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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.

Sam
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    Start with $$ \sum\limits_{r = 0}^n {\binom{n}{r}\cos (r\theta )} = \Re \sum\limits_{r = 0}^n {\binom{n}{r}e^{ir\theta } } = \Re (1 + e^{i\theta } )^n = \Re (1 + \cos \theta + i\sin \theta )^n . $$ – Gary Nov 03 '20 at 09:15
  • The result given by Mathematica is the same as yours $$\sum _{r=0}^n \binom{n}{r} \cos (r t)=2^n \cos ^n\left(\frac{t}{2}\right) \cos \left(\frac{n t}{2}\right)$$ I don't believe it can be simplified further. ($t$ is $\theta$) – Raffaele Nov 03 '20 at 09:17

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

Shubham Johri
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  • thank you! we did not cover binomial theorem in my methods class and i have only briefly touched on it in my own time so i never thought to use it – Sam Nov 03 '20 at 10:34
  • @Sam You are welcome, if this answered your question satisfactorily consider accepting my answer by clicking the tick mark button next to it. – Shubham Johri Nov 03 '20 at 10:51