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Mathematics Question on Inverse Trigonometric Functions

Find the value of tan⁡−1[2cos⁡(2sin⁡−112)]\tan ^{-1}\bigg[2\cos\Big(2\sin^{-1}\frac{1}{2}\Big)\bigg]

Answer

tan⁡−1[2cos⁡(2sin⁡−112)]\tan ^{-1}\bigg[2\cos\Big(2\sin^{-1}\frac{1}{2}\Big)\bigg]
Let sin⁡−112=x.\sin^{-1}\frac{1}{2}=x.

Then, sin⁡x=12=sin⁡(π6)\sin x=\frac{1}{2}=\sin(\frac{\pi}{6})

∴sin⁡−112=π6\therefore\sin^{-1}\frac{1}{2}=\frac{\pi}{6}

tan⁡−1[2cos⁡(2sin⁡−112)]=tan⁡−1[2cos⁡(2xπ6)]\tan ^{-1}\bigg[2\cos\Big(2\sin^{-1}\frac{1}{2}\Big)\bigg]=\tan^{-1}\bigg[2\cos\Big(\frac{2x\pi}{6}\Big)\bigg]

=tan⁡−1[2cos⁡π3]=tan⁡−1[2∗12]\tan^{-1}\bigg[2\cos\frac{\pi}{3}\bigg]=\tan^{-1}\bigg[2*\frac{1}{2}\bigg]

=tan⁡−11=π4\tan^{-1}1=\frac{\pi}{4}