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Sketching y = a sin bx and y = a cos bx

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1 Sketching y = a sin bx and y = a cos bx

2 y = sin x y = cos x x 0 30 45 60 90 120 135 150 180 210 225
240 270 300 315 330 360 sin x .5 .71 .87 1 –.5 –.71 –.87 –1 90 180 270 360 1 –1 0.5 –0.5 45 135 225 315 y = cos x x 0 30 45 60 90 120 135 150 180 210 225 240 270 300 315 330 360 cos x 1 .87 .71 .5 –.5 –.71 –.87 –1 90 180 270 360 1 –1 0.5 –0.5 45 135 225 315

3 y = sin x y = cos x x –2 –3/2 – –/2 /2  3/2 2 5/2 3 7/2 4
/2 3/2 2 5/2 3 7/2 4 sin x y = cos x x –2 –3/2 – –/2 /2 3/2 2 5/2 3 7/2 4 cos x

4 y = sin x y = –sin x y = cos x y = –cos x x /2  3/2 2 sin x x /2
/2 3/2 2 sin x x /2 3/2 2 –sin x y = cos x y = –cos x x /2 3/2 2 cos x x /2 3/2 2 –cos x

5 Sketching y = a sin x y = 2 sin x y = ½ sin x y = –3 sin x
/2 3/2 2 sin x 2 sin x x /2 3/2 2 sin x 2 sin x x /2 3/2 2 sin x 2 sin x Sketching y = a cos x y = –2 cos x y = ¾ cos x y = –4 cos x x /2 3/2 2 cos x –2 cos x x /2 3/2 2 cos x ¾ cos x x /2 3/2 2 cos x –4 cos x

6 To be replaced by Ryder 1

7 Sketching y = a sin bx and y = a cos bx:
|a| = amplitude = ½(ymax – ymin) i.e., the “height” from the middle to the top |b| = frequency = number of cycles in 2 units P = period = the least number of positive units it takes to complete one cycle Plot the two functions f(x) and g(x) on the same graph on 0  x  2. For each function, identify the amplitude, frequency, period and answer the number of times they intersect. f(x) = 2 cos (3x) Amplitude: ____ Frequency: ____ Period: ____ g(x) = –2 sin (¾x) Amplitude: ____ Frequency: ____ Period: ____ 2 1 2 3 –1 –2 –3 How many times do they intersect on the interval [0, 2]? ___


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