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# Draw the graph of Graph y = 1/2 sin(x pi/2)

- The graph of y = A sin Bx has the property
- (1). amplitude = |A|
- (2). period = 2pi/B
- For y = 1/2 sin [(pi/2) x],
- since A = 1/2, so its amplitude = |1/2|
- since B = pi/2, so its period = 2pi/B = 2pi ÷ pi/2 = 2pi × 2/pi = 4
- Thus, its amplitude is 1/2 and period is 4

- Find the five points in one period
- one period is 4, half period is 2, quarter period is 1
- divide the five points equally in a period [0, 4]
- The five points in x-axis are: x
_{1}= 0, x_{2}= 1, x_{3}= 2, x_{4}= 3, x_{5}= 4 - so the five points in xy-plane are (0, ?), (1, ?), (2, ?), (3, ?), (4, ?)

- Now find the values of the function y = 1/2sin(x pi/2) in the five points
- when x = 0, y = 1/2 sin[(pi/2) × 0] = 1/2 sin(0) = 0, so the point is (0, 0)
- when x = 1, y = 1/2 sin[(pi/2) × 1] = 1/2 sin(pi/2) = 1/2, so the point is (1, 1/2)
- when x = 2, y = 1/2 sin[(pi/2) × 2] = 1/2 sin(pi) = 0, so the point is (2, 0)
- when x = 3, y = 1/2 sin[(pi/2) × 3] = 1/2 sin(3pi/2) = - 1/2, so the point is (3, - 1/2)
- when x = 4, y = 1/2 sin[(pi/2) × 4] = 1/2 sin(2 pi) = 0, so the point is (4, 0)
- The five points are (0, 0), (1, 1/2), (2, 0), (3, -1/2), (4, 0)

- Draw the graph of y = 1/2sin(x pi/2) based on the five points

- Note the values of the Sine function at the special angles are:
- sin(0) = 0
- sin(pi/2) = 1
- sin(pi) = 0
- sin(3pi/2) = -1
- sin(2pi) = 0

- Analysis of the graph:
- x = 0, y = 0.
- x = 2 is its half period, at this point, its value of y is 0.
- x = 1 is its quarter period, at this point, its value of y is 1/2 which is maximum.
- x = 3 is its three-fourth period, at this point, its value of y is -1/2 which is minimum.
- x = 4 is its end point of the first period, at this point, its value of y is 0.
- The curve of y = 1/2 sin (x pi/2) is continuous, it will repeat at period of 4.