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**1. The problem statement, all variables and given/known data**

I have encountered this problem from the book

*The Physics of Waves*and in the end of chapter six, it asks me to prove the following identity as part of the operation to prove that as the limit of ##W## tends to infinity, the series becomes an integral. The series involved is as follows:

##\sum_{n=1}^{W} sin(\frac {nk\pi} {W+1}) sin(\frac {n'k\pi} {W+1})##

I need to prove that:

##\sum_{n=1}^{W} sin(\frac {nk\pi} {W+1}) sin(\frac {n'k\pi} {W+1})=b## when ##n=n'## and ##=0## when ##n \neq n'## where ##n, n'## are positive integers.

**2. Relevant equations**

Trigonometric Identities?

**3. The attempt at a solution**

I used the identities to convert the series into the following:

##\sum_{n=1}^{W} \frac 1 2 (cos( \frac {(n-n')k \pi} {W+1}) - cos( \frac {(n+n')k \pi} {W+1}))##

By writing out the terms and regrouping, I concluded that:

##\sum_{n=1}^{W} cos( \frac {(n-n')k \pi} {W+1}) = \sum_{n=1}^{\frac 1 2 W} cos( \frac {(n-n')k \pi} {W+1}) + cos( \frac {(n-n') (W+1-k) \pi} {W+1})## for even ##W##

and:

##\sum_{n=1}^{W} cos( \frac {(n-n')k \pi} {W+1}) = \sum_{n=1}^{\frac {2W+1} {2}} cos( \frac {(n-n')k \pi} {W+1}) + cos( \frac {(n-n') (W+1-k) \pi} {W+1})## for odd ##W##.

In both cases,

##\sum_{n=1}^{W} cos( \frac {(n-n')k \pi} {W+1}) = 2 cos( \frac {\pi} {2} (n-n')) cos( \frac {(n-n') (W+1-2k) \pi} {2(W+1)})## which will be zero for all odd ##(n-n')##.

The same procedure goes for

##\sum_{n=1}^{W} (cos( \frac {(n+n')k \pi} {W+1})##, whereas the series become ##2 cos( \frac {\pi} {2} (n+n')) cos( \frac {(n+n') (W+1-2k) \pi} {2(W+1)})## which will be zero for all odd ##(n+n')## since if ##(n-n')## is odd, ##(n+n')## is also odd. Here I appear to have proved the identity that the series is zero for all ##(n-n')## is odd; however, I could not find any method to prove it for cases when ##(n-n')## is even. I really need help with this. Can someone point me at a certain direction?