Euler's Forumla
$ e^{ix} = \cos x + i * \sin x $
Proof
Using Taylor Series:
$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $
$ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots $
$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots $
$ e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \cdots $
Expanding the complex terms yield:
$ e^{ix} = 1 + ix - \frac{x^2}{2!} - \frac{(ix)^3}{3!} + \frac{x^4}{4!} + \frac{(ix)^5}{5!} - \cdots $
Rearranging the terms into even and odd exponents and factoring out the imaginary number:
$ e^{ix} = (1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots) + i(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots) $
Which is equivalent to:
$ e^{ix} = \cos x + i\sin x $
