Inverse Trigonometric Functions — Class 12 Maths Solution

Let ${\sin ^{ - 1}}\left( { - \frac{1}{2}} \right) = x \Rightarrow \sin x = - \frac{1}{2}$
As we know that the range of the principal value branch of ${\sin ^{ - 1}}$ is $\left[ { - \frac{\pi }{2},\;\;\frac{\pi }{2}} \right].$

Then, $\sin \left( { - \frac{\pi }{6}} \right) = - \frac{1}{2},\,\;where\;\; - \frac{\pi }{6} \in \left[ { - \frac{\pi }{2},\;\;\frac{\pi }{2}} \right]$

Hence, the principal value of ${\sin ^{ - 1}}\left( { - \frac{1}{2}} \right)\;\;is\;\; - \frac{\pi }{6}$