Proof of the formula for $\sin \theta + 2\sin 2\theta +\cdots + n\sin n\theta$

Question

I'm looking to show that

$$\sin \theta + 2\sin 2\theta +\cdots + n\sin n\theta = \frac14\left(\;(n+1)\sin (n\theta) - n \sin((n+1)\theta)\;\right)\csc^2\left(\frac{\theta}{2}\right)$$

So far, I believe that the answer is given by

$$\operatorname{Im}\left({{z+nz^{n+2} - (n+1)z^{n+1}}\over(1-z)^2}\right) \quad\text{where}\;z = \cos \theta + i\sin \theta$$

(from the power sum formula: http://mathworld.wolfram.com/PowerSum.html). But I am struggling to substitute in the values to get the desired result.

Any insights would be appreciated.

Answer 1

Hint: \begin{align} \sin(\theta) + 2\sin(2\theta) + ... + n\sin (n\theta)&=-(\cos\theta+\cos2\theta+\dots+\cos n\theta)'\\&=-\operatorname{Re}\bigl(\mathrm e^{i\theta}+\mathrm e^{2i\theta}+\dots +\mathrm e^{ni\theta}\bigr)'. \end{align}