What kind of expansion is this?

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 1

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, $$\eqalign{\cos(m\theta) &=\Re(\cos m\theta+i\sin m\theta)\cr &=\Re((\cos\theta+i\sin\theta)^m)\cr &=\Re\biggl(\cos^m\theta+\binom m1\cos^{m-1}\theta(i\sin\theta) +\binom m2\cos^{m-2}\theta(i\sin\theta)^2+\cdots\biggr)\cr &=\cos^m\theta-\binom m2\cos^{m-2}\theta\sin^2\theta+\cdots\cr &=\cos^m\theta-\binom m2\cos^{m-2}\theta(1-\cos^2\theta)+\cdots\ .\cr}$$

Answer 2

Hint: Use De Moivre's 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.