Question: In a paper I came across an expansion like this:
$$\cos(m\theta) = C\_m^0\cos^m(\theta) - C\_m^2\cos^{m-2}(\theta)(1-\cos^2(\theta)) + C\_m^4\cos^{m-4}(\theta)(1-\cos^2(\theta))^2 + ... 
 (-1)^nC\_m^{2n}\cos^{m-2n}(\theta)(1-\cos^2(\theta))^n + ...$$

What kind of expansion is this and how do you calculate those $C\_m^i$'s?

Answer: 
The coefficients are binomial coefficients
$$C^k\_m=\frac{m!}{k!\,(m-k)!}\ ,$$
though you should **note** that the notation used is slightly unusual, many people (AFAIK most people) write
$$C^m\_k\quad\hbox{or}\quad {}^mC\_k\quad\hbox{or}\quad \binom mk\ .$$
The formula comes from de Moivre's Theorem and the binomial expansion,
$$\eqalign{\cos(m\theta)
 &=\Re(\cos m\theta+i\sin m\theta)\cr
 &=\Re((\cos\theta+i\sin\theta)^m)\cr
 &=\Re\biggl(\cos^m\theta+\binom m1\cos^{m-1}\theta(i\sin\theta)
 +\binom m2\cos^{m-2}\theta(i\sin\theta)^2+\cdots\biggr)\cr
 &=\cos^m\theta-\binom m2\cos^{m-2}\theta\sin^2\theta+\cdots\cr
 &=\cos^m\theta-\binom m2\cos^{m-2}\theta(1-\cos^2\theta)+\cdots\ .\cr}$$
**Hint:** Use [De Moivre's formula](https://en.wikipedia.org/wiki/De_Moivre%27s_formula)
$$(\cos\theta+i\sin\theta)^m=\cos m\theta+i\sin m\theta$$
write the expansion of the left side and then pick up the real part of both sides. You will find the desired formula.