Prove that: ^32x+3 2x=4( ^6x- ^6x) - Vidyadip

Question

Prove that:
$\cos^32\text{x}+3\cos2\text{x}=4(\cos^6\text{x}-\sin^6\text{x})$

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Answer

$\cos^32\text{x}+3\cos2\text{x}=4(\cos^6\text{x}-\sin^6\text{x})$
$\text{RHS}=4\Big[(\cos^2\theta)^3-(\sin^2\theta)^3\Big]$
$=4(\cos^2\theta-\sin^2\theta)\Big[\cos^4\theta+\sin^4\theta+\sin^2\theta+\cos^2\theta\Big]$
$=4\cos^2\theta\Big[(\cos^2\theta-\sin^2\theta)+2\sin^2\theta+\sin^2\theta+\sin^2\theta+\cos^2\theta\Big]$
$=4\cos^2\theta[\cos^22\theta+3\sin^2\theta\cos^2\theta]$
$=4\cos^2\theta\Big[\cos^22\theta+3\Big(\frac{1-\cos^2\theta}{2}\Big)\Big(\frac{1+\cos^2\theta}{2}\Big)\Big]$
$=4\cos^2\theta\Big(\cos^22\theta+\frac{3}{4}(1-\cos^22\theta)\Big)$
$=\cos^2\theta\big[4\cos^22\theta+3-3\cos^22\theta\big]$
$=\cos^2\theta[\cos^22\theta+3]$
$=\cos^32\theta+3\cos^2\theta\ \text{LHS}$
$\text{LHS=RHS}$

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