It is striking that in lists of trigonometric identities like Wikipedia's there is a product-to-sum rule for a product of arbitrary number of cosines, namely $\prod_{k=1}^n\cos\left(\theta_k\right)=\frac{1}{2^n}\sum_{\{e_k = \pm 1\}}\cos\left(\sum_{k=1}^{n} e_k \theta_k\right)$.
However there is no counterpart for a sum of arbitrary number of cosines. The $\{\theta_k\}$ in here are assumed to be arbitrary, i.e. with no special relation between them.
I conjecture that this should be a (sum of)-products rule, in the form of
$$ \sum_{k=1}^n\cos\left(\theta_k\right)=\sum_{\{e_k = \pm 1\}}\prod_{a_{k,\{e_k\}}}\cos\left(\sum_{k=1}^{n} a_{k,\{e_k\}} \theta_k\right) $$
The notation is to say that for any permutation $\{e_k\}$ there is a product with a new set of coefficients $a_{k,\{e_k\}}$ inside the $\cos$-functions. It is desired that not for all permutations such products are present; indeed the fewer the better.
Examples:
For $n=2$ there is the well-known rule: $\cos(A)+\cos(B)=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$.
For $n=3$ one can write (see e.g. here): $\cos(A)+\cos(B)+\cos(C)=\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)+\cos\left(\frac{B+C}{2}\right)\cos\left(\frac{B-C}{2}\right)+\cos\left(\frac{A+C}{2}\right)\cos\left(\frac{A-C}{2}\right)$.
The last example already illustrates the above statement that a small number of products is desired, since here one can argue that the RHS with three more complicated terms than the three terms on the LHS doesn't produce a great improvement. I know of no more compact form for $n=3$.
Question: is there a formula which transforms a sum of arbitrary number of cosines into a convenient (sum of)-products of cosines ?
