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Ok, let me explain why (I said something is missing) by some special cases:
1. a=0. This is an extreme case. Let's assume N=pi/(2ΔΘ). Then, N->Infinity when ΔΘ->0. In this case, since cos(a*sin nΔΘ)=cos(0)=1, your sum will be (let me denote that sum by SUM)
SUM=N, which tends to Infinity when ΔΘ->0.
2. When a>0 (similar for a<0) and N is large (or ΔΘ small), sin(nΔΘ) is very small for not big n's. Thefore, cos(a*sin nΔΘ) will be close to 1. That means your SUM will tend to big.
So, my conclusion is that whatever the SUM is, it'll tend to Infinity when ΔΘ->0. This is why I said your formula is missing a factor ΔΘ, which would have balanced your SUM down to a finite value.
Yes, whatever the SUM is, it still makes sense to sum it over to a simpler form. Sorry, this is the part that I missed in the last post. Here is the reason that we probably can't make it simpler. Assume that we can, and let me denote the simplified form by F(N). In other words,
Multiplying the two sides by ΔΘ (=pi/(2N)) produces
The left hand side will tend to the 0th order bessel function as I indicated last time, and we all know that the 0th order bessel function cannot be expressed in a close form (1/2 order can, by the way). Now, look at the right hand side. When N->Infinity (ΔΘ->0), it's just a limit and will tend to a close form (because only derivatives are possibly involved). Therefore, we end up with a conflict.