# Prove the following$3\sin^{-1}x=\sin^{-1}(3x-4x^3)\;x\in\bigg[-\frac{1}{2},\frac{1}{2}\bigg]$

Toolbox:
• $$sin\: 3A=3\: sinA-4sin^3A$$
• $$sin^{-1}sinx=x\: \: \: \: if\: x \: \in\: \bigg[ \large\frac{-\pi}{2} \large\frac{\pi}{2} \bigg]$$
Let $sin^{-1}x = A, \rightarrow x = sin A$.
R.H.S.: $sin^{-1}(3x - 4x^3)$
Substituting for $x = sinA$, R.H.S. = $$sin^{-1} \bigg[ 3sinA-4sin^3A \bigg]$$
Substituting the value of $$sin\: 3A=3\: sinA-4sin^3A$$, we get
R.H.S.:$sin^{-1}(sin 3A) \rightarrow 3A$.
L.H.S. $3 sin^{-1} x$
Substituting the value of $sin^{-1}x = A$, L.H.S.= $3A$.
Therefore, L.H.S. = R.H.S.

answered Feb 22, 2013
edited Mar 15, 2013