Find $\displaystyle\sum_{r=0}^n\binom nr\cos(r\theta)$using$1+\cos(\theta)+i\sin(\theta)=2\cos(\frac\theta2)(\cos(\frac\theta2)+i\sin(\frac\theta2))$

Question

I have proven the identity shown as the textbook requests, and made attempts on manipulating the identity. The most I have procured is $$(1+(\cos(\theta)+i\sin(\theta))^n=2^n\cos^n(\frac\theta2)(\cos(n\frac \theta2))$$Recognising that $$2^n=\sum_{r=0}^n \binom {n}{r}$$ But this is as far as I can get as I cannot recognise any geometric series or other techniques I could use. The book provides an answer but if someone could provide a nudge in the right direction or processes/techniques I could use it would be greatly appreciated.

Answer 1

$\cos(r\theta)=\mathfrak{Re}(e^{ir\theta})$ so you want to find$$\begin{align*}\mathfrak{Re}\left(\sum_{r=0}^n\binom nre^{ir\theta}\right)&=\mathfrak{Re}[(1+e^{i\theta})^n]\\&=\mathfrak{Re}[(1+\cos\theta+i\sin\theta)^n]\\&=\mathfrak{Re}[2^n\cos^n(\theta/2)e^{in\theta/2}]\\&=2^n\cos^n(\theta/2)\cos(n\theta/2)\end{align*}$$