Question

Find the inverse of the following functions:

$f\left(x\right)={\sin }^{-1}\left(\frac{x}{3}\right),x\in [-3,3]$
then $f^{-1}\left(\pi /2\right)$

Answer 1

$f\left(x\right)={\sin }^{-1}\left(\frac{x}{3}\right)=y$ (say)
$\Rightarrow \frac{x}{3}=\sin y\Rightarrow x=3\sin y$
$\therefore f^{-1}\left(x\right)=f\left(y\right)=3\sin y$
Hence $f^{-1}\left(\pi /2\right)=3$