Find the value of \(\tan ^{-1}\bigg[2\cos\Big(2\sin^{-1}\frac{1}{2}\Big)\bigg]\)
\(\tan ^{-1}\bigg[2\cos\Big(2\sin^{-1}\frac{1}{2}\Big)\bigg]\)
Let \(\sin^{-1}\frac{1}{2}=x.\)
Then, \(\sin x=\frac{1}{2}=\sin(\frac{\pi}{6})\)
\(\therefore\sin^{-1}\frac{1}{2}=\frac{\pi}{6}\)
\(\tan ^{-1}\bigg[2\cos\Big(2\sin^{-1}\frac{1}{2}\Big)\bigg]=\tan^{-1}\bigg[2\cos\Big(\frac{2x\pi}{6}\Big)\bigg]\)
=\(\tan^{-1}\bigg[2\cos\frac{\pi}{3}\bigg]=\tan^{-1}\bigg[2*\frac{1}{2}\bigg]\)
=\(\tan^{-1}1=\frac{\pi}{4}\)
The elementary properties of inverse trigonometric functions will help to solve problems. Here are a few important properties related to inverse trigonometric functions:
Tanβ1x + Tanβ1y = Ο + tanβ1 (x+y/ 1-xy), if xy > 1
Tanβ1x + Tanβ1y = tanβ1 (x+y/ 1-xy), if xy < 1
Tanβ1x + Tanβ1y = tanβ1 (x+y/ 1-xy), if xy < 1
Tanβ1x + Tanβ1y = -Ο + tanβ1 (x+y/ 1-xy), if xy > 1
= x, if xβ[βΟ/2, Ο/2]
= Οβx, if xβ[Ο/2, 3Ο/2]
=β2Ο+x, if xβ[3Ο/2, 5Ο/2] And so on.
= βx, β[βΟ,0]
= x, β[0,Ο]
= 2Οβx, β[Ο,2Ο]
=β2Ο+x, β[2Ο,3Ο]
= x, (βΟ/2, Ο/2)
= xβΟ, (Ο/2, 3Ο/2)
= xβ2Ο, (3Ο/2, 5Ο/2)
