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Mathematics Question on Properties of Inverse Trigonometric Functions

The value of sin⁡−1(cos⁡ 53π5)\sin^{-1}\left ( \cos\, \frac { 53 \pi}{5}\right) is

A

3π5\frac {3 \pi}{5}

B

−3π5\frac {-3 \pi}{5}

C

π10\frac {\pi}{10}

D

−π10\frac {- \pi}{10}

Answer

−π10\frac {- \pi}{10}

Explanation

Solution

The correct answer is D:−π10\frac{-\pi}{10}
We have, sin⁡−1(cos⁡53π5)=sin⁡−1[cos⁡(10π+3π5)]\sin ^{-1}\left(\cos \frac{53 \pi}{5}\right)=\sin ^{-1}\left[\cos \left(10 \pi+\frac{3 \pi}{5}\right)\right]
=sin⁡−1[cos⁡(3π5)]=\sin ^{-1}\left[\cos \left(\frac{3 \pi}{5}\right)\right]
[∵cos⁡(2nπ±θ)=cos⁡θ][\because \cos (2 n \pi \pm \theta)=\cos \theta]
=sin⁡−1[sin⁡(π2−3π5)]=\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-\frac{3 \pi}{5}\right)\right]
=sin⁡−1[sin⁡(5π−6π10)]=\sin ^{-1}\left[\sin \left(\frac{5 \pi-6 \pi}{10}\right)\right]
=sin⁡−1[sin⁡(−π10)]=−π10=\sin ^{-1}\left[\sin \left(\frac{-\pi}{10}\right)\right]=\frac{-\pi}{10}