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$ e $ and Trigonometry

The Taylor series of $ e^x $ is

                $ e^x = \sum^{\infty}_{n=0}{\frac{x^n}{n!}} = 1 + x + \frac{x^2}2 + \frac{x^3}6 + \cdots $

Using this equation, it is possible to relate $ e $ to the seemingly unrelated worlds of trigonometry and the complex numbers by simply plugging in a complex number, $ ix $ for example. This yields:

                $ \begin{align} e^ix &= \sum^{\infty}_{n=0}{\frac{(ix)^n}{n!}}\\ &= \sum^{\infty}_{n=0}{\frac{i^nx^n}{n!}}\\ &= 1 + ix - \frac{x^2}{2!} - i\frac{x^3}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!}\cdots \end{align} $

But by rearranging this, one gets the identity

                $ e^ix = \sum^{\infty}_{n=0}{\frac{(-x)^{2n}}{(2n)!}} + i\sum^{\infty}_{n=0}{\frac{(-x)^{2n+1}}{(2n+1)!}} $


References:
(Reference 1)
(Reference 2)

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