\(3cos^{-1x}=cos^{-1}(4x^3-3x)\,x\in\bigg[-\frac{1}{2},1\bigg]\) prove.
\(3cos^{-1x}=cos^{-1}(4x^3-3x)\,x\in[-\frac{1}{2},1]\)
To prove:
Let x = cosθ. Then, \(cos^{-1}x=\theta\).
We have,
RHS= \(cos^{-1}(4x^3-3x)\)
= \(cos^{-1}(4cos2\theta-3cos\theta)\)
= \(cos^{-1}(cos3\theta)\)
= 3θ
= \(3cos^{-1x}\)
\(=L.H.S.\)
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)