Question

Find the value of ${\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{3}} \right)$

Answer 1

Hint: To solve this question, we need to know the basic theory related to the Inverse Trigonometric Functions. Here we have two terms ${\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right)$ and ${\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{3}} \right)$. So first we find the calculated value of both terms and then we simply add these two to get our result.

Complete step-by-step solution:
First term of the expression:
${\cos ^{ - 1}}\left[ {\cos \left( {\dfrac{{5\pi }}{3}} \right)} \right]$
The above expression can be written as:
${\cos ^{ - 1}}\left\{ {\cos \left( {2\pi - \dfrac{\pi }{3}} \right)} \right\}$
We know that Cos($2\pi - \theta $)=Cos$\theta $
So, we can write
${\cos ^{ - 1}}\left\{ {\cos \left( {2\pi - \dfrac{\pi }{3}} \right)} \right\}$=${\cos ^{ - 1}}\left\{ {\cos \dfrac{\pi }{3}} \right\}$.
Now, we know that
${\cos ^{ - 1}}\left( {\cos x} \right) = x$ , for $0 \leqslant x \leqslant \pi $ .
So, we have:
${\cos ^{ - 1}}\left\{ {\cos \dfrac{\pi }{3}} \right\}$= $\dfrac{\pi }{3}$
Second term of the expression:
${\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{3}} \right)$
The above expression can be written as:
\[{\sin ^{ - 1}}\left\{ {\sin \left( {2\pi - \dfrac{\pi }{3}} \right)} \right\}\]
We know that Sin($2\pi - \theta $)= -Sin$\theta $
So, we can write
\[{\sin ^{ - 1}}\left\{ {\sin \left( {2\pi - \dfrac{\pi }{3}} \right)} \right\}\]= \[{\sin ^{ - 1}}\left( { - \sin \dfrac{\pi }{3}} \right)\]
We know that $\operatorname{Sin} ( - \theta ) = - \operatorname{Sin} \theta $
So , we have:
\[{\sin ^{ - 1}}\left( { - \sin \dfrac{\pi }{3}} \right)\]= \[{\sin ^{ - 1}}\left( {\sin ( - \dfrac{\pi }{3})} \right)\]
Now, we know that
\[{\operatorname{Sin} ^{ - 1}}\left( {\sin x} \right) = x\] , for $\dfrac{{ - \pi }}{2} \leqslant x \leqslant \dfrac{\pi }{2}$
So, we have:
\[{\sin ^{ - 1}}\left({\sin ( - \dfrac{\pi }{3}} \right)\] = $ - \dfrac{\pi }{3}$
Now, as we need to calculated-
${\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{3}} \right)$
Put the value of both term, which we already calculated-
= $\dfrac{\pi }{3} - \dfrac{\pi }{3}$
= 0
Therefore, the value of ${\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{3}} \right)$ is 0.

Note: The important basic step is to split the given angle in terms of $2\pi \pm \theta $. So, we should remember the formula related to trigonometric function of the allied angle.
Sin ($2\pi - \theta $)= -Sin$\theta $
Sin ($2\pi + \theta $)= Sin$\theta $.
 And also, we need to remember some formula related to this question.
${\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta $ where, $0 \leqslant \theta \leqslant \pi $
${\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta $ where, $ - \dfrac{\pi }{2} \leqslant \theta \leqslant \dfrac{\pi }{2}$