No worries - I was going to make you work it out for yourself since we are not supposed to do your work for you.
Jahaan spoiled my cunning plan - need more weasels!

Euler relation -> trig.
Also work out [itex]e^{-i\theta}[/itex] as per the Euler relation.
Gives you the other half of a pair of simultaneous equations that you can solve for sin or cos in terms of complex exponentials.

To see how we get the math to look so pretty, just click "quote" at the bottom of a post and compare the stuff in the "tex" or "itex" tags with what you see.

Start with [itex]e^{i\theta} = \cos\theta + i\sin\theta[/itex] and [itex]e^{-i\theta} = \cos\theta - i\sin\theta[/itex]

Then you add the equations and get[tex]e^{i\theta} + e^{-i\theta} = 2\cos\theta \Longrightarrow \cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}[/tex]

That's really really helpful thanks for clearing this up. Revising for a maths retake at the moment, just done my first year of a physics degree. Think i will be visiting this forum a lot more!

The Taylor-series representation of ##e^z## is a very powerful trick for this sort of thing, especially when combined with the formulas other people mentioned.

This series converges for any real or complex ##z##. It can also be used to define trig functions and hyperbolic trig functions for real or complex numbers:

##\cosh(z)## = the even terms of ##\exp(z)## (including 1 as the 0th term)
##\sinh(z)## = the odd terms of ##\exp(z)##

##\cos(z)## = the even terms of ##\exp(\imath z)## (including 1 as the 0th term)
##\imath \sin(z)## = the odd terms of ##\exp(\imath z)##

For example, ##\cos(\imath)## = the even terms of ##\exp(\imath^2)##
##
= 1 + \tfrac{1}{2}(-1)^2 + \tfrac{1}{4!}(-1)^4 + \cdots
= 1 + \tfrac{1}{2} + \tfrac{1}{4!} + \cdots = \cosh(1)
##