Prove that cos^2 x + cos^2 3x + cos^2 5x + ... to n terms = (1/2)[n + ((sin4nx)/(2sin2x))]


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Question Number 13658 by Tinkutara last updated on 22/May/17

$Prove that$ $\cos^{2} x + \cos^{2} 3 x + \cos^{2} 5 x + . . . to n terms$ $= \frac{1}{2} [n + \frac{\sin 4 n x}{2 \sin 2 x}]$
	Answered by ajfour last updated on 22/May/17

	$2 S = (1 + \cos 2 x) + (1 + \cos 6 x)$ $+ (1 + \cos 10 z) + . . . . + (1 + \cos [(4 n - 2) x])$ $2 S = n + \frac{1}{2 \sin 2 x} {2 \cos 2 x \sin 2 x + 2 \cos 6 x \sin 2 x + 2 \cos 10 x \sin 2 x + . . . .$ $. . . . . + 2 \cos [4 (n - 2) x] \sin 2 x}$ $2 S = n + \frac{1}{2 \sin 2 x} {\sin 4 x +$ $\sin 8 x - \sin 4 x$ $+ \sin 12 x - \sin 8 x$ $+ . . . . . . . . . - . . . . . .$ $+ \sin [(4 n - 4) x] - \sin [(4 n - 8) x]$ $+ \sin 4 n x - \sin [(4 n - 4) x]}$ $2 S = n + \frac{\sin 4 n x}{2 \sin 2 x} .$ $S = \frac{1}{2} [n + \frac{\sin 4 n x}{2 \sin 2 x}] .$
	Commented by Tinkutara last updated on 23/May/17

	$Thanks!$
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