⎛ ⎞ 1 y 2 (ω ) 51. Comparing ( 1) = 2sin ⎜ ⎟ to = sin , y x y A x –– –, (–3– 1) ( , 0) ⎝ ⎠ –, 4 2 we find ω = 2 and = 1 . Therefore, the A 2 (– , 0) (0, 0) = 4π . – 2 amplitude is 2 = 2 and the period is 2π 1/ 2 2 Because the amplitude is 2, the graph of x 2 (– , 1) – 4 = 2sin ⎛ 1 y ⎞ will lie between −2 and 2 on the x ⎝ 2 ⎠ 50. Since cosine is an even function, we can plot the Because the period is 4π , one cycle will y-axis. = 4π . We divide equivalent form x begin at = x (2 ) 0 and end at = cos . y x ] into four subintervals, each Comparing [0, the interval 4π (ω = cos y A x ) (2 ) , we = cos to y x ω = 2 . Therefore, the amplitude find by finding the following = 1 and A π of length 4π = 4 = = π is 1 1 and the period is 2π . Because values: the π 0, , 2π , 3π , and 4π 2 amplitude is 1, the graph of These values of x (2 ) determine the = cos will lie y x coordinates of between −1 and 1 on the Because the y-axis. π = the five key points on the graph. To obtain the period is , one cycle will begin at 0 and y- x [0, ] coordinates of the five key points for π π = end at . We divide the interval x = ⎛ ⎞ 2sin 1 , we multiply the of y x y-coordinates into ⎜ ⎟ 2 the five key points for π four subintervals, each of length by finding five key points are 4 = 2 . The A the following values: ⎝ ⎠ = sin by y x π π π 0, , , 3π , and (0, (π (2π (3π (4π 0) , ,2) , , 0) , , −2) , , 0) 4 2 4 the five key points on the graph. To obtain These values of determine the of x x-coordinates the coordinates of the five key points for y-
y ⎛ 1 ⎞ 2 ( –– –, 1) 4 (–3– –, 1) ( , 0) 51. Comparing y = 2sin ⎜ x ⎟ to ⎝ ⎠ y = A sin (ω x ) , we find A = 2 and ω = 1 . Therefore, the 2 2 (0, 0) 2 x (– – , 0) 2 amplitude is 2 = 2 and the period is 2π 1/ 2 Because the amplitude is 2, the graph of = 4π . 2 (– – , 1) 4 y = 2sin ⎛ 1 ⎝ 2 x ⎞ will lie between −2 and 2 on the ⎠ 50. Since cosine is an even function, we can plot the y-axis. Because the period is 4π , one cycle will equivalent form y = cos (2 x ) . begin at x = 0 and end at x = 4π . We divide Comparing y = cos (2 x ) to y = A cos (ω x ) , we the interval [0, 4π ] into four subintervals, each find A = 1 and ω = 2 . Therefore, the amplitude of length 4π = π 4 by finding the following is 1 = 1 and the period is 2π = π . Because the 2 values: 0, π , 2π , 3π , and 4π amplitude is 1, the graph of y = cos (2 x ) will lie These values of x determine the of between −1 and 1 on the y-axis. Because the period is π , one cycle will begin at x = 0 and end at x = π . We divide the interval [0, π ] into the five key points on the graph. To obtain the y- coordinates of the five key points for y = 2sin ⎛ 1 x ⎞ , we multiply the y-coordinates of ⎜ 2 ⎟ four subintervals, each of length π by finding 4 the following values: ⎝ ⎠ the five key points for five key points are y = sin x by A = 2 . The 0, π , π , 3π , and π (0, 0) , (π ,2) , (2π , 0) , (3π , −2) , (4π , 0) 4 2 4 These values of x determine the x-coordinates of the five key points on the graph. To obtain the y- coordinates of the five key points for coordinates
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