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

Toolbox:
• $cos3A=4cos^3A-3cosA$
• $cos^{-1}cosx=x \: \: \: if\: x \in [ 0\: \pi ]$
Let $x = cosA$ $\Rightarrow A = cos^{-1}x$.

Given, R.H.S.: = $cos^{-1}(4x^3-3x)$, substituting for $x = cosA$, we get:
R.H.S.: = $cos^{-1} [ 4\: cos^3A-3\: cosA]$

Substituting $cos3A=4cos^3A-3cosA$, we get:

R.H.S. = $= cos^{-1}\: cos3A$ = 3A.

Substituting for $A = cos^{-1}x$, we get:

R.H.S. = $3 cos^{-1}x =$ L.H.S.

edited Mar 15, 2013