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

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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.

 

answered Feb 22, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
 

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