How to find principle value of Inverse trigonometric function?

Inverse trigonometric function

• ${\sin }^{–1}x$ should not be confused with ${\sin }^{–1}x$. In fact ${\sin }^{–1}x={\displaystyle \frac{1}{\sin x}}$ and similarly for other trigonometric functions.

• Whenever no branch of an inverse trigonometric functions is mentioned, we mean the principle value branch of that function.

• The value of an inverse trigonometric functions which lies in the range of principle branch is called the principle value of that inverse trigonometric functions.

Inverse Trigonometric Functions	Domain	Range
${\sin }^{-1}$	$[-1,1]$	$[-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}]$
${\cos }^{-1}$	$[-1,1]$	$[0,\pi ]$
${\mathrm{cosec}}^{-1}$	$R-(-1,1)$	$[-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}]-\{{\displaystyle \frac{\pi }{2}}\}$
${\sec }^{-1}$	$R-(-1,1)$	$(0,\pi )-\{{\displaystyle \frac{\pi }{2}}\}$
${\tan }^{-1}$	$R$	$(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}})$
${\cot }^{-1}$	$R$	$(0,\pi )$

Example 1:

Find the principle value of ${\sin }^{-1}{\displaystyle \frac{1}{\sqrt{2}}}$.

Solution:

Let ${\sin }^{-1}({\displaystyle \frac{1}{\sqrt{2}}})=y$.Then, $\sin y={\displaystyle \frac{1}{\sqrt{2}}}$

we know the range of principle value of ${\sin }^{-1}$ is 

$[-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}]$ and $\sin [{\displaystyle \frac{\pi }{4}}]={\displaystyle \frac{1}{\sqrt{2}}}$

Therefore,principle value of ${\sin }^1({\displaystyle \frac{1}{\sqrt{2}}})$ is ${\displaystyle \frac{\pi }{4}}$

Example 2:

Find the principle value of ${\cot }^{-1}{\displaystyle \frac{-1}{\sqrt{3}}}$.

Solution:

Let ${\cot }^{-1}({\displaystyle \frac{1}{\sqrt{3}}})=y$.Then, 

$\cot y={\displaystyle \frac{-1}{\sqrt{3}}}=-\cot {\displaystyle \frac{\pi }{3}}$$=\cot (\pi -{\displaystyle \frac{\pi }{3}})=\cot ({\displaystyle \frac{2\pi }{3}})$

we know the range of principle value of ${\cot }^{-1}$ is $(0,\pi )$ and $\cot ({\displaystyle \frac{2\pi }{3}})={\displaystyle \frac{-1}{\sqrt{3}}}$

Hence the principle value of ${\cot }^{-1}{\displaystyle \frac{1}{\sqrt{2}}}$ is ${\displaystyle \frac{2\pi }{3}}$.