I found two triangle identities about the sum of cos/sine.
$$ \sum_{n=0}^{N-1} {\cos \left( \frac{2\pi \Delta}{N} n \right)} = 0, {\quad \rm for\ }\Delta = 1, 2,\cdots$$ $$ \sum_{n=0}^{N-1} {\sin \left( \frac{2\pi \Delta}{N} n\right)} = 0, {\quad \rm for\ }\Delta = 0, 1,\cdots$$
For the second identity, I feel much easier to accept in an intuitive way when examing the special case $\sin (\frac{2\pi}{N}n) = -\sin (\frac{2\pi}{N}(N-n))$ (i.e., when $\Delta = 1$). For the first identity and the second one in a more generalised condition (any $\Delta$), I didn't have a clue.
What you can use is Euler's formula and deal with a sum of exponentials instead, like so :
$$e^{i \frac{2 \pi n \Delta}{N}} = \cos{\frac{2 \pi n \Delta}{N}} + i \cdot \sin{\frac{2 \pi n \Delta}{N}}$$
Then your sums become: $$Re(\sum_{n = 0}^{N-1} e^{i \frac{2 \pi n \Delta}{N}}) = \sum_{n=0}^{N-1} {\cos \left( \frac{2\pi \Delta}{N} n \right)}$$ and $$Im(\sum_{n = 0}^{N-1} e^{i \frac{2 \pi n \Delta}{N}}) = \sum_{n=0}^{N-1} {\sin \left( \frac{2\pi \Delta}{N} n \right)}$$ respectively. The sums are now easy to deal with geometric sums.