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

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 $$\frac{\cos(n+1)\vartheta\sin n\vartheta}{\sin\vartheta}$$?

\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 $\varphi=\theta/2$.
• I have a doubt. You used the geometric series sum formula. However I fail to see how $e^{i\theta}$. Could you clarify it please? Sep 14, 2018 at 21:35
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 $\frac\theta2$.