Draw the graph of `y = sin^(3)x.`

Explanation

We have `y = f(x) = sin^(3)x`
Clearly `f(x + 2pi) = f(x)`
Hence y = f(x) has period `2pi`.
So first we draw the graph for `x in [0, 2pi].`
Let `x in [0, pi]`
`f(0) = f(pi) = 0 " and " f(pi//2) = 1`
Hence the graphs of `y = sin x " and " y = sin^(3)x` coincide for `x = 0, pi//2 " and " pi.`
For `x in (0, pi//2) uu (pi//2, pi), sin x in (0,1)`
`therefore` `sin^(3)x lt sin x`
So the graph of `y = sin^(3)x` lies below the graph of y = sin x (even below the graph of `y = sin^(2)x`).
So graphs of the functions are as shown in the following figure.

For `x in (pi, 2pi), -1 lt sin x lt 0`
`therefore` `-1 lt sin^(3)x lt 0`
Also `sin^(3)x gt sin x`
So the graph of `y = sin^(3) x` lies above the graph of y = sin x as shown in the following figure.

Since the period of `y = sin^(3)x " is " 2pi`, the graph of the function can be repeated as shown in the following figure.