I was just wondering why for each $k \in \mathbb{N}$, it holds that
$$\sum_{i=1}^k \sin\left(i \dfrac{2 \pi}{k}\right)=0$$
Is it some simple reason I am overseeing?
Asked
Active
Viewed 47 times
0
-
Here's a more-general question: "How can we sum up sin and cos series when the angles are in arithmetic progression?". – Blue Oct 03 '20 at 18:58
3 Answers
2
That is because there's a well-known formula for the sum of sines (and another for the sum of cosines) of arcs in arithmetic progression: $$\sum_{i=1}^k\sin i\theta=\frac{\sin\dfrac{(k+1)\theta}2}{\sin\dfrac{\theta}2} \:\sin\frac{k\theta}2.$$
Bernard
- 175,478
-
-
I learnt it on my last year in high-school. It is easy to prove using complex exponentials, as the imaginary part of the sum of terms of a geometric sequence. – Bernard Oct 03 '20 at 19:15
-
Without complex, use $\sin(a)\sin(b) = (\cos(a-b)-\cos(a+b))/2$ with $a = \sin(i\theta), b = \sin(\theta/2)$ and it telescopes. – marty cohen Oct 03 '20 at 21:01
-
-
Multiply both sides by $\sin\frac\theta 2$ and you get the product of sines that you need for marty cohen comment. – Somos Oct 03 '20 at 21:59
-
0
Geometric sums gives you \begin{align*} \sum_{j=1}^ne^{i\frac{2\pi j}{n}} &=\frac{e^{i\frac{2\pi(n+1)}{n}}-e^{i\frac{2\pi n}{n}}}{e^{i\frac{2\pi}{n}}-1}-1\\ &=\frac{e^{i\frac{2\pi}{n}}-1}{e^{i\frac{2\pi}{n}}-1}-1\\ &=1-1=0 \end{align*}
and since
$$ \sin x=\Im (e^{ix}) $$ you conclude by linearity of $\Im$.
Note that since $\cos x=\Re(e^{ix})$ you get the same conclusion for sums of cosines.
Joe
- 11,745
0
Cannot be simpler:
Consider a unit circle with center at the origin and points on the circle defining angles measured from x-axis. Your sum includes $k$ angles defined by $k$ equidistant points on the unit circle. It's easy to see that, no matter if $k$ is odd or even, points are symmetric with respect to $x$-axis. In other words, for each point above $x$-axis with distance $y$ from $x$-axis, there is a symmetric point with the same distance below the axis. These distances are actually sines of respective angles. When you sum all these distances with proper signs, they will eventually cancel each other. So the total sum must be zero.
Saša
- 15,906