Question 11 Mark
Find the principal value of $\sin ^{-1}\left(-\frac{1}{2}\right)$.
Answer
View full question & answer→$\text { Let } \sin ^{-1}\left(-\frac{1}{2}\right)=x$
$\therefore \sin x=-\frac{1}{2}=-\sin \frac{\pi}{6}=\sin \left(-\frac{\pi}{6}\right)$
$ \text { Here }-\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \text { and } x \in(-1,1)$
$\therefore$ Principal value of $\sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}$.
$\therefore \sin x=-\frac{1}{2}=-\sin \frac{\pi}{6}=\sin \left(-\frac{\pi}{6}\right)$
$ \text { Here }-\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \text { and } x \in(-1,1)$
$\therefore$ Principal value of $\sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}$.