Question

Find the principal value of ${\sin }^{-1}\left(-\frac{1}{2}\right)$

Answer 1

Solve using trigonometric rules:

The given inverse trigonometric function is ${\sin }^{-1}\left(-\frac{1}{2}\right)$.

Let, ${\sin }^{-1}\left(-\frac{1}{2}\right)=y$

$\Rightarrow$ $\left(-\frac{1}{2}\right)=\sin y$ $\left[\because {\sin }^{-1}x=y\Rightarrow x=\sin y\right]$

$\Rightarrow$ $\sin y=-\frac{1}{2}$

As we know, the range of principal value of ${\sin }^{-1}\mathrm{x}$ lies in $\left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$

$\Rightarrow$ $\sin y=\sin \left(-\frac{\mathrm{\pi }}{6}\right)$

$\Rightarrow$ $y=-\frac{\mathrm{\pi }}{6}$

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

Hence, the required principal value is $-\frac{\mathrm{\pi }}{6}$.