Find the value of sin^-1 ≤ft( sin (3π)/5 ).

The correct answer is Option (2) → $\frac{2\pi}{5}$ ##

We know that $\sin^{-1}(\sin x) = x$. Therefore, $\sin^{-1} \left( \sin \frac{3\pi}{5} \right) = \frac{3\pi}{5}$

But $\frac{3\pi}{5} \notin \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$, which is the principal branch of $\sin^{-1} x$

However $\sin \left( \frac{3\pi}{5} \right) = \sin \left( \pi - \frac{3\pi}{5} \right) = \sin \frac{2\pi}{5}$ and $\frac{2\pi}{5} \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$

Therefore $\sin^{-1} \left( \sin \frac{3\pi}{5} \right) = \sin^{-1} \left( \sin \frac{2\pi}{5} \right) = \frac{2\pi}{5}$