How do I evaluate this finite sum using simple techniques?

Question

I am trying to calculate the definitive integral by definition (with Riemann sum). $$\int_{\frac{-\pi}{2}}^{\frac{3\pi}{2}} (2\sin{(2x+\frac{3\pi}{2})}) \ dx$$

But during the process of calculating it I have troubles evaluating this finite sum:

$$\sum_{i=1}^{n} \sin({\frac{\pi}{2}+\frac{4 \pi i}{n}})$$

I would greatly appreciate if you could help me figure out a nice expression for this using high school techniques.

Answer 1

Consider the set of points

$$[\sin(\pi/2 + 4 \pi i/n),\cos(\pi/2 + 4 \pi i /n)] \subset \mathbf{R}^2, \ i = 1,\ldots n.$$

if $n$ is odd, these are the vertices of a regular $n$-gon centered at the origin, and if $n$ is even, they are the vertices of a regulat $n/2$-gon centered at the origin which each point occuring with multiplicity two.

In either case, the set of points is invariant by the rotation by $4 \pi/n$. Thus the average of the points is also invariant by this rotation. But the only point invariant by a non-trivial rotation is zero. Hence, if $n \ne 1,2$, the sum of all the points is zero, and thus the sum is zero for $n > 2$.

Answer 2

Euler's formula is $e^{it} = \cos t + i \sin t$ so $\sin t = \operatorname{im} e^{it}$.

The Riemann sum with a uniform partition is ${2 \over n} \sum_{k=0}^{n-1} \sin (-{\pi \over 2}+ 2 \pi {k \over n} ) = {2 \over n} \operatorname{im}\left( \sum_{k=0}^{n-1} e^{i(-{\pi \over 2}+ 2 \pi {k \over n} )} \right)$. The advantage of this is the summation becomes a geometric series.

\begin{eqnarray} \sum_{k=0}^{n-1} e^{i(-{\pi \over 2}+ 2 \pi {k \over n} )} &=& e^{-i{\pi \over 2}} \sum_{k=0}^{n-1} (e^{i {2 \pi \over n} })^k \\ &=& e^{-i{\pi \over 2}} { (e^{i {2 \pi \over n} })^n -1 \over e^{i {2 \pi \over n} } -1 } \\ &=& 0 \end{eqnarray} If you do not want to use the Euler formula, you could note that if you split the sum into two appropriate parts, the second part cancels the first exactly (a glance at a plot will show this). You need to deal with odd & even $n$, but it is straightforward. (This is what the other answer is getting at, albeit in a more complicated manner.)