 2 5π 12 + πk = cos 5π = cos 5π 6 √ = − 3 2 6 + 2πk (the period of cosine is 2π) 1See the comments at the beginning of Section 10.5 for a review of this concept. 2Don’t forget to divide the 2πk by 3 as well! 858 Foundations of Trigonometry 12 + πk = cos 7π √ Similarly, we ﬁnd cos 2 7π 3 2. To determine which of our solutions lie in [0, 2π), we substitute integer values for k. The solutions we keep come from the values of k = 0 and k = 1 and are x = 5π 12, 17π 12. Using a √ 3
calculator, we graph y = cos(2x) and y = − 2 over [0, 2π) and examine where these two graphs intersect. We see that the x-coordinates of the intersection points correspond to the decimal representations of our exact answers. 6 + 2πk = cos 7π 12 and 19π = − 12, 7π 6 2. Since this equation has the form csc(u) = 2, we rewrite this as sin(u) = u = π 4 + 2πk or u = 3π 4 + 2πk for integers k. Since the argument of cosecant here is 1 √ 2 2 and ﬁnd 3 x − π + 2πk or 1 3 x − π = 3π 4 + 2πk To solve 1 3 x − π = π 4 + 2πk, we ﬁrst add π to both sides 1 3 x = π 4 + 2πk + π A common error is to treat the ‘2πk’ and ‘π’ terms as ‘like’ terms and try to combine them when they are not.3 We can, however, combine the ‘π’ and ‘ π 4 ’ terms to get We now ﬁnish by multiplying both sides by 3 to get 1 3 x = 5π 4 + 2πk x = 3 5π 4 + 2πk = 15π 4 + 6πk Solving the other equation, 1 check the ﬁrst family of answers, we substitute, combine line terms, and simplify. 4 + 2πk produces x = 21π 3 x − π = 3π 4 + 6πk for integers k. To csc 1 3 15π 4 + 6πk − π = csc 5π = csc π = csc π √ 4 2 = 4 + 2πk − π 4 + 2πk (the period of cosecant is 2π) The family x = 21π 4 + 6πk checks similarly. Despite having inﬁnitely many solutions, we ﬁnd that none of them lie in [0, 2π). To verify this graphically, we use a reciprocal identity to 2 do not intersect at rewrite the cosecant as a sine and we ﬁnd that y = and y = �
� 1 sin( 1 3 x−π) all over the interval [0, 2π). 3Do you see why? 10.7 Trigonometric Equations and Inequalities 859 y = cos(2x) and y = − √ 3 2 y = 1 sin( 1 3 x−π) and y = √ 2 3. Since cot(3x) = 0 has the form cot(u) = 0, we know u = π 2 + πk, so, in this case, 3x = π 2 + πk for integers k. Solving for x yields x = π 3 k. Checking our answers, we get 6 + π 2 + πk cot 3 π 6 + π 3 k = cot π = cot π 2 = 0 (the period of cotangent is π) As k runs through the integers, we obtain six answers, corresponding to k = 0 through k = 5, which lie in [0, 2π): x = π 2 and 11π 6. To conﬁrm these graphically, we must be careful. On many calculators, there is no function button for cotangent. We choose4 to use sin(3x). Graphing y = cos(3x) the quotient identity cot(3x) = cos(3x) sin(3x) and y = 0 (the x-axis), we see that the x-coordinates of the intersection points approximately match our solutions. 6, 3π 6, 7π 2, 5π 6, π 4. The complication in solving an equation like sec2(x) = 4 comes not from the argument of secant, which is just x, but rather, the fact the secant is being squared. To get this equation to look like one of the forms listed on page 857, we extract square roots to get sec(x) = ±2. Converting to cosines, we have cos(x) = ± 1 3 + 2πk or x = 5π 3 + 2πk for integers k. For cos(x) = − 1 3 + 2πk for integers k. If we take a step back and think of these families of solutions geometrically, we see we are ﬁnding the measures of all angles with a reference angle of π 3
. As a result, these solutions can be combined and we may write our solutions as x = π 3 + πk for integers k. To check the ﬁrst family of solutions, we note that, depending on the integer. However, it is true that for all integers k, k, sec π sec π 3 = ±2. (Can you show this?) As a result, 3 + πk doesn’t always equal sec π 2. For cos(x) = 1 2, we get x = 2π 2, we get x = π 3 + 2πk or x = 4π 3 + πk and x = 2π 3 + πk = ± sec π 3 sec2 π 3 + πk = ± sec π 3 2 = (±2)2 = 4 The same holds for the family x = 2π the values k = 0 and k = 1, namely x = π 3 + πk. The solutions which lie in [0, 2π) come from 3. To conﬁrm graphically, we use 3 and 5π 3, 4π 3, 2π 4The reader is encouraged to see what happens if we had chosen the reciprocal identity cot(3x) = 1 tan(3x) instead. The graph on the calculator appears identical, but what happens when you try to ﬁnd the intersection points? 860 Foundations of Trigonometry a reciprocal identity to rewrite the secant as cosine. The x-coordinates of the intersection points of y = (cos(x))2 and y = 4 verify our answers. 1 y = cos(3x) sin(3x) and y = 0 y = 1 cos2(x) and y = 4 5. The equation tan x 2 Hence, x = −3 has the form tan(u) = −3, whose solution is u = arctan(−3)+πk. 2 = arctan(−3) + πk, so x = 2 arctan(−3) + 2πk for integers k. To check, we note tan 2 arctan(−3)+2πk 2 = tan (arctan(−3) + πk) = tan (arctan(−3)) = −3 (the period of tangent is π) (See Theorem 10.
27) To determine which of our answers lie in the interval [0, 2π), we ﬁrst need to get an idea of the value of 2 arctan(−3). While we could easily ﬁnd an approximation using a calculator,5 we proceed analytically. Since −3 < 0, it follows that − π 2 < arctan(−3) < 0. Multiplying through by 2 gives −π < 2 arctan(−3) < 0. We are now in a position to argue which of the solutions x = 2 arctan(−3) + 2πk lie in [0, 2π). For k = 0, we get x = 2 arctan(−3) < 0, so we discard this answer and all answers x = 2 arctan(−3) + 2πk where k < 0. Next, we turn our attention to k = 1 and get x = 2 arctan(−3) + 2π. Starting with the inequality −π < 2 arctan(−3) < 0, we add 2π and get π < 2 arctan(−3) + 2π < 2π. This means x = 2 arctan(−3) + 2π lies in [0, 2π). Advancing k to 2 produces x = 2 arctan(−3) + 4π. Once again, we get from −π < 2 arctan(−3) < 0 that 3π < 2 arctan(−3) + 4π < 4π. Since this is outside the interval [0, 2π), we discard x = 2 arctan(−3) + 4π and all solutions of the form x = 2 arctan(−3) + 2πk for k > 2. Graphically, we see y = tan x and y = −3 intersect only 2 once on [0, 2π) at x = 2 arctan(−3) + 2π ≈ 3.7851. 6. To solve sin(2x) = 0.87, we ﬁrst note that it has the form sin(u) = 0.87, which has the family of solutions u = arcsin(0.87) + 2πk or u = π − arcsin(0.87) + 2πk for integers k. Since
the argument of sine here is 2x, we get 2x = arcsin(0.87) + 2πk or 2x = π − arcsin(0.87) + 2πk 2 − 1 which gives x = 1 2 arcsin(0.87) + πk for integers k. To check, 2 arcsin(0.87) + πk or x = π 5Your instructor will let you know if you should abandon the analytic route at this point and use your calculator. But seriously, what fun would that be? 10.7 Trigonometric Equations and Inequalities 861 sin 2 1 2 arcsin(0.87) + πk = sin (arcsin(0.87) + 2πk) = sin (arcsin(0.87)) (the period of sine is 2π) = 0.87 (See Theorem 10.26) For the family x = π 2 − 1 2 arcsin(0.87) + πk, we get sin 2 π 2 − 1 2 arcsin(0.87) + πk = sin (π − arcsin(0.87) + 2πk) = sin (π − arcsin(0.87)) = sin (arcsin(0.87)) = 0.87 (the period of sine is 2π) (sin(π − t) = sin(t)) (See Theorem 10.26) 2 arcsin(0.87) < π 2 so that multiplying through by 1 4. Starting with the family of solutions x = 1 To determine which of these solutions lie in [0, 2π), we ﬁrst need to get an idea of the value of x = 1 2 arcsin(0.87). Once again, we could use the calculator, but we adopt an analytic route here. By deﬁnition, 0 < arcsin(0.87) < π 2 gives us 0 < 1 2 arcsin(0.87) + πk, we use the same kind of arguments as in our solution to number 5 above and ﬁnd only the solutions corresponding to k = 0 and k = 1 lie in [0, 2π): x = 1 2 arcsin(0.87) + π. Next, we move to the family x = π 2 arcsin(0.87)
+ πk for integers k. Here, we need to get a better estimate of π 2 arcsin(0.87) < π 4, we ﬁrst multiply through by −1 and then add π 4, or 4 < π π 2. Proceeding with the usual arguments, we ﬁnd the only solutions which lie in [0, 2π) correspond to k = 0 and k = 1, namely x = π 2 arcsin(0.87) and x = 3π 2 arcsin(0.87). All told, we have found four solutions to sin(2x) = 0.87 in [0, 2π): x = 1 2 arcsin(0.87). By graphing y = sin(2x) and y = 0.87, we conﬁrm our results. 2 arcsin(0.87). From the inequality 0 < 1 2 − 1 2 arcsin(0.87), x = 1 2 arcsin(0.87) and x = 3π 2 arcsin(0.87) + π, x = π 2 arcsin(0.87) and x = 1 2 arcsin(0.87) > π 2 arcsin(0.87) < π 2 to get = tan x 2 and y = −3 y = sin(2x) and y = 0.87 862 Foundations of Trigonometry Each of the problems in Example 10.7.1 featured one trigonometric function. If an equation involves two diﬀerent trigonometric functions or if the equation contains the same trigonometric function but with diﬀerent arguments, we will need to use identities and Algebra to reduce the equation to the same form as those given on page 857. Example 10.7.2. Solve the following equations and list the solutions which lie in the interval [0, 2π). Verify your solutions on [0, 2π) graphically. 1. 3 sin3(x) = sin2(x) 3. cos(2x) = 3 cos(x) − 2 5. cos(3x) = cos(5x) 7. sin(x) cos x 2 + cos(x) sin x 2 = 1 Solution. 2. sec2(x) = tan(x) + 3 4. cos(3x) = 2 − cos
(x) 6. sin(2x) = √ 3 cos(x) 8. cos(x) − √ 3 sin(x) = 2 1. We resist the temptation to divide both sides of 3 sin3(x) = sin2(x) by sin2(x) (What goes wrong if you do?) and instead gather all of the terms to one side of the equation and factor. 3 sin3(x) = sin2(x) 3 sin3(x) − sin2(x) = 0 sin2(x)(3 sin(x) − 1) = 0 Factor out sin2(x) from both terms. We get sin2(x) = 0 or 3 sin(x) − 1 = 0. Solving for sin(x), we ﬁnd sin(x) = 0 or sin(x) = 1 3. The solution to the ﬁrst equation is x = πk, with x = 0 and x = π being the two solutions +2πk which lie in [0, 2π). To solve sin(x) = 1 + 2πk for integers k. We ﬁnd the two solutions here which lie in [0, 2π) or x = π − arcsin 1 3. To check graphically, we plot y = 3(sin(x))3 and and x = π − arcsin 1 to be x = arcsin 1 3 3 y = (sin(x))2 and ﬁnd the x-coordinates of the intersection points of these two curves. Some extra zooming is required near x = 0 and x = π to verify that these two curves do in fact intersect four times.6 3, we use the arcsine function to get x = arcsin 1 3 2. Analysis of sec2(x) = tan(x) + 3 reveals two diﬀerent trigonometric functions, so an identity is in order. Since sec2(x) = 1 + tan2(x), we get sec2(x) = tan(x) + 3 1 + tan2(x) = tan(x) + 3 (Since sec2(x) = 1 + tan2(x).) tan2(x) − tan(x) − 2 = 0 u2 − u − 2 = 0 (u + 1)(u − 2) = 0 Let u = tan
(x). 6Note that we are not counting the point (2π, 0) in our solution set since x = 2π is not in the interval [0, 2π). In the forthcoming solutions, remember that while x = 2π may be a solution to the equation, it isn’t counted among the solutions in [0, 2π). 10.7 Trigonometric Equations and Inequalities 863 This gives u = −1 or u = 2. Since u = tan(x), we have tan(x) = −1 or tan(x) = 2. From tan(x) = −1, we get x = − π 4 + πk for integers k. To solve tan(x) = 2, we employ the arctangent function and get x = arctan(2) + πk for integers k. From the ﬁrst set of solutions, we get x = 3π 4 as our answers which lie in [0, 2π). Using the same sort of argument we saw in Example 10.7.1, we get x = arctan(2) and x = π + arctan(2) as answers from our second set of solutions which lie in [0, 2π). Using a reciprocal identity, we rewrite the secant as a cosine and graph y = (cos(x))2 and y = tan(x) + 3 to ﬁnd the x-values of the points where they intersect. 4 and x = 7π 1 y = 3(sin(x))3 and y = (sin(x))2 y = 1 (cos(x))2 and y = tan(x) + 3 3. In the equation cos(2x) = 3 cos(x) − 2, we have the same circular function, namely cosine, on both sides but the arguments diﬀer. Using the identity cos(2x) = 2 cos2(x) − 1, we obtain a ‘quadratic in disguise’ and proceed as we have done in the past. cos(2x) = 3 cos(x) − 2 2 cos2(x) − 1 = 3 cos(x) − 2 (Since cos(2x) = 2 cos2(x) − 1.) 2 cos2(x) − 3 cos(x) + 1 = 0 2u2 − 3u +
1 = 0 (2u − 1)(u − 1) = 0 Let u = cos(x). 2, we get x = π 2 or u = 1. Since u = cos(x), we get cos(x) = 1 This gives u = 1 2 or cos(x) = 1. Solving cos(x) = 1 3 + 2πk or x = 5π 3 + 2πk for integers k. From cos(x) = 1, we get x = 2πk for integers k. The answers which lie in [0, 2π) are x = 0, π 3. Graphing y = cos(2x) and y = 3 cos(x) − 2, we ﬁnd, after a little extra eﬀort, that the curves intersect in three places on [0, 2π), and the x-coordinates of these points conﬁrm our results. 3, and 5π 4. To solve cos(3x) = 2 − cos(x), we use the same technique as in the previous problem. From Example 10.4.3, number 4, we know that cos(3x) = 4 cos3(x) − 3 cos(x). This transforms the equation into a polynomial in terms of cos(x). cos(3x) = 2 − cos(x) 4 cos3(x) − 3 cos(x) = 2 − cos(x) 2 cos3(x) − 2 cos(x) − 2 = 0 4u3 − 2u − 2 = 0 Let u = cos(x). 864 Foundations of Trigonometry To solve 4u3 − 2u − 2 = 0, we need the techniques in Chapter 3 to factor 4u3 − 2u − 2 into (u − 1) 4u2 + 4u + 2. We get either u − 1 = 0 or 4u2 + 2u + 2 = 0, and since the discriminant of the latter is negative, the only real solution to 4u3 − 2u − 2 = 0 is u = 1. Since u = cos(x), we get cos(x) = 1, so x = 2πk for integers k. The only solution which lies in [0, 2π) is x = 0. Graphing y = cos(3x) and y = 2 − cos(x) on the
same set of axes over [0, 2π) shows that the graphs intersect at what appears to be (0, 1), as required. y = cos(2x) and y = 3 cos(x) − 2 y = cos(3x) and y = 2 − cos(x) 5. While we could approach cos(3x) = cos(5x) in the same manner as we did the previous two problems, we choose instead to showcase the utility of the Sum to Product Identities. From cos(3x) = cos(5x), we get cos(5x) − cos(3x) = 0, and it is the presence of 0 on the right hand side that indicates a switch to a product would be a good move.7 Using Theorem 10.21, we have that cos(5x) − cos(3x) = −2 sin 5x+3x = −2 sin(4x) sin(x). Hence, 2 the equation cos(5x) = cos(3x) is equivalent to −2 sin(4x) sin(x) = 0. From this, we get sin(4x) = 0 or sin(x) = 0. Solving sin(4x) = 0 gives x = π 4 k for integers k, and the solution to sin(x) = 0 is x = πk for integers k. The second set of solutions is contained in the ﬁrst set of solutions,8 so our ﬁnal solution to cos(5x) = cos(3x) is x = π 4 k for integers k. There are eight of these answers which lie in [0, 2π): x = 0, π 4, 3π 4, π, 5π 4. Our plot of the graphs of y = cos(3x) and y = cos(5x) below (after some careful zooming) bears this out. sin 5x−3x 2 and 7π 2, 3π 4, π 2 √ 6. In examining the equation sin(2x) = 3 cos(x), not only do we have diﬀerent circular functions involved, namely sine and cosine, we also have diﬀerent arguments to contend with, namely 2x and x. Using the identity sin(2x) = 2 sin(x) cos(x) makes all of
the arguments the same and we proceed as we would solving any nonlinear equation – gather all of the nonzero terms on one side of the equation and factor. sin(2x) = 2 sin(x) cos(x) = √ 2 sin(x) cos(x) − cos(x)(2 sin(x) − 3 cos(x) = 0 3) = 0 √ √ √ 3 cos(x) 3 cos(x) (Since sin(2x) = 2 sin(x) cos(x).) from which we get cos(x) = 0 or sin(x) = 2, we get x = π integers k. From sin(x) = √ 3 3 2. From cos(x) = 0, we obtain x = π 2 + πk for 3 +2πk for integers k. The answers 3 +2πk or x = 2π √ 7As always, experience is the greatest teacher here! 8As always, when in doubt, write it out! 10.7 Trigonometric Equations and Inequalities which lie in [0, 2π) are x = π after some careful zooming, verify our answers. 3 and 2π 2, 3π 2, π 3. We graph y = sin(2x) and y = 865 √ 3 cos(x) and, y = cos(3x) and y = cos(5x) y = sin(2x) and y = √ 3 cos(x) 7. Unlike the previous problem, there seems to be no quick way to get the circular functions or their arguments to match in the equation sin(x) cos x = 1. If we stare at 2 it long enough, however, we realize that the left hand side is the expanded form of the sum formula for sin x + x 2 x = 1. Solving, 2 we ﬁnd x = π 3 and x = 5π 3. Graphing y = sin(x) cos x 2. Hence, our original equation is equivalent to sin 3 3 k for integers k. Two of these solutions lie in [0, 2π): x = π and y = 1 validates our solutions. + cos(x) sin x 2 + cos(x) sin x 2 3 + 4π 8. With the absence of double angles or squares, there doesn’t seem to be
