Chapter 5 The Next Wave: MORE MODELING AND TRIGONOMETRY

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1 ANSWERS Mathematics (Mathematical Analysis) page 1 Chapter The Next Wave: MORE MODELING AND TRIGONOMETRY NW-1. TI-8, points; Casio, points a) An infinite number of them. b) 17p, - 7p c) Add p n to p, p d) p, 9p e) - 1p, - p NW-. a) The y-coordinates of the points are. NW-. a) b) : 0, p c) 0 + p n, p + p n NW-. a) p, p b) p + p n, p + p n NW-. a) The x and y are interchanged. b) Reflection about the line y = x. c) no NW-7. a) p + p n b) p, p d) p + p n, p + p n, p + p n c) p, p, p NW-8. (a) and (b) are the graphs whose inverses are also functions. NW cos a 1 NW-10. a) p, p b) p + p n, p + p n NW-11. a) b) - 1 Â (k - 1) + 7 k=1 NW-1. a) b = b) y = 0. c) sin z = 1, p + p n NW-1. x = a + b, y = b - a NW-1.. feet NW-1. a) p, 7p b) p + p n, 7p + p n NW-1. Domain = (-, 7) and Range = [-, ] NW-17. b) Because inverses are symmetric about the line y = x. c) No, because the vertical line test for functions fails.

2 ANSWERS Mathematics (Mathematical Analysis) page NW-18. a) [- p, p ] b) [ p, p ] c) Domain is [-1, 1] and Range is [- p, p ] NW-19. a) 1.07 b) It is not in the range of y = sin 1 (x). sin 1 selects only one of the infinitely many solutions to the equation. c) x = p + p norp + p n d) You have to use the unit circle or a wave. NW-0. a) No, it does not pass the vertical line test b) [0, p ] and [p, p ] c) [ 0, p ] d) Domain = -1 x 1, Range = 0 x p NW-1. y = sin -1 x: D: [-1, 1], R: [- p, p ]; y = cos-1 x: D: [-1, 1], R: [0, p ] NW-. a) It is not in the range of y = cos -1 (x). cos -1 selects only one of the infinitely many solutions to the equation. b) x = p + p n, p + p n c) You have to draw and think. NW- 0, p, p NW-. a) The equation cos x = -0. will have multiple solutions. b) Sylvie needs to include all the solutions, which she can get using a graph or unit circle. She needs to add multiples of p, and include the negative values. x = ± p n, where n is an integer. NW-. a) - p b) p c) p d) p NW-. - NW-7. x = - ±i NW-8. a) x ( x- 1 ) y b) x + h NW-9. a) feet b).0 feet NW-0. sin 1 (1) = p ª sin 1 = It is confusing because sin (x) = (sin x), but sin 1 (x) π (sin x) 1. NW-1. a) - b) 1 c) 1 d) e) impossible f)

3 ANSWERS Mathematics (Mathematical Analysis) page NW-. at x = 0, -, NW-. Answers vary. NW-. a) vertical asymptotes: x =- p,- p, p, p b) roots: -p, -p, 0, p, p NW-. a) x π np, where n is any odd integer b) all real numbers c) y = 0, x = ±p n, n is any integer d) x = np, where n is any odd integer e) increasing; odd NW-. a) Restrict the domain b) (- p, p ) c) Domain: all real numbers, Range: (- p, p ) NW-7. a) sin - 1 ( )= 0.97 b) x = 0.97 and.1 c) p n,.1 + p n, n is an integer. NW-8. a) 1.18,.1 b) no solution c) p n,. + p n d) p n,.0 + p n NW-9. tan q = y x NW-. a) p, p c) p + p n, p NW-0. 8., q= 19.7 NW-1..7 b) p + p n, n is an integer. + p n d) no solutions NW-. Yes, the first is the inverse function, the second the reciprocal function of y = cos x NW-. It is false. For example, take a = p, b = p NW-. a = 8, b = NW i NW-7. A =, B = - NW8. Amplitude =, Horizontal shift = to the right, Vertical shift = 1 up, Period = NW-9. - NW-0. Laurel is. Hardy's equation only shifts the graph p to the right. NW-1. y = sin ( (x p ) ) +

4 ANSWERS Mathematics (Mathematical Analysis) page NW-. a), p to the right, down, p = p b) y = sin (x p ) NW-. a) (0., ) and (., ) b) period =., amplitude = 10, horizontal shift =. or -1., vertical shift = c) One possible answer is h = 10 cos (( p 9 )(t 0.)) +. NW-. a) 8 b) c) - 9 = NW-7. a) The range of sine and cosine is -1 y 1. b) A fraction can equal 7 without the numerator being and the denominator being 7. For example, = 7. c) 0.0 or. NW-8. x =± ab ae- 1 NW-9 a) 7, - b), - c) -, d), - NW-0 1 NW-1. y 91 = ±0.(x 8) NW-. a). b) p NW-. a) -,0.9 b) +,0.9 c) +,0.9 d) -,0.9 NW NW-8. a) sin q b) cos q NW-9. a) - b) - 7 c) d) NW-70. cos (p q ) = -cos q NW-71. sin (p q ) = sin q NW-7. a) p, p b) p, 7p c) p, p, p, 7p d) p +pn, p + p n, 7p + p n NW-7. b) f(x) = x, g(x) = 1 x NW-7. a) x y z (x + z ) b) (x 1) y (y + 8z) NW-7 7 sin z NW-77. a) z =, b) z = 1. NW-78. y = 8 cos ( ( x + p ))- NW-79. q= p

