63. Since sine is an odd function, we can plot the 64. Since cosine is an even function, we consider the equivalent form equivalent form = 9 cos ⎛ 3π y ⎞ . Comparing 5 sin ⎛ 2π ⎞ . y x x ⎝ ⎠ 3 3 5 ⎝ 2 ⎠ Comparing (ω ) = cos , we find y A x ⎛ ⎞ = ⎛ ⎞ 5 sin 2π to 9 cos 3π y x y x to ⎝ ⎠ 3 3 5 ⎝ 2 ⎠ 3 3 (ω ) 3π . Therefore, the amplitude is = sin , we y A x 2 find and = 9 ω A and 5 5 ω = =− A = 2π . / 2 3 and the = 4 . Therefore, the amplitude is and the 9 = 9 − = period 5 5 5 5 3 3 is 2π period is 2π = 3 . Because the amplitude is 2π / 3 the amplitude is 9 , the graph of 5 5 , the graph of and 9 5 sin ⎛ 2π ⎞ will lie = 9 cos ⎛ 3π ⎞ will lie between − y x y x 9 3 3 ⎝ 3 ⎠ 5 ⎝ 2 ⎠ 5 5 between − 5 3 on the Because the y-axis. and 5 on the Because the period is 4 , one y-axis. 3 3 = period is 3 , one cycle will begin at 0 and x [0, = end at 3 . We divide the interval 3] into x cycle will begin at = 0 and end at = 4 . We x x 3 four subintervals, each of length 3 by finding into four subintervals, divide the interval ⎡0, 4 ⎤ 4 ⎢ 3 ⎥ ⎣ ⎦ the following values: by finding the following 0, 3 , 3 , 9 , and = each of length 4/3 1 3 4 3
63. Since sine is an odd function, we can plot the 64. Since cosine is an even function, we consider the equivalent form y 5 sin ⎛ 2π x ⎞ . equivalent form y = 9 cos ⎛ 3π x ⎞ . Comparing 3 ⎝ 3 ⎠ 5 ⎝ 2 ⎠ Comparing y 5 sin ⎛ 2π x ⎞ to y = 9 cos ⎛ 3π x ⎞ to y = A cos (ω x ) , we find 3 ⎝ 3 ⎠ 5 ⎝ 2 ⎠ y = A sin (ω x ) , we find 5 A =− 3 and ω = 2π . 3 A = 9 5 and ω = 3π . Therefore, the amplitude is 2 Therefore, the amplitude is − 5 = 5 3 3 and the 9 = 9 5 5 and the period is 2π = 4 . / 2 3 period is 2π = 3 . Because the amplitude is 2π / 3 the amplitude is 9 , the graph of 5 5 , the graph of y 5 sin ⎛ 2π x ⎞ will lie y = 9 cos ⎛ 3π x ⎞ will lie between − 9 and 9 3 3 ⎝ 3 ⎠ 5 ⎝ 2 ⎠ 5 5 between − 5 3 and 5 3 on the y-axis. Because the on the y-axis. Because the period is 4 , one 3 period is 3 , one cycle will begin at x = 0 and end at x = 3 . We divide the interval [0, 3] into cycle will begin at x = 0 and end at x = 4 . We 3 four subintervals, each of length 3 by finding divide the interval ⎡0, 4 ⎤ into four subintervals, 4 ⎢ 3 ⎥ ⎣ ⎦ the following values: 0, 3 , 3 , 9 , and 3 each of length 4/3 = 1 4 3 by finding the following
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