It is shown that the product-to-sum identities are given by:
$\prod _{k=1}^n \cos \left(\theta _k\right)=\frac{1}{2^n}\sum _{\text{e$\epsilon $S}} \text{cos}\left(e_1 \theta _1+\text{...}+e_n \theta _n\right)$ (1)
where $\left(e_1,+\text{...}+,e_n\right)\,\epsilon \,\,S=\{1,-1\}^n$
An example of $\cos(\theta)\cos(\phi)$ is given.
I would like to know how to apply this to say for example
i) $\cos(\theta)\cos(\phi)\cos(\omega)$
or
ii) $\cos(\theta)\cos(\phi)\cos(\omega)\cos(\zeta)$
I'm not sure that I understand how to apply (1) to these as I do not understand
$\left(e_1,+\text{...}+,e_n\right)\,\epsilon \,\,S=\{1,-1\}^n$
so a worked example would be good.
If you have $n$ factors in the product, you have $2^n$ terms in the sum since you have two choices ($+1$ or $-1$) for $e_1, e_2$ etc...
In the case $n =3$ you thus have $8$ terms which starts to become a bit cumbersome to write :
\begin{align} \cos(\theta)\cos(\phi)\cos(\omega) &= (\cos(\theta+\phi+\omega) \\ &+ \cos(\theta+\phi-\omega) \\ &+ \cos(\theta-\phi+\omega) \\ &+ \cos(\theta-\phi-\omega) \\ &+ \cos(-\theta+\phi+\omega) \\ &+ \cos(-\theta+\phi-\omega) \\ &+ \cos(-\theta-\phi+\omega) \\ &+ \cos(-\theta-\phi-\omega))/8 \\ \end{align}
This corresponds to the following vectors $e$ (in the same order) : $$[1, 1, 1] \\ [1, 1, -1] \\ [1, -1, 1] \\ [1, -1, -1] \\ [-1, 1, 1] \\ [-1, 1, -1] \\ [-1, -1, 1] \\ [-1, -1, -1] \\$$
Due to cosine being an even function, some of these terms can be simplified (for example the first and last term of the sum are equal). In fact, with the enumeration order I chosed, there is a kind of symmetry as terms 1 and 8 are equal, 2 and 7, 3 and 6, 4 and 5 are equal.