$Re(e^{i\theta}+e^{i\theta}+...+e^{ni\theta})=\frac{\cos(n+1)\vartheta\sin n\varphi}{\sin\varphi}$

Question

For any $0<\theta<\pi$ and the integer $n\geqslant 1$ show that:

$$\sin\theta+\frac{\sin 2\theta}{2}+...+\frac{\sin n\theta}{n}>0$$

Denote by $s_n(\theta)$ the left-hand side of the inequality to be shown. Put $\vartheta=\frac{\theta}{2}$ for brevity.

Since $s'_n(\theta)=Re(e^{i\theta}+e^{i\theta}+...+e^{ni\theta})=\frac{\cos(n+1)\vartheta\sin n\vartheta}{\sin\vartheta}$

I have been trying to understand how the author gets from here $s'_n(\theta)=Re(e^{i\theta}+e^{i\theta}+...+e^{ni\theta})$to this expression$\frac{\cos(n+1)\vartheta\sin n\varphi}{\sin\varphi}$.

I think $s'_n(\theta)=-n\cos(\theta)$ so I thought of using the sum of geometric series $\frac{a-ar^n}{1-r}$(since $0<\theta<\pi$) to obtain $\frac{\cos(n+1)\vartheta\sin n\vartheta}{\sin\vartheta}$. However I got nowhere.

Question:

How did the author derive $\frac{\cos(n+1)\vartheta\sin n\vartheta}{\sin\vartheta}$?

Thanks in advance!

Answer 1

\begin{align} e^{i\theta}+e^{i\theta}+...+e^{ni\theta} &= \dfrac{e^{i\theta}(1-e^{in\theta})}{1-e^{i\theta}} \\ &= \dfrac{e^{i\theta}-e^{i(n+1)\theta}}{1-e^{i\theta}} \\ &= \dfrac{(e^{i\theta}-e^{i(n+1)\theta})(1-e^{-i\theta})}{|1-e^{i\theta}|^2} \\ &= \dfrac{e^{i\theta}-e^{i(n+1)\theta}-1+e^{in\theta}}{|1-e^{i\theta}|^2} \end{align} \begin{align} 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)\vartheta\sin n\varphi}{\sin\varphi} \end{align} where $\varphi=\theta/2$.

Answer 2

Your notation is messed up: you've got two symbols $\vartheta$ and $\varphi$ for the same thing. I'll use $\varphi$ for both. To prove $$ \Re(e^{i\varphi}+e^{i2\varphi}+...+e^{ni\varphi})=\frac{\cos(n+1)\varphi\sin n\varphi}{\sin\varphi}, $$ you can consult the solution to this question, remembering that $\varphi$ here is an abbreviation for $\frac\theta2$.