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

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Lynne Trussell · Accepted Answer

That is because there&#039;s a well-known formula for the sum of sines (and another for the sum of cosines) of arcs in arithmetic progression:∑i=1ksin⁡iθ=sin⁡(k+1)θ2sin⁡θ2sin⁡kθ2

Orlando Paz · Accepted Answer

Geometric sums gives you  ∑j=1nei2πjn=ei2π(n+1)n−ei2πnnei2πn−1−1  =ei2πn−1ei2πn−1−1   =1−1=0   and since  sin⁡x=J(eix)  you conclude by linearity of J  Note that since cos⁡x=R(eix) you get the same conclusion for sums of cosines.

Vasquez · Accepted Answer

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&#039;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.

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

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