Summation of $\cos(2n-1)\theta$ - Extended

Question

This question is an extension to this question(Summation of cos (2n-1) theta):

Deduce that $\sum_{n=1}^{N}(2 n-1) \sin \left[\frac{(2 n-1) \pi}{N}\right]=-N \csc \frac{\pi}{N}$

Answer 1

a typical method

$$\begin{aligned} \sum_{n=1}^{N} {\cos((2n-1)\theta)} & = \frac1{\sin\theta} \sum_{n=1}^{N} {\cos((2n-1)\theta)\sin\theta}\\ & = \frac1{2\sin\theta} \sum_{n=1}^{N} {(\sin(2n\theta) - \sin((2n-2)\theta))}\\ & = \frac{\sin(2N\theta)}{2\sin\theta} \end{aligned}$$

where notice $2\cos A\sin B=\sin(A+B)-\sin(A-B)$

so your extended is

$$\begin{aligned} -\frac{\mathrm{d}}{\mathrm{d}\theta} \left(\sum_{n=1}^{N} {\cos((2n-1)\theta)}\right) & = -\frac{\mathrm{d}}{\mathrm{d}\theta} \left(\frac{\sin(2N\theta)}{2\sin\theta}\right)\\ & = -N\cos(2N\theta)\csc\theta + \frac1{2}\cot\theta\csc\theta\sin(2N\theta) \end{aligned}$$

plug-in $\theta=\pi/N$ and you will find your answer.