The first part of the question requires me to show that the sum of $$cos(2n - 1)x = \frac{sin(2Nx)}{2sin(x)}$$ from $n = 1$ to $N$.
This I have done by considering the real part of the geometric series $(e^{ix} + e^{3ix} + e^{5ix}+\dots)$ up to $N$ terms.
The second part then asks me to deduce that the sum of $$(2n - 1)sin((2n-1)(\pi/N))\ \text{from}\ n=1\ \text{to}\ N\ \text{is}\ -N(cosec(\pi/N).$$
I have considered opening the bracket and then deriving the result by subtracting the sum of $sin(2n-1)x$ from twice the sum of $(n)(sin(2n - 1)x$ and then substituting $x = \frac{\pi}{N},$ but I cannot sum the latter part. Any hint in that regard will be greatly appreciated.