Sum of trigonometric function with argument in arithmetic progresion

Posted January 19th, 2008 by Structure

Let

$c_n=1+\cos x+\cos 2x+\cos3x..+\cos nx$

and

$s_n=\sin x+\sin 2x+\sin 3x+...+\sin nx$

We have

$c_n+is_n=1+\cos x+\cos 2x+\cos3x..+\cos nx+i(\sin x+\sin 2x+\sin 3x+...+\sin nx)=$

$1+(\cos x+i\sin x)+(\cos 2x++i\sin 2x)+(\cos3x+\sin 3x)+..+(\cos nx+i \sin nx)$

$=1+(\cos x+i\sin x)+(\cos x+i\sin x)^2+(\cos x+i\sin x)^3+...(\cos x+i\sin x)^n=\frac{1-(\cos x+i\sin x)^{n+1}}{1-(\cos x+i\sin x)}=$

$\frac{1-\cos (n+1)x-i\sin (n+1)x}{1-\cos x-i\sin x}=\frac{2\sin^2 \frac{(n+1)x}{2}-2i\sin \frac{(n+1)x}{2}\cos \frac{(n+1)x}{2}}{2\sin^2 \frac{x}{2}-i2\sin \frac{x}{2}\cos \frac{x}{2}}=$

$=\frac{-i^2 2\sin^2 \frac{(n+1)x}{2}-2i\sin \frac{(n+1)x}{2}\cos \frac{(n+1)x}{2}}{-i^22\sin^2 \frac{x}{2}-i2\sin \frac{x}{2}\cos \frac{x}{2}}=\frac{-2i\sin \frac{(n+1)x}{2}(\cos \frac{(n+1)x}{2}+i\sin \frac{(n+1)x}{2})}{-2i\sin \frac{x}{2}(\cos \frac{x}{2}+i\sin \frac{x}{2})}=$