About a binomial expansion of complex numbers

Question

Prove that $$1+{n \choose 1}\cos x + {n \choose 2}\cos 2x+... \cos nx=(2 \cos\frac{x}{2})^n(\cos\frac{nx}{2})$$ given that $$(1+\cos x+i\sin x)^n=(2\cos\frac{x}{2})^n(\cos\frac{nx}{2}+i\sin\frac{nx}{2})$$

I'm not so sure how to even start with this, any hints?

Answer 1

Hint: We have $$\sum_{k=0}^{n}\dbinom{n}{k}e^{kix}=\left(1+e^{ix}\right)^{n}=\left(1+\cos\left(x\right)+i\sin\left(x\right)\right)^{n} $$ then $$\sum_{k=0}^{n}\dbinom{n}{k}\cos\left(kx\right)=\textrm{Re}\left(\sum_{k=0}^{n}\dbinom{n}{k}e^{kix}\right)=\textrm{Re}\left(\left(1+e^{ix}\right)^{n}\right) $$ now use the given identity.

Answer 2

HINT...Consider the real part of the binomial expansion of $(1+\cos x+i\sin x)^n=(1+e^{ix})^n$