$$ \sum_{n=1}^N\cos(2n-1)\theta=\dfrac{\sin(2N\theta)}{2\sin\theta}, $$ where $\sin\theta\neq0.$ Deduce that $$ \sum_{n=1}^N (2n-1)\sin\left[\dfrac {(2n-1)\pi}N\right]=-N\operatorname{cosec}\dfrac\pi N. $$
Please help me in this problem , I know how to show the first part...then in the next part they say to deduce that, how do we do that part ? :/
Hint Calculate the derivative of the first part with respect to $\theta$ and then choose $\theta$ according to your question.