2 (4, 2) ( 4, 2) 1 –– 2 x y –– 2 8 4 4 8 (0, 1) (8, 1) ( 8, 1) 2 66. We begin by considering ⎛ π ⎞ 1 sin . y x 2 65. We begin by considering π 1 ⎛ ⎞ π ⎛ ⎞ 3 cos y x . ⎝ 8 ⎠ 2 (ω x π 3 cos ⎛ ⎞ to y x 1 ) ⎟ to x = , y A Comparing ⎠ sin ⎜ sin π ⎝ ⎠ 4 ⎝ 8 2 =− y ⎝ ⎠ 2 4 ω = − and = . Therefore, the A we find 2 8 (ω ) = cos , we y A x find ω π 3 and = . 1 1 A − = 3 3 − = and the Therefore, the amplitude is is period is 4 2 2 2π 2 and the 1 2 , the = 16 . Because the amplitude = period is 2π 8 . Because the amplitude is 3 2 π , /8 is 2 2 graph of π /4 π 1 sin ⎛ ⎞ will lie between − 1 y x the graph of ⎝ ⎠ 2 8 2 1 ⎛ π ⎞ 3 cos will lie between y x 2 ⎝ 4 ⎠ on the Because the period is 16, y-axis. and 2 − 3 and 3 on the Because the period is y-axis. 2 2 = = one cycle will begin at 0 and end at 16 . x x = 8, one cycle will begin at 0 and end at x [0,16] We divide the interval into four [0, = 8 . We divide the interval 8] into four x subintervals, each of length 16 = 4 by finding 4 subintervals, each of length 8 = 2 by finding the 4 the following values: following values: 0, 2, 4, 6, and 8 0, 4, 8, 12, and 16 These values of determine the of x x-coordinates the five key points on the graph. To obtain the y-
( 4, 2) y –– 2 2 (4, 2) 1 –– 2 x 8 4 4 8 ( 8, 1) (0, 1) (8, 1) 2 66. We begin by considering y 1 sin ⎛ π x ⎞ . 65. We begin by considering y 3 cos ⎛ π x ⎞ . 2 1 ⎛ π ⎞ ⎝ 8 ⎠ Comparing 2 y 3 cos ⎛ π x ⎞ to ⎝ 4 ⎠ y =− 2 1 sin ⎜ ⎝ 8 x ⎟ to ⎠ π y = A sin (ω x ) , 2 ⎝ 4 ⎠ we find A = − and ω = . Therefore, the 2 8 y = A cos (ω x ) , we find A 3 and ω = π . 1 1 Therefore, the amplitude is 2 − 3 = 3 4 and the is 2π − = and the period is 2 2 1 period is 2π 2 2 = 8 . Because the amplitude is 3 , π /8 = 16 . Because the amplitude is , the 2 π /4 2 graph of y 1 sin ⎛ π x ⎞ will lie between − 1 the graph of y 3 cos ⎛ π x ⎞ will lie between 2 ⎝ 8 ⎠ 2 1 2 ⎝ 4 ⎠ and on the y-axis. Because the period is 16, 2 − 3 and 3 on the y-axis. Because the period is 2 2 8, one cycle will begin at x = 0 and end at x = 8 . We divide the interval [0, 8] into four one cycle will begin at x = 0 and end at x = 16 . We divide the interval [0,16] into four subintervals, each of length 16 = 4 by finding 4 subintervals, each of length 8 = 2 by finding the 4 following values: 0, 2, 4, 6, and 8 These values of x determine the x-coordinates of the five key points on the graph. To obtain the y- the following values: 0, 4, 8, 12, and 16
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