One part of a STEP-question from 1991 is
Prove that
$$1 + m \cos 2\theta + \binom {m} {2}\cos 4\theta + \cdots + \binom {m}{r}\cos 2r\theta + \cdots + \cos 2m\theta ~=~ 2^m \cos^m \theta ~\cos m\theta$$
I have tried a few things, proof by induction being one of the first. I tried adding
$$\sum_{i=0}^{m} \binom{m}{i} \cos 2(i+1)\theta $$
On both sides as an inductive step, making use of $\binom{n}{m} + \binom{n}{m+1} = \binom{n+1}{m+1}$, yielding (after some hopefully correct rearrangement)
$$\sum_{i=0}^{m+1} \binom{m+1}{i} \cos 2i\theta = 2^m \cos^m \theta \cos m\theta + \cos 2\theta \sum_{i=0}^m \binom{m}{i} \cos 2i\theta - \sin 2\theta \sum_{i=0}^m \binom{m}{i} \sin 2i\theta $$
However, it doesn't seem as if this leads to anything. One could go about it in another way, namely fiddle with the equation to make the right-hand side of the equation $2^{m+1} \cos^{m+1} \theta \cos (m+1)\theta$, though I can't quite see how that would be done.
$$\sum_{k=0}^m\binom{m}{k}\cos(2k\theta)=Re\sum_{k=0}^m\binom{m}{k}e^{2ik\theta}=Re(1+e^{2i\theta})^m=Re\left( 2^m\left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right)^me^{im\theta}\right)=2^m\cos^m\theta\cos m\theta$$