algebra precalculus - Evaluating $sum_{n=1}^{99}sin(n)$

I'm looking for a trick, or a quick way to evaluate the sum $\displaystyle{\sum_{n=1}^{99}\sin(n)}$. I was thinking of applying a sum to product formula, but that doesn't seem to help the situation. Any help would be appreciated.

Answer

Hint: compute $\sum_{n=0}^{99} (\cos(n) + i\sin(n))$.