Question

Найдите сумму cos^2 x + cos^2 2x + cos^2 3x + ⋯ + cos^2 nx

Answer 1

$S = \cos^2x + \cos^22x + \cos^23x + \cdots + \cos^2nx$

$\cos^2(\alpha) = \frac{1+\cos 2\alpha}{2}$

$S = \frac{1+\cos2x}{2} + \frac{1+\cos4x}{2} + \frac{1+\cos6x}{2} +\cdots +\frac{1+\cos(2nx)}{2} =$

$= \frac{1}{2}\cdot(n+S_1)$

$S_1 = \cos(2x)+\cos(4x)+\cos(6x)+\cdots +\cos(2nx) =$

$= \frac{1}{2\sin(x)}\cdot(2\sin(x)\cos(2x) + 2\sin(x)\cos(4x)+2\sin(x)\cos(6x)+\cdots +$

$+ 2\sin(x)\cos(2nx)) = V$

$2\sin(\alpha)\cos(\beta) = \sin(\alpha+\beta) + \sin(\alpha - \beta) =$

$= \sin(\beta + \alpha) - \sin(\beta - \alpha)$

$V = \frac{1}{2\sin(x)}\cdot( (\sin(2x+x) - \sin(2x - x) )+ (\sin(4x+x) - \sin(4x-x)) +$

$+ (\sin(6x+x) - \sin(6x-x)) + \cdots + (\sin(2nx + x) - \sin(2nx - x)) ) =$

$= \frac{1}{2\sin(x)}\cdot( (\sin(3x) - \sin(x)) + (\sin(5x) - \sin(3x)) +$

$+ (\sin(7x) - \sin(5x)) + \cdots + (\sin((2n+1)x) - \sin((2n-1)x)) ) =$

$= \frac{1}{2\sin(x)}\cdot( \sin((2n+1)x) - \sin(x) ) = W$

$\sin(\alpha) - \sin(\beta) = 2\cdot\sin(\frac{\alpha - \beta}{2})\cdot\cos(\frac{\alpha + \beta}{2})$

$W = \frac{1}{2\sin(x)}\cdot( 2\sin(\frac{2nx+x - x}{2})\cos(\frac{2nx+x+x}{2})) =$

$= \frac{1}{\sin(x)}\cdot( \sin(nx)\cos((n+1)x) ) = \frac{\sin(nx)\cos((n+1)x)}{\sin(x)}$

$S = \frac{1}{2}\cdot(n+\frac{\sin(nx)\cos((n+1)x)}{\sin(x)}) =$

$= \frac{n}{2} + \frac{\sin(nx)\cos((n+1)x)}{2\sin(x)}$