Let $A_n=\sum\limits_{k=1}^n \sin k $ , show that there exists $M>0$ , $|A_n|<M $  for every $n$ . 
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. 
