In this question the user ask to prove the next identity:
$$1 + 4\cos{2\theta} + 6 \cos{4\theta} +4\cos{6\theta}+ \cos {8\theta} =16\cos{4\theta} \cos^4 \theta$$
I realized the terms in the left side are expression of a more general term:
$$\sum_{k=0}^n {n \choose k} \cos{2k\theta} $$
(the expression of the OP is $n=4$).
Question there is a way to calculate that sum?
Obviously using the binomial theorem you can show:
$$\sum_{k=0}^n {n \choose k} \frac{e^{2ik\theta}+e^{-2ik\theta}}{2}=\frac{1}{2}(1+e^{2i\theta})^n+ \frac{1}{2}(1+e^{-2i\theta})^n$$ $$=\frac{1}{2}e^{2i\theta n}(1+e^{-2i\theta})^n+ \frac{1}{2}(1+e^{-2i\theta})^n$$ $$=\frac{1}{2}(e^{2i\theta n}+1)(1+e^{-2i\theta})^n$$
But how show an expression in terms of $\cos$?
$$(e^{2i\theta n}+1)(1+e^{-2i\theta})^n = (e^{in\theta}+e^{-in\theta})(e^{i\theta}+e^{-i\theta})^n = 2^{n+1}\cos n\theta \cos^n \theta$$