$\cos(c) = \cos(a)\cos(b) + \sin(a)\sin(b)\cos(C)$
Prove that if $a$, $b$, and $c$ is approximately $0$, then $c^2 = a^2 + b^2 - 2ab~\cos(C)$.
I wasn't sure how to prove this. One thought I had was to use the series definition of cosine and sine, and go from there.
When you see "approximately zero" you should think Taylor series-they are wonderful in the neighborhood of a solution. Plug in the Taylor series up to your favorite order. To zero order, $\cos=1, \sin=0$ and it is easy. Try for the next orders and see where it works and where it doesn't.