Why does \sum_{i=1}^k \sin (i \frac{2\pi}{k})=0 for integers k

zeotropojd

zeotropojd

Answered question

2021-12-30

Why does i=1ksin(i2πk)=0 for integers k

Answer & Explanation

Lynne Trussell

Lynne Trussell

Beginner2021-12-31Added 32 answers

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:
i=1ksiniθ=sin(k+1)θ2sinθ2sinkθ2
Orlando Paz

Orlando Paz

Beginner2022-01-01Added 42 answers

Geometric sums gives you
j=1nei2πjn=ei2π(n+1)nei2πnnei2πn11
=ei2πn1ei2πn11
=11=0
and since
sinx=J(eix)
you conclude by linearity of J
Note that since cosx=R(eix) you get the same conclusion for sums of cosines.
Vasquez

Vasquez

Expert2022-01-09Added 669 answers

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.

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