Draw the graph `y = sin^(2)x.`

Explanation

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

Since the period of the function is `pi`, the graph is repeated as shown in the following figure.