much we can do. However, since the arguments of the cosine and sine are the same, we can rewrite the left hand side of this equation as a sinusoid.9 To ﬁt f (x) = cos(x) − 3 sin(x) to the form A sin(ωt + φ) + B, we use what we learned in Example 10.5.3 and ﬁnd A = 2, B = 0, ω = 1 = 2, 3 sin(x) = 2 as 2 sin x + 5π and φ = 5π 6 or sin x + 5π 3 + 2πk for integers k. Only one of 6 these solutions, x = 5π 3, which corresponds to k = 1, lies in [0, 2π). Geometrically, we see that y = cos(x) − 6. Hence, we can rewrite the equation cos(x) − = 1. Solving the latter, we get x = − π 3 sin(x) and y = 2 intersect just once, supporting our answer. √ √ √ y = sin(x) cos x 2 + cos(x) sin x 2 and y = 1 y = cos(x) − √ 3 sin(x) and y = 2 We repeat here the advice given when solving systems of nonlinear equations in section 8.7 – when it comes to solving equations involving the trigonometric functions, it helps to just try something. 9We are essentially ‘undoing’ the sum / diﬀerence formula for cosine or sine, depending on which form we use, so this problem is actually closely related to the previous one! 866 Foundations of Trigonometry Next, we focus on solving inequalities involving the trigonometric functions. Since these functions are continuous on their domains, we may use the sign diagram technique we’ve used in the past to solve the inequalities.10 Example 10.7.3. Solve the following inequalities on [0, 2π). Express your answers using interval notation and verify your answers graphically. 1. 2 sin(x) ≤ 1 2. sin(2x) > cos(x) 3. tan(x) ≥ 3 Solution. 1. We begin solving 2 sin(x) ≤ 1 by collecting all of the terms on one side of the equation and zero on the other to get 2 sin(x) −
1 ≤ 0. Next, we let f (x) = 2 sin(x) − 1 and note that our original inequality is equivalent to solving f (x) ≤ 0. We now look to see where, if ever, f is undeﬁned and where f (x) = 0. Since the domain of f is all real numbers, we can immediately set about ﬁnding the zeros of f. Solving f (x) = 0, we have 2 sin(x) − 1 = 0 or sin(x) = 1 2. The solutions here are x = π 6 + 2πk for integers k. Since we are restricting our attention to [0, 2π), only x = π 6 are of concern to us. Next, we choose test values in [0, 2π) other than the zeros and determine if f is positive or negative there. For = 1 and for x = π we get f (π) = −1. 2 we get f π x = 0 we have f (0) = −1, for x = π Since our original inequality is equivalent to f (x) ≤ 0, we are looking for where the function is negative (−) or 0, and we get the intervals 0, π 6, 2π. We can conﬁrm our answer 6 graphically by seeing where the graph of y = 2 sin(x) crosses or is below the graph of y = 1. 6 + 2πk and x = 5π 6 and x = 5π ∪ 5π 2 (−) 0 (+) π 6 0 (−) 5π 6 2π 0 y = 2 sin(x) and y = 1 2. We ﬁrst rewrite sin(2x) > cos(x) as sin(2x) − cos(x) > 0 and let f (x) = sin(2x) − cos(x). Our original inequality is thus equivalent to f (x) > 0. The domain of f is all real numbers, so we can advance to ﬁnding the zeros of f. Setting f (x) = 0 yields sin(2x) − cos(x) = 0, which, by way of the double angle identity for sine, becomes 2 sin(x) cos(x) − cos(x) = 0 or 2 +πk for integers k of which only x = �
� cos(x)(2 sin(x)−1) = 0. From cos(x) = 0, we get x = π 2 and x = 3π 2 which gives x = π 6 + 2πk or x = 5π 6 lie in [0, 2π). Next, we choose 2 lie in [0, 2π). For 2 sin(x) − 1 = 0, we get sin(x) = 1 6 + 2πk for integers k. Of those, only x = π 6 and x = 5π 10See page 214, Example 3.1.5, page 321, page 399, Example 6.3.2 and Example 6.4.2 for discussion of this technique. 10.7 Trigonometric Equations and Inequalities 867 4 we get f 3π our test values. For x = 0 we ﬁnd f (0) = −1; when x = π = −1 + for x = 3π ; when x = π we have f (π) = 1, and lastly, for √, so this is 2 x = 7π 6, π our answer. We can use the calculator to check that the graph of y = sin(2x) is indeed above the graph of y = cos(x) on those intervals. 2−2 2. We see f (x) > 0 on π 2 √ 2 2 = 2 = −2− 4 we get f 7π = −1 − 2 = 2− ∪ 5π 6, 3π 4 we get ; √ √ (−) 0 (+) π 6 0 (−) π 2 0 (+) 5π 6 0 (−) 3π 2 2π 0 y = sin(2x) and y = cos(x) 2 and 3π 3. Proceeding as in the last two problems, we rewrite tan(x) ≥ 3 as tan(x) − 3 ≥ 0 and let f (x) = tan(x) − 3. We note that on [0, 2π), f is undeﬁned at x = π 2, so those values will need the usual disclaimer on the sign diagram.11 Moving along to zeros, solving f (x) = tan(x) − 3 = 0 requires the arctangent function. We ﬁnd x = arctan(3) + πk
for integers k and of these, only x = arctan(3) and x = arctan(3) + π lie in [0, 2π). Since 3 > 0, we know 0 < arctan(3) < π 2 which allows us to position these zeros correctly on the sign diagram. To choose test values, we begin with x = 0 and ﬁnd f (0) = −3. Finding a is a bit more challenging. Keep in mind convenient test value in the interval arctan(3), π 2 that the arctangent function is increasing and is bounded above by π 2. This means that the number x = arctan(117) is guaranteed12 to lie between arctan(3) and π 2. We see that f (arctan(117)) = tan(arctan(117)) − 3 = 114. For our next test value, we take x = π and ﬁnd f (π) = −3. To ﬁnd our next test value, we note that since arctan(3) < arctan(117) < π 2, it follows13 that arctan(3) + π < arctan(117) + π < 3π 2. Evaluating f at x = arctan(117) + π yields f (arctan(117) + π) = tan(arctan(117) + π) − 3 = tan(arctan(117)) − 3 = 114. We = −4. Since we want f (x) ≥ 0, we 4 and ﬁnd f 7π choose our last test value to be x = 7π. Using the graphs of y = tan(x) ∪ arctan(3) + π, 3π see that our answer is arctan(3), π 2 2 and y = 3, we see when the graph of the former is above (or meets) the graph of the latter. 4 11See page 321 for a discussion of the non-standard character known as the interrobang. 12We could have chosen any value arctan(t) where t > 3. 13... by adding π through the inequality... 868 Foundations of Trigonometry (−) 0 (+) 0 arctan(3) (−) �
� 2 (arctan(3) + π) 0 (+) (−) 3π 2 2π y = tan(x) and y = 3 Our next example puts solving equations and inequalities to good use – ﬁnding domains of functions. Example 10.7.4. Express the domain of the following functions using extended interval notation.14 1. f (x) = csc 2x + π 3 2. f (x) = sin(x) 2 cos(x) − 1 3. f (x) = 1 − cot(x) Solution. 1. To ﬁnd the domain of f (x) = csc 2x + π 3, we rewrite f in terms of sine as f (x) = 1 sin(2x+ π 3 ) Since the sine function is deﬁned everywhere, our only concern comes from zeros in the denominator. Solving sin 2x + π 2 k for integers k. In set-builder notation, 3 our domain is x : x = − π 2 k for integers k. To help visualize the domain, we follow the 6,..., old mantra ‘When in doubt, write it out!’ We get x : x = − π where we have kept the denominators 6 throughout to help see the pattern. Graphing the situation on a numberline, we have = 0, we get x = − π 6 + π 6, − 7π 6, − 4π 6 + π 6, 5π 6, 8π 6, 2π. − 7π 6 − 4π 6 − π 6 2π 6 5π 6 8π 6 Proceeding as we did on page 756 in Section 10.3.1, we let xk denote the kth number excluded from the domain and we have xk = − π for integers k. The intervals which comprise the domain are of the form (xk, xk + 1) = integers. Using extended interval notation, we have that the domain is as k runs through the 2 k = (3k−1)π 6 (3k−1)π 6, (3k+2)π 6 6 + π ∞ k=−∞ (3k − 1)π 6, (3k + 2)π 6 We can check our answer by substituting in values of k to
see that it matches our diagram. 14See page 756 for details about this notation. 10.7 Trigonometric Equations and Inequalities 869 2. Since the domains of sin(x) and cos(x) are all real numbers, the only concern when ﬁnding 2 cos(x)−1 is division by zero so we set the denominator equal to zero and 3 +2πk for integers 3 + 2πk for integers k, the domain of f (x) = sin(x) solve. From 2 cos(x)−1 = 0 we get cos(x) = 1 k. Using set-builder notation, the domain is x : x = π or x : x = ± π 3, ± 11π 3,..., so we have 3 + 2πk and x = 5π 3 +2πk or x = 5π 2 so that x = π 3, ± 7π 3, ± 5π − 11π 3 − 7π 3 − 5π 3 − π 3 π 3 5π 3 7π 3 11π 3 Unlike the previous example, we have two diﬀerent families of points to consider, and we present two ways of dealing with this kind of situation. One way is to generalize what we did in the previous example and use the formulas we found in our domain work to describe the intervals. To that end, we let ak = π for integers k. The goal now is to write the domain in terms of the a’s an b’s. We ﬁnd a0 = π 3, a1 = 7π 3 and b−2 = − 7π 3. Hence, in terms of the a’s and b’s, our domain is 3 + 2πk = (6k+5)π 3 + 2πk = (6k+1)π 3, a−2 = − 11π 3, a−1 = − 5π 3, b−1 = − π 3, a2 = 13π 3, b1 = 11π 3, b0 = 5π 3, b2 = 17π and bk = 5π 3 3... (a−2, b−2) ∪ (b−2, a−1) ∪ (a−1, b−1) ∪ (b−1, a0) �
� (a0, b0) ∪ (b0, a1) ∪ (a1, b1) ∪... If we group these intervals in pairs, (a−2, b−2)∪(b−2, a−1), (a−1, b−1)∪(b−1, a0), (a0, b0)∪(b0, a1) and so forth, we see a pattern emerge of the form (ak, bk) ∪ (bk, ak + 1) for integers k so that our domain can be written as ∞ k=−∞ (ak, bk) ∪ (bk, ak + 1) = ∞ k=−∞ (6k + 1)π 3, (6k + 5)π 3 ∪ (6k + 5)π 3, (6k + 7)π 3 A second approach to the problem exploits the periodic nature of f. Since cos(x) and sin(x) have period 2π, it’s not too diﬃcult to show the function f repeats itself every 2π units.15 This means if we can ﬁnd a formula for the domain on an interval of length 2π, we can express the entire domain by translating our answer left and right on the x-axis by adding integer multiples of 2π. One such interval that arises from our domain work is π. The portion. Adding integer multiples of 2π, we get the family of of the domain here is π 3, 7π 3 + 2πk for integers k. We leave it to the reader intervals π 3 + 2πk, 7π to show that getting common denominators leads to our previous answer. ∪ 5π 3 + 2πk ∪ 5π 3 + 2πk, 5π 3, 5π 3, 7π 3 3 3 15This doesn’t necessarily mean the period of f is 2π. The tangent function is comprised of cos(x) and sin(x), but its period is half theirs. The reader is invited to investigate the period of f. 870 Foundations of Trigonometry 3. To ﬁnd the domain of f (x) = 1 − cot(x), we ﬁrst note that, due to the presence of the cot(x
) term, x = πk for integers k. Next, we recall that for the square root to be deﬁned, we need 1−cot(x) ≥ 0. Unlike the inequalities we solved in Example 10.7.3, we are not restricted here to a given interval. Our strategy is to solve this inequality over (0, π) (the same interval which generates a fundamental cycle of cotangent) and then add integer multiples of the period, in this case, π. We let g(x) = 1 − cot(x) and set about making a sign diagram for g over the interval (0, π) to ﬁnd where g(x) ≥ 0. We note that g is undeﬁned for x = πk for integers k, in particular, at the endpoints of our interval x = 0 and x = π. Next, we look for the zeros of g. Solving g(x) = 0, we get cot(x) = 1 or x = π 4 + πk for integers k and 6 and x = π only one of these, x = π 2, we get √ 3, and g π g π 2 6 4, lies in (0, π). Choosing the test values x = π = 1. = 1 − (−) 0 0 (+) π 4 π We ﬁnd g(x) ≥ 0 on π the intervals π express our ﬁnal answer as 4 + πk, π + πk = 4, π. Adding multiples of the period we get our solution to consist of. Using extended interval notation, we, (k + 1)π (4k+1)π 4 ∞ k=−∞ (4k + 1)π 4, (k + 1)π We close this section with an example which demonstrates how to solve equations and inequalities involving the inverse trigonometric functions. Example 10.7.5. Solve the following equations and inequalities analytically. Check your answers using a graphing utility. 1. arcsin(2x) = π 3 2. 4 arccos(x) − 3π = 0 3. 3 arcsec(2x − 1) + π = 2π 4. 4 arctan2(x) − 3
π arctan(x) − π2 = 0 5. π2 − 4 arccos2(x) < 0 6. 4 arccot(3x) > π Solution. 1. To solve arcsin(2x) = π 3 is in the range of the arcsine function (so a solution exists!) Next, we exploit the inverse property of sine and arcsine from Theorem 10.26 3, we ﬁrst note that π 10.7 Trigonometric Equations and Inequalities 871 arcsin(2x) = π 3 sin (arcsin(2x)) = sin π 3 2x = x = √ 3 2 √ 3 4 Since sin(arcsin(u)) = u Graphing y = arcsin(2x) and the horizontal line y = π √ 3 4 ≈ 0.4430. 2. Our ﬁrst step in solving 4 arccos(x) − 3π = 0 is to isolate the arccosine. Doing so, we get 3, we see they intersect at arccos(x) = 3π 4. Since 3π 4 is in the range of arccosine, we may apply Theorem 10.26 arccos(x) = 3π 4 cos (arccos(x)) = cos 3π 4 x = − √ 2 2 Since cos(arccos(u)) = u The calculator conﬁrms y = 4 arccos(x) − 3π crosses y = 0 (the x-axis) at − √ 2 2 ≈ −0.7071. y = arcsin(2x) and y = π 3 y = 4 arccos(x) − 3π 3. From 3 arcsec(2x − 1) + π = 2π, we get arcsec(2x − 1) = π 3. As we saw in Section 10.6, there are two possible ranges for the arcsecant function. Fortunately, both ranges contain π 3. Applying Theorem 10.28 / 10.29, we get arcsec(2x − 1) = π 3 sec(arcsec(2x − 1)) = sec π 3 2x − 1 = 2 x = 3 2 Since sec(arcsec(u)) = u To check using our
calculator, we need to graph y = 3 arcsec(2x − 1) + π. To do so, we make from Theorems 10.28 and 10.29.16 We see the graph use of the identity arcsec(u) = arccos 1 u of y = 3 arccos + π and the horizontal line y = 2π intersect at 3 1 2 = 1.5. 2x−1 16Since we are checking for solutions where arcsecant is positive, we know u = 2x − 1 ≥ 1, and so the identity applies in both cases. 872 Foundations of Trigonometry 4. With the presence of both arctan2(x) ( = (arctan(x))2) and arctan(x), we substitute u = arctan(x). The equation 4 arctan2(x) − 3π arctan(x) − π2 = 0 becomes 4u2 − 3πu − π2 = 0. Factoring,17 we get (4u + π)(u − π) = 0, so u = arctan(x) = − π 4 or u = arctan(x) = π. Since − π 4 is in the range of arctangent, but π is not, we only get solutions from the ﬁrst equation. Using Theorem 10.27, we get arctan(x) = − π 4 tan(arctan(x)) = tan − π 4 x = −1 Since tan(arctan(u)) = u. The calculator veriﬁes our result. y = 3 arcsec(2x − 1) + π and y = 2π y = 4 arctan2(x) − 3π arctan(x) − π2 5. Since the inverse trigonometric functions are continuous on their domains, we can solve inequalities featuring these functions using sign diagrams. Since all of the nonzero terms of π2 − 4 arccos2(x) < 0 are on one side of the inequality, we let f (x) = π2 − 4 arccos2(x) and note the domain of f is limited by the arccos(x) to [−1, 1]. Next, we ﬁnd the zeros of f by
setting f (x) = π2 − 4 arccos2(x) = 0. We get arccos(x) = ± π 2, and since the range of arccosine is = 0 [0, π], we focus our attention on arccos(x) = π as our only zero. Hence, we have two test intervals, [−1, 0) and (0, 1]. Choosing test values x = ±1, we get f (−1) = −3π2 < 0 and f (1) = π2 > 0. Since we are looking for where f (x) = π2 − 4 arccos2(x) < 0, our answer is [−1, 0). The calculator conﬁrms that for these values of x, the graph of y = π2 − 4 arccos2(x) is below y = 0 (the x-axis.) 2. Using Theorem 10.26, we get x = cos π 2 17It’s not as bad as it looks... don’t let the π throw you! 10.7 Trigonometric Equations and Inequalities 873 (−) (+) 0 0 1 −1 y = π2 − 4 arccos2(x) 6. To begin, we rewrite 4 arccot(3x) > π as 4 arccot(3x) − π > 0. We let f (x) = 4 arccot(3x) − π, and note the domain of f is all real numbers, (−∞, ∞). To ﬁnd the zeros of f, we set f (x) = 4 arccot(3x) − π = 0 and solve. We get arccot(3x) = π 4 is in the range of arccotangent, we may apply Theorem 10.27 and solve 4, and since π arccot(3x) = π 4 cot(arccot(3x)) = cot π 4 3x = 1 x = 1 3 Since cot(arccot(u)) = u. 3, we have two test intervals, −∞, 1 Next, we make a sign diagram for f. Since the domain of f is all real numbers, and there is 3,
∞. Ideally, we wish only one zero of f, x = 1 to ﬁnd test values x in these intervals so that arccot(4x) corresponds to one of our oft-used ‘common’ angles. After a bit of computation,18 we choose x = 0 for x < 1 3, we √ 3 = − π choose x = 3 < 0. Since we are looking for where 3. We ﬁnd f (0) = π > 0 and f. To check graphically, we use the f (x) = 4 arccot(3x) − π > 0, we get our answer −∞, 1 3 technique in number 2c of Example 10.6.5 in Section 10.6 to graph y = 4 arccot(3x) and we see it is above the horizontal line y = π on −∞, 1 3 3 and for x > 1 = −∞, 0.3. and 1 √ 3 3 3 (−) (+) 0 1 3 y = 4 arccot(3x) and y = π 18Set 3x equal to the cotangents of the ‘common angles’ and choose accordingly. 874 Foundations of Trigonometry 10.7.1 Exercises In Exercises 1 - 18, ﬁnd all of the exact solutions of the equation and then list those solutions which are in the interval [0, 2π). 1. sin (5x) = 0 2. cos (3x) = 1 2 4. tan (6x) = 1 5. csc (4x) = −1 √ 3 3 = 0 7. cot (2x) = − 10. cos x + 5π 6 13. csc(x) = 0 16. sec2 (x) = 4 3 8. cos (9x) = 9 11. sin 2x − π 3 = − 1 2 14. tan (2x − π) = 1 17. cos2 (x. sin (−2x) = 6. sec (3x) = 9. sin = x 3 12. 2 cos x + √ 3 = 7π 4 15. tan2 (x) = 3 18. sin2 (x) = 3 4 In Exercises 19 - 42, solve the equation, giving the exact solutions which lie in [0,
