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 equation \[ 2 \cos^{-1} x = \sin^{-1} \left( 2 \sqrt{1 - x^2} \right) \] is valid for all values of \(x\) satisfying:
The elementary properties of inverse trigonometric functions will help to solve problems. Here are a few important properties related to inverse trigonometric functions:
Tan−1x + Tan−1y = π + tan−1 (x+y/ 1-xy), if xy > 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = -π + tan−1 (x+y/ 1-xy), if xy > 1
= x, if x∈[−π/2, π/2]
= π−x, if x∈[π/2, 3π/2]
=−2π+x, if x∈[3π/2, 5π/2] And so on.
= −x, ∈[−π,0]
= x, ∈[0,π]
= 2π−x, ∈[π,2π]
=−2π+x, ∈[2π,3π]
= x, (−π/2, π/2)
= x−π, (π/2, 3π/2)
= x−2π, (3π/2, 5π/2)