Find the values of $sin^{-1} \bigg( sin \frac{2\pi}{3} \bigg)$

Toolbox:
• $sin(\pi - x) = sinx$
• The range of the principal value of $\sin^{-1}x$ is $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$
Given $sin^{-1} \bigg( sin \frac{2\pi}{3} \bigg)$
$$sin\frac{2\pi}{3}=sin(\pi-\frac{\pi}{3})$$
We know that $sin(\pi - x) = sinx \Rightarrow sin \frac{2\pi}{3} = sin \frac{\pi}{3}$
The given expression becomes $sin^{-1} sin \frac{\pi}{3}$ which is $= \frac{\pi}{3}$
edited Mar 14, 2013