2π) 19. sin (x) = cos (x) 21. sin (2x) = cos (x) 23. cos (2x) = cos (x) 20. sin (2x) = sin (x) 22. cos (2x) = sin (x) 24. cos(2x) = 2 − 5 cos(x) 25. 3 cos(2x) + cos(x) + 2 = 0 26. cos(2x) = 5 sin(x) − 2 27. 3 cos(2x) = sin(x) + 2 29. tan2(x) = 1 − sec(x) 31. sec(x) = 2 csc(x) 33. sin(2x) = tan(x) 35. cos(2x) + csc2(x) = 0 37. tan2 (x) = 3 2 sec (x) 39. tan(2x) − 2 cos(x) = 0 28. 2 sec2(x) = 3 − tan(x) 30. cot2(x) = 3 csc(x) − 3 32. cos(x) csc(x) cot(x) = 6 − cot2(x) 34. cot4(x) = 4 csc2(x) − 7 36. tan3 (x) = 3 tan (x) 38. cos3 (x) = − cos (x) 40. csc3(x) + csc2(x) = 4 csc(x) + 4 41. 2 tan(x) = 1 − tan2(x) 42. tan (x) = sec (x) 10.7 Trigonometric Equations and Inequalities 875 In Exercises 43 - 58, solve the equation, giving the exact solutions which lie in [0, 2π) 43. sin(6x) cos(x) = − cos(6x) sin(x) 44. sin(3x) cos(x) = cos(3x) sin(x) 45. cos(2x) cos(x) + sin(2x) sin(x) = 1 46. cos(5x) cos(3x) − sin(5x) sin(3x) = √ 3 2 47. sin(x) + cos(x) = 1 √ 49. 2 cos
(x) − √ 2 sin(x) = 1 51. cos(2x) − √ 3 sin(2x) = √ 2 48. sin(x) + √ 3 cos(x) = 1 √ 50. 3 sin(2x) + cos(2x) = 1 √ 52. 3 3 sin(3x) − 3 cos(3x) = 3 √ 3 53. cos(3x) = cos(5x) 54. cos(4x) = cos(2x) 55. sin(5x) = sin(3x) 56. cos(5x) = − cos(2x) 57. sin(6x) + sin(x) = 0 58. tan(x) = cos(x) In Exercises 59 - 68, solve the equation. 59. arccos(2x) = π 60. π − 2 arcsin(x) = 2π 61. 4 arctan(3x − 1) − π = 0 62. 6 arccot(2x) − 5π = 0 63. 4 arcsec x 2 = π 64. 12 arccsc x 3 = 2π 65. 9 arcsin2(x) − π2 = 0 66. 9 arccos2(x) − π2 = 0 67. 8 arccot2(x) + 3π2 = 10π arccot(x) 68. 6 arctan(x)2 = π arctan(x) + π2 In Exercises 69 - 80, solve the inequality. Express the exact answer in interval notation, restricting your attention to 0 ≤ x ≤ 2π. 69. sin (x) ≤ 0 72. cos2 (x) > 75. cot2 (x) ≥ 1 2 1 3 78. cos(3x) ≤ 1 70. tan (x) ≥ √ 3 73. cos (2x) ≤ 0 76. 2 cos(x) ≥ 1 79. sec(x) ≤ √ 2 71. sec2 (x) ≤ 4 74. sin x + π 3 > 1 2 77. sin(5x) ≥ 5 80. cot(x) ≤ 4 876 Foundations of Trigonometry In Exercises 81 - 86, solve the inequality. Express the exact answer in interval notation, restricting your
attention to −π ≤ x ≤ π. 81. cos (x) > 84. sin2 (x) < √ 3 2 3 4 82. sin(x) > 1 3 83. sec (x) ≤ 2 85. cot (x) ≥ −1 86. cos(x) ≥ sin(x) In Exercises 87 - 92, solve the inequality. Express the exact answer in interval notation, restricting your attention to −2π ≤ x ≤ 2π. 87. csc (x) > 1 90. tan2 (x) ≥ 1 88. cos(x) ≤ 5 3 89. cot(x) ≥ 5 91. sin(2x) ≥ sin(x) 92. cos(2x) ≤ sin(x) In Exercises 93 - 98, solve the given inequality. 93. arcsin(2x) > 0 94. 3 arccos(x) ≤ π 95. 6 arccot(7x) ≥ π 96. π > 2 arctan(x) 97. 2 arcsin(x)2 > π arcsin(x) 98. 12 arccos(x)2 + 2π2 > 11π arccos(x) In Exercises 99 - 107, express the domain of the function using the extended interval notation. (See page 756 in Section 10.3.1 for details.) 99. f (x) = 1 cos(x) − 1 100. f (x) = cos(x) sin(x) + 1 101. f (x) = tan2(x) − 1 102. f (x) = 2 − sec(x) 103. f (x) = csc(2x) 104. f (x) = sin(x) 2 + cos(x) 105. f (x) = 3 csc(x) + 4 sec(x) 106. f (x) = ln (\| cos(x)\|) 107. f (x) = arcsin(tan(x)) 108. With the help of your classmates, determine the number of solutions to sin(x) = 1 2 and sin(4x) = 1 2 in [0, 2π). Then ﬁnd the number of solutions to sin(2x) = 1 2 in [0, 2π). A pattern should emerge. Explain how this pattern would help you solve equations
like sin(11x) = 1 2. What do you ﬁnd? 2, sin(3x) = 1 2 and sin 5x 2, sin 3x = 1 = 1 2 2 = 1 2. Now consider sin x with −1 and repeat the whole exploration. 2 Replace 1 2 10.7 Trigonometric Equations and Inequalities 877 10.7.2 Answers 1. x = πk 5 ; x = 0, π 5, 2π 5, 3π 5, 4π 5, π, 6π 5, 7π 5, 8π 5, 9π 5 2. x = π 9 + 2πk 3 or x = + πk or x = 5π 9 5π 6 + 2πk 3 ; x = π 9, 5π 9, 7π 9, 11π 9, 13π 9, 17π 9 + πk; x = 2π 3, 5π 6, 5π 3, 11π 6 + + πk 6 πk 2 ; x = π 24, 5π 24, 3π 8, 13π 24, 17π 24, 7π 8, 25π 24, 29π 24, 11π 8, 37π 24, 41π 24, 15π 8 ; x = 3π 8, 7π 8, 11π 8, 15π 8 + 2πk 3 or x = 7π 12 + 2πk 3 ; x = π 12, 7π 12, 3π 4, 5π 4, 17π 12, 23π 12 3. x = 4. x = 5. x = 6. x = 2π 3 π 24 3π 8 π 12 7. x = π 3 + πk 2 ; x = π 3, 5π 6, 4π 3, 11π 6 8. No solution 9. x = 3π 4 10. x = − π 3 11. x = 3π 4 + 6πk or x = 9π 4 + 6πk; x = 3π 4 + πk; x = 2π 3, 5π 3 + πk or x = 13π 12 + πk; x = π 12, 3π 4, 13π 12, 7π 4 12. x = − 19π 12 + 2πk or x = π 12 + 2πk; x = π 12,
5π 12 13. No solution 14. x = 5π 8 + πk 2 ; x = π 8, 5π 8, 9π 8, 13π 8 15. x = 16. x = 17. x = 18 + πk or x = + πk or x = 2π 3 5π 6 + πk; x = + πk; x = π 3 π 6,, 2π 3 5π 6,, 4π 3 7π 6,, 5π 3 11π 6 + πk 2 ; x = π 4, 3π 4, 5π 4, 7π 4 + πk or x = 2π 3 + πk; x = π 3, 2π 3, 4π 3, 5π 3 878 19. x = 21. x = π 4 π 6 23. x = 0,,, 5π 4 π 2 2π 3,, 3π 2, 5π 6 4π 3 25. x = 27. x = 2π 3 7π 6,, 4π 3, arccos 11π 6, arcsin 1 3 1 3, 2π − arccos, π − arcsin 1 3 1 3 29. x = 0, 2π 3, 4π 3 31. x = arctan(2), π + arctan(2) Foundations of Trigonometry 20. x = 0, 22. x = 24. x = 26, 5π 3 3π 2, π 3 5π 6 5π 3 5π 6,,, 28. x = 3π 4, 7π 4, arctan 1 2, π + arctan 1 2 30. x = 32. x = 34. x = π 6 π 6 π 6 36. x = 0, 38. x = 40. x = π 2 π 6,,,,,, 5π 6 7π 6 π 4 π 3 3π 2 5π 6,,, π 2 5π 6 3π 4 2π 3,, 11π 6 5π 6 4π 3,, 7π 6 5π 3,, π,, 7π 6, 3π 2, 11π 6 5π 4, 7π 4, 11π 6 33. x = 0, π,, 3π 4, 5π 4,
7π 4 π 4 3π 2 5π 3 π 2 5π 8, 35. x = 37. x = 39. x = 41,,,,, 5π 6 9π 8, 3π 2 13π 8 4π 7,,, 3π 7 3π 2 13π 48, 43. x = 0, 44. x = 0, π 7 π 2, 2π 7, π, 46. x = π 48, 11π 48 47. x = 0, π 2 49. x = 51. x =, π 12 17π 24 17π 12 41π 24,, 23π 24, 47π 24 42. No solution, 5π 7, 6π 7, π, 8π 7, 9π 7, 10π 7, 11π 7, 12π 7, 13π 7 45. x = 0, 23π 48, 25π 48, 35π 48, 37π 48, 47π 48, 49π 48, 59π 48, 48. x =, 71π 48, 73π 48, 83π 48, 85π 48, 95π 48 61π 48 π 2, 11π 6 π 3 5π 18,, 50. x = 0, π, 52. x = π 6, 4π 3 5π 6, 17π 18, 3π 2, 29π 18 10.7 Trigonometric Equations and Inequalities 879 53. x = 0, 55. x = 0 3π 8 3π 4 5π 8,, π,, 5π 4 7π 8, 3π 2 9π 8 7π 4 11π 8,, 13π 8, 15π 8 54. x = 0, π 3, 2π 3, π, 4π 3, 5π 3,, π, 56. x = π 7, π 3, 3π 7, 5π 7, π, 9π 7, 11π 7, 5π 3, 13π 7 57. x = 0, 2π 7, 58. x = arcsin 59. x = − 1 2 61. x = 2 3 63. x = 2 √ 2 65. x = ± √ 3 2 67. x = −1, 0, 6π 7, 4π 7 −1 + 2 8π 7 √ 5, 10π 7, 12π 7, π 5
, 3π 5, π,, 7π 5 ≈ 0.6662, π − arcsin 9π 5 −1 + 2 √ 5 ≈ 2.4754 60. x = −1 62. x = − √ 3 2 64. x = 6 66. x = 1 2 √ 68. x = − 3 69. [π, 2π] 71. 73. 0, π 3 ∪ π 4, 3π 4, 2π 3 5π 4 ∪, 7π 4 4π 3 ∪ 5π 3, 2π 0, 75. π 3 ∪ 2π 3, π ∪ π, 4π 3 ∪ 5π 3, 2π 70. 72. 74. 76. π 3, π 2 ∪ 0, 0, 0,, 4π 3 3π 4 11π 6 5π 3, 2π 3π 2 5π 4 ∪ 7π 4, 2π, 2π 77. No solution 78. [0, 2π] 0, 79. π 4 ∪ π 2, 3π 2 ∪ 7π 4, 2π 81. − π 6, π 6 −π, − 83 80. [arccot(4), π) ∪ [π + arccot(4), 2π) 82. arcsin 1 3, π − arcsin 1 3 84. − 2π 3, − π 3 ∪ π 3, 2π 3 880 Foundations of Trigonometry −π, − 85. π 4 ∪ 0, 3π 4 −2π, − 87. 3π 2 ∪ − 3π 2, −π ∪ 0, π 2 ∪ π 2, π 86. − 3π 4, π 4 88. [−2π, 2π] 89. (−2π, arccot(5) − 2π] ∪ (−π, arccot(5) − π] ∪ (0, arccot(5)] ∪ (π, π + arccot(5)] − 3π 2, − 5π 4 − ∪ 3π 4, − −π, − 0, 3π 4 ∪ 5π 4, 3π 2 ∪ 3π 2, 7π 4 ∪ π, 5π 3 90. −
7π 4, − 3π 2 −2π, − 91. 5π 3 ∪ 92. − 93. 0, 1 2, − 7π 6 11π 6 ∪ ∪ π 6, 5π 6 ∪,, − π 3π 2 2 2, 1 94. 1 96. (−∞, ∞) 97. [−1, 0) 99. ∞ k=−∞ (2kπ, (2k + 2)π) 100. ∞ k=−∞ (4k − 1)π 2, (4k + 3)π 2 −∞, √ 3 7 95. 98. −14k + 1)π k=−∞ 4 ∞ (6k − 1)π 3 (2k + 1)π 2 ∪ (2k + 1)π 2, (4k + 3)π 4 (6k + 1)π 3 ∪ (4k + 1)π 2, (4k + 3)π 2,, kπ 2 kπ 2,, (k + 1)π 2 (k + 1)π 2 (4k − 1)π 4, (4k + 1)π 4 104. (−∞, ∞) 106. ∞ k=−∞ (2k − 1)π 2, (2k + 1)π 2 101. 102. 103. 105. 107. k=−∞ ∞ k=−∞ ∞ k=−∞ ∞ k=−∞ Chapter 11 Applications of Trigonometry 11.1 Applications of Sinusoids In the same way exponential functions can be used to model a wide variety of phenomena in nature,1 the cosine and sine functions can be used to model their fair share of natural behaviors. In section 10.5, we introduced the concept of a sinusoid as a function which can be written either in the form C(x) = A cos(ωx+φ)+B for ω > 0 or equivalently, in the form S(x) = A sin(ωx+φ)+B for ω > 0. At the time, we remained undecided as to which form we preferred, but the time for such indecision is over. For clarity of exposition we focus on the sine function2 in this section and switch to the independent variable t, since the applications in this
section are time-dependent. We reintroduce and summarize all of the important facts and deﬁnitions about this form of the sinusoid below. Properties of the Sinusoid S(t) = A sin(ωt + φ) + B The amplitude is \|A\| The angular frequency is ω and the ordinary frequency is f = ω 2π The period is T = 1 f = 2π ω The phase is φ and the phase shift is − φ ω The vertical shift or baseline is B Along with knowing these formulas, it is helpful to remember what these quantities mean in context. The amplitude measures the maximum displacement of the sine wave from its baseline (determined by the vertical shift), the period is the length of time it takes to complete one cycle of the sinusoid, the angular frequency tells how many cycles are completed over an interval of length 2π, and the ordinary frequency measures how many cycles occur per unit of time. The phase indicates what 1See Section 6.5. 2Sine haters can use the co-function identity cos π 2 − θ = sin(θ) to turn all of the sines into cosines. 882 Applications of Trigonometry angle φ corresponds to t = 0, and the phase shift represents how much of a ‘head start’ the sinusoid has over the un-shifted sine function. The ﬁgure below is repeated from Section 10.5. amplitude baseline period In Section 10.1.1, we introduced the concept of circular motion and in Section 10.2.1, we developed formulas for circular motion. Our ﬁrst foray into sinusoidal motion puts these notions to good use. Example 11.1.1. Recall from Exercise 55 in Section 10.1 that The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height 136 feet. It completes two revolutions in 2 minutes and 7 seconds. Assuming that the riders are at the edge of the circle, ﬁnd a sinusoid which describes the height of the passengers above the ground t seconds after they pass the point on the wheel closest to the ground. Solution. We sketch the problem situation below and assume a counter-clockwise rotation.3 θ Q h P O 3Otherwise, we could just observe the motion of the wheel from the other side. 11.1 Applications
of Sinusoids 883 We know from the equations given on page 732 in Section 10.2.1 that the y-coordinate for counterclockwise motion on a circle of radius r centered at the origin with constant angular velocity (frequency) ω is given by y = r sin(ωt). Here, t = 0 corresponds to the point (r, 0) so that θ, the angle measuring the amount of rotation, is in standard position. In our case, the diameter of the wheel is 128 feet, so the radius is r = 64 feet. Since the wheel completes two revolutions in 2 minutes and 7 seconds (which is 127 seconds) the period T = 1 seconds. Hence, the angular frequency is ω = 2π T = 4π 127 radians per second. Putting these two pieces of information together, we have that y = 64 sin 4π 127 t describes the y-coordinate on the Giant Wheel after t seconds, assuming it is centered at (0, 0) with t = 0 corresponding to the point Q. In order to ﬁnd an expression for h, we take the point O in the ﬁgure as the origin. Since the base of the Giant Wheel ride is 8 feet above the ground and the Giant Wheel itself has a radius of 64 feet, its center is 72 feet above the ground. To account for this vertical shift upward,4 we add 72 to our formula for y to obtain the new formula h = y + 72 = 64 sin 4π 127 t + 72. Next, we need to adjust things so that t = 0 corresponds to the point P instead of the point Q. This is where the phase comes into play. Geometrically, we need to shift the angle θ in the ﬁgure back π 2 radians. From Section 10.2.1, we know θ = ωt = 4π 127 t, so we (temporarily) write the height in terms of θ as h = 64 sin (θ) + 72. + 72. We 127 t − π Subtracting π 2. can check the reasonableness of our answer by graphing y = h(t) over the interval 0, 127 2 2 from θ gives the ﬁnal answer h(t) = 64 sin θ − π + 72 = 64 sin 4π 2 (127) = 127 2 2 y 136 72 8 t
127 2 A few remarks about Example 11.1.1 are in order. First, note that the amplitude of 64 in our answer corresponds to the radius of the Giant Wheel. This means that passengers on the Giant Wheel never stray more than 64 feet vertically from the center of the Wheel, which makes sense. 8 = 15.875. This represents the Second, the phase shift of our answer works out to be ‘time delay’ (in seconds) we introduce by starting the motion at the point P as opposed to the point Q. Said diﬀerently, passengers which ‘start’ at P take 15.875 seconds to ‘catch up’ to the point Q. 4π/127 = 127 π/2 Our next example revisits the daylight data ﬁrst introduced in Section 2.5, Exercise 6b. 4We are readjusting our ‘baseline’ from y = 0 to y = 72. 884 Applications of Trigonometry Example 11.1.2. According to the U.S. Naval Observatory website, the number of hours H of daylight that Fairbanks, Alaska received on the 21st day of the nth month of 2009 is given below. Here t = 1 represents January 21, 2009, t = 2 represents February 21, 2009, and so on. Month Number Hours of Daylight 1 2 3 4 5 6 7 8 9 10 11 12 5.8 9.3 12.4 15.9 19.4 21.8 19.4 15.6 12.4 9.1 5.6 3.3 1. Find a sinusoid which models these data and use a graphing utility to graph your answer along with the data. 2. Compare your answer to part 1 to one obtained using the regression feature of a calculator. Solution. 1. To get a feel for the data, we plot it below. H 22 20 18 16 14 12 10 10 11 12 6 The data certainly appear sinusoidal,5 but when it comes down to it, ﬁtting a sinusoid to data manually is not an exact science. We do our best to ﬁnd the constants A, ω, φ and B so that the function H(t) = A sin(ωt + φ) + B closely matches the data. We ﬁrst go after the vertical shift B whose value determines the baseline. In a typical sinusoid, the value
of B is the average of the maximum and minimum values. So here we take B = 3.3+21.8 = 12.55. Next is the amplitude A which is the displacement from the baseline to the maximum (and minimum) values. We ﬁnd A = 21.8 − 12.55 = 12.55 − 3.3 = 9.25. At this point, we have H(t) = 9.25 sin(ωt + φ) + 12.55. Next, we go after the angular frequency ω. Since the data collected is over the span of a year (12 months), we take the period T = 12 months.6 This 2 5Okay, it appears to be the ‘∧’ shape we saw in some of the graphs in Section 2.2. Just humor us. 6Even though the data collected lies in the interval [1, 12], which has a length of 11, we need to think of the data point at t = 1 as a representative sample of the amount of daylight for every day in January. That is, it represents H(t) over the interval [0, 1]. Similarly, t = 2 is a sample of H(t) over [1, 2], and so forth. 11.1 Applications of Sinusoids 885 12 = π T = 2π means ω = 2π 6. The last quantity to ﬁnd is the phase φ. Unlike the previous example, it is easier in this case to ﬁnd the phase shift − φ ω. Since we picked A > 0, the phase shift corresponds to the ﬁrst value of t with H(t) = 12.55 (the baseline value).7 Here, we choose t = 3, since its corresponding H value of 12.4 is closer to 12.55 than the next value, 15.9, which corresponds to t = 4. Hence, − φ 2. We have H(t) = 9.25 sin π + 12.55. Below is a graph of our data with the curve y = H(t). ω = 3, so φ = −3ω = −. Using the ‘SinReg’ command, we graph the calculator’s regression below. While both models seem to be reasonable ﬁts to the data, the calculator model is possibly the better ﬁt. The calculator does not
give us an r2 value like it did for linear regressions in Section 2.5, nor does it give us an R2 value like it did for quadratic, cubic and quartic regressions as in Section 3.1. The reason for this, much like the reason for the absence of R2 for the logistic model in Section 6.5, is beyond the scope of this course. We’ll just have to use our own good judgment when choosing the best sinusoid model. 11.1.1 Harmonic Motion One of the major applications of sinusoids in Science and Engineering is the study of harmonic motion. The equations for harmonic motion can be used to describe a wide range of phenomena, from the motion of an object on a spring, to the response of an electronic circuit. In this subsection, we restrict our attention to modeling a simple spring system. Before we jump into the Mathematics, there are some Physics terms and concepts we need to discuss. In Physics, ‘mass’ is deﬁned as a measure of an object’s resistance to straight-line motion whereas ‘weight’ is the amount of force (pull) gravity exerts on an object. An object’s mass cannot change,8 while its weight could change. 7See the ﬁgure on page 882. 8Well, assuming the object isn’t subjected to relativistic speeds... 886 Applications of Trigonometry An object which weighs 6 pounds on the surface of the Earth would weigh 1 pound on the surface of the Moon, but its mass is the same in both places. In the English system of units, ‘pounds’ (lbs.) is a measure of force (weight), and the corresponding unit of mass is the ‘slug’. In the SI system, the unit of force is ‘Newtons’ (N) and the associated unit of mass is the ‘kilogram’ (kg). We convert between mass and weight using the formula9 w = mg. Here, w is the weight of the object, m is the mass and g is the acceleration due to gravity. In the English system, g = 32 feet second2, and in the SI system, g = 9.8 meters second2. Hence, on Earth a mass of 1 slug weighs 32 lbs. and a mass of 1 kg weighs 9.8 N.10 Suppose we attach an object with mass m
