Show that $A_n=\sum\limits_{k=1}^n \sin k $ is bounded? Let $A_n=\sum\limits_{k=1}^n \sin k $ , show that there exists $M>0$ , $|A_n|<M $  for every $n$ . 
 A: 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: $\sin\alpha+\sin\beta=2\sin(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2})$.
A: Hint: Since $$\sin x = \frac{e^{ix}-e^{-ix}}{2i}$$ we can rewrite $$\sum_{n=1}^{K} \sin n = \frac {1}{2i}\sum_{n=1}^K (e^i)^n-\frac {1}{2i}\sum_{n=1}^K (e^{-i})^n$$ and both of these are geometric series.
A: Note that 
$$2(\sin k)(\sin(0.5))=\cos(k-0.5)-\cos(k+0.5).$$
This is obtained by using the ordinary expression for the cosine of a sum.
Add up, $k=1$ to $n$.  On the right, there is mass cancellation. We get 
$$\cos(0.5)-\cos(n+0.5).$$
Thus our sum of sines is 
$$\frac{\cos(0.5)-\cos(n+0.5)}{2\sin(0.5)}.$$
We can now obtain the desired bound for $|A_n|$.  For example, $2$ works, but not by much.
We could modify the appearance of the above formula by using the fact that 
$\cos(0.5)-\cos(n+0.5)=2\sin(n/2)\sin(n/2+0.5)$. 
Generalization: The same idea can be used to find a closed form for
$$\sum_{k=0}^{n-1} \sin(\alpha +k\delta).$$
Sums of cosines can be handled in a similar way.
Comment:  This answer was written up because the OP, in a comment, asked for a solution that only uses real functions.  However, summing complex exponentials, as in the solution by @Eric Naslund, is the right way to handle the problem. 
