Section 8.5 AN63 61. Suppose 0 . θ π < < Then, by the Law of Cosines, d r r r r d r r 2 cos 4 1 cos 2 2 1 cos 2 2 sin 2 . 2 2 2 2 2 θ θ θ θ ( ) = + − = − ⇒ = − = Since, for any angle in d 0, , π ( ) is strictly less than the length of the arc subtended by ,θ that is, d r ,θ < then r r 2 sin 2 , θ θ < or 2sin 2 . θ θ < Since cos 2 1, θ < then, for 0 , sin 2sin 2 cos 2 2sin 2 . θ π θ θ θ θ θ < < = < < Thus sinθ θ < for 0 . θ π < < 63. ( ) ( )( ) ( )( ) ( )( ) = − = − + − = − − + = − − = + − + − = − − = − − C C a b c ab ab a b c ab c a b ab c a b c b a ab s b s a ab s a s b ab sin 2 1 cos 2 1 2 2 2 4 4 4 2 2 2 2 4 2 2 2 2 2 2 2 2 70. y x 8 5 22 x = 3 y = 2 (4, 9) 1 2 , 0 2 1 3 0, 2 11 2 5, 71. ln3 ln4 ln3 { } − 72. sin 2 6 7 ; csc 7 6 12 ; sec 7 5 ; cot 5 6 12 θ θ θ θ = = = − = − 73. y x 3sin 4( ) = − 74. f x A x x 2 5 1( ) = − − 75. 2 3 76. x x 4 3 ln3 1 2 x 3 2 ( ) ⋅ ⋅ − 77. , 2 5 4 3 , ( ) −∞ ⎤ ⎦ ⎥ ∪ ⎡ ∞ ⎢ ⎣ 78. 8 15 π 79. Quadrant II; 2 x-intercepts 8.4 Assess Your Understanding (page 580) 3. ab C 1 2 sin 4. s s a s b s c a b c ; 1 2 ( )( )( ) ( ) − − − + + 5. 6 6. T 7. c 8. c 9. 2.83 11. 17.46 13. 13.42 15. 9.56 17. 4.60 19. 3.86 21. 2.72 23. 210 25. 6.93 27. 74.15 29. K ab C a C a B A a B C A 1 2 sin 1 2 sin sin sin sin sin 2sin 2 = = ⋅ = 31. 0.92 33. 2.27 35. 5.44 37. 9.03 sq ft 39. $544,638 41. 18.18 m2 43. The area of home plate is about 216.5 in.2 45. K r 1 2 sin 2 θ θ ( ) = + 47. The ground area is 7517.4 ft .2 49. Letting d 0 = gives K s a s b s c s abc s s a s b s c s a b c a b c 0 0 cos where 1 2 0 1 2 2 θ ( )( )( )( ) ( )( )( ) ( ) ( ) = − − − − − ⋅ ⋅ = − − − = + + + = + + 51. (a) Area OAC OC AC OC AC 1 2 1 2 1 1 1 2 sin cos α α Δ = = ⋅ ⋅ = (b) Area OCB BC OC OB BC OB OC OB OB 1 2 1 2 1 2 sin cos 2 2 β β Δ = = ⋅ = (c) Area OAB BD OA OB BD OB OB 1 2 1 2 1 2 sin α β ( ) Δ = = = + (d) OC OC OB OB cos cos 1 α β = = (e) OAB OAC OCB Area Area Area Δ = Δ + Δ OB OB 1 2 sin 1 2 sin cos 1 2 sin cos 2 α β α α β β ( ) + = + OB OB sin 1 sin cos sin cos α β α α β β ( ) + = + sin cos cos sin cos cos cos sin cos α β β α α α α β β β ( ) + = + sin sin cos cos sin α β α β α β ( ) + = + 53. 31,145 ft2 55. (a) The perimeter and area are both 36. (b) The perimeter and area are both 60. 57. K ah ab C h b C a B C A 1 2 1 2 sin sin sin sin sin = = ⇒ = = 59. POQ A B C C 180 2 2 180 1 2 180 90 2 , ( ) ( ) ∠ =°−+=°− °−=°+ and C C sin 90 2 cos 2 . ( ) ° + = So, r c A B C c A B C sin 2 sin 2 sin 90 2 sin 2 sin 2 cos 2 . ( ) = ° + = 61. A B C s a r s b r s c r s a b c r s s r s r cot 2 cot 2 cot 2 3 3 2 ( ) + + = − + − + − = − + + = − = 66. Maximum value; 17 67. , 3 1, 3 ( ) [ ) −∞ − ∪ − 68. t t t t t t sin 2 3 , cos 7 3 , tan 14 7 , csc 3 2 2 , sec 3 7 7 , cot 14 2 = = − = − = = − = − 69. csc sin 1 sin sin 1 sin sin cos sin cos cos sin cos cot 2 2 θ θ θ θ θ θ θ θ θ θ θ θ θ − = − = − = = ⋅ = 70. , 5 5, ( ) ( ) −∞ − ∪ ∞ 71. P w w w 2 2 144 2 ( ) = + − 72. 1 2 , 3 2 , 1, 2, 3, 6 ± ± ± ± ± ± 73. 2.39, 2.41 [ ] 74. 2, 9 { } − 75. y x2 = − 8.5 Assess Your Understanding (page 589) 4. simple harmonic; amplitude 5. simple harmonic; damped 6. T 7. d t t 5cos π ( ) ( ) = − 9. d t t 7cos 2 5( ) ( ) = − 11. d t t 5sin π ( ) ( ) = − 13. d t t 7sin 2 5( ) ( ) = − 15. (a) Simple harmonic (b) 5 m (c) 2 3 sec π (d) 3 2 oscillation sec π
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