to a spring as depicted below. The weight of the object will stretch the spring. The system is said to be in ‘equilibrium’ when the weight of the object is perfectly balanced with the restorative force of the spring. How far the spring stretches to reach equilibrium depends on the spring’s ‘spring constant’. Usually denoted by the letter k, the spring constant relates the force F applied to the spring to the amount d the spring stretches in accordance with Hooke’s Law11 F = kd. If the object is released above or below the equilibrium position, or if the object is released with an upward or downward velocity, the object will bounce up and down on the end of the spring until some external force stops it. If we let x(t) denote the object’s displacement from the equilibrium position at time t, then x(t) = 0 means the object is at the equilibrium position, x(t) < 0 means the object is above the equilibrium position, and x(t) > 0 means the object is below the equilibrium position. The function x(t) is called the ‘equation of motion’ of the object.12 x(t) = 0 at the equilibrium position x(t) < 0 above the equilibrium position x(t) > 0 below the equilibrium position If we ignore all other inﬂuences on the system except gravity and the spring force, then Physics tells us that gravity and the spring force will battle each other forever and the object will oscillate indeﬁnitely. In this case, we describe the motion as ‘free’ (meaning there is no external force causing the motion) and ‘undamped’ (meaning we ignore friction caused by surrounding medium, which in our case is air). The following theorem, which comes from Diﬀerential Equations, gives x(t) as a function of the mass m of the object, the spring constant k, the initial displacement x0 of the 9This is a consequence of Newton’s Second Law of Motion F = ma where F is force, m is mass and a is acceleration. In our present setting, the force involved is weight which is caused by the acceleration due to gravity. 10Note that 1 pound = 1 slug foot 11Look familiar? We saw Hooke’s Law in Section 4.3.1. 12To keep units compatible, if we are using the English system
, we use feet (ft.) to measure displacement. If we second2 and 1 Newton = 1 kg meter second2. are in the SI system, we measure displacement in meters (m). Time is always measured in seconds (s). 11.1 Applications of Sinusoids 887 object and initial velocity v0 of the object. As with x(t), x0 = 0 means the object is released from the equilibrium position, x0 < 0 means the object is released above the equilibrium position and x0 > 0 means the object is released below the equilibrium position. As far as the initial velocity v0 is concerned, v0 = 0 means the object is released ‘from rest,’ v0 < 0 means the object is heading upwards and v0 > 0 means the object is heading downwards.13 Theorem 11.1. Equation for Free Undamped Harmonic Motion: Suppose an object of mass m is suspended from a spring with spring constant k. If the initial displacement from the equilibrium position is x0 and the initial velocity of the object is v0, then the displacement x from the equilibrium position at time t is given by x(t) = A sin(ωt + φ) where ω = k m and A = x2 0 + 2 v0 ω A sin(φ) = x0 and Aω cos(φ) = v0. It is a great exercise in ‘dimensional analysis’ to verify that the formulas given in Theorem 11.1 work out so that ω has units 1 s and A has units ft. or m, depending on which system we choose. Example 11.1.3. Suppose an object weighing 64 pounds stretches a spring 8 feet. 1. If the object is attached to the spring and released 3 feet below the equilibrium position from rest, ﬁnd the equation of motion of the object, x(t). When does the object ﬁrst pass through the equilibrium position? Is the object heading upwards or downwards at this instant? 2. If the object is attached to the spring and released 3 feet below the equilibrium position with an upward velocity of 8 feet per second, ﬁnd the equation of motion of the object, x(t). What is the longest distance the object travels above the equilibrium position? When does this ﬁrst happen? Conﬁrm your result using a graphing utility. Solution. In order to use the formulas in Theorem
11.1, we ﬁrst need to determine the spring constant k and the mass of the object m. To ﬁnd k, we use Hooke’s Law F = kd. We know the object weighs 64 lbs. and stretches the spring 8 ft.. Using F = 64 and d = 8, we get 64 = k · 8, or k = 8 lbs. s2. We get m = 2 slugs. We can now proceed to apply Theorem 11.1. ft.. To ﬁnd m, we use w = mg with w = 64 lbs. and g = 32 ft. k 8 1. With k = 8 and m = 2, we get ω = 2 = 2. We are told that the object is released 3 feet below the equilibrium position ‘from rest.’ This means x0 = 3 and v0 = 0. Therefore, 32 + 02 = 3. To determine the phase φ, we have A sin(φ) = x0, 2 and angles coterminal to it A = which in this case gives 3 sin(φ) = 3 so sin(φ) = 1. Only φ = π 0 + v0 x2 m = 2 √ = ω 13The sign conventions here are carried over from Physics. If not for the spring, the object would fall towards the ground, which is the ‘natural’ or ‘positive’ direction. Since the spring force acts in direct opposition to gravity, any movement upwards is considered ‘negative’. 888 Applications of Trigonometry = 0. Going through the usual analysis we ﬁnd t = − π satisfy this condition, so we pick14 the phase to be φ = π 2. Hence, the equation of motion is x(t) = 3 sin 2t + π. To ﬁnd when the object passes through the equilibrium position we 2 solve x(t) = 3 sin 2t + π 2 k for 2 integers k. Since we are interested in the ﬁrst time the object passes through the equilibrium position, we look for the smallest positive t value which in this case is t = π 4 ≈ 0.78 seconds after the start of the motion. Common sense suggests that if we release the object below the equilibrium position, the object should be traveling upwards when it ﬁrst passes through it
. To check this answer, we graph one cycle of x(t). Since our applied domain in this situation is t ≥ 0, and the period of x(t) is T = 2π 2 = π, we graph x(t) over the interval [0, π]. Remembering that x(t) > 0 means the object is below the equilibrium position and x(t) < 0 means the object is above the equilibrium position, the fact our graph is crossing through the t-axis from positive x to negative x at t = π ω = 2π 4 + π 4 conﬁrms our answer. ω = 2 0 + v0 x2 5. From Aω cos(φ) = v0, we get 10 cos(φ) = −8, or cos(φ) = − 4 2. The only diﬀerence between this problem and the previous problem is that we now release the object with an upward velocity of 8 ft s. We still have ω = 2 and x0 = 3, but now we have v0 = −8, the negative indicating the velocity is directed upwards. Here, we get 32 + (−4)2 = 5. From A sin(φ) = x0, we get 5 sin(φ) = 3 which gives A = sin(φ) = 3 5. This means that φ is a Quadrant II angle which we can describe in terms of either arcsine or arccosine. Since x(t) is expressed in terms of sine, we choose to express φ = π − arcsin 3. Hence, 5. Since the amplitude of x(t) is 5, the object will travel x(t) = 5 sin 2t + π − arcsin 3 5 at most 5 feet above the equilibrium position. To ﬁnd when this happens, we solve the equation x(t) = 5 sin 2t + π − arcsin 3 = −5, the negative once again signifying that 5 the object is above the equilibrium position. Going through the usual machinations, we get 2 arcsin 3 t = 1 4 + πk for integers k. The smallest of these values occurs when k = 0, 5 that is, t = 1 4 ≈ 1.107 seconds after the start of the motion. To check our answer using the calculator, we graph y = 5 sin 2x
+ π − arcsin 3 on a graphing utility 5 and conﬁrm the coordinates of the ﬁrst relative minimum to be approximately (1.107, −5). + π 2 arcsin 3 + π 5 x π 4 π 2 3π 4 π t 3 2 1 −1 −2 −3 x(t) = 3 sin 2t + π 2 y = 5 sin 2x + π − arcsin 3 5 It is possible, though beyond the scope of this course, to model the eﬀects of friction and other external forces acting on the system.15 While we may not have the Physics and Calculus background 14For conﬁrmation, we note that Aω cos(φ) = v0, which in this case reduces to 6 cos(φ) = 0. 15Take a good Diﬀerential Equations class to see this! 11.1 Applications of Sinusoids 889 to derive equations of motion for these scenarios, we can certainly analyze them. We examine three cases in the following example. Example 11.1.4. 1. Write x(t) = 5e−t/5 cos(t) + 5e−t/5 using a graphing utility. √ 3 sin(t) in the form x(t) = A(t) sin(ωt + φ). Graph x(t) 2. Write x(t) = (t + 3) √ 2 cos(2t) + (t + 3) √ x(t) using a graphing utility. 2 sin(2t) in the form x(t) = A(t) sin(ωt + φ). Graph 3. Find the period of x(t) = 5 sin(6t) − 5 sin (8t). Graph x(t) using a graphing utility. Solution. √ √ √ 1. We start rewriting x(t) = 5e−t/5 cos(t) + 5e−t/5 3 sin(t) by factoring out 5e−t/5 from both terms to get x(t) = 5e−t/5 cos(t) + 3 sin(t). We convert what’s left in parentheses to the required form using the formulas introduced in Exercise 36 from Section 10.5. We ﬁnd
so that x(t) = 10e−t/5 sin t + π cos(t) +. Graphing this on the 3 sin(t) = 2 sin t + π 3 3 calculator as y = 10e−x/5 sin x + π reveals some interesting behavior. The sinusoidal nature 3 continues indeﬁnitely, but it is being attenuated. In the sinusoid A sin(ωx + φ), the coeﬃcient, we can think A of the sine function is the amplitude. In the case of y = 10e−x/5 sin x + π 3 of the function A(x) = 10e−x/5 as the amplitude. As x → ∞, 10e−x/5 → 0 which means the amplitude continues to shrink towards zero. Indeed, if we graph y = ±10e−x/5 along with y = 10e−x/5 sin x + π, we see this attenuation taking place. This equation corresponds to 3 the motion of an object on a spring where there is a slight force which acts to ‘damp’, or slow the motion. An example of this kind of force would be the friction of the object against the air. In this model, the object oscillates forever, but with smaller and smaller amplitude. y = 10e−x/5 sin x + π 3 y = 10e−x/5 sin x + π 3, y = ±10e−x/5 √ 2. Proceeding as in the ﬁrst example, we factor out (t + 3) 2 from each term in the function 2(cos(2t) + sin(2t)). We ﬁnd x(t) = (t + 3). Graphing this on the (cos(2t) + sin(2t)) = calculator as y = 2(x + 3) sin 2x + π, we ﬁnd the sinusoid’s amplitude growing. Since our 4 amplitude function here is A(x) = 2(x + 3) = 2x + 6, which continues to grow without bound, so x(t) = 2(t + 3) sin 2t + π 4 √ 2 sin 2t + π 4 2 sin(2t) to get x(t) = (
t + 3) 2 cos(2t) + (t + 3) √ √ √ 890 Applications of Trigonometry as x → ∞, this is hardly surprising. The phenomenon illustrated here is ‘forced’ motion. That is, we imagine that the entire apparatus on which the spring is attached is oscillating as well. In this case, we are witnessing a ‘resonance’ eﬀect – the frequency of the external oscillation matches the frequency of the motion of the object on the spring.16 y = 2(x + 3) sin 2x + π 4 y = 2(x + 3) sin 2x + π 4 y = ±2(x + 3) 8 = 3 3. Last, but not least, we come to x(t) = 5 sin(6t) − 5 sin(8t). To ﬁnd the period of this function, we need to determine the length of the smallest interval on which both f (t) = 5 sin(6t) and g(t) = 5 sin(8t) complete a whole number of cycles. To do this, we take the ratio of their frequencies and reduce to lowest terms: 6 4. This tells us that for every 3 cycles f makes, In other words, the period of x(t) is three times the period of f (t) (which is g makes 4. four times the period of g(t)), or π. We graph y = 5 sin(6x) − 5 sin(8x) over [0, π] on the calculator to check this. This equation of motion also results from ‘forced’ motion, but here the frequency of the external oscillation is diﬀerent than that of the object on the spring. Since the sinusoids here have diﬀerent frequencies, they are ‘out of sync’ and do not amplify each other as in the previous example. Taking things a step further, we can use a sum to product identity to rewrite x(t) = 5 sin(6t) − 5 sin(8t) as x(t) = −10 sin(t) cos(7t). The lower frequency factor in this expression, −10 sin(t), plays an interesting role in the graph of x(t). Below we graph y = 5 sin(6x) − 5 sin(8x)
and y = ±10 sin(x) over [0, 2π]. This is an example of the ‘beat’ phenomena, and the curious reader is invited to explore this concept as well.17 y = 5 sin(6x) − 5 sin(8x) over [0, π] y = 5 sin(6x) − 5 sin(8x) and y = ±10 sin(x) over [0, 2π] 16The reader is invited to investigate the destructive implications of resonance. 17A good place to start is this article on beats. 11.1 Applications of Sinusoids 891 11.1.2 Exercises 1. The sounds we hear are made up of mechanical waves. The note ‘A’ above the note ‘middle second. Find a sinusoid which C’ is a sound wave with ordinary frequency f = 440 Hertz = 440 cycles models this note, assuming that the amplitude is 1 and the phase shift is 0. 2. The voltage V in an alternating current source has amplitude 220 √ 2 and ordinary frequency f = 60 Hertz. Find a sinusoid which models this voltage. Assume that the phase is 0. 3. The London Eye is a popular tourist attraction in London, England and is one of the largest Ferris Wheels in the world. It has a diameter of 135 meters and makes one revolution (counterclockwise) every 30 minutes. It is constructed so that the lowest part of the Eye reaches ground level, enabling passengers to simply walk on to, and oﬀ of, the ride. Find a sinsuoid which models the height h of the passenger above the ground in meters t minutes after they board the Eye at ground level. 4. On page 732 in Section 10.2.1, we found the x-coordinate of counter-clockwise motion on a circle of radius r with angular frequency ω to be x = r cos(ωt), where t = 0 corresponds to the point (r, 0). Suppose we are in the situation of Exercise 3 above. Find a sinsusoid which models the horizontal displacement x of the passenger from the center of the Eye in meters t minutes after they board the Eye. Here we take x(t) > 0 to mean the passenger is to the right of the center, while x(t) < 0 means the passenger is to the left of the center. 5. In Exercise 52 in Section 10
.1, we introduced the yo-yo trick ‘Around the World’ in which a yo-yo is thrown so it sweeps out a vertical circle. As in that exercise, suppose the yo-yo string is 28 inches and it completes one revolution in 3 seconds. If the closest the yo-yo ever gets to the ground is 2 inches, ﬁnd a sinsuoid which models the height h of the yo-yo above the ground in inches t seconds after it leaves its lowest point. 6. Suppose an object weighing 10 pounds is suspended from the ceiling by a spring which stretches 2 feet to its equilibrium position when the object is attached. (a) Find the spring constant k in lbs. (b) Find the equation of motion of the object if it is released from 1 foot below the equilibrium position from rest. When is the ﬁrst time the object passes through the equilibrium position? In which direction is it heading? ft. and the mass of the object in slugs. (c) Find the equation of motion of the object if it is released from 6 inches above the equilibrium position with a downward velocity of 2 feet per second. Find when the object passes through the equilibrium position heading downwards for the third time. 892 Applications of Trigonometry 7. Consider the pendulum below. Ignoring air resistance, the angular displacement of the pen- dulum from the vertical position, θ, can be modeled as a sinusoid.18 θ The amplitude of the sinusoid is the same as the initial angular displacement, θ0, of the pendulum and the period of the motion is given by T = 2π l g where l is the length of the pendulum and g is the acceleration due to gravity. (a) Find a sinusoid which gives the angular displacement θ as a function of time, t. Arrange things so θ(0) = θ0. (b) In Exercise 40 section 5.3, you found the length of the pendulum needed in Jeﬀ’s antique Seth-Thomas clock to ensure the period of the pendulum is 1 2 of a second. Assuming the initial displacement of the pendulum is 15◦, ﬁnd a sinusoid which models the displacement of the pendulum θ as a function of time, t, in seconds. 8. The table below lists the average temperature of Lake Erie as measured in Cleveland, Ohio on the ﬁr
