What is ${\sin }^{-1}(\sin \frac{3\pi }{5})$ equal to?

We know that ${\sin }^{-1}(\sin \theta )=\theta$ only if $\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$.Here, $\theta =\frac{3\pi }{5}=108^{\circ }$, which is outside the principal value range $[-90^{\circ },90^{\circ }]$.We use the identity $\sin \theta =\sin (\pi -\theta )$:
$$\sin \frac{3\pi }{5}=\sin (\pi -\frac{3\pi }{5})=\sin \frac{2\pi }{5}$$
Now, $\frac{2\pi }{5}=72^{\circ }$, which is within the range $[-\frac{\pi }{2},\frac{\pi }{2}]$.Therefore:
$${\sin }^{-1}(\sin \frac{3\pi }{5})={\sin }^{-1}(\sin \frac{2\pi }{5})=\frac{2\pi }{5}$$