Find the values of \(\sin^{-1}(\sin^2\frac{\pi}{3})\)
\(\sin^{-1}(\sin^2\frac{\pi}{3})\)
We know that \(\sin^{-1}\)(sin x) = x if x∈ \([\frac{\pi}{2},\frac{\pi}{2}]\), which is the principal value branch of \(\sin^{-1}\) x.
Here, \(2 \frac{\pi}{3}\) ∉ [\(-\frac{\pi}{2},\frac{\pi}{2}\) ]
Now,\(\sin^{-1}(\sin^2\frac{\pi}{3})\)can be written as:
\(\sin^{-1}(\sin^2\frac{\pi}{3})=\sin^{-1}[(\sin\frac{\pi-2\pi}{3})]=\sin^{-1}(sin\frac{\pi}{3})\) where \(\frac{\pi}{3}\in[-\frac{\pi}{2},\frac{\pi}{2}]\)
\(\therefore \sin^{-1}(\sin^2\frac{\pi}{3}=\sin^{-1}(\sin\frac{\pi}{3})=\frac{\pi}{3}\)
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)