Moivre formula

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

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