635 6.4 Translations of the Graphs of the Sine and Cosine Functions We join the corresponding points with a smooth curve to obtain the solid blue graph shown in Figure 33. The period is 2p, and the amplitude is 3. Method 2 Write y = 3 cos Ax + p 4 B in the form y = a cos1x - d2. y = 3 cos ax + p 4b , or y = 3 cos c x - a- p 4b d Rewrite to subtract - p 4 . This result shows that d = - p 4 . Because - p 4 is negative, the phase shift is to the left 0 - p 4 0 = p 4 unit. The graph is the same as that of y = 3 cos x (the red graph in the calculator screen shown in the margin), except that it is translated to the left p 4 unit (the blue graph). S Now Try Exercise 43. EXAMPLE 2 Graphing y =a cos 1x −d2 Graph y = 3 cos Ax + p 4 B over one period. SOLUTION Method 1 We first solve the following three-part inequality. 0 … x + p 4 … 2p Three-part inequality - p 4 … x … 7p 4 Subtract p 4 from each part. Dividing this interval into four equal parts gives the following x-values. We use these x-values to make a table of key points. - p 4 , p 4 , 3p 4 , 5p 4 , 7p 4 Key x-values x - p 4 p 4 3p 4 5p 4 7p 4 x +P 4 0 p 2 p 3p 2 2p cos 1x +P 4 2 1 0 -1 0 1 3 cos 1x +P 4 2 3 0 -3 0 3 These x-values lead to maximum points, minimum points, and x-intercepts. y y = 3 cos x 0 x –3 3 y = 3 cos (x + ) 4 P 3p 4 5p 4 7p 4 p 4 p 4 – Figure 33 y2 = 3 cosx −4 4 11p 4 11p 4 − EXAMPLE 3 Graphing y =a cos 3 b 1x −d2 4 Graph y = -2 cos13x + p2 over two periods. SOLUTION Method 1 We first solve the three-part inequality 0 … 3x + p … 2p to find the interval C - p 3 , p 3 D . Dividing this interval into four equal parts gives the points A - p 3 , -2B , A - p 6 , 0B , 10, 22, A p 6 , 0B , and A p 3 , -2B . We plot these points and join them with a smooth curve. By graphing an additional half period to the left and to the right, we obtain the graph shown in Figure 34. Method 2 First write the equation in the form y = a cos 3b1x - d24. y = -2 cos13x + p2, or y = -2 cos c 3 ax + p 3 b d Rewrite by factoring out 3. Then a = -2, b = 3, and d = - p 3 . The amplitude is -2 = 2, and the period is 2p 3 (because the value of b is 3). The phase shift is to the left 0 - p 3 0 = p 3 units compared to the graph of y = -2 cos 3x. Again, see Figure 34. S Now Try Exercise 49. y x 0 –2 2 y = –2 cos (3x + P) –p p – – 2p 3 2p 3 p 3 p 3 –p 6 p 6 Figure 34
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