Solution We know that \(sin^{-1}(sin x) = x\). Therefore, \(sin^{-1}(sin\frac{3\pi}{5}) =
\frac{3\pi}{5}\). But \(\frac{3\pi}{5}
\notin [-\frac{\pi}{2}, \frac{\pi}{2}]\), which is the principal
branch of \(sin^{-1}x\). However, \(sin(\frac{3\pi}{5}) = sin(\pi - \frac{2\pi}{5}) =
sin\frac{2\pi}{5}\) and \(\frac{2\pi}{5} \in [-\frac{\pi}{2},
\frac{\pi}{2}]\). Therefore, \(sin^{-1}(sin\frac{3\pi}{5}) =
sin^{-1}(sin\frac{2\pi}{5}) = \frac{2\pi}{5}\).