It depends where you want to start.
If you know the Taylor series for #e^z#, #sin theta# and #cos theta#, together with the basic properties of #i = sqrt(-1)#, then you can easily find that:
#e^(itheta) = cos theta + i sin theta#
Then:
#cos 2theta + isin 2theta = e^(2itheta) = (e^(itheta))^2#
#= (cos theta + isin theta)^2#
#= cos^2theta + 2i cos theta sin theta + i^2 sin^2 theta#
#= cos^2theta + 2i cos theta sin theta - sin^2 theta#
#= (cos^2theta - sin^2 theta) + i (2cos theta sin theta)#
Comparing the real and imaginary parts we get two formulae for the price of one:
#cos 2theta = cos^2 theta - sin^2 theta#
#sin 2theta = 2cos theta sin theta#
