Answer
$sin^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}$
Work Step by Step
The domain of $sin^{-1}(x)$ is $[-1, 1]$ and the range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. $sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$, so $sin^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}$.
