Moivre formula

Submitted by Structure on Sat, 01/19/2008 - 19:06.

We shall prove

$(\cos x+i\sin x)^n=\cos nx+i\sin nx$

We have

$(\cos x+i\sin x)(\cos y+i\sin y)=\cos x \cos y-\sin x\sin y +i(\sin x\cos y+\sin y \cos x)=$

$=\cos(x+y)+i\sin(x+y)$

Then for x=y we have

$(\cos x+i\sin x)^2=\cos 2x+i\sin 2x$

Now we prove by induction

$(\cos x+i\sin x)^{n+1}=(\cos x+i\sin x)^{n}(\cos x+i\sin x)=(\cos nx+i\sin nx)(\cos x+i\sin x)=$

$\cos (n+1)x+i\sin (n+1)x$

From

$(\cos x+i\sin x)^2=\cos^2x-\sin^2x+2i\sin x\cos x=\cos 2x+i\sin 2x$

we have

$\cos 2x=\cos^2x-\sin^2x\:$

$\sin 2x=2\sin x\cos x$