2cos x - cos 3x - cos 5x = | vedclass

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(A) Given expression: $2\cos x - (\cos 5x + \cos 3x)$Using the sum-to-product formula $\cos C + \cos D = 2\cos\left(\frac{C+D}{2}\right)\cos\left(\fra

Vedclass · Accepted Answer

$16\cos^3 x \sin^2 x$

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(A) Given expression: $2\cos x - (\cos 5x + \cos 3x)$Using the sum-to-product formula $\cos C + \cos D = 2\cos\left(\frac{C+D}{2}ight)\cos\left(\frac{C-D}{2}ight)$:$\cos 5x + \cos 3x = 2\cos\left(\frac{5x+3x}{2}ight)\cos\left(\frac{5x-3x}{2}ight) = 2\cos 4x \cos x$Substituting this back into the expression:$2\cos x - 2\cos 4x \cos x = 2\cos x(1 - \cos 4x)$Using the identity $1 - \cos 2	heta = 2\sin^2 	heta$,where $2	heta = 4x$ (so $	heta = 2x$):$2\cos x(2\sin^2 2x) = 4\cos x \sin^2 2x$Using the double angle identity $\sin 2x = 2\sin x \cos x$:$4\cos x (2\sin x \cos x)^2 = 4\cos x (4\sin^2 x \cos^2 x) = 16\sin^2 x \cos^3 x$.

$2\cos x - \cos 3x - \cos 5x = $

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