st of the month for each month during the years 1971 – 2000.19 For example, t = 3 represents the average of the temperatures recorded for Lake Erie on every March 1 for the years 1971 through 2000. Month Number, t Temperature (◦ F), 10 11 12 36 33 34 38 47 57 67 74 73 67 56 46 (a) Using the techniques discussed in Example 11.1.2, ﬁt a sinusoid to these data. (b) Using a graphing utility, graph your model along with the data set to judge the reason- ableness of the ﬁt. 18Provided θ is kept ‘small.’ Carl remembers the ‘Rule of Thumb’ as being 20◦ or less. Check with your friendly neighborhood physicist to make sure. 19See this website: http://www.erh.noaa.gov/cle/climate/cle/normals/laketempcle.html. 11.1 Applications of Sinusoids 893 (c) Use the model you found in part 8a to predict the average temperature recorded for Lake Erie on April 15th and September 15th during the years 1971–2000.20 (d) Compare your results to those obtained using a graphing utility. 9. The fraction of the moon illuminated at midnight Eastern Standard Time on the tth day of June, 2009 is given in the table below.21 Day of June, t Fraction Illuminated, F 3 6 9 12 15 18 21 24 27 30 0.81 0.98 0.98 0.83 0.57 0.27 0.04 0.03 0.26 0.58 (a) Using the techniques discussed in Example 11.1.2, ﬁt a sinusoid to these data.22 (b) Using a graphing utility, graph your model along with the data set to judge the reason- ableness of the ﬁt. (c) Use the model you found in part 9a to predict the fraction of the moon illuminated on June 1, 2009. 23 (d) Compare your results to those obtained using a graphing utility. 10. With the help of your classmates, research the phenomena mentioned in Example 11.1.4, namely resonance and beats. 11. With the help of your classmates, research Amplitude Modulation and Frequency Modulation. 12. What other things in the world might be roughly sinusoidal? Look to see what models you
can ﬁnd for them and share your results with your class. 20The computed average is 41◦F for April 15th and 71◦F for September 15th. 21See this website: http://www.usno.navy.mil/USNO/astronomical-applications/data-services/frac-moon-ill. 22You may want to plot the data before you ﬁnd the phase shift. 23The listed fraction is 0.62. 894 Applications of Trigonometry 11.1.3 Answers 1. S(t) = sin (880πt) 2. V (t) = 220 √ 2 sin (120πt) 3. h(t) = 67.5 sin π 15 t − π 2 + 67.5 4. x(t) = 67.5 cos π 15 t − π 2 = 67.5 sin π 15 t 5. h(t) = 28 sin 2π + 30 2 3 t − π ft. and m = 5 16 slugs 6. (a) k = 5 lbs. (b) x(t) = sin 4t + π 2. The object ﬁrst passes through the equilibrium point when t = π 8 ≈ 0.39 seconds after the motion starts. At this time, the object is heading upwards. (c) x(t) = √ 2 2 sin 4t + 7π 4. The object passes through the equilibrium point heading down- wards for the third time when t = 17π 16 ≈ 3.34 seconds. 7. (a) θ(t) = θ0 sin g l t + π 2 (b) θ(t) = π 12 sin 4πt + π 2 8. (a) T (t) = 20.5 sin π (b) Our function and the data set are graphed below. The sinusoid seems to be shifted to 6 t − π + 53.5 the right of our data. (c) The average temperature on April 15th is approximately T (4.5) ≈ 39.00◦F and the average temperature on September 15th is approximately T (9.5) ≈ 73.38◦F. (d) Using a graphing calculator, we get the following This model predicts the average temperature for April 15th to be approximately
42.43◦F and the average temperature on September 15th to be approximately 70.05◦F. This model appears to be more accurate. 11.1 Applications of Sinusoids 895 9. (a) Based on the shape of the data, we either choose A < 0 or we ﬁnd the second value of t which closely approximates the ‘baseline’ value, F = 0.505. We choose the latter to obtain F (t) = 0.475 sin π 15 t − 2π + 0.505 = 0.475 sin π 15 t + 0.505 (b) Our function and the data set are graphed below. It’s a pretty good ﬁt. (c) The fraction of the moon illuminated on June 1st, 2009 is approximately F (1) ≈ 0.60 (d) Using a graphing calculator, we get the following. This model predicts that the fraction of the moon illuminated on June 1st, 2009 is approximately 0.59. This appears to be a better ﬁt to the data than our ﬁrst model. 896 Applications of Trigonometry 11.2 The Law of Sines Trigonometry literally means ‘measuring triangles’ and with Chapter 10 under our belts, we are more than prepared to do just that. The main goal of this section and the next is to develop theorems which allow us to ‘solve’ triangles – that is, ﬁnd the length of each side of a triangle and the measure of each of its angles. In Sections 10.2, 10.3 and 10.6, we’ve had some experience solving right triangles. The following example reviews what we know. Example 11.2.1. Given a right triangle with a hypotenuse of length 7 units and one leg of length 4 units, ﬁnd the length of the remaining side and the measures of the remaining angles. Express the angles in decimal degrees, rounded to the nearest hundreth of a degree. Solution. For deﬁnitiveness, we label the triangle below √ To ﬁnd the length of the missing side a, we use the Pythagorean Theorem to get a2 + 42 = 72 which then yields a = 33 units. Now that all three sides of the triangle are known, there are several ways we can ﬁnd α using the
inverse trigonometric functions. To decrease the chances of propagating error, however, we stick to using the data given to us in the problem. In this case, the lengths 4 and 7 were given, so we want to relate these to α. According to Theorem 10.4, cos(α) = 4 7. Since α is an acute angle, α = arccos 4 radians. Converting to degrees, we ﬁnd α ≈ 55.15◦. Now 7 that we have the measure of angle α, we could ﬁnd the measure of angle β using the fact that α and β are complements so α + β = 90◦. Once again, we opt to use the data given to us in the radians and we problem. According to Theorem 10.4, we have that sin(β) = 4 have β ≈ 34.85◦. 7 so β = arcsin 4 7 A few remarks about Example 11.2.1 are in order. First, we adhere to the convention that a lower case Greek letter denotes an angle1 and the corresponding lowercase English letter represents the side2 opposite that angle. Thus, a is the side opposite α, b is the side opposite β and c is the side opposite γ. Taken together, the pairs (α, a), (β, b) and (γ, c) are called angle-side opposite pairs. Second, as mentioned earlier, we will strive to solve for quantities using the original data given in the problem whenever possible. While this is not always the easiest or fastest way to proceed, it 1as well as the measure of said angle 2as well as the length of said side 11.2 The Law of Sines 897 minimizes the chances of propagated error.3 Third, since many of the applications which require solving triangles ‘in the wild’ rely on degree measure, we shall adopt this convention for the time being.4 The Pythagorean Theorem along with Theorems 10.4 and 10.10 allow us to easily handle any given right triangle problem, but what if the triangle isn’t a right triangle? In certain cases, we can use the Law of Sines to help. Theorem 11.2. The Law of Sines: Given a triangle with angle-side opposite pairs (α, a), (β, b) and (γ, c), the following ratios hold or, equivalently,
sin(α) a = sin(β) b = sin(γ) c a sin(α) = b sin(β) = c sin(γ) The proof of the Law of Sines can be broken into three cases. For our ﬁrst case, consider the triangle ABC below, all of whose angles are acute, with angle-side opposite pairs (α, a), (β, b) and (γ, c). If we drop an altitude from vertex B, we divide the triangle into two right triangles: ABQ and BCQ. If we call the length of the altitude h (for height), we get from Theorem 10.4 that sin(α) = h a so that h = c sin(α) = a sin(γ). After some rearrangement of the last equation, we get sin(α). If we drop an altitude from vertex A, we can proceed as above using the triangles ABQ and ACQ to get sin(β) B, completing the proof for this case. c and sin(γ) = h a = sin(γ) b = sin(γ For our next case consider the triangle ABC below with obtuse angle α. Extending an altitude from vertex A gives two right triangles, as in the previous case: ABQ and ACQ. Proceeding as before, we get h = b sin(γ) and h = c sin(β) so that sin(β). b = sin(γ 3Your Science teachers should thank us for this. 4Don’t worry! Radians will be back before you know it! 898 Applications of Trigonometry Dropping an altitude from vertex B also generates two right triangles, ABQ and BCQ. We know that sin(α) = h c so that h = c sin(α). Since α = 180◦ − α, sin(α) = sin(α), so in fact, we have h = c sin(α). Proceeding to BCQ, we get sin(γ) = h a so h = a sin(γ). Putting this together with the previous equation, we get sin(γ), and we are ﬁnished with this case. c = sin(α The remaining case is when ABC is a right triangle. In this case, the Law of Sines reduces to the formulas given in Theorem 10.4 and is left to the reader. In order to use the Law of
Sines to solve a triangle, we need at least one angle-side opposite pair. The next example showcases some of the power, and the pitfalls, of the Law of Sines. Example 11.2.2. Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. 1. α = 120◦, a = 7 units, β = 45◦ 2. α = 85◦, β = 30◦, c = 5.25 units 3. α = 30◦, a = 1 units, c = 4 units 4. α = 30◦, a = 2 units, c = 4 units 5. α = 30◦, a = 3 units, c = 4 units 6. α = 30◦, a = 4 units, c = 4 units Solution. b sin(45◦) = 1. Knowing an angle-side opposite pair, namely α and a, we may proceed in using the Law of √ 6 Sines. Since β = 45◦, we use 3 ≈ 5.72 units. Now that we have two angle-side pairs, it is time to ﬁnd the third. To ﬁnd γ, we use the fact that the sum of the measures of the angles in a triangle is 180◦. Hence, γ = 180◦ − 120◦ − 45◦ = 15◦. To ﬁnd c, we have no choice but to used the derived value γ = 15◦, yet we can minimize the propagation of error here by using the given angle-side opposite pair (α, a). The Law of Sines gives us sin(120◦) so b = 7 sin(45◦) sin(120◦) so that c = 7 sin(15◦) sin(120◦) ≈ 2.09 units.5 sin(120◦) = 7 c sin(15◦) = 7 7 2. In this example, we are not immediately given an angle-side opposite pair, but as we have the measures of α and β, we can solve for γ since γ = 180◦ − 85◦ − 30◦ = 65◦. As in the previous example, we are forced to use a derived value in our computations since the only 5The exact value of sin(15◦) could be
found using the diﬀerence identity for sine or a half-angle formula, but that becomes unnecessarily messy for the discussion at hand. Thus “exact” here means 7 sin(15◦) sin(120◦). 11.2 The Law of Sines 899 angle-side pair available is (γ, c). The Law of Sines gives rearrangement, we get a = 5.25 sin(85◦) which yields sin(65◦). After the usual sin(65◦) ≈ 5.77 units. To ﬁnd b we use the angle-side pair (γ, c) sin(65◦) hence b = 5.25 sin(30◦) sin(65◦) ≈ 2.90 units. sin(30◦) = 5.25 b a sin(85◦) = 5.25 β = 45◦ a = 7 c ≈ 2.09 α = 120◦ γ = 15◦ b ≈ 5.72 Triangle for number 1 β = 30◦ c = 5.25 a ≈ 5.77 α = 85◦ γ = 65◦ b ≈ 2.90 Triangle for number 2 1 4 = sin(30◦) 3. Since we are given (α, a) and c, we use the Law of Sines to ﬁnd the measure of γ. We start with sin(γ) and get sin(γ) = 4 sin (30◦) = 2. Since the range of the sine function is [−1, 1], there is no real number with sin(γ) = 2. Geometrically, we see that side a is just too short to make a triangle. The next three examples keep the same values for the measure of α and the length of c while varying the length of a. We will discuss this case in more detail after we see what happens in those examples. 2 4 = sin(30◦) 4. In this case, we have the measure of α = 30◦, a = 2 and c = 4. Using the Law of Sines, we get sin(γ) so sin(γ) = 2 sin (30◦) = 1. Now γ is an angle in a triangle which also contains α = 30◦. This means that γ must measure between 0◦
and 150◦ in order to ﬁt inside the triangle with α. The only angle that satisﬁes this requirement and has sin(γ) = 1 is γ = 90◦. In other words, we have a right triangle. We ﬁnd the measure of β to be β = 180◦ − 30◦ − 90◦ = 60◦ and then determine b using the Law of Sines. We ﬁnd b = 2 sin(60◦) 3 ≈ 3.46 units. In this case, the side a is precisely long enough to form a sin(30◦) = 2 unique right triangle. √ c = 4 a = 1 α = 30◦ c = 4 β = 60◦ a = 2 α = 30◦ b ≈ 3.46 Diagram for number 3 Triangle for number 4 5. Proceeding as we have in the previous two examples, we use the Law of Sines to ﬁnd γ. In this 3. Since γ lies in a triangle with α = 30◦, or sin(γ) = 4 sin(30◦) case, we have sin(γ) 4 = sin(30◦) = 2 3 3 900 Applications of Trigonometry 3 3 : γ = arcsin 2 we must have that 0◦ < γ < 150◦. There are two angles γ that fall in this range and have radians ≈ 138.19◦. At radians ≈ 41.81◦ and γ = π − arcsin 2 sin(γ) = 2 3 this point, we pause to see if it makes sense that we actually have two viable cases to consider. As we have discussed, both candidates for γ are ‘compatible’ with the given angle-side pair (α, a) = (30◦, 3) in that both choices for γ can ﬁt in a triangle with α and both have a sine of 2 3. The only other given piece of information is that c = 4 units. Since c > a, it must be true that γ, which is opposite c, has greater measure than α which is opposite a. In both cases, γ > α, so both candidates for γ are compatible with this last piece of given information as radians ≈ 41.81◦, we well. Thus have
two triangles on our hands. In the case γ = arcsin 2 3 ﬁnd6 β ≈ 180◦ − 30◦ − 41.81◦ = 108.19◦. Using the Law of Sines with the angle-side opposite radians pair (α, a) and β, we ﬁnd b ≈ 3 sin(108.19◦) ≈ 138.19◦, we repeat the exact same steps and ﬁnd β ≈ 11.81◦ and b ≈ 1.23 units.7 Both triangles are drawn below. sin(30◦) ≈ 5.70 units. In the case γ = π − arcsin 2 3 β ≈ 11.81◦ c = 4 β ≈ 108.19◦ a = 3 α = 30◦ γ ≈ 41.81◦ α = 30◦ c = 4 a = 3 γ ≈ 138.19◦ b ≈ 5.70 b ≈ 1.23 6. For this last problem, we repeat the usual Law of Sines routine to ﬁnd that sin(γ) so that sin(γ) = 1 2. Since γ must inhabit a triangle with α = 30◦, we must have 0◦ < γ < 150◦. Since the measure of γ must be strictly less than 150◦, there is just one angle which satisﬁes both required conditions, namely γ = 30◦. So β = 180◦ − 30◦ − 30◦ = 120◦ and, using the √ Law of Sines one last time, b = 4 sin(120◦) 3 ≈ 6.93 units. 4 = sin(30◦) 4 sin(30◦) = 4 c = 4 β = 120◦ a = 4 α = 30◦ γ = 30◦ b ≈ 6.93 Some remarks about Example 11.2.2 are in order. We ﬁrst note that if we are given the measures of two of the angles in a triangle, say α and β, the measure of the third angle γ is uniquely 6 radians, γ = arcsin 2 6To ﬁnd an exact expression for β, we convert everything back to radians: α = 30◦ = π 3 radians and
180◦ = π radians. Hence, β = π − π 7An exact answer for β in this case is β = arcsin 2 6 − arcsin 2 − π 3 6 − arcsin 2 = 5π 6 radians ≈ 11.81◦. 3 3 radians ≈ 108.19◦. 11.2 The Law of Sines 901 determined using the equation γ = 180◦ − α − β. Knowing the measures of all three angles of a triangle completely determines its shape. If in addition we are given the length of one of the sides of the triangle, we can then use the Law of Sines to ﬁnd the lengths of the remaining two sides to determine the size of the triangle. Such is the case in numbers 1 and 2 above. In number 1, the given side is adjacent to just one of the angles – this is called the ‘Angle-Angle-Side’ (AAS) case.8 In number 2, the given side is adjacent to both angles which means we are in the so-called ‘Angle-Side-Angle’ (ASA) case. If, on the other hand, we are given the measure of just one of the angles in the triangle along with the length of two sides, only one of which is adjacent to the given angle, we are in the ‘Angle-Side-Side’ (ASS) case.9 In number 3, the length of the one given side a was too short to even form a triangle; in number 4, the length of a was just long enough to form a right triangle; in 5, a was long enough, but not too long, so that two triangles were possible; and in number 6, side a was long enough to form a triangle but too long to swing back and form two. These four cases exemplify all of the possibilities in the Angle-Side-Side case which are summarized in the following theorem. Theorem 11.3. Suppose (α, a) and (γ, c) are intended to be angle-side pairs in a triangle where α, a and c are given. Let h = c sin(α) If a < h, then no triangle exists which satisﬁes the given criteria. If a = h, then γ = 90◦ so exactly one (right) triangle exists which satisﬁes the criteria. If
