Product of sines to sum

I stumbled across the following identity in a system I'm considering: \begin{align} \prod_{j=1}^N \sin a_j \end{align} which I need to rewrite as a sum (or otherwise) of sines or other expressions. $$a_j$$ in my context must be treated as completely general real variables, but with distinct numerical values. (At a certain point I am required to take a sum over $$p$$ when $$a_j\mapsto pb_j+c_j$$. If the expressions that arise remain simple that would be a tremendous plus. It is fine if we get products of rational functions, as these can be simplified using Heaviside's decomposition.)

So I know of the double angle formula, which yes, in principle allows for rewriting this expression as a sum: \begin{align} \prod_{j=1}^N \sin a_j &= \frac1{2^{N-1}} \sum_{\sigma_2,\cdots,\sigma_N=\pm} \left[\prod_{j=2}^N \sigma_j\right]\sin\left[a_1 + \sum_{j=2}^N \sigma_j a_j + \frac{\pi(1-N)}2\right] \end{align} however it is not clear what the meaning is (in the analytical sense) of the resulting expressions. In my system $$N$$ will get very large, and it is not clear how to proceed with the expressions that follow (what is an infinite-fold signed sum of sines?! No idea either).

What I was considering is using Euler's reflection. Intuitively this should lead to an identity containing a Barnes $$G$$ function. I would like to consider expressions of the form \begin{align} \prod_1^N \Gamma(a_j) \end{align} but I have not found general identities relating this to Barnes $$G$$, and it is also not clear how to sum Barnes $$G$$ functions given my context above. I'd here like to discuss alternatives options and hope Math.SE can offer insight that I could not find.

EDIT: a potential option would be to proceed with the expression and use the following result of the binomial theorem in reverse: \begin{align} \prod_{j=1}^N f_j &= \frac1{2^{N-1}N!}\sum_{\sigma_{2,\dots,N} = \pm} \left(\prod_{j=2}^N \sigma_j \right) \left(f_1+\sum_{i=2}^N \sigma_j f_j\right)^N \end{align}

This would require us to find a function $$f$$ that satisfies it.

EDIT 2: Possibly useful:

Using the above identity, one could for example perform a subset of summations like so: \begin{align} \sum_{p=-n}^n \prod_{j=1}^N \sin (pb_j+c_j) &= \frac1{2^{N-1}} \sum_{\sigma_2,\cdots,\sigma_N=\pm} \left[\prod_{j=2}^N \sigma_j\right] \sum_{p=-n}^n\sin\left[pb_j+c_j + p\sum_{j=2}^N \sigma_j b_j + \sum_{j=2}^N c_j + \frac{\pi(1-N)}2\right]\\ &= \frac1{i2^{N}} \sum_{\sigma_2,\cdots,\sigma_N=\pm} \left[\prod_{j=2}^N \sigma_j\right] \left[\exp\left[ic_j + i\sum_{j=2}^N c_j + \frac{i\pi(1-N)}2 \right] \left(\sum_{p=-n}^n \exp\left[ipb_j + ip\sum_{j=2}^N \sigma_j b_j\right]\right) - \exp\left[-ic_j - i\sum_{j=2}^N c_j + \frac{i\pi(N-1)}2 \right] \left(\sum_{p=-n}^n \exp\left[-ipb_j - ip\sum_{j=2}^N \sigma_j b_j\right]\right)\right] \end{align} then (heuristically) using symmetricity in dummy index $$p$$ to perform the summation, then perhaps using Cauchy products (in reverse) in the finite case to extract dirichlet kernels from the summation.