Use De Moivre's theorem to show that
$$\sin (2m+1) \theta = (\sin^{2m+1} \theta) \cdot P_m (\cot^2 \theta)$$ for $0 < \theta < \pi/2$, where
$$P_m(x) = \sum_{k=0}^{m} (-1)^k C^{2m+1}_{2k+1} x^{m-k}$$ with the usual $C^n_r = \frac{n!}{(n-r)!r!}$. Hence, deduce that
$$\sum_{k=1}^{m} \cot^2 \frac{k \pi}{2m+1} = \frac{m(2m - 1)}{3}.$$
I've managed to do the first part, but can't quite get the 'Hence, deduce' part. My thinking so far was to substitute $\theta = \frac{k \pi}{2m+1}$ for $k \in \mathbb{N}$ into the top equation to get:
$$P_m \left( \cot^2 \left(\frac{k\pi}{2m+1} \right) \right) = 0$$
implying that $$\sum_{k=0}^{m} (-1)^k C^{2m+1}_{2k+1} \left( \cot^2 \left(\frac{k\pi}{2m+1} \right) \right)^{m-k} = 0,$$
and I simply cannot figure out where to go from here. I certainly don't need a full proof but if anyone has any hints/guide as to whether this is the right approach and/or where to go from here, it would be greatly appreciated. Thanks.
