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

The value of cos⁡−1(cos⁡5π3)+sin⁡−1(cos⁡5π3){\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) is
A.π2\dfrac{\pi }{2}
B.5π3\dfrac{{5\pi }}{3}
C.10π3\dfrac{{10\pi }}{3}
D.0

Explanation

Solution

We will first simplify the given expression using the properties of inverse trigonometry such as cos⁡−1(cos⁡x)=x{\cos ^{ - 1}}\left( {\cos x} \right) = x, when x∈[0,π]x \in \left[ {0,\pi } \right] and sin⁡−1(sin⁡x)=x{\sin ^{ - 1}}\left( {\sin x} \right) = x when x∈[−π2,π2]x \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]. Then, we will substitute and simplify the values to get the required answer.

Complete step by step answer:

We have to find the value of cos⁡−1(cos⁡5π3)+sin⁡−1(cos⁡5π3){\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right)
It is known that cos⁡−1(cos⁡x)=x{\cos ^{ - 1}}\left( {\cos x} \right) = x, when x∈[0,π]x \in \left[ {0,\pi } \right] and sin⁡−1(sin⁡x)=x{\sin ^{ - 1}}\left( {\sin x} \right) = x when x∈[−π2,π2]x \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]
We will rewrite the angles using the formulas of trigonometry.
We can rewrite the angle ascos⁡(5π3)=cos⁡(2π−π3)\cos \left( {\dfrac{{5\pi }}{3}} \right) = \cos \left( {2\pi - \dfrac{\pi }{3}} \right)
Also, cos⁡(2π−θ)=cos⁡θ\cos \left( {2\pi - \theta } \right) = \cos \theta
Therefore, cos⁡(2π−π3)=cos⁡π3\cos \left( {2\pi - \dfrac{\pi }{3}} \right) = \cos \dfrac{\pi }{3}
On substituting the values in the given expression, we get,
cos⁡−1(cos⁡π3)+sin⁡−1(cos⁡π3){\cos ^{ - 1}}\left( {\cos \dfrac{\pi }{3}} \right) + {\sin ^{ - 1}}\left( {\cos \dfrac{\pi }{3}} \right)
Now, we know that cos⁡θ=sin⁡(π2−θ)\cos \theta = \sin \left( {\dfrac{\pi }{2} - \theta } \right)
Hence,
cos⁡π3=sin⁡(π2−π3) ⇒cos⁡π3=sin⁡(π6)  \cos \dfrac{\pi }{3} = \sin \left( {\dfrac{\pi }{2} - \dfrac{\pi }{3}} \right) \\\ \Rightarrow \cos \dfrac{\pi }{3} = \sin \left( {\dfrac{\pi }{6}} \right) \\\
We can now write the given expression as,
cos⁡−1(cos⁡π3)+sin⁡−1(sin⁡π6){\cos ^{ - 1}}\left( {\cos \dfrac{\pi }{3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{\pi }{6}} \right)
We will use the property cos⁡−1(cos⁡x)=x{\cos ^{ - 1}}\left( {\cos x} \right) = x, when x∈[0,π]x \in \left[ {0,\pi } \right] and sin⁡−1(sin⁡x)=x{\sin ^{ - 1}}\left( {\sin x} \right) = x when x∈[−π2,π2]x \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] to further simplify.
⇒π3+π6=π2\Rightarrow\dfrac{\pi }{3} + \dfrac{\pi }{6} = \dfrac{\pi }{2}
Hence,option A is correct.

Note: We cannot directly write cos⁡−1(cos⁡5π3){\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) as 5π3\dfrac{{5\pi }}{3} because the range of cos⁡−1x{\cos ^{ - 1}}x is [0,π]\left[ {0,\pi } \right] and 5π3>π\dfrac{{5\pi }}{3} > \pi . Similarly, the range of sin⁡−1x{\sin ^{ - 1}}x is [−π2,π2]\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]. Many students make mistakes by simply writing the values without considering the range.