Question

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

Answer 1

Hint: In this question we are asked to find the principal value of inverse trigonometric function. To solve this question, first we need to know the range of principal values of inverse trigonometric functions where domain is all possible values of $\theta $ and range is all possible values of angles.

Complete step-by-step answer:
To know the range of inverse trigonometric functions we can take the help of the following table.

	Range	Positive	Negative
${{\sin }^{-1}}$	$\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$	$\theta $	$-\theta $
${{\cos }^{-1}}$	\[\left[ 0,\pi \right]\]	$\theta $	$\pi -\theta $
${{\tan }^{-1}}$	$\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$	$\theta $	$-\theta $

The range of ${{\sin }^{-1}}$ is between $\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$ . Equate the expression to $'\theta '$ , then locate the value in the basic trigonometric table and find the angle. Simplify and get the principal value.
A principal value of a function is the value selected at a point in the domain of multiple valued function, chosen so that the function has a single value at the point. The principal value of ${{\sin }^{-1}}x$ for \[x>0\] , is the length of the arc of a unit circle centred at the origin which subtends an angle at the centre whose sine is x.
 The principal value of the ${{\sin }^{-1}}$ will be between the range $\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$.We have been given the function,
 ${{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)$ , for which we need to find the principal value ,
 let us take ${{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}\left( x \right)$
So, we will get ${{\sin }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)=-{{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$ .
We can write it as –
$\sin \theta =\left( \dfrac{\sqrt{3}}{2} \right)$ .
We know that $\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}$ .
$\theta ={{60}^{\circ }}={{60}^{\circ }}\times \dfrac{\pi }{180}$ .
=$\dfrac{\pi }{3}$ .
Thus, we got the principal value of inverse sine function as $-\dfrac{\pi }{3}$ , which lies between the range $\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$ .
Hence, the principal value of ${{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)$ is $\dfrac{-\pi }{3}$ .
Note: Students should be careful while taking the range of inverse trigonometric function, range of principle value of ${{\sin }^{-1}}$ is between $\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$ .They should be careful about the sign. If the given value is positive then the principal value is $'\theta '$ , if it is negative then the principal value will be$'-\theta '$ . Otherwise they may lead to getting the wrong answer.