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Q.
The principal value of $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$ is
Inverse Trigonometric Functions
Solution:
Let $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=y$. Then, $\sin y=\frac{1}{\sqrt{2}}$.
We know that, the range of the principal value branch of $\sin ^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ and $\sin \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$. Therefore, principal value of $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$ is $\frac{\pi}{4}$.