Question

Find the principal value of ${\sin ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)$.

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 ${\sin ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)$. We know that for sine function to be negative, the angle should be negative, that is less than $0^o$. We know that $\sin {60^o} = \dfrac{{\sqrt 3 }}{2}$. This means that the value of ${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = {60^o}$.So the value of the given expression can be calculated as-
${\sin ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right) = - {\left( {60} \right)^o} = - {60^o}$
We know that ${{\pi }}$ rad = $180^o$, so
${\sin ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right) = - \dfrac{{{\pi }}}{3}$
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 function 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. One common mistake is that the students often neglect the negative sign in the function in a hurry, and often write the wrong answer.