# Sum of Product of Sines

• August 23rd 2012, 08:46 PM
sumofsines
Sum of Product of Sines
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?
• August 23rd 2012, 09:12 PM
MaxJasper
Re: Sum of Product of Sines
Your sum is equal to zero if y<>k and:

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