Show that $A_n=\sum\limits_{k=1}^n \sin k $ is bounded?

Solution 1:

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.

Solution 2:

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.