624 CHAPTER 6 The Circular Functions and Their Graphs While the coefficient a in y = a sin x or y = a cos x affects the amplitude of the graph, the coefficient of x in the argument affects the period. Consider y = sin 2x. We can complete a table of values for the interval 30, 2p4. x 0 p 4 p 2 3p 4 p 5p 4 3p 2 7p 4 2p sin 2 x 0 1 0 -1 0 1 0 -1 0 Note that one complete cycle occurs in p units, not 2p units. Therefore, the period here is p, which equals 2p 2 . Now consider y = sin 4x. Look at the next table. x 0 p 8 p 4 3p 8 p 2 5p 8 3p 4 7p 8 p sin 4 x 0 1 0 -1 0 1 0 -1 0 Divide each part by the positive number b. These values suggest that one complete cycle is achieved in p 2 or 2p 4 units, which is reasonable because sin a4 # p 2b = sin 2p= 0. In general, the graph of a function of the form y =sin bx or y =cos bx, for b +0, will have a period different from 2P when b 31. To see why this is so, remember that the values of sin bx or cos bx will take on all possible values as bx ranges from 0 to 2p. Therefore, to find the period of either of these functions, we must solve the following three-part inequality. 0 … bx … 2p bx ranges from 0 to 2p. 0 … x … 2p b Thus, the period is 2P b . By dividing the interval C 0, 2p b D into four equal parts, we obtain the values for which sin bx or cos bx is -1, 0, or 1. These values will give minimum points, x-intercepts, and maximum points on the graph. (If a function has b 60, then identities can be used to rewrite the function so that b 70.) NOTE One method to divide an interval into four equal parts is as follows. Step 1 Find the midpoint of the interval by adding the x-values of the endpoints and dividing by 2. Step 2 Find the quarter points (the midpoints of the two intervals found in Step 1) using the same procedure. EXAMPLE 2 Graphing y =sin bx Graph y = sin 2x, and compare to the graph of y = sin x. SOLUTION In this function the coefficient of x is 2, so b = 2 and the period is 2p 2 = p. Therefore, the graph will complete one period over the interval 30, p4. We can divide the interval 30, p4 into four equal parts by first finding its midpoint: 1 2 10 + p2 = p 2 . The quarter points are found next by determining the midpoints of the two intervals C 0, p 2 D and C p 2 , pD . 1 2 a0 + p 2b = p 4 and 1 2 a p 2 + pb = 3p 4 Quarter points 1 2 A p 2 + pB = 1 2 A 3p 2 B = 3p 4
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