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Question

Mathematics Question on Inverse Trigonometric Functions

The value of sin⁡−1(223)+sin⁡−1(13)\sin ^{-1} (\frac {2 \sqrt {2}}{3})+\sin^{-1} (\frac {1}{3}) is equal to

A

π6\frac {\pi} {6}

B

π2\frac {\pi} {2}

C

π4\frac {\pi} {4}

D

2π4\frac {2\pi} {4}

Answer

π2\frac {\pi} {2}

Explanation

Solution

sin⁡−1(223)+sin⁡−113\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\sin ^{-1} \frac{1}{3}
=sin⁡−1(223)+cos⁡−11−(13)2=\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\cos ^{-1} \sqrt{1-\left(\frac{1}{3}\right)^{2}}
=sin⁡−1(223)+cos⁡−1(223)=\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\cos ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)
=π2[∵sin⁡−1x+cos⁡−1x=π2]=\frac{\pi}{2} \left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]