77. The graph is a reflected cosine graph with = 1 = −3 t ⎛ − 1 ⎞ amplitude 6 and period 8. = − cos 1 ; t 3 = ⎜ ⎟ 8 3 ω Find : 2π ⎝ ⎠ ω ⎛ 2 2 ⎞ 8ω = 2π 2 2 ⎛ 3 ⎞ ⎜ 3 ⎟ tan = ⎝ ⎠ = ⋅ ⎜ − t ⎟ = −2 2 ; ω = 2π = ⎛ − 1 ⎞ ⎝ ⎠ π 3 1 8 4 ⎜ 3 ⎟ 1 2 2 = − ⎛ π ⎞ 6 cos y x The equation is: . ⎝ ⎠ ⎜ ⎟ 4 4 cot = ⋅ = − t is on a circle of radius ⎝ ⎠ 84. The point = (−2, 5) P 78. The graph is a reflected sine graph with amplitude 7 and period 8. −2 2 2 ω Find : = + = 29 4 25 r with the (−2)2 8 = 2π center + 52 = ω at the origin. So, we have = −2 , = 5 , and x y 8ω = 2π 5 5 29 y = 29 . Thus, sin = = = ; r t ω = 2π = π 8 4 29 29 r −2 2 29 5 x y The equation is: cos = = = ; tan = =− t t r x − . = −7 sin ⎛ π ⎞ . y x 29 29 2 ⎜ ⎟ 4 ⎝ ⎠ π The domain of ⎫ 85. ⎧ π ≈ 79. Set the calculator to radian mode: sin 0.38 . = sec is y x 8 ⎨ ≠ odd multiple of ⎬ . x x ⎩ ⎭ 2 The range of {y = sec is ≤ −1 or ≥ 1}. y x y y The period is 2π . = 1 ≈ 1.02 . cos10o Set the calculator to degree mode: 80.
77. The graph is a reflected cosine graph with amplitude 6 and period 8. cos t = − 1 ; 3 t = 1 = −3 ⎛ − 1 ⎞ Find ω : 8 = 2π ⎜ 3 ⎟ ⎝ ⎠ ω ⎛ 2 2 ⎞ 8ω = 2π ⎜ 3 ⎟ 2 2 ⎛ 3 ⎞ tan t = ⎝ ⎠ = ⋅ ⎜ − ⎟ = −2 2 ; ω = 2π = π ⎛ − 1 ⎞ 3 ⎝ 1 ⎠ 8 4 ⎜ 3 ⎟ The equation is: y = − 6 cos ⎛ π x ⎞ . ⎝ ⎠ 1 2 2 ⎜ 4 ⎟ cot t = ⋅ = − 4 ⎝ ⎠ 78. The graph is a reflected sine graph with amplitude 7 and period 8. −2 2 2 84. The point P = (−2, 5) is on a circle of radius Find ω : 8 = 2π r = (−2)2 + 52 = 4 + 25 = 29 with the center ω at the origin. So, we have x = −2 , y = 5 , and 8ω = 2π y 5 5 29 r = 29 . Thus, sin t = = = ; ω = 2π = π 8 4 r 29 29 x −2 2 29 y 5 The equation is: y = −7 sin ⎛ π x ⎞ . cos t = = r = − 29 29 ; tan t = x =− . 2 ⎜ 4 ⎟ ⎝ ⎠ 79. Set the calculator to radian mode: sin π ≈ 0.38 . 85. The domain of ⎧ y = sec x is π ⎫ 8 ⎨ x x ≠ odd multiple of ⎬ . ⎩ 2 ⎭ The range of y = sec x is {y y ≤ −1 or y ≥ 1}. The period is 2π . 80. Set the calculator to degree mode: = 1 ≈ 1.02 . cos10o
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