^2 2 x- ^2 6 x= 4 x 8 x - Vidyadip

Question

$\cos ^2 2 x-\cos ^2 6 x=\sin 4 x \sin 8 x$

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Answer

$\begin{aligned}
\text { L.H.S. } & =\cos ^2 2 x-\cos ^2 6 x \\
& =(\cos 2 x)^2-(\cos 6 x)^2 \\
& =(\cos 2 x+\cos 6 x)(\cos 2 x-\cos 6 x) \\
& =\left[2 \cos \left(\frac{2 x+6 x}{2}\right) \cos \left(\frac{2 x-6 x}{2}\right)\right] \\
& \cdot\left[2 \sin \left(\frac{2 x+6 x}{2}\right) \sin \left(\frac{6 x-2 x}{2}\right)\right] \\
& =[2 \cos 4 x \cos (-2 x)][2 \sin 4 x \sin 2 x] \\
& =(2 \cos 4 x \cos 2 x)(2 \sin 4 x \sin 2 x) \\
& =(2 \sin 2 x \cos 2 x)(2 \sin 4 x \cos 4 x) \\
& =\sin 4 x \sin 8 x \\
& =\text { R.H.S.}
\end{aligned}$

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