I am trying to understand this derivation provided in another answer:$$s'_n(\theta) = {\bf Re}(e^{i\theta}+e^{i\theta}+...+e^{ni\theta}) \\ = \dfrac{-1+\cos\theta+\cos n\theta-\cos(n+1)\theta}{2(1-\cos\theta)} \\ = \dfrac{-2\sin^2\theta/2+2\sin(n+\theta/2)\sin\theta/2}{4\sin^2\theta/2} \\ = \frac{\cos(n+1)\varphi\sin n\varphi}{\sin\varphi}$$
Note:$\frac{\theta}{2}=\varphi$
I am stuck on the following point:
I used $\cos\theta=\cos^2(\frac{\theta}{2})-\sin^2(\frac{\theta}{2})=1-2\sin^2{\frac{\theta}{2}}$
After repacing I got:$\dfrac{-1+\cos\theta+\cos n\theta-\cos(n+1)\theta}{2(1-\cos\theta)}=\dfrac{-2\sin^2(\frac{\theta}{2})-2\sin^2(\frac{n\theta}{2})+2\sin^2(\frac{(n+1)\theta}{2})}{4\sin^2(\frac{\theta}{2})} $
However, after some search I did not find any identity that replaced would yield me the expression $$\frac{\cos(n+1)\varphi\sin n\varphi}{\sin\varphi}.$$
Question:
How can I go from $\dfrac{-2\sin^2(\frac{\theta}{2})-2\sin^2(\frac{n\theta}{2})+2\sin^2(\frac{(n+1)\theta}{2})}{4\sin^2(\frac{\theta}{2})}$ to $\dfrac{\cos(n+1)\vartheta\sin n\varphi}{\sin\varphi}$
Thanks in advance!
Without considering real or imaginary versions it can be seen that: \begin{align} \sum_{k=1}^{n} e^{i k \theta} &= \frac{e^{i \theta} (1 - e^{i n \theta})}{1 - e^{i \theta}} = \frac{e^{i \theta} (1 - e^{- i \theta}) (1 - e^{i n \theta})}{(1 - e^{i \theta})(1 - e^{- i \theta})} \\ &= \frac{(e^{i \theta} - 1)(1 - e^{i n \theta})}{2(1- \cos\theta)} \\ &= - \frac{e^{i(n+1)\theta/2} (e^{i \theta/2} - e^{- i \theta/2})(e^{i n\theta/2} - e^{- i n\theta/2})}{4 \sin^{2}(\theta/2)} \\ &= - \frac{(2 i)^2 \sin(\theta/2) \sin(n \theta/2)}{4 \sin^{2}(\theta/2)} \, e^{i (n+1)\theta/2} \\ &= \frac{\sin(n \theta/2)}{\sin(\theta/2)} \, e^{i (n+1)\theta/2}. \end{align} Now take the real and imaginary components to obtain \begin{align} Re\left(\sum_{k=1}^{n} e^{i k \theta}\right) &= \frac{\sin(n \theta/2) \cos((n+1)\theta/2)}{\sin(\theta/2)} \\ Im\left( \sum_{k=1}^{n} e^{i k \theta} \right) &= \frac{\sin(n \theta/2) \sin((n+1)\theta/2)}{\sin(\theta/2)} \end{align}
For the case of taking the Re component first: \begin{align} Re\left(\sum_{k=1}^{n} e^{i k \theta} \right) &= Re\left( \frac{(e^{i \theta} - 1)(1 - e^{i n \theta})}{2(1- \cos\theta)} \right) \\ &= \frac{\cos\theta - \cos(n+1)\theta + \cos n\theta -1}{2(1-\cos\theta)} \\ &= \frac{\cos(n \theta) - \cos((n+1)\theta) - 2 \sin^{2}(\theta/2)}{4 \sin^{2}(\theta/2)} \\ &= \frac{\sin(n\theta + \theta/2) - \sin(\theta/2)}{2 \sin(\theta/2)} \\ &= \frac{\sin(n\theta/2) \cos((n+1)\theta/2)}{\sin(\theta/2)}, \end{align} where \begin{align} \cos\theta - \cos\phi &= - 2 \sin\left(\frac{\theta + \phi}{2}\right) \sin\left(\frac{\theta - \phi}{2}\right) \\ \sin\theta - \sin\phi &= 2 \sin\left(\frac{\theta - \phi}{2}\right) \cos\left(\frac{\theta + \phi}{2}\right) \end{align} was used.