h < a < c, then two distinct triangles exist which satisfy the given criteria. If a ≥ c, then γ is acute and exactly one triangle exists which satisﬁes the given criteria Theorem 11.3 is proved on a case-by-case basis. If a < h, then a < c sin(α). If a triangle were c = sin(α) to exist, the Law of Sines would have sin(γ) a = 1, which is impossible. In the ﬁgure below, we see geometrically why this is the case. so that sin(γ) = c sin(α) > a a a a h = c sin(α sin(α) a < h, no triangle a = h, γ = 90◦ Simply put, if a < h the side a is too short to connect to form a triangle. This means if a ≥ h, we are always guaranteed to have at least one triangle, and the remaining parts of the theorem 8If this sounds familiar, it should. From high school Geometry, we know there are four congruence conditions for triangles: Angle-Angle-Side (AAS), Angle-Side-Angle (ASA), Side-Angle-Side (SAS) and Side-Side-Side (SSS). If we are given information about a triangle that meets one of these four criteria, then we are guaranteed that exactly one triangle exists which satisﬁes the given criteria. 9In more reputable books, this is called the ‘Side-Side-Angle’ or SSA case. 902 Applications of Trigonometry c a a = a a = sin(γ) < 1 which means there are two solutions to sin(γ) = c sin(α) tell us what kind and how many triangles to expect in each case. If a = h, then a = c sin(α) and so that sin(γ) = c sin(α) the Law of Sines gives sin(α) a = 1. Here, γ = 90◦ as required. Moving along, now suppose h < a < c. As before, the Law of Sines10 gives sin(γ) = c sin(α). Since h < a, c sin(α) < a or c sin(α) : an acute angle which we’ll call γ0, and its supplement, 180
◦ − γ0. We need to argue that each of these angles ‘ﬁt’ into a triangle with α. Since (α, a) and (γ0, c) are angle-side opposite pairs, the assumption c > a in this case gives us γ0 > α. Since γ0 is acute, we must have that α is acute as well. This means one triangle can contain both α and γ0, giving us one of the triangles promised in the theorem. If we manipulate the inequality γ0 > α a bit, we have 180◦ −γ0 < 180◦ −α which gives (180◦ − γ0) + α < 180◦. This proves a triangle can contain both of the angles α and (180◦ − γ0), giving us the second triangle predicted in the theorem. To prove the last case in the theorem, we assume a ≥ c. Then α ≥ γ, which forces γ to be an acute angle. Hence, we get only one triangle in this case, completing the proof. a a c a a h α γ0 γ0 h < a < c, two triangles c α h a γ a ≥ c, one triangle One last comment before we use the Law of Sines to solve an application problem. In the AngleSide-Side case, if you are given an obtuse angle to begin with then it is impossible to have the two triangle case. Think about this before reading further. Example 11.2.3. Sasquatch Island lies oﬀ the coast of Ippizuti Lake. Two sightings, taken 5 miles apart, are made to the island. The angle between the shore and the island at the ﬁrst observation point is 30◦ and at the second point the angle is 45◦. Assuming a straight coastline, ﬁnd the distance from the second observation point to the island. What point on the shore is closest to the island? How far is the island from this point? Solution. We sketch the problem below with the ﬁrst observation point labeled as P and the second as Q. In order to use the Law of Sines to ﬁnd the distance d from Q to the island, we ﬁrst need to ﬁnd the measure of β which is the angle opposite the side of length 5 miles. To that end, we note
that the angles γ and 45◦ are supplemental, so that γ = 180◦ − 45◦ = 135◦. We can now 5 ﬁnd β = 180◦ − 30◦ − γ = 180◦ − 30◦ − 135◦ = 15◦. By the Law of Sines, we have sin(15◦) which gives d = 5 sin(30◦) sin(15◦) ≈ 9.66 miles. Next, to ﬁnd the point on the coast closest to the island, which we’ve labeled as C, we need to ﬁnd the perpendicular distance from the island to the coast.11 d sin(30◦) = 10Remember, we have already argued that a triangle exists in this case! 11Do you see why C must lie to the right of Q? 11.2 The Law of Sines 903 Let x denote the distance from the second observation point Q to the point C and let y denote the distance from C to the island. Using Theorem 10.4, we get sin (45◦) = y d. After some rearranging, we ﬁnd y = d sin (45◦) ≈ 9.66 ≈ 6.83 miles. Hence, the island is approximately 6.83 miles from the coast. To ﬁnd the distance from Q to C, we note that β = 180◦ − 90◦ − 45◦ = 45◦ so by symmetry,12 we get x = y ≈ 6.83 miles. Hence, the point on the shore closest to the island is approximately 6.83 miles down the coast from the second observation point. √ 2 2 Sasquatch Island Sasquatch Island β β d ≈ 9.66 miles d ≈ 9.66 miles y miles γ Q 30◦ 5 miles P 45◦ Shoreline 45◦ Q C x miles We close this section with a new formula to compute the area enclosed by a triangle. Its proof uses the same cases and diagrams as the proof of the Law of Sines and is left as an exercise. Theorem 11.4. Suppose (α, a), (β, b) and (γ, c) are the angle-side opposite pairs of a triangle. Then the area A enclosed by the triangle is given by A = 1 2 bc sin(α) = 1 2
ac sin(β) = 1 2 ab sin(γ) Example 11.2.4. Find the area of the triangle in Example 11.2.2 number 1. Solution. From our work in Example 11.2.2 number 1, we have all three angles and all three sides to work with. However, to minimize propagated error, we choose A = 1 2 ac sin(β) from Theorem 11.4 because it uses the most pieces of given information. We are given a = 7 and β = 45◦, and we calculated c = 7 sin(15◦) sin (45◦) =≈ 5.18 square units. The reader is encouraged to check this answer against the results obtained using the other formulas in Theorem 11.4. sin(120◦). Using these values, we ﬁnd A = 1 7 sin(15◦) sin(120◦) 2 (7) 12Or by Theorem 10.4 again... 904 Applications of Trigonometry 11.2.1 Exercises In Exercises 1 - 20, solve for the remaining side(s) and angle(s) if possible. As in the text, (α, a), (β, b) and (γ, c) are angle-side opposite pairs. 1. α = 13◦, β = 17◦, a = 5 2. α = 73.2◦, β = 54.1◦, a = 117 3. α = 95◦, β = 85◦, a = 33.33 4. α = 95◦, β = 62◦, a = 33.33 5. α = 117◦, a = 35, b = 42 6. α = 117◦, a = 45, b = 42 7. α = 68.7◦, a = 88, b = 92 8. α = 42◦, a = 17, b = 23.5 9. α = 68.7◦, a = 70, b = 90 10. α = 30◦, a = 7, b = 14 11. α = 42◦, a = 39, b = 23.5 12. γ = 53◦, α = 53◦, c = 28.01 13. α = 6◦, a = 57, b = 100 14. γ = 74.6◦
, c = 3, a = 3.05 15. β = 102◦, b = 16.75, c = 13 16. β = 102◦, b = 16.75, c = 18 17. β = 102◦, γ = 35◦, b = 16.75 18. β = 29.13◦, γ = 83.95◦, b = 314.15 19. γ = 120◦, β = 61◦, c = 4 20. α = 50◦, a = 25, b = 12.5 21. Find the area of the triangles given in Exercises 1, 12 and 20 above. (Another Classic Application: Grade of a Road) The grade of a road is much like the pitch of a roof (See Example 10.6.6) in that it expresses the ratio of rise/run. In the case of a road, this ratio is always positive because it is measured going uphill and it is usually given as a percentage. For example, a road which rises 7 feet for every 100 feet of (horizontal) forward progress is said to have a 7% grade. However, if we want to apply any Trigonometry to a story problem involving roads going uphill or downhill, we need to view the grade as an angle with respect to the horizontal. In Exercises 22 - 24, we ﬁrst have you change road grades into angles and then use the Law of Sines in an application. 22. Using a right triangle with a horizontal leg of length 100 and vertical leg with length 7, show that a 7% grade means that the road (hypotenuse) makes about a 4◦ angle with the horizontal. (It will not be exactly 4◦, but it’s pretty close.) 23. What grade is given by a 9.65◦ angle made by the road and the horizontal?13 13I have friends who live in Paciﬁca, CA and their road is actually this steep. It’s not a nice road to drive. 11.2 The Law of Sines 905 24. Along a long, straight stretch of mountain road with a 7% grade, you see a tall tree standing perfectly plumb alongside the road.14 From a point 500 feet downhill from the tree, the angle of inclination from the road to the top of the tree is 6◦. Use the Law of Sines to �
��nd the height of the tree. (Hint: First show that the tree makes a 94◦ angle with the road.) (Another Classic Application: Bearings) In the next several exercises we introduce and work with the navigation tool known as bearings. Simply put, a bearing is the direction you are heading according to a compass. The classic nomenclature for bearings, however, is not given as an angle in standard position, so we must ﬁrst understand the notation. A bearing is given as an acute angle of rotation (to the east or to the west) away from the north-south (up and down) line of a compass rose. For example, N40◦E (read “40◦ east of north”) is a bearing which is rotated clockwise 40◦ from due north. If we imagine standing at the origin in the Cartesian Plane, this bearing would have us heading into Quadrant I along the terminal side of θ = 50◦. Similarly, S50◦W would point into Quadrant III along the terminal side of θ = 220◦ because we started out pointing due south (along θ = 270◦) and rotated clockwise 50◦ back to 220◦. Counter-clockwise rotations would be found in the bearings N60◦W (which is on the terminal side of θ = 150◦) and S27◦E (which lies along the terminal side of θ = 297◦). These four bearings are drawn in the plane below. N N40◦E N60◦W 60◦ 40◦ W E 50◦ 27◦ S50◦W S27◦E S The cardinal directions north, south, east and west are usually not given as bearings in the fashion described above, but rather, one just refers to them as ‘due north’, ‘due south’, ‘due east’ and ‘due west’, respectively, and it is assumed that you know which quadrantal angle goes with each cardinal direction. (Hint: Look at the diagram above.) 25. Find the angle θ in standard position with 0◦ ≤ θ < 360◦ which corresponds to each of the bearings given below. (a) due west (b) S83◦E (c) N5.5◦E (d) due south 14
The word ‘plumb’ here means that the tree is perpendicular to the horizontal. 906 Applications of Trigonometry (e) N31.25◦W (f) S72◦4112W15 (g) N45◦E (h) S45◦W 26. The Colonel spots a campﬁre at a of bearing N42◦E from his current position. Sarge, who is positioned 3000 feet due east of the Colonel, reckons the bearing to the ﬁre to be N20◦W from his current position. Determine the distance from the campﬁre to each man, rounded to the nearest foot. 27. A hiker starts walking due west from Sasquatch Point and gets to the Chupacabra Trailhead before she realizes that she hasn’t reset her pedometer. From the Chupacabra Trailhead she hikes for 5 miles along a bearing of N53◦W which brings her to the Muﬃn Ridge Observatory. From there, she knows a bearing of S65◦E will take her straight back to Sasquatch Point. How far will she have to walk to get from the Muﬃn Ridge Observatory to Sasquach Point? What is the distance between Sasquatch Point and the Chupacabra Trailhead? 28. The captain of the SS Bigfoot sees a signal ﬂare at a bearing of N15◦E from her current location. From his position, the captain of the HMS Sasquatch ﬁnds the signal ﬂare to be at a bearing of N75◦W. If the SS Bigfoot is 5 miles from the HMS Sasquatch and the bearing from the SS Bigfoot to the HMS Sasquatch is N50◦E, ﬁnd the distances from the ﬂare to each vessel, rounded to the nearest tenth of a mile. 29. Carl spies a potential Sasquatch nest at a bearing of N10◦E and radios Jeﬀ, who is at a bearing of N50◦E from Carl’s position. From Jeﬀ’s position, the nest is at a bearing of S70◦W. If Jeﬀ and Carl are 500 feet apart, how far is Jeﬀ from the Sasquatch nest? Round your answer to the nearest foot. 30
. A hiker determines the bearing to a lodge from her current position is S40◦W. She proceeds to hike 2 miles at a bearing of S20◦E at which point she determines the bearing to the lodge is S75◦W. How far is she from the lodge at this point? Round your answer to the nearest hundredth of a mile. 31. A watchtower spots a ship oﬀ shore at a bearing of N70◦E. A second tower, which is 50 miles from the ﬁrst at a bearing of S80◦E from the ﬁrst tower, determines the bearing to the ship to be N25◦W. How far is the boat from the second tower? Round your answer to the nearest tenth of a mile. 32. Skippy and Sally decide to hunt UFOs. One night, they position themselves 2 miles apart on an abandoned stretch of desert runway. An hour into their investigation, Skippy spies a UFO hovering over a spot on the runway directly between him and Sally. He records the angle of inclination from the ground to the craft to be 75◦ and radios Sally immediately to ﬁnd the angle of inclination from her position to the craft is 50◦. How high oﬀ the ground is the UFO at this point? Round your answer to the nearest foot. (Recall: 1 mile is 5280 feet.) 15See Example 10.1.1 in Section 10.1 for a review of the DMS system. 11.2 The Law of Sines 907 33. The angle of depression from an observer in an apartment complex to a gargoyle on the building next door is 55◦. From a point ﬁve stories below the original observer, the angle of inclination to the gargoyle is 20◦. Find the distance from each observer to the gargoyle and the distance from the gargoyle to the apartment complex. Round your answers to the nearest foot. (Use the rule of thumb that one story of a building is 9 feet.) 34. Prove that the Law of Sines holds when ABC is a right triangle. 35. Discuss with your classmates why knowing only the three angles of a triangle is not enough to determine any of the sides. 36. Discuss with your classmates why the Law of Sines cannot be used to ﬁnd the angles in the triangle when only the three sides are given. Also discuss what happens if only two sides and the
angle between them are given. (Said another way, explain why the Law of Sines cannot be used in the SSS and SAS cases.) 37. Given α = 30◦ and b = 10, choose four diﬀerent values for a so that (a) the information yields no triangle (b) the information yields exactly one right triangle (c) the information yields two distinct triangles (d) the information yields exactly one obtuse triangle Explain why you cannot choose a in such a way as to have α = 30◦, b = 10 and your choice of a yield only one triangle where that unique triangle has three acute angles. 38. Use the cases and diagrams in the proof of the Law of Sines (Theorem 11.2) to prove the area formulas given in Theorem 11.4. Why do those formulas yield square units when four quantities are being multiplied together? 908 Applications of Trigonometry 11.2.2 Answers 1. 3. 5. 7. α = 13◦ β = 17◦ a = 5 γ = 150◦ b ≈ 6.50 c ≈ 11.11 Information does not produce a triangle Information does not produce a triangle α = 68.7◦ β ≈ 76.9◦ γ ≈ 34.4◦ c ≈ 53.36 a = 88 α = 68.7◦ β ≈ 103.1◦ γ ≈ 8.2◦ c ≈ 13.47 a = 88 b = 92 b = 92 9. Information does not produce a triangle 11. 13. 15. 17. 19. α = 42◦ β ≈ 23.78◦ γ ≈ 114.22◦ a = 39 c ≈ 53.15 b = 23.5 α = 6◦ β ≈ 169.43◦ γ ≈ 4.57◦ c ≈ 43.45 a = 57 b = 100 α = 6◦ β ≈ 10.57◦ γ ≈ 163.43◦ c ≈ 155.51 a = 57 b = 100 α ≈ 28.61◦ β = 102◦ a ≈ 8.20 b = 16.75 c = 13 γ ≈ 49.39◦ α = 43◦ γ = 35◦ β = 102◦ a ≈ 11.68 b = 16.75 c ≈ 9.82
Information does not produce a triangle 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. α = 73.2◦ β = 54.1◦ γ = 52.7◦ c ≈ 97.22 b ≈ 99.00 a = 117 β = 62◦ α = 95◦ a = 33.33 b ≈ 29.54 c ≈ 13.07 γ = 23◦ α = 117◦ β ≈ 56.3◦ γ ≈ 6.7◦ c ≈ 5.89 b = 42 a = 45 b = 23.5 α = 42◦ β ≈ 67.66◦ γ ≈ 70.34◦ c ≈ 23.93 a = 17 α = 42◦ β ≈ 112.34◦ γ ≈ 25.66◦ c ≈ 11.00 a = 17 b = 23.5 α = 30◦ β = 90◦ γ = 60◦ √ 3 b = 14 a = 7 c = 7 α = 53◦ β = 74◦ a = 28.01 b ≈ 33.71 c = 28.01 γ = 53◦ α ≈ 78.59◦ β ≈ 26.81◦ γ = 74.6◦ b ≈ 1.40 a = 3.05 α ≈ 101.41◦ β ≈ 3.99◦ γ = 74.6◦ b ≈ 0.217 a = 3.05 c = 3 c = 3 Information does not produce a triangle α = 66.92◦ β = 29.13◦ γ = 83.95◦ c ≈ 641.75 a ≈ 593.69 b = 314.15 α = 50◦ β ≈ 22.52◦ γ ≈ 107.48◦ a = 25 c ≈ 31.13 b = 12.5 21. The area of the triangle from Exercise 1 is about 8.1 square units. The area of the triangle from Exercise 12 is about 377.1 square units. The area of the triangle from Exercise 20 is about 149 square units. ≈ 0.699 radians, which is equivalent to 4.004◦ 22. arctan 7 100 23. About 17% 24. About 53 feet 11.2 The Law of Sines
