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Question: The value of \({\sin ^{ - 1}}\left( {\cos \dfrac{{53\pi }}{5}} \right)\) is A.\(\dfrac{{3\pi }}{5...

The value of sin⁡−1(cos⁡53π5){\sin ^{ - 1}}\left( {\cos \dfrac{{53\pi }}{5}} \right) is
A.3π5\dfrac{{3\pi }}{5}
B.−3π5\dfrac{{ - 3\pi }}{5}
C.π10\dfrac{\pi }{{10}}
D.−π10\dfrac{{ - \pi }}{{10}}

Explanation

Solution

Hint : Try to break angles in general angles.

We know,
cos⁡(2kπ+θ)=cos⁡θ   \cos (2k\pi + \theta ) = \cos \theta \\\ \\\
So, on comparing the above equation with question we get,
cos⁡53π5=\cos \dfrac{{53\pi }}{5} = cos⁡(10π+3π5)=cos⁡3π5\cos \left( {10\pi + \dfrac{{3\pi }}{5}} \right) = \cos \dfrac{{3\pi }}{5} ……(i)
We know,
3π5=π2+π10\dfrac{{3\pi }}{5} = \dfrac{\pi }{2} + \dfrac{\pi }{{10}} ……(ii)
And we also know,
cos⁡(π2+π10)=\cos \left( {\dfrac{\pi }{2} + \dfrac{\pi }{{10}}} \right) = −sin⁡π10 - \sin \dfrac{\pi }{{10}}= cos⁡3π5\cos \dfrac{{3\pi }}{5} ……(iii) (From i & ii )
We have to find the value of
sin⁡−1(cos⁡53π5){\sin ^{ - 1}}\left( {\cos \dfrac{{53\pi }}{5}} \right)
We know the value of (cos⁡53π5)\left( {\cos \dfrac{{53\pi }}{5}} \right) is −sin⁡π10 - \sin \dfrac{\pi }{{10}}
So, sin⁡−1(cos⁡53π5){\sin ^{ - 1}}\left( {\cos \dfrac{{53\pi }}{5}} \right)=sin⁡−1(−sin⁡π10){\sin ^{ - 1}}\left( { - \sin \dfrac{\pi }{{10}}} \right)=−π10 - \dfrac{\pi }{{10}}
As we know(sin⁡−1sin⁡a=a{\sin ^{ - 1}}\sin a = a)
Hence the correct option is (D).

Note : In these types of problems of finding value of trigonometry we have to use the quadrant rule of finding angle and also use some of the properties of inverse trigonometric functions as shown above.