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.
 A: Hint: compute $\sum_{n=0}^{99} (\cos(n) + i\sin(n))$.
A: As in other people's answers, you can do this by complex methods.  However if (as suggested by the algebra-precalculus tag) you have not yet met complex numbers...
Let $S$ be the sum.  Using a "products to sums" formula,
$$\eqalign{(2\sin{\textstyle\frac{1}{2}})S
  &=\sum_{n=1}^{99}2\sin n\sin{\textstyle\frac{1}{2}}\cr
  &=\sum_{n=1}^{99}\bigl(\cos(n-{\textstyle\frac{1}{2}})-\cos(n+{\textstyle\frac{1}{2}})\bigr)\cr
  &=(\cos{\textstyle\frac{1}{2}}+\cos(1{\textstyle\frac{1}{2}})+\cdots
    +\cos(98{\textstyle\frac{1}{2}}))\cr
  &\qquad\qquad{}-(\cos(1{\textstyle\frac{1}{2}})+\cdots
    +\cos(98{\textstyle\frac{1}{2}})+\cos(99{\textstyle\frac{1}{2}}))\cr
  &=\cos{\textstyle\frac{1}{2}}-\cos(99{\textstyle\frac{1}{2}})\cr}$$
and so
$$S=\frac{\cos{\textstyle\frac{1}{2}}-\cos(99{\textstyle\frac{1}{2}})}{2\sin{\textstyle\frac{1}{2}}}\ .$$
Comment.  For alternative answers you can follow the same basic idea starting with $(2\sin1)S$, $(2\cos1)S$ etc.
