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Suppose $\sin \frac{\theta}{2} \neq 0$ . Prove that

$$\frac{1}{2}+ \sum_{k=1}^n \cos (k\theta) = \frac{\sin[(n+ \frac{1}{2})\theta]}{2\sin\frac{\theta}{2}}$$

The question also give the hint, $$z=\cos\theta + i\sin\theta = e^{i\theta}$$ $$\sum_{k=1}^n z^k = z + z^2 + \cdots + z^n = \frac{z(z^n-1)}{z-1}$$

What I did was to change z into polar form and applied double angle formula to remove $\cos\theta$, but I have no idea what to do after that and it is still $\sum_{k=1}^n z^k = \sum_{k=1}^n \cos(k\theta)+i\sin(k\theta)$, not exactly $\sum_{k=1}^n \cos(k\theta)$ that I am supposed to prove

Michael Hardy
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Steve
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    https://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro, https://math.stackexchange.com/questions/773131/prove-that-sin-frac-pi14-is-a-root-of-8x3-4x2-4x-1-0/773960#773960 and many other questions – Jean-Claude Arbaut Aug 18 '18 at 18:11

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\begin{equation} \cos k \theta = \frac{e^{jk\theta} + e^{-jk\theta}}{2} \end{equation} Let's use $\sum\limits_{k=0}^{N-1 }r^k= \frac{1-r^N}{1-r} $ \begin{equation} \sum \cos k \theta = \frac{1}{2} \sum\limits_{k=1}^{N} e^{jk\theta} + \frac{1}{2} \sum\limits_{k=1}^{N} e^{-jk\theta} = \frac{1}{2} \big( \frac{1 - e^{-j (N+1) \theta}}{1 - e^{j \theta}}-1 + \frac{1 - e^{j (N+1) \theta}}{1 - e^{-j \theta}}-1 \big) \end{equation} So \begin{equation} \frac{1}{2} + \sum \cos k \theta = \frac{1}{2} \big( -1+ \frac{1 - e^{j (N+1) \theta}}{1 - e^{j \theta}} + \frac{1 - e^{-j (N+1) \theta}}{1 - e^{-j \theta}} \big) \end{equation} So \begin{equation} \frac{1}{2} + \sum \cos k \theta = \frac{1}{2} \big( \frac{-(1-e^{j \theta})(1-e^{-j \theta}) + (1 - e^{j (N+1) \theta})(1-e^{-j \theta}) + (1 - e^{-j (N+1) \theta})(1-e^{j \theta}) } {(1-e^{j \theta})(1-e^{-j \theta})} \big) \end{equation} that is \begin{equation} \frac{1}{2} + \sum \cos k \theta = \frac{1}{2} \big( \frac{ -2 + e^{j \theta} + e^{- j\theta} + 1 - e^{-j \theta} - e^{j (N+1) \theta} + e^{j N \theta} + 1 - e^{j \theta} - e^{- j (N+1) \theta }+ e^{-jN \theta} } {(1-e^{j \theta})(1-e^{-j \theta})} \big) \end{equation} that is \begin{equation} \frac{1}{2} + \sum \cos k \theta = \frac{1}{2} \big( \frac{ - (e^{j (N+1) \theta} +e^{- j (N+1) \theta })+ (e^{j N \theta} + e^{-jN \theta}) } {(1-e^{j \theta})(1-e^{-j \theta})} \big) \end{equation} i.e. \begin{equation} \frac{1}{2} + \sum \cos k \theta = \frac{1}{2} \big( \frac{ - 2 \cos (N+1) \theta+ 2 \cos N \theta } {4 \sin^2 \frac{\theta}{2}} \big) \end{equation} Using $\cos a - \cos b = 2 \sin \frac{1}{2}(a+b) \sin \frac{1}{2} (b-a)$, we get \begin{equation} \frac{1}{2} + \sum \cos k \theta = \frac{1}{2} \big( \frac{ 4 \sin \frac{ \theta}{2} \sin \frac{ \theta}{2} (2N + 1) } {4 \sin^2 \frac{\theta}{2}} \big) = \frac{1}{2} \frac{\sin \frac{ \theta}{2} (2N + 1) }{\sin \frac{ \theta}{2}} \end{equation}

Ahmad Bazzi
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