5 ANSWERS Mathematics (Mathematical Analysis) page NW-80. y = 0 cos p ( x - )+ a) 9.18 inches b) at approximately 9:9 AM NW-81. 1) h = cos p (t 1.) + a) 9. cm b) 0. sec and 1.97 sec ) h = cos ( p ( x±) )+ a) 7.77 feet b) 1.8 sec and.7 sec ) d = 9 sin ( p ( t±.) )+ a) 8. cm b).7 seconds ) h = cos ( 8p ( x±0.1) )+ 8 a) 0.0 cm b) 0.07 seconds ) F = 19 sin ( 1 p ( t±10) )+ 8 a) b) 1.1 hours after noon or about 1:10 PM ) A = 1.1 sin ( p ( t±.) ) a) 1.7 liters b).01 seconds 7) h = 1. sin ( p ( t±.) )+. a). cm b) 0.10 seconds 8) h = 1 sin (p (t.)) + 71 a) 10 cm b) seconds 9) h = cos ( p ( t±) )+ 1 a) 7.77 cm b).99 seconds NW-8. a) a =± 1, ± i 1 b) x = 1±i 7 c) y = 7 d) x = 9, -1 NW-8. a) y = + sin x; other answers are possible. b) y = + cos (x p ); other answers are possible. NW-8. a) d) b) - 1 e) -0.9 c) - 1 NW-8. sin a=-, tana =,csca =-, seca =-, and cot a= NW-8. x = 10 NW-87. x = -., y = or x = 1., y = -1 NW-88. ab b- a NW

6 ANSWERS Mathematics (Mathematical Analysis) page NW-90. a) p b) c) 10 d) 10 NW-91. -sin a NW-9. p +pn, p +pn NW-9. a) sin (a + a) or sin a cos a + sin a cos a NW-9. a) cos (a + a) or cos a cos a sin a sin a NW-9. a) sin x b) 1 c) cos 80 d) cos (y 10)e) sin 70 f) cos w NW-97. a = q NW-99. a) sin a=-, cosa =-, tana = b) csca =-, seca =-, cot a= c) sin a = 1 17, cos a =- 8 17, sin a = 78+(1 ) NW-100.a) x = 0, 1,, b) x = 99 NW-101.a) + x y + y x b) xy NW-10.a) + i b) a = -, b = 1 c) a = - + i, b = i NW-10. sin x =, cosx = 7, sin x = 10, cos x = NW-10.x ª.17 NW-10. a) x π, - b) x =, - NW-10.a) sin 10p b) - sin p 1 NW ,.80 NW-109.After using the sum and difference rules, find the product. Use the Pythagorean Identity to express a in terms of sine, b in terms of cosine. NW-110. d = h h- 1. NW-111.a) (a b) b) a 810a b a b 70a b + 0ab b NW-11.9 NW-11. a) y = p x b) 00x p x

7 ANSWERS Mathematics (Mathematical Analysis) page 7 NW-11.The total length is 1.79 cm Teacher s Solution: Draw a line through B parallel to CD meeting AC at E. Then AE = 0 cm, AB = 100 cm, and ABE is a right triangle. Hence BE = CD = 80. Let q be the central BAC. Then cos q = 0., so qª 0.97 radians. Thus the wire length around the large log is 80(p (0.97)) cm. The wire around the small log is 0((0.97) cm in length and the wire between the logs is (80) cm. Thus, the total length is p- 10 cos - 1 (0.) ª 1.79 NW-11.a) b) NW cos ( p ( x±) )+. NW-117. cos x =-, tan x =, csc x =-, sec x =-, cot x = NW-118.a) sin x±sin x - = 0 c) p NW-119.a) c = 0, 1 b) p NW-10. a) p, p, p, p NW-11. np NW-1.c) p + p n NW-1. a) x = 0, p,p b) x =- p c) p, p, p, p, all + p n d) x = p, p NW-1.a) cosine, secant b) sine, cosecant, tangent, cotangent NW-1.A =, B = - NW-1. x = i NW-19. x =, y = NW-10. a) x π 7 b) y goes to c) y goes to - d) It approaches y = e) It approaches y = NW-11. a) q= 0, p, p, p,p b) q= p, p, p, 7p NW-1.a) 80 seconds b) y = 7 sin ( p x) c) y = p sin - 1 x ( 7 ) NW-1.a).0 feet b) at 0.70 and.9 seconds NW-1.a) 9. b) in 0 seconds c) q = tan 1 1 ( 0-11t ) NW-17.a) 0, p,p, p, p c) 0, p,p, p, p b) 0, p, p, p, all + p n d) 0, p, p, p, all + p n

8 ANSWERS Mathematics (Mathematical Analysis) page 8 NW-19.sin -1 : [- p, p ], cos-1 : [0, p ], tan -1 : (- p, p ) NW-11.a) 1± 11 b) 1,- 1 c) d) -, NW-1. 1, 1, 0 NW-1.a) p, p, both + p n b) p, p, 7p, 11p, all + p n NW-1.y + = - 9 ( x±) NW-1. x ( a±b) ( a + b) ( a + b )