Sum of Product of Sines

**sumofsines** · August 23rd 2012, 08:46 PM

Hello,

I'm trying to show that $\sum _{j=1}^{N-1} \sin \left(\frac{j \pi y}{N}\right)\sin\left(\frac{j \pi k}{N}\right)=0$ , if y does not equal k (with y and k both being integers). I've tried using the trig identity for the product of sines. That gives $\sum _{j=1}^{N-1} \left [\cos\left (\frac{j \pi (y-k)}{N}\right ) - \cos\left (\frac{j \pi (y+k)}{N}\right )\right ]$ , but I'm still stuck there.

Any ideas?

**MaxJasper** · August 23rd 2012, 09:12 PM

Your sum is equal to zero if y<>k and:

$\left\{\left\{\cos \left(\pi \left(-\frac{k}{N}+k-y\right)\right)\to -\cos \left(\pi \left(k-\frac{y}{N}+y\right)\right)+\cos \left(\pi \left(-\frac{k}{N}+k+y\right)\right)+\cos \left(\pi \left(k+\left(\frac{1}{N}-1\right) y\right)\right)\right\}\right\}$

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