e^{ix} = cos(x) + *i*sin(x)

Here is a proof of a trigonometric identity commonly used in representing complex numbers; it provides a simple relationship between polar and exponential form.
So here goes...

We know the following Taylor Series:

x^{2} x^{3} x^{4}
e^{x} = 1 + x + ---- + ---- + ---- + ... ...
2! 3! 4!
x^{3} x^{5}
sin(x) = x - ---- + ---- - ... ...
3! 5!
x^{4} x^{6}
cos(x) = 1 - ---- + ---- - ... ...
4! 6!

And we know that the imaginary number *i* represents the square root of -1, so that *i*^{2} = -1.

So if we raise **e** to the power *i*x:

(*i*x)^{2} (*i*x)^{3} (*i*x)^{4}
e^{ix} = 1 + ix + ------- + ------- + ------- + ... ...
2! 3! 4!

Seperating out the powers gives:

*i*^{2}x^{2} *i*^{3}x^{3} *i*^{4}x^{4}
e^{ix} = 1 + ix + ------ + ------ + ------ + ... ...
2! 3! 4!

Evaluating the powers of *i* simplifies things a bit:

x^{2} *i*x^{3} x^{4}
e^{ix} = 1 + *i*x - ---- - ----- + ---- + ... ...
2! 3! 4!

And finally, seperate the odd- and even- powered terms, and factorize out the *i*, to give:

/ x^{2} x^{4} \ / x^{3} x^{5} \
e^{ix} = | 1 - ---- + ---- - ... ... | + *i* | x - ---- + ---- ... ... |
\ 2! 4! / \ 3! 5! /

Which, given the Taylor Identities given above, is equal to:

e^{ix} = cos(x) + *i*sin(x)

This has been an essay in the craft of the `<PRE>` tag.