909 25. (a) θ = 180◦ (b) θ = 353◦ (c) θ = 84.5◦ (d) θ = 270◦ (e) θ = 121.25◦ (f) θ = 197◦1848 (g) θ = 45◦ (h) θ = 225◦ 26. The Colonel is about 3193 feet from the campﬁre. Sarge is about 2525 feet to the campﬁre. 27. The distance from the Muﬃn Ridge Observatory to Sasquach Point is about 7.12 miles. The distance from Sasquatch Point to the Chupacabra Trailhead is about 2.46 miles. 28. The SS Bigfoot is about 4.1 miles from the ﬂare. The HMS Sasquatch is about 2.9 miles from the ﬂare. 29. Jeﬀ is about 371 feet from the nest. 30. She is about 3.02 miles from the lodge 31. The boat is about 25.1 miles from the second tower. 32. The UFO is hovering about 9539 feet above the ground. 33. The gargoyle is about 44 feet from the observer on the upper ﬂoor. The gargoyle is about 27 feet from the observer on the lower ﬂoor. The gargoyle is about 25 feet from the other building. 910 Applications of Trigonometry 11.3 The Law of Cosines In Section 11.2, we developed the Law of Sines (Theorem 11.2) to enable us to solve triangles in the ‘Angle-Angle-Side’ (AAS), the ‘Angle-Side-Angle’ (ASA) and the ambiguous ‘Angle-Side-Side’ (ASS) cases. In this section, we develop the Law of Cosines which handles solving triangles in the ‘Side-Angle-Side’ (SAS) and ‘Side-Side-Side’ (SSS) cases.1 We state and prove the theorem below. Theorem 11.5. Law of Cosines: Given a triangle with angle-side opposite pairs (α, a), (β, b) and (γ, c), the following equations hold a2 = b2 + c2 − 2bc cos(α) b2 = a2
+ c2 − 2ac cos(β) c2 = a2 + b2 − 2ab cos(γ) or, solving for the cosine in each equation, we have cos(α) = b2 + c2 − a2 2bc cos(β) = a2 + c2 − b2 2ac cos(γ) = a2 + b2 − c2 2ab To prove the theorem, we consider a generic triangle with the vertex of angle α at the origin with side b positioned along the positive x-axis. B = (c cos(α), c sin(α)) c α a A = (0, 0) b C = (b, 0) From this set-up, we immediately ﬁnd that the coordinates of A and C are A(0, 0) and C(b, 0). From Theorem 10.3, we know that since the point B(x, y) lies on a circle of radius c, the coordinates 1Here, ‘Side-Angle-Side’ means that we are given two sides and the ‘included’ angle - that is, the given angle is adjacent to both of the given sides. 11.3 The Law of Cosines 911 of B are B(x, y) = B(c cos(α), c sin(α)). (This would be true even if α were an obtuse or right angle so although we have drawn the case when α is acute, the following computations hold for any angle α drawn in standard position where 0 < α < 180◦.) We note that the distance between the points B and C is none other than the length of side a. Using the distance formula, Equation 1.1, we get a = (c cos(α) − b)2 + (c sin(α) − 0)2 2 (c cos(α) − b)2 + c2 sin2(α) a2 = a2 = (c cos(α) − b)2 + c2 sin2(α) a2 = c2 cos2(α) − 2bc cos(α) + b2 + c2 sin2(α) a2 = c2 cos2(α) + sin2(α) + b2 − 2bc cos(α) a2 = c2(1) + b2 − 2bc cos(α) a2 = c2 + b
2 − 2bc cos(α) Since cos2(α) + sin2(α) = 1 The remaining formulas given in Theorem 11.5 can be shown by simply reorienting the triangle to place a diﬀerent vertex at the origin. We leave these details to the reader. What’s important about a and α in the above proof is that (α, a) is an angle-side opposite pair and b and c are the sides adjacent to α – the same can be said of any other angle-side opposite pair in the triangle. Notice that the proof of the Law of Cosines relies on the distance formula which has its roots in the Pythagorean Theorem. That being said, the Law of Cosines can be thought of as a generalization of the Pythagorean Theorem. If we have a triangle in which γ = 90◦, then cos(γ) = cos (90◦) = 0 so we get the familiar relationship c2 = a2 + b2. What this means is that in the larger mathematical sense, the Law of Cosines and the Pythagorean Theorem amount to pretty much the same thing.2 Example 11.3.1. Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. 1. β = 50◦, a = 7 units, c = 2 units 2. a = 4 units, b = 7 units, c = 5 units Solution. 1. We are given the lengths of two sides, a = 7 and c = 2, and the measure of the included angle, β = 50◦. With no angle-side opposite pair to use, we apply the Law of Cosines. We get b2 = 72 + 22 − 2(7)(2) cos (50◦) which yields b = In order to determine the measures of the remaining angles α and γ, we are forced to used the derived value for b. There are two ways to proceed at this point. We could use the Law of Cosines again, or, since we have the angle-side opposite pair (β, b) we could use the Law of Sines. The advantage to using the Law of Cosines over the Law of Sines in cases like this is that unlike the sine function, the cosine function distinguishes between acute and obtuse angles. The cosine of an acute is positive,
whereas the cosine of an obtuse angle is negative. Since the sine of both acute and obtuse angles are positive, the sine of an angle alone is not 53 − 28 cos (50◦) ≈ 5.92 units. 2This shouldn’t come as too much of a shock. All of the theorems in Trigonometry can ultimately be traced back to the deﬁnition of the circular functions along with the distance formula and hence, the Pythagorean Theorem. 912 Applications of Trigonometry enough to determine if the angle in question is acute or obtuse. Since both authors of the textbook prefer the Law of Cosines, we proceed with this method ﬁrst. When using the Law of Cosines, it’s always best to ﬁnd the measure of the largest unknown angle ﬁrst, since this will give us the obtuse angle of the triangle if there is one. Since the largest angle is opposite the longest side, we choose to ﬁnd α ﬁrst. To that end, we use the formula cos(α) = b2+c2−a2 53 − 28 cos (50◦) and c = 2. We get3 and substitute a = 7, b = 2bc cos(α) = 2 − 7 cos (50◦) 53 − 28 cos (50◦) Since α is an angle in a triangle, we know the radian measure of α must lie between 0 and π radians. This matches the range of the arccosine function, so we have α = arccos 2 − 7 cos (50◦) 53 − 28 cos (50◦) radians ≈ 114.99◦ At this point, we could ﬁnd γ using γ = 180◦ − α − β ≈ 180◦ − 114.99◦ − 50◦ = 15.01◦, that is if we trust our approximation for α. To minimize propagation of error, however, we could use the Law of Cosines again,4 in this case using cos(γ) = a2+b2−c2. Plugging in a = 7, b = 53 − 28 cos (50◦) and c = 2, we get γ = arccos sketch the triangle below. √ 7−2 cos(50�
�) 53−28 cos(50◦) 2ab radians ≈ 15.01◦. We β = 50◦ a = 7 c = 2 α ≈ 114.99◦ γ ≈ 15.01◦ b ≈ 5.92 As we mentioned earlier, once we’ve determined b it is possible to use the Law of Sines to ﬁnd the remaining angles. Here, however, we must proceed with caution as we are in the ambiguous (ASS) case. It is advisable to ﬁrst ﬁnd the smallest of the unknown angles, since we are guaranteed it will be acute.5 In this case, we would ﬁnd γ since the side opposite γ is smaller than the side opposite the other unknown angle, α. Using the angle-side opposite pair (β, b), we get sin(γ). The usual calculations produces γ ≈ 15.01◦ and α = 180◦ − β − γ ≈ 180◦ − 50◦ − 15.01◦ = 114.99◦. sin(50◦) 53−28 cos(50◦) 2 = √ 2. Since all three sides and no angles are given, we are forced to use the Law of Cosines. Following our discussion in the previous problem, we ﬁnd β ﬁrst, since it is opposite the longest side, radians ≈ 101.54◦. As in b. We get cos(β) = a2+c2−b2 5, so we get β = arccos − 1 = − 1 2ac 5 3after simplifying... 4Your instructor will let you know which procedure to use. It all boils down to how much you trust your calculator. 5There can only be one obtuse angle in the triangle, and if there is one, it must be the largest. 11.3 The Law of Cosines 913 the previous problem, now that we have obtained an angle-side opposite pair (β, b), we could proceed using the Law of Sines. The Law of Cosines, however, oﬀers us a rare opportunity to ﬁnd the remaining angles using only the data given to us in the statement of the problem. Using this, we get γ = arccos 5 7 radians ≈ 44.42◦ and
α = arccos 29 35 radians ≈ 34.05◦. β ≈ 101.54◦ c = 5 a = 4 α ≈ 34.05◦ γ ≈ 44.42◦ b = 7 We note that, depending on how many decimal places are carried through successive calculations, and depending on which approach is used to solve the problem, the approximate answers you obtain may diﬀer slightly from those the authors obtain in the Examples and the Exercises. A great example of this is number 2 in Example 11.3.1, where the approximate values we record for the measures of the angles sum to 180.01◦, which is geometrically impossible. Next, we have an application of the Law of Cosines. Example 11.3.2. A researcher wishes to determine the width of a vernal pond as drawn below. From a point P, he ﬁnds the distance to the eastern-most point of the pond to be 950 feet, while the distance to the western-most point of the pond from P is 1000 feet. If the angle between the two lines of sight is 60◦, ﬁnd the width of the pond. 1000 feet 950 feet 60◦ P Solution. We are given the lengths of two sides and the measure of an included angle, so we may apply the Law of Cosines to ﬁnd the length of the missing side opposite the given angle. Calling this length w (for width), we get w2 = 9502 + 10002 − 2(950)(1000) cos (60◦) = 952500 from which we get w = 952500 ≈ 976 feet. √ 914 Applications of Trigonometry In Section 11.2, we used the proof of the Law of Sines to develop Theorem 11.4 as an alternate formula for the area enclosed by a triangle. In this section, we use the Law of Cosines to derive another such formula - Heron’s Formula. Theorem 11.6. Heron’s Formula: Suppose a, b and c denote the lengths of the three sides of a triangle. Let s be the semiperimeter of the triangle, that is, let s = 1 2 (a + b + c). Then the area A enclosed by the triangle is given by A = s(s − a)(s − b)(s − c) We prove Theorem 11.
6 using Theorem 11.4. Using the convention that the angle γ is opposite the side c, we have A = 1 2 ab sin(γ) from Theorem 11.4. In order to simplify computations, we start by manipulating the expression for A2. A2 = 2 ab sin(γ) 1 2 a2b2 sin2(γ) = = 1 4 a2b2 4 1 − cos2(γ) since sin2(γ) = 1 − cos2(γ). The Law of Cosines tells us cos(γ) = a2+b2−c2 2ab, so substituting this into our equation for A2 gives A2 = = = = = = = = 1 − 1 − 2 1 − cos2(γ) a2 + b2 − c2 2ab a2 + b2 − c22 4a2b2 4a2b2 − a2 + b2 − c22 4a2b2 a2b2 4 a2b2 4 a2b2 4 a2b2 4 4a2b2 − a2 + b2 − c22 16 (2ab)2 − a2 + b2 − c22 16 2ab − a2 + b2 − c2 2ab + a2 + b2 − c2 16 c2 − a2 + 2ab − b2 a2 + 2ab + b2 − c2 16 diﬀerence of squares. 11.3 The Law of Cosines 915 A2 = = = = = c2 − a2 − 2ab + b2 a2 + 2ab + b2 − c2 16 c2 − (a − b)2 (a + b)2 − c2 16 (c − (a − b))(c + (a − b))((a + b) − c)((a + b) + c) 16 perfect square trinomials. diﬀerence of squares. (b + c − a)(a + c − b)(a + b − c)(a + b + c) 16 (a + c − b) 2 (b + c − a) 2 (a + b − c) 2 · · · (a + b + c) 2 At this stage, we recognize the last factor as the semiperimeter, s = 1 complete the proof, we note that 2 (a + b + c) =
a+b+c 2. To (s − a − 2a 2 = b + c − a 2 Similarly, we ﬁnd (s − b) = a+c−b 2 and (s − c) = a+b−c 2. Hence, we get A2 = (b + c − a) 2 · (a + c − b) 2 · (a + b − c) 2 · (a + b + c) 2 = (s − a)(s − b)(s − c)s so that A = s(s − a)(s − b)(s − c) as required. We close with an example of Heron’s Formula. Example 11.3.3. Find the area enclosed of the triangle in Example 11.3.1 number 2. Solution. We are given a = 4, b = 7 and c = 5. Using these values, we ﬁnd s = 1 2 (4 + 7 + 5) = 8, (s − a) = 8 − 4 = 4, (s − b) = 8 − 7 = 1 and (s − c) = 8 − 5 = 3. Using Heron’s Formula, we get √ A = s(s − a)(s − b)(s − c) = 6 ≈ 9.80 square units. (8)(4)(1)(3) = 96 = 4 √ 916 Applications of Trigonometry 11.3.1 Exercises In Exercises 1 - 10, use the Law of Cosines to ﬁnd the remaining side(s) and angle(s) if possible. 1. a = 7, b = 12, γ = 59.3◦ 2. α = 104◦, b = 25, c = 37 3. a = 153, β = 8.2◦, c = 153 4. a = 3, b = 4, γ = 90◦ 5. α = 120◦, b = 3, c = 4 6. a = 7, b = 10, c = 13 7. a = 1, b = 2, c = 5 8. a = 300, b = 302, c = 48 9. a = 5, b = 5, c = 5 10. a = 5, b = 12, ; c = 13 In Exercises 11 - 16, solve for the remaining side(s)
and angle(s), if possible, using any appropriate technique. 11. a = 18, α = 63◦, b = 20 12. a = 37, b = 45, c = 26 13. a = 16, α = 63◦, b = 20 14. a = 22, α = 63◦, b = 20 15. α = 42◦, b = 117, c = 88 16. β = 7◦, γ = 170◦, c = 98.6 17. Find the area of the triangles given in Exercises 6, 8 and 10 above. 18. The hour hand on my antique Seth Thomas schoolhouse clock in 4 inches long and the minute hand is 5.5 inches long. Find the distance between the ends of the hands when the clock reads four o’clock. Round your answer to the nearest hundredth of an inch. 19. A geologist wants to measure the diameter of a crater. From her camp, it is 4 miles to the northern-most point of the crater and 2 miles to the southern-most point. If the angle between the two lines of sight is 117◦, what is the diameter of the crater? Round your answer to the nearest hundredth of a mile. 20. From the Pedimaxus International Airport a tour helicopter can ﬂy to Cliﬀs of Insanity Point by following a bearing of N8.2◦E for 192 miles and it can ﬂy to Bigfoot Falls by following a bearing of S68.5◦E for 207 miles.6 Find the distance between Cliﬀs of Insanity Point and Bigfoot Falls. Round your answer to the nearest mile. 21. Cliﬀs of Insanity Point and Bigfoot Falls from Exericse 20 above both lie on a straight stretch of the Great Sasquatch Canyon. What bearing would the tour helicopter need to follow to go directly from Bigfoot Falls to Cliﬀs of Insanity Point? Round your angle to the nearest tenth of a degree. 6Please refer to Page 905 in Section 11.2 for an introduction to bearings. 11.3 The Law of Cosines 917 22. A naturalist sets oﬀ on a hike from a lodge on a bearing of S80◦W. After 1.5 miles, she changes her bearing to S17◦W and continues hiking for 3 miles. Find her distance
from the lodge at this point. Round your answer to the nearest hundredth of a mile. What bearing should she follow to return to the lodge? Round your angle to the nearest degree. 23. The HMS Sasquatch leaves port on a bearing of N23◦E and travels for 5 miles. It then changes course and follows a heading of S41◦E for 2 miles. How far is it from port? Round your answer to the nearest hundredth of a mile. What is its bearing to port? Round your angle to the nearest degree. 24. The SS Bigfoot leaves a harbor bound for Nessie Island which is 300 miles away at a bearing of N32◦E. A storm moves in and after 100 miles, the captain of the Bigfoot ﬁnds he has drifted oﬀ course. If his bearing to the harbor is now S70◦W, how far is the SS Bigfoot from Nessie Island? Round your answer to the nearest hundredth of a mile. What course should the captain set to head to the island? Round your angle to the nearest tenth of a degree. 25. From a point 300 feet above level ground in a ﬁretower, a ranger spots two ﬁres in the Yeti National Forest. The angle of depression7 made by the line of sight from the ranger to the ﬁrst ﬁre is 2.5◦ and the angle of depression made by line of sight from the ranger to the second ﬁre is 1.3◦. The angle formed by the two lines of sight is 117◦. Find the distance between the two ﬁres. Round your answer to the nearest foot. (Hint: In order to use the 117◦ angle between the lines of sight, you will ﬁrst need to use right angle Trigonometry to ﬁnd the lengths of the lines of sight. This will give you a Side-Angle-Side case in which to apply the Law of Cosines.) ﬁre 117◦ ﬁre ﬁretower 26. If you apply the Law of Cosines to the ambiguous Angle-Side-Side (ASS) case, the result is a quadratic equation whose variable is that of the missing side. If the equation has no positive real zeros then the information given does not yield a triangle. If the equation has only one positive real zero then exactly one triangle
is formed and if the equation has two distinct positive real zeros then two distinct triangles are formed. Apply the Law of Cosines to Exercises 11, 13 and 14 above in order to demonstrate this result. 27. Discuss with your classmates why Heron’s Formula yields an area in square units even though four lengths are being multiplied together. 7See Exercise 78 in Section 10.3 for the deﬁnition of this angle. 918 Applications of Trigonometry 11.3.2 Answers 1. 3. 5. 7. 9. 11. 13. 15. α ≈ 35.54◦ β ≈ 85.16◦ γ = 59.3◦ c ≈ 10.36 a = 7 b = 12 α ≈ 85.90◦ β = 8.2◦ a = 153 b ≈ 21.88 c = 153 γ ≈ 85.90◦ α = 120◦ β ≈ 25.28◦ γ ≈ 34.72◦ a = c = 4 b = 3 37 √ Information does not produce a triangle α = 60◦ β = 60◦ γ = 60 = 20 α = 63◦ β ≈ 98.11◦ γ ≈ 18.89◦ a = 18 α = 63◦ β ≈ 81.89◦ γ ≈ 35.11◦ c ≈ 11.62 a = 18 c ≈ 6.54 b = 20 Information does not produce a triangle α = 42◦ a ≈ 78.30 b = 117 β ≈ 89.23◦ γ ≈ 48.77◦ c = 88 α = 104◦ a ≈ 49.41 b = 25 β ≈ 29.40◦ γ ≈ 46.60◦ c = 37 α ≈ 36.87◦ β ≈ 53.13◦ γ = 90 ≈ 32.31◦ β ≈ 49.58◦ γ ≈ 98.21◦ a = 7 c = 13 b = 10 α ≈ 83.05◦ β ≈ 87.81◦ γ ≈ 9.14◦ b = 302 a = 300 c = 48 α ≈ 22.62◦ β ≈ 67.38◦ γ = 90◦ c = 13 b = 12 a = 5 α ≈ 55.30
