TITLE: Show that $A_n=\sum\limits_{k=1}^n \sin k $ is bounded?
QUESTION [17 upvotes]: Let $A_n=\sum\limits_{k=1}^n \sin k $ , show that there exists $M>0$ , $|A_n|<M $  for every $n$ .

REPLY [5 votes]: A more general formula would be:
$$A_n=\sum_{k=1}^{n} \sin k\theta = \frac{\sin\theta+\sin n\theta-\sin(n+1)\theta}{2(1-\cos\theta)}$$
So $A_n$ is clearly bounded (just simply check the case where $\theta=1$).
The formula can be proved by induction using the trig identity: $\sin\alpha+\sin\beta=2\sin(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2})$.