Draw the graph of y = sin(x/3 )

The graph of y = A sin Bx has the property
(1). amplitude = |A|
(2). period = 2pi/B
For y = sin(x/3),
since A = 1, so its amplitude is |1|
since B = 1/3, so its period = 2pi/B = 2pi ÷ 1/3 = 2pi × 3 = 6pi
Thus, its amplitude is 1 and period is 6pi
Find the five points in one period
one period is 6pi, half period is 3pi, quarter period is 3pi/2
divide the five points equally in a period [0, 6pi]
so the five points in x-axis are:
x1 = 0
x2 = 3pi/2
x3 = 3pi
x4 = 3pi + 3pi/2 = 6pi/2 + 3pi/2 = 9pi/2
x5 = 9pi/2 + 3pi/2 = 12pi/2 = 6pi
so the five points in xy-plane are (0, ?), (3pi/2, ?), (3pi, ?), (9pi/2, ?), (6pi, ?)
Now find the values of the function y = sin(x/3) in the five points
when x = 0, y = sin(x/3) = sin[(1/3) × 0] = sin(0) = 0, so the point is (0, 0)
when x = 3pi/2, y = sin(x/3) = sin[(1/3) × 3pi/2] = sin(pi/2) = 1, so the point is (3pi/2, 1)
when x = 3pi, y = sin(x/3) = sin[(1/3) × 3pi] = sin(pi) = 0, so the point is (3pi, 0)
when x = 9pi/2, y = sin(x/3) = sin[(1/3) × 9pi/2] = sin(3pi/2) = -1, so the point is (9pi/2, -1)
when x = 6pi, y = sin(x/3) = sin[(1/3) × 6pi] = sin(2pi) = 0, so the point is (6pi, 0)
The five points are (0, 0), (3pi/2, 1), (3pi, 0), (9pi/2, -1), (6pi, 0)
Draw the graph of y = sin(x/3) based on the five points
draw the graph of y = sin(x/3)
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 = 3 pi is its half period, at this point, its value of y is 0.
x = (3/2)pi is its quarter period, at this point, its value of y is 1 which is maximum.
x = (9/2)pi is its three-fourth period, at this point, its value of y is -1 which is minimum.
x = 6 pi is its end point of one period, at this point, its value of y is 0.
The curve of y = sin (x/3) is continuous,
its second period is from x = 6 pi to x = 6 pi + 6 pi = 12 pi,
its third period is from 12 pi to 12 pi + 6 pi = 18 pi,
its fourth period is from 18 pi to 18 pi + 6 pi = 24 pi, so on.