Section 6.4 AN47 47. θ θ θ θ θ = − = − = = − = − cos 12 13 ; tan 5 12 ; csc 13 5 ; sec 13 12 ; cot 12 5 49. θ θ θ θ θ = = − = = − = − sin 2 2 3 ; tan 2 2; csc 3 2 4 ; sec 3; cot 2 4 51. θ θ θ θ = − = − = = − cos 5 3 ; tan 2 5 5 ; csc 3 2 ; sec 3 5 5 ; θ = − cot 5 2 53. θ θ θ = − = = − sin 3 2 ; cos 1 2 ; tan 3; θ θ = − = − csc 2 3 3 ; cot 3 3 55. θ θ = − = − sin 3 5 ; cos 4 5 ; θ θ θ = − = − = csc 5 3 ; sec 5 4 ; cot 4 3 57. θ θ = = − sin 10 10 ; cos 3 10 10 ; θ θ θ = = − = − csc 10; sec 10 3 ; cot 3 59. − 3 2 61. − 3 3 63. 2 65. −1 67. −1 69. 2 2 71. 0 73. − 2 75. 2 3 3 77. 1 79. 1 81. 0 83. 1 85. −1 87. 0 89. 0.9 91. 9 93. 0 95. All real numbers 97. Odd multiples of π 2 99. Odd multiples of π 2 101. { } − ≤ ≤ y y 1 1 103. All real numbers 105. { } ≥ y y 1 107. Odd; yes; origin 109. Odd; yes; origin 111. Even; yes; y-axis 113. (a) − 1 3 (b) 1 115. (a) −2 (b) 6 117. (a) −4 (b) −12 119. ≈15.81 min 121. 130, 90, 70 123. Let a be a real number and ( ) = P x y , be the point on the unit circle that corresponds to t. Consider the equation = = t y x a tan . Then = y ax. But + = x y 1, 2 2 so + = x a x 1. 2 2 2 So = ± + x a 1 1 2 and = ± + y a a 1 ; 2 that is, for any real number a, there is a point ( ) = P x y , on the unit circle for which = t a tan . In other words, the range of the tangent function is the set of all real numbers. 125. Suppose that there is a number π < < p p , 0 2 , for which θ θ ( ) + = p sin sin for all θ. If θ = 0, then ( ) + = = = p p sin 0 sin sin0 0, so π = p . If θ π = 2 , then π π ( ) ( ) + = p sin 2 sin 2 . But π = p . Thus, π π ( ) ( ) = − = = sin 3 2 1 sin 2 1. This is impossible. Therefore, the smallest positive number p for which θ θ ( ) + = p sin sin for all θ is π2 . 127. θ θ = sec 1 cos ; since θ cos has period π2 , so does θ sec . 129. If ( ) = P x y , is the point on the unit circle corresponding to θ, then ( ) = − − Q x y , is the point on the unit circle corresponding to θ π + . Thus, θ π θ ( ) + = − − = = y x y x tan tan . Suppose that there exists a number π < < p p , 0 , for which θ θ ( ) + = p tan tan for all θ. Then, if θ = 0, then = = p tan tan0 0. But this means that p is a multiple of π. Since no multiple of π exists in the interval π ( ) 0, , this is a contradiction. Therefore, the period of θ θ ( ) = f tan is π. 131. Let ( ) = P x y , be the point on the unit circle corresponding to θ. Then θ θ θ θ = = = = y x csc 1 1 sin ; sec 1 1 cos ; θ θ = = = x y y x cot 1 1 tan . 133. sin cos sin sin cos sin cos sin sin cos 2 2 2 2 2 2 2 2 θ φ θ φ θ θ φ θ φ θ ( ) ( ) + + = + + sin cos sin cos sin cos 1 2 2 2 2 2 2 θ φ φ θ θ θ ( ) = + + = + = 135. 7 5 137. 7 13 − 143. f g x x x 14 46 2 ( )( ) = − + 144. (2, 3) (3, 5) (4, 3) y 6 6x x 5 3 26 26 Vertex: 3, 5 ( ) axis of symmetry: x 3 = 145. ln6 4 { } + 146. 9 147. 89 16 { } 148. 8 and 3 − − 149. 21 { } 150. c 35 = 151. 0, 3 , 5, 0 ( ) ( ) − 152. x h 3 5 3 2 − + 6.4 Assess Your Understanding (page 436) 3. (a) 1 (b) 0 (c) 1− (d) 0 (e) 1, 1 [ ] − (f) origin; odd (g) 1 2 ; 2 2 ; 1− 4. (a) 1− (b) 0; 1 (c) 1, 1 [ ] − (d) y-axis; even (e) 3 2 ; 2 2 ; 0 5. 1; 2 π 6. 3; π 7. 3; 3 π 8. T 9. F 10. T 11. d 12. d 13. (a) 0 (b) x 2 2 π π − ≤ ≤ (c) 1 (d) 0, , 2 π π (e) f x x f x x 1 for 3 2 , 2 ; 1 for 2 , 3 2 π π π π ( ) ( ) = = − = − = − (f) 5 6 , 6 , 7 6 , 11 6 π π π π − − (g) x x k k , an integer π { } = 15. Amplitude 5; period 2π = = 17. Amplitude 3; period 2 π = = 19. Amplitude 6; period 2 = = 21. Amplitude 1 7 ; period 4 7 π = = 23. Amplitude 10 9 ; period 5 = = 25. F 27. A 29. H 31. C 33. J 35. 5P 2 5P 2 P 2 , 0 3P 2 , 0 y 5 x (0, 4) (22P, 4) (2P, 24) (2P, 4) (P, 24) 2 Domain: , ( ) −∞ ∞ Range: 4, 4 [ ] − 37. 3P 2 , 24 3P 2 , 4 P 2 , 24 P 2 , 4 2 y 5 25 x 22P 2P (2P, 0) (0, 0) 2 Domain: , ( ) −∞∞ Range: 4, 4 [ ] − 39. 3P 8 , 0 3P 8 , 0 P 2 , 1 P 8 , 0 P 4 , 21 P 4 , 21 2 P 2 , 1 2 P 8 , 0 y 2.5 22.5 x 2 2 (0, 1) P 2 P 2 2 Domain: , ( ) −∞∞ Range: 1, 1 [ ] − 41. 3P 4 , 1 P 2 , 0 P 4 , 1 2 P 4 , 21 y 1.25 x (0, 0) 22P 2P (P, 0) Domain: , ( ) −∞∞ Range: 1, 1 [ ] − 43. y 2.5 x 5P (2P, 0) (0, 0) (23P, 2) (P, 2) (4P, 0) (2P, 22) (3P, 22) Domain: , ( ) −∞ ∞ Range: 2, 2 [ ] − 45. 5P 4 , 0 3P 4 , 0 P 4 , 0 P 2 1 2 , 1 2 2p, 1 2 0, y 2.5 x 22P 2P 2 2 Domain: , ( ) −∞∞ Range: 1 2 , 1 2 ⎡ − ⎣ ⎢ ⎤ ⎦ ⎥ 47. P 2 , 5 P 2 , 1 2 3P 2 , 5 2 3P 2 , 1 y 6 24 x (0, 3) 22P 2P (P, 3) (2P, 3) Domain: , ( ) −∞∞ Range: 1, 5 [ ] 49. 3 2 2 , 23 3 2 , 23 1 2 , 23 y 3 29 x (0, 2) (2, 2) (22, 2) (1, 28) 22 2 Domain: , ( ) −∞∞ [ ] − Range: 8, 2
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