Answer
$0$
Work Step by Step
Suppose $\theta =\sin (\pi)$
This gives: $\sin^{-1} \theta=\sin^{-1}[\sin(\pi)]$
The inverse property states that $\sin^{-1}(\sin a)=a $ satisfies for every $ a $ in the range $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]$ .
But the value $ a=\pi $ does not lie in the range of $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]$.
Therefore, we have $ \sin^{-1}[\sin(\pi)]=0$
