Draw the graph of `f(x) = sqrt(sin x)`.

Explanation

We have `y = f(x) = sqrt(sin x)`
f(x) is defined if `sin ge 0`
Clearly, the domain of the function is `x in underset(n in I)uuu [2npi, pi + 2npi]`
Also the period of the function is `2pi`.
So we need to draw the graph for `x in [0, pi]` only.
Now `0 lt sin x lt 1 " for " x in (0, pi)`
`therefore` `sqrt(sin x) le sin x` for `x in (0, pi)`
`sqrt(sin 0) = sin 0 = 0, sqrt(sin pi) = sin pi = 0 " and " sqrt("sin"(pi)/(2)) = "sin" (pi)/(2) = 1`
Graphs of y = sin x and `y = sqrt(sin x)` meet at `x = 0, (pi)/(2)` and `pi`.
For other values of x, the graph of `y = sqrt(sin x)` lies above the graph of y = sin x.
In `3^(rd)` and `4^(th)` quadrants, `y = sqrt(sin x)` does not exist as `sin x lt 0` here. Thus the graph of `y = sqrt(sin x)` is as shown in the following figure.