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: x₁ = 0, x₂ = 1, x₃ = 2, x₄ = 3, x₅ = 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)

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.