1 + cos 2x + cos 4x + cos 6x = | vedclass

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Question

(C) Given expression: $1 + \cos 2x + \cos 4x + \cos 6x$Group the terms to use the sum-to-product formulas:$= (1 + \cos 6x) + (\cos 2x + \cos 4x)$Using

Vedclass · Accepted Answer

$4\cos x \cos 2x \cos 3x$

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(C) Given expression: $1 + \cos 2x + \cos 4x + \cos 6x$Group the terms to use the sum-to-product formulas:$= (1 + \cos 6x) + (\cos 2x + \cos 4x)$Using the identities $1 + \cos 2	heta = 2\cos^2 	heta$ and $\cos A + \cos B = 2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$:$= 2\cos^2 3x + 2\cos(\frac{4x+2x}{2})\cos(\frac{4x-2x}{2})$$= 2\cos^2 3x + 2\cos 3x \cos x$Factor out $2\cos 3x$:$= 2\cos 3x (\cos 3x + \cos x)$Apply the sum-to-product formula again for $(\cos 3x + \cos x)$:$= 2\cos 3x [2\cos(\frac{3x+x}{2})\cos(\frac{3x-x}{2})]$$= 2\cos 3x [2\cos 2x \cos x]$$= 4\cos x \cos 2x \cos 3x$.

$1 + \cos 2x + \cos 4x + \cos 6x = $

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