Question

$3sin^{-1}(3x-4x^3), x \in[-\frac {1}{2},\frac{1}{2}]$ prove.

Answer 1

$3\,sin^{-1}(3x-4x^3), x \in[-\frac {1}{2},\frac{1}{2}]$
To prove:
Let x = sinθ. Then, $sin^{-1x}=\theta$
We have,
R.H.S. = $sin^{-1}(3x-4x^3)=sin^{-1}(2sin\theta-4sin3\theta)$
=$sin^{-1}(sin3\theta)$
= 3θ
= $3\,sin^{-1x}$
= L.H.S