Question

What is the ${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$ $?$

Answer 1

Hint : The inverse of the trigonometric function must be used to determine the measure of the angle. The inverse of the tangent function is read tangent inverse and is also called the arctangent relation. The inverse of the cosine function is read cosine inverse and is also called the arccosine relation. The inverse of the sine function is read sine inverse and is also called the arcsine relation.

Complete step by step solution:
Given ${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$ -----(1)
We know that $\sin \left( {\dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}$ ------(2)
Taking ${\sin ^{ - 1}}$ on both sides of the equation (2). Then the equation (2) becomes
${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = \dfrac{\pi }{3}$ -------(3)
Since $\sin x$ is a periodic function with period $\pi$ . By definition of a periodic function, there exist any integer $n$ , such that
$\sin \left( {2n\pi + \dfrac{\pi }{3}} \right) = \sin \left( {\dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}$ --(4) for any integer $n$ .
Since the range of ${\sin ^{ - 1}}\left( x \right)$ lie in the range $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ .
From the equation (4) only $\dfrac{\pi }{3}$ lies in the closed interval $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ .
Hence, the value of ${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$ is $\dfrac{{\sqrt 3 }}{2}$ .
So, the correct answer is “$\dfrac{\pi }{3}$”.

Note : Note that the domain of ${\sin ^{ - 1}}\left( x \right)$ is $\left[ { - 1,1} \right]$ . The principal value denoted ${\tan ^{ - 1}}$ is chosen to lie in the range $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ . Hence the exact value of ${\tan ^{ - 1}}\left( { - x} \right)$ for any value of $x$ lies in the $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ .Also note that $\sin ( - x) = - \sin (x)$ , $\cos ( - x) = \cos (x)$ and $\tan ( - x) = - \tan (x)$ .