Question

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

Answer 1

Solve using trigonometric rules:

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

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

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

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

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

$\Rightarrow$ $\sin y=\sin \frac{\mathrm{\pi }}{4}$

$\Rightarrow$ $y=\frac{\mathrm{\pi }}{4}$

So, the principal value of ${\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)$ is $\frac{\mathrm{\pi }}{4}$.

Hence, the required principal value is $\frac{\mathrm{\pi }}{4}$.