Question

How do you evaluate ${\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right)$?

Answer 1

Hint: In this question we have to find the value of the given function which is a trigonometric function, we will make use of inverse trigonometric formulas, first as the given angle will not be range of sin which is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$, so write the angle $5\pi $as the difference of two angles as $6\pi - \pi $, and the angle becomes $\pi - \dfrac{\pi }{6}$, and now using the fact $\sin \left( {\pi - \theta } \right) = \sin \theta $simplify the expression, and then by using the fact ${\sin ^{ - 1}}\left( {\sin x} \right) = x$we will get the required result.

Complete step by step solution:
Given function is ${\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right)$,
First as the given angle $\dfrac{{5\pi }}{6}$ is not in the range of sin which is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$, we will write the angle as the difference of two angles, by writing we get,
$ \Rightarrow {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right) = {\sin ^{ - 1}}\left( {\sin \dfrac{{6\pi - \pi }}{6}} \right)$,
Now simplifying the expression we get,
$ \Rightarrow {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right) = {\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{6\pi }}{6} - \dfrac{\pi }{6}} \right)} \right)$,
Now again simplifying by eliminating the like terms, we get,
$ \Rightarrow {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right) = {\sin ^{ - 1}}\left( {\sin \left( {\pi - \dfrac{\pi }{6}} \right)} \right)$,
Now using the trigonometric equality fact that is $\sin \left( {\pi - \theta } \right) = \sin \theta $, we get,
$ \Rightarrow {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right) = {\sin ^{ - 1}}\left( {\sin \dfrac{\pi }{6}} \right)$,
Now using the inverse trigonometric formula of sine i.e., ${\sin ^{ - 1}}\left( {\sin x} \right) = x$,
Here $x = \dfrac{\pi }{6}$, by substituting the value we get,
$ \Rightarrow {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right) = \dfrac{\pi }{6}$,
So, the value of the given function is $\dfrac{\pi }{6}$.

$\therefore $The value of the given trigonometric function ${\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right)$ will be equal to $\dfrac{\pi }{6}$.

Note: The inverse trigonometric functions are used to find the unknown measure of an angle of a right triangle when two side lengths are known. Basic inverse trigonometric functions are ${\sin ^{ - 1}}x$, ${\cos ^{ - 1}}x$ and ${\tan ^{ - 1}}x$.Specifically, they are the inverses of the sine, cosine, tangent. Some important formulas related to Inverse trigonometry are given below:
${\sin ^{ - 1}}\left( {\sin x} \right) = x$,
${\cos ^{ - 1}}\left( {\cos x} \right) = x$,
${\tan ^{ - 1}}\left( {\tan x} \right) = x$,
${\sec ^{ - 1}}\left( {\sec x} \right) = x$,
${\csc ^{ - 1}}\left( {\csc x} \right) = x$,
${\cot ^{ - 1}}\left( {\cot x} \right) = x$,
${\sin ^{ - 1}}x = {\csc ^{ - 1}}\dfrac{1}{x}$,
${\cos ^{ - 1}}x = {\sec ^{ - 1}}\dfrac{1}{x}$,
${\tan ^{ - 1}}x = {\cot ^{ - 1}}\dfrac{1}{x}$,
${\sin ^{ - 1}}x = {\cos ^{ - 1}}\sqrt {1 - {x^2}} = {\tan ^{ - 1}}\left( {\dfrac{x}{{\sqrt {1 - {x^2}} }}} \right)$,
${\cos ^{ - 1}}x = {\sin ^{ - 1}}\sqrt {1 - {x^2}} = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 - {x^2}} }}{x}} \right)$,
${\tan ^{ - 1}}x = {\sin ^{ - 1}}\left( {\dfrac{x}{{\sqrt {1 + {x^2}} }}} \right) = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt {1 + {x^2}} }}} \right)$.