◦ β ≈ 89.40◦ γ ≈ 35.30◦ a = 37 c = 26 b = 45 α = 63◦ β ≈ 54.1◦ γ ≈ 62.9◦ c ≈ 21.98 a = 22 b = 20 α ≈ 3◦ γ = 170◦ β = 7◦ a ≈ 29.72 b ≈ 69.2 c = 98.6 √ 2. 4. 6. 8. 10. 12. 14. 16. √ √ 17. The area of the triangle given in Exercise 6 is The area of the triangle given in Exercise 8 is The area of the triangle given in Exercise 10 is exactly 30 square units. 1200 = 20 51764375 ≈ 7194.75 square units. 3 ≈ 34.64 square units. 18. The distance between the ends of the hands at four o’clock is about 8.26 inches. 19. The diameter of the crater is about 5.22 miles. 20. About 313 miles 21. N31.8◦W 22. She is about 3.92 miles from the lodge and her bearing to the lodge is N37◦E. 23. It is about 4.50 miles from port and its heading to port is S47◦W. 24. It is about 229.61 miles from the island and the captain should set a course of N16.4◦E to reach the island. 25. The ﬁres are about 17456 feet apart. (Try to avoid rounding errors.) 11.4 Polar Coordinates 919 11.4 Polar Coordinates In Section 1.1, we introduced the Cartesian coordinates of a point in the plane as a means of assigning ordered pairs of numbers to points in the plane. We deﬁned the Cartesian coordinate plane using two number lines – one horizontal and one vertical – which intersect at right angles at a point we called the ‘origin’. To plot a point, say P (−3, 4), we start at the origin, travel horizontally to the left 3 units, then up 4 units. Alternatively, we could start at the origin, travel up 4 units, then to the left 3 units and arrive at the same location. For the most part, the ‘motions’ of the Cartesian system (over and up) describe a rectangle, and most points can be thought of as the
corner diagonally across the rectangle from the origin.1 For this reason, the Cartesian coordinates of a point are often called ‘rectangular’ coordinates. In this section, we introduce a new system for assigning coordinates to points in the plane – polar coordinates. We start with an origin point, called the pole, and a ray called the polar axis. We then locate a point P using two coordinates, (r, θ), where r represents a directed distance from the pole2 and θ is a measure of rotation from the polar axis. Roughly speaking, the polar coordinates (r, θ) of a point measure ‘how far out’ the point is from the pole (that’s r), and ‘how far to rotate’ from the polar axis, (that’s θ). y P (−3, 4) P (r, θ) 3 2 1 r θ −4 −3 −2 −1 −1 1 2 3 4 x Pole r Polar Axis −2 −3 −4 For example, if we wished to plot the point P with polar coordinates 4, 5π 6 move out along the polar axis 4 units, then rotate 5π, we’d start at the pole, 6 radians counter-clockwise. P 4, 5π 6 r = 4 Pole θ = 5π 6 Pole Pole We may also visualize this process by thinking of the rotation ﬁrst.3 To plot P 4, 5π 6 we rotate 5π this way, 6 counter-clockwise from the polar axis, then move outwards from the pole 4 units. 1Excluding, of course, the points in which one or both coordinates are 0. 2We will explain more about this momentarily. 3As with anything in Mathematics, the more ways you have to look at something, the better. The authors encourage the reader to take time to think about both approaches to plotting points given in polar coordinates. 920 Applications of Trigonometry Essentially we are locating a point on the terminal side of 5π 6 which is 4 units away from the pole. θ = 5π 6 Pole θ = 5π 6 Pole P 4, 5π 6 Pole If r < 0, we begin by moving in the opposite direction on the polar axis from the pole. For example, to plot Q −3.5, π 4 we have r = −3.5 Pole θ = π 4 Pole Pole Q −3
.5, π 4 If we interpret the angle ﬁrst, we rotate π Here we are locating a point 3.5 units away from the pole on the terminal side of 5π 4 radians, then move back through the pole 3.5 units. 4, not π 4. θ = π 4 Pole θ = π 4 Pole Pole Q −3.5, π 4 As you may have guessed, θ < 0 means the rotation away from the polar axis is clockwise instead of counter-clockwise. Hence, to plot R 3.5, − 3π 4 we have the following. r = 3.5 Pole Pole θ = − 3π 4 Pole R 3.5, − 3π 4 From an ‘angles ﬁrst’ approach, we rotate − 3π R is the point on the terminal side of θ = − 3π 4 then move out 3.5 units from the pole. We see that 4 which is 3.5 units from the pole. Pole θ = − 3π 4 Pole θ = − 3π 4 Pole R 3.5, − 3π 4 11.4 Polar Coordinates 921 The points Q and R above are, in fact, the same point despite the fact that their polar coordinate representations are diﬀerent. Unlike Cartesian coordinates where (a, b) and (c, d) represent the same point if and only if a = c and b = d, a point can be represented by inﬁnitely many polar coordinate pairs. We explore this notion more in the following example. Example 11.4.1. For each point in polar coordinates given below plot the point and then give two additional expressions for the point, one of which has r > 0 and the other with r < 0. 4. P −3, − π 4 3. P 117, − 5π 2 2. P −4, 7π 6 1. P (2, 240◦) Solution. 1. Whether we move 2 units along the polar axis and then rotate 240◦ or rotate 240◦ then move out 2 units from the pole, we plot P (2, 240◦) below. θ = 240◦ Pole Pole P (2, 240◦) We now set about ﬁnding alternate descriptions (r, θ) for the point P. Since P is 2 units from the
pole, r = ±2. Next, we choose angles θ for each of the r values. The given representation for P is (2, 240◦) so the angle θ we choose for the r = 2 case must be coterminal with 240◦. (Can you see why?) One such angle is θ = −120◦ so one answer for this case is (2, −120◦). For the case r = −2, we visualize our rotation starting 2 units to the left of the pole. From this position, we need only to rotate θ = 60◦ to arrive at location coterminal with 240◦. Hence, our answer here is (−2, 60◦). We check our answers by plotting them. Pole θ = −120◦ θ = 60◦ Pole P (2, −120◦) P (−2, 60◦) 2. We plot −4, 7π 6 6 radians. Since r = −4 < 0, we ﬁnd our point lies 4 units from the pole on the terminal side of π 6. by ﬁrst moving 4 units to the left of the pole and then rotating 7π P −4, 7π 6 Pole Pole θ = 7π 6 922 Applications of Trigonometry To ﬁnd alternate descriptions for P, we note that the distance from P to the pole is 4 units, so any representation (r, θ) for P must have r = ±4. As we noted above, P lies on the terminal as one of our answers. To ﬁnd a diﬀerent side of π representation for P with r = −4, we may choose any angle coterminal with the angle in the as our second answer. original representation of P −4, 7π 6 6, so this, coupled with r = 4, gives us 4, π. We pick − 5π 6 and get −4, − 5π 6 6 P 4, π 6 θ = π 6 Pole θ = − 5π 6 Pole P −4, − 5π 6, we move along the polar axis 117 units from the pole and rotate 3. To plot P 117, − 5π 2 clockwise 5π 2 radians as illustrated below. Pole θ = − 5π 2 Pole P 117, − 5π 2 Since P is 117 units
from the pole, any representation (r, θ) for P satisﬁes r = ±117. For the r = 117 case, we can take θ to be any angle coterminal with − 5π 2. In this case, we choose as one answer. For the r = −117 case, we visualize moving left 117 θ = 3π units from the pole and then rotating through an angle θ to reach P. We ﬁnd that θ = π 2 satisﬁes this requirement, so our second answer is −117, π 2 2, and get 117, 3π. 2 Pole θ = 3π 2 Pole θ = π 2 P 117, 3π 2 P −117, π 2 11.4 Polar Coordinates 923 4. We move three units to the left of the pole and follow up with a clockwise rotation of π 4 radians to plot P −3, − π 4. We see that P lies on the terminal side of 3π 4. P −3, − π 4 θ = − π 4 Pole Pole. To Since P lies on the terminal side of 3π ﬁnd a diﬀerent representation for P with r = −3, we may choose any angle coterminal with − π 4, one alternative representation for P is 3, 3π 4 for our ﬁnal answer −3, 7π 4. We choose θ = 7π. 4 4 P 3, 3π 4 P −3, 7π 4 θ = 3π 4 Pole θ = 7π 4 Pole Now that we have had some practice with plotting points in polar coordinates, it should come as no surprise that any given point expressed in polar coordinates has inﬁnitely many other representations in polar coordinates. The following result characterizes when two sets of polar coordinates determine the same point in the plane. It could be considered as a deﬁnition or a theorem, depending on your point of view. We state it as a property of the polar coordinate system. Equivalent Representations of Points in Polar Coordinates Suppose (r, θ) and (r, θ) are polar coordinates where r = 0, r = 0 and the angles are measured in radians. Then (r, θ) and (r, θ) determine the same point P if and only
if one of the following is true: r = r and θ = θ + 2πk for some integer k r = −r and θ = θ + (2k + 1)π for some integer k All polar coordinates of the form (0, θ) represent the pole regardless of the value of θ. The key to understanding this result, and indeed the whole polar coordinate system, is to keep in mind that (r, θ) means (directed distance from pole, angle of rotation). If r = 0, then no matter how much rotation is performed, the point never leaves the pole. Thus (0, θ) is the pole for all 924 Applications of Trigonometry values of θ. Now let’s assume that neither r nor r is zero. If (r, θ) and (r, θ) determine the same point P then the (non-zero) distance from P to the pole in each case must be the same. Since this distance is controlled by the ﬁrst coordinate, we have that either r = r or r = −r. If r = r, then when plotting (r, θ) and (r, θ), the angles θ and θ have the same initial side. Hence, if (r, θ) and (r, θ) determine the same point, we must have that θ is coterminal with θ. We know that this means θ = θ + 2πk for some integer k, as required. If, on the other hand, r = −r, then when plotting (r, θ) and (r, θ), the initial side of θ is rotated π radians away from the initial side of θ. In this case, θ must be coterminal with π + θ. Hence, θ = π + θ + 2πk which we rewrite as θ = θ + (2k + 1)π for some integer k. Conversely, if r = r and θ = θ + 2πk for some integer k, then the points P (r, θ) and P (r, θ) lie the same (directed) distance from the pole on the terminal sides of coterminal angles, and hence are the same point. Now suppose r = −r and θ = θ + (
2k + 1)π for some integer k. To plot P, we ﬁrst move a directed distance r from the pole; to plot P, our ﬁrst step is to move the same distance from the pole as P, but in the opposite direction. At this intermediate stage, we have two points equidistant from the pole rotated exactly π radians apart. Since θ = θ + (2k + 1)π = (θ + π) + 2πk for some integer k, we see that θ is coterminal to (θ + π) and it is this extra π radians of rotation which aligns the points P and P. Next, we marry the polar coordinate system with the Cartesian (rectangular) coordinate system. To do so, we identify the pole and polar axis in the polar system to the origin and positive x-axis, respectively, in the rectangular system. We get the following result. Theorem 11.7. Conversion Between Rectangular and Polar Coordinates: Suppose P is represented in rectangular coordinates as (x, y) and in polar coordinates as (r, θ). Then x = r cos(θ) and y = r sin(θ) x2 + y2 = r2 and tan(θ) = y x (provided x = 0) In the case r > 0, Theorem 11.7 is an immediate consequence of Theorem 10.3 along with the quotient identity tan(θ) = sin(θ) If r < 0, then we know an alternate representation for (r, θ) cos(θ). is (−r, θ + π). Since cos(θ + π) = − cos(θ) and sin(θ + π) = − sin(θ), applying the theorem to (−r, θ + π) gives x = (−r) cos(θ + π) = (−r)(− cos(θ)) = r cos(θ) and y = (−r) sin(θ + π) = (−r)(− sin(θ)) = r sin(θ). Moreover, x2 + y2 = (−r)2 = r2, and y x = tan(θ + π) = tan(θ), so the theorem is true in this case, too
. The remaining case is r = 0, in which case (r, θ) = (0, θ) is the pole. Since the pole is identiﬁed with the origin (0, 0) in rectangular coordinates, the theorem in this case amounts to checking ‘0 = 0.’ The following example puts Theorem 11.7 to good use. Example 11.4.2. Convert each point in rectangular coordinates given below into polar coordinates with r ≥ 0 and 0 ≤ θ < 2π. Use exact values if possible and round any approximate values to two decimal places. Check your answer by converting them back to rectangular coordinates. 1. P 2, −2 √ 3 2. Q(−3, −3) 3. R(0, −3) 4. S(−3, 4) 11.4 Polar Coordinates 925 Solution. 1. Even though we are not explicitly told to do so, we can avoid many common mistakes by taking 3 shows that the time to plot the points before we do any calculations. Plotting P 2, −2 √ it lies in Quadrant IV. With x = 2 and y = −2 = 4 + 12 = 16 so r = ±4. Since we are asked for r ≥ 0, we choose r = 4. To ﬁnd θ, we have that tan(θ) = y 3, and since P lies in Quadrant IV, we know θ is a Quadrant IV angle. We are asked to have 0 ≤ θ < 2π,. To check, we convert (r, θ) = 4, 5π 3. Hence, our answer is 4, 5π so we choose θ = 5π 3 back to rectangular coordinates and we ﬁnd x = r cos(θ) = 4 cos 5π = 2 and 3 y = r sin(θ) = 4 sin 5π 3 3. This tells us θ has a reference angle of π 3, we get r2 = x2 + y2 = (2)2 + −, as required. x = −2 = 4 = −2 32 √ √ 18 = ±3 2. The point Q(−3, −3) lies in Quadrant III. Using x = y = −3, we get r2 = (−3)2 + (−3)2 = 18 2. We ﬁnd 4. Since
Q lies in Quadrant III, 4, which satisﬁes the requirement that 0 ≤ θ < 2π. Our ﬁnal answer is. To check, we ﬁnd x = r cos(θ) = (3 = −3 so r = ± tan(θ) = −3 we choose θ = 5π (r, θ) = 3 −3 = 1, which means θ has a reference angle of π 2. Since we are asked for r ≥ 0, we choose r = 3 = (3 2) √ √ √ √ − √ 2) cos 5π 4 2, 5π 4 2 2 and y = r sin(θ) = (3 √ 2) sin 5π 4 √ = (3 − √ 2 2 2) = −3, so we are done. y y θ = 5π 3 x θ = 5π 4 x P Q P has rectangular coordinates (2, −2 P has polar coordinates 4, 5π 3 √ 3) Q has rectangular coordinates (−3, −3) Q has polar coordinates 3 √ 2, 5π 4 3. The point R(0, −3) lies along the negative y-axis. While we could go through the usual computations4 to ﬁnd the polar form of R, in this case we can ﬁnd the polar coordinates of R using the deﬁnition. Since the pole is identiﬁed with the origin, we can easily tell the point R is 3 units from the pole, which means in the polar representation (r, θ) of R we know r = ±3. Since we require r ≥ 0, we choose r = 3. Concerning θ, the angle θ = 3π 2 satisﬁes 0 ≤ θ < 2π 4Since x = 0, we would have to determine θ geometrically. 926 Applications of Trigonometry with its terminal side along the negative y-axis, so our answer is 3, 3π 2 x = r cos(θ) = 3 cos 3π 2 = (3)(0) = 0 and y = r sin(θ) = 3 sin 3π 2 = 3(−1) = −3.. To check, we note −3 = − 4 4. The point S(−3
, 4) lies in Quadrant II. With x = −3 and y = 4, we get r2 = (−3)2 +(4)2 = 25 so r = ±5. As usual, we choose r = 5 ≥ 0 and proceed to determine θ. We have tan(θ) = y x = 4 3, and since this isn’t the tangent of one the common angles, we resort to using the arctangent function. Since θ lies in Quadrant II and must satisfy 0 ≤ θ < 2π, we choose θ = π − arctan 4 ≈ (5, 2.21). To radians. Hence, our answer is (r, θ) = 5, π − arctan 4 3 3 check our answers requires a bit of tenacity since we need to simplify expressions of the form: cos π − arctan 4. These are good review exercises and are hence 3 left to the reader. We ﬁnd cos π − arctan 4 5, so that 3 x = r cos(θ) = (5) − 3 5 = − 3 = −3 and y = r sin(θ) = (5) 4 5 = 4 which conﬁrms our answer. and sin π − arctan 4 3 5 and sin π − arctan 4 = 4 3 y y S θ = 3π 2 x R θ = π − arctan 4 3 x R has rectangular coordinates (0, −3) R has polar coordinates 3, 3π 2 S has rectangular coordinates (−3, 4) S has polar coordinates 5, π − arctan 4 3 Now that we’ve had practice converting representations of points between the rectangular and polar coordinate systems, we now set about converting equations from one system to another. Just as we’ve used equations in x and y to represent relations in rectangular coordinates, equations in the variables r and θ represent relations in polar coordinates. We convert equations between the two systems using Theorem 11.7 as the next example illustrates. Example 11.4.3. 1. Convert each equation in rectangular coordinates into an equation in polar coordinates. (a) (x − 3)2 + y2 = 9 (b) y = −x (c) y = x2 2. Convert each equation in polar coordinates into an equation in rectangular coordinates.

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