Question

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

Answer 1

Hint: The principal value of an inverse trigonometric function at a point x is the value of the inverse function at the point x, which lies in the range of the principal branch. The principal value of the sine function is from $\left[ { - \dfrac{{{\pi }}}{2},\dfrac{{{\pi }}}{2}} \right]$. From the table we will find the value of the function within the principal range, to get our final answer.

Complete step-by-step answer:

The values of the sine functions are-

Function	$0^o$	$30^o$	$45^o$	$60^o$	$90^o$
sin	0	\[\dfrac{1}{2}\]	\[\dfrac{1}{{\sqrt 2 }}\]	\[\dfrac{{\sqrt 3 }}{2}\]	1

In the given question we need to find the principal value of ${\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right)$. We know that for sine function to be positive, the angle should be acute, that is less than $90^o$. We know that $\sin {45^o} = \dfrac{1}{{\sqrt 2 }}$. This means that-

${\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right) = {45^o}$
We know that ${{\pi }}$ rad = $180^o$, so
${\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right) = \dfrac{{{\pi }}}{4}$
This is the required value.

Note: In such types of questions, we need to strictly follow the range of the principal values that have been specified. The principal value of sine in the question should always be between $-90^o$ and $90^o$. This is because there can be infinite values of any inverse trigonometric functions. For example, the value of $sin^{-1}0$ can be $0^{o}, 180^{o}, 360^{o}$ and so on. Hence, we need to write only that value which is inside the principal range.