Answer
$one- to- one$,
$ domain$,
$[-\frac{\pi}{2}, \frac{\pi}{2}]$
Work Step by Step
Inverse function is defined for one-to-one functions only. Now Sine being trigonometric will not be one-to-one for all domains hence we have to restrict it to interval on which it is one-to-one. Now sine is one to one in interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$, so this solves our problem.
Thus
For a function to have an inverse, it must be $one- to- one$.
To define the inverse sine function, we restrict the $ domain$ of the sine function to the $[-\frac{\pi}{2}, \frac{\pi}{2}]$ interval .
