Question

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

Answer 1

Hint:In this question first we have to let the given function equal to $y$. Then using inverse trigonometric functions we have to find that its possible value will lie in the principal value that function or not. If not then check for another possible value.

Complete step-by-step answer:
Let $y = {\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$
Now, on taking on both sides. we get
$\Rightarrow \sin y = \sin \\{ {\sin ^{ - 1}}\sin \left( {\dfrac{{2\pi }}{3}} \right)\\} \\\ \Rightarrow \sin y = \sin \left( {\dfrac{{2\pi }}{3}} \right){\text{ eq}}{\text{.1}} \\\$
But range of principal value of ${\sin ^{ - 1}}{\text{ is }}\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$.Therefore, $y = \dfrac{{2\pi }}{3}$is not possible.
We know that $\sin x$is positive in the first and second quadrant and negative in the third and fourth quadrant.
$\Rightarrow \sin (\pi - \theta ) = \sin \theta$ eq.2
Now, again consider the eq.1
$\Rightarrow \sin y = \sin \left( {\dfrac{{2\pi }}{3}} \right){\text{ }} \\\ \Rightarrow \sin y = \sin \left( {\pi - \dfrac{{2\pi }}{3}} \right){\text{ \\{ from eq}}{\text{.2\\} }} \\\ \Rightarrow \sin y = \sin \left( {\dfrac{\pi }{3}} \right){\text{ }} \\\ \Rightarrow {\text{y = }}\dfrac{\pi }{3} \\\$
Which is in range of principal value of ${\sin ^{ - 1}}{\text{ i}}{\text{.e}}{\text{. }}\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$.
Hence,
${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$=$\dfrac{\pi }{3}$

Note:Whenever you get this type of question the key concept to solve this is to learn the principal values of inverse trigonometric functions. And properties of trigonometric functions like in this question we need the property of $\sin x$that it is positive in the first and second quadrant and negative in rest.