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

Find the value of cos⁡−1(cos⁡5π3)+sin⁡−1(sin⁡5π3){\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{3}} \right)

Explanation

Solution

To solve this question, we need to know the basic theory related to the Inverse Trigonometric Functions. Here we have two terms cos⁡−1(cos⁡5π3){\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) and sin⁡−1(sin⁡5π3){\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[cos⁡(5π3)]{\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(2π−θ2\pi - \theta )=Cosθ\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(cos⁡x)=x{\cos ^{ - 1}}\left( {\cos x} \right) = x , for 0⩽x⩽π0 \leqslant x \leqslant \pi .
So, we have:
{\cos ^{ - 1}}\left\\{ {\cos \dfrac{\pi }{3}} \right\\}= π3\dfrac{\pi }{3}
Second term of the expression:
sin⁡−1(sin⁡5π3){\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(2π−θ2\pi - \theta )= -Sinθ\theta
So, we can write
{\sin ^{ - 1}}\left\\{ {\sin \left( {2\pi - \dfrac{\pi }{3}} \right)} \right\\}= sin⁡−1(−sin⁡π3){\sin ^{ - 1}}\left( { - \sin \dfrac{\pi }{3}} \right)
We know that Sin⁡(−θ)=−Sin⁡θ\operatorname{Sin} ( - \theta ) = - \operatorname{Sin} \theta
So , we have:
sin⁡−1(−sin⁡π3){\sin ^{ - 1}}\left( { - \sin \dfrac{\pi }{3}} \right)= sin⁡−1(sin⁡(−π3)){\sin ^{ - 1}}\left( {\sin ( - \dfrac{\pi }{3})} \right)
Now, we know that
Sin⁡−1(sin⁡x)=x{\operatorname{Sin} ^{ - 1}}\left( {\sin x} \right) = x , for −π2⩽x⩽π2\dfrac{{ - \pi }}{2} \leqslant x \leqslant \dfrac{\pi }{2}
So, we have:
sin⁡−1(sin⁡(−π3){\sin ^{ - 1}}\left({\sin ( - \dfrac{\pi }{3}} \right) = −π3 - \dfrac{\pi }{3}
Now, as we need to calculated-
cos⁡−1(cos⁡5π3)+sin⁡−1(sin⁡5π3){\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-
= π3−π3\dfrac{\pi }{3} - \dfrac{\pi }{3}
= 0
Therefore, the value of cos⁡−1(cos⁡5π3)+sin⁡−1(sin⁡5π3){\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 2π±θ2\pi \pm \theta . So, we should remember the formula related to trigonometric function of the allied angle.
Sin (2π−θ2\pi - \theta )= -Sinθ\theta
Sin (2π+θ2\pi + \theta )= Sinθ\theta .
And also, we need to remember some formula related to this question.
cos⁡−1(cos⁡θ)=θ{\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta where, 0⩽θ⩽π0 \leqslant \theta \leqslant \pi
sin⁡−1(sin⁡θ)=θ{\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta where, −π2⩽θ⩽π2 - \dfrac{\pi }{2} \leqslant \theta \leqslant \dfrac{\pi }{2}