Michael, that looks like the same explanation you have on Wikipedia, which I have seen, as well as your post on this site about a related topic. While it's a correct formula for $cos(\displaystyle\sum_{n=0}^m a_{n})$, as I said in a previous comment, I am looking for a much more elementary formula that doesn't ask one to "take every permutation of a subset of natural numbers of order k" in the index without an explicit formula. If you'll bear with the indices, I think I've figured it out to my satisfaction.

In order to find

$(A_{0}+B_{0})(A_{1}+B_{1})...(A_{m}+B_{m})$

we'll need to relabel A and B and instead use $\alpha_{0}$ in place of $A$ and $\alpha_{1}$ in place of $B$. Rewriting, we need to find:

$(\alpha_{00}+\alpha_{10})(\alpha_{01}+\alpha_{11})...(\alpha_{0m}+\alpha_{1m})$

Since each term will be formed by choosing one term from each binomial, every term will be of the form $\displaystyle\prod_{k=0}^m\alpha_{j_{k}k}$ where $j_{k} \in ${$0,1$}

Thus, we may write $\displaystyle\prod_{n=0}^m (\alpha_{0n}+\alpha_{1n})$ as an 'm-sum', viz.

$\displaystyle\prod_{n=0}^m (\alpha_{0n}+\alpha_{1n}) = \displaystyle\sum_{j_{0}=0}^1\displaystyle\sum_{j_{1}=0}^1\displaystyle\sum_{j_{2}=0}^1...\displaystyle\sum_{j_{m}=0}^1(\displaystyle\prod_{k=0}^m\alpha_{j_{k}k})$

The product basically sets up the term, and the sums take you through every permutation.

As an example, in the case where m=1:

$(A_{0}+B_{0})(A_{1}+B_{1})$
$= (\alpha_{00}+\alpha_{10})(\alpha_{01}+\alpha_{11})$
$= \displaystyle\prod_{n=0}^1 (\alpha_{0n}+\alpha_{1n})$
$= \displaystyle\sum_{i_{0}=0}^1\displaystyle\sum_{i_{1}=0}^1(\displaystyle\prod_{k=0}^1\alpha_{i_{k}k})$
$= \displaystyle\sum_{i_{0}=0}^1\displaystyle\sum_{i_{1}=0}^1 \alpha_{i_{0}0}\alpha_{i_{1}1}$
$= \alpha_{00}\alpha_{01}+\alpha_{00}\alpha_{11}+\alpha_{10}\alpha_{01}+\alpha_{10}\alpha_{11}$
$= A_{0}A_{1}+A_{0}B_{1}+B_{0}A_{1}+B_{0}B_{1}$

as promised.

Define $\beta_{0n}:=cos(a_{n})$ and $\beta_{1n}:=isin(a_{n})$. Using the result from above, we see that:

$\displaystyle\prod_{n=0}^m[cos(a_{n})+isin(a_{n})]=$
$\displaystyle\sum_{j_{0}=0}^1\displaystyle\sum_{j_{1}=0}^1\displaystyle\sum_{j_{2}=0}^1...\displaystyle\sum_{j_{m}=0}^1(\displaystyle\prod_{k=0}^m\beta_{j_{k}k})$

Similarly, define $\gamma_{0n}:=cos(a_{n})$ and $\gamma_{1n}:=-isin(a_{n})$, then:

$\displaystyle\prod_{n=0}^m[cos(a_{n})-isin(a_{n})]=$
$\displaystyle\sum_{j_{0}=0}^1\displaystyle\sum_{j_{1}=0}^1\displaystyle\sum_{j_{2}=0}^1...\displaystyle\sum_{j_{m}=0}^1(\displaystyle\prod_{k=0}^m\gamma_{j_{k}k})$

Thus,

$cos(\displaystyle\sum_{n=0}^ma_{n})=\frac{1}{2}[\displaystyle\sum_{j_{0}=0}^1\displaystyle\sum_{j_{1}=0}^1\displaystyle\sum_{j_{2}=0}^1...\displaystyle\sum_{j_{m}=0}^1(\displaystyle\prod_{k=0}^m\beta_{j_{k}k})+\displaystyle\sum_{j_{0}=0}^1\displaystyle\sum_{j_{1}=0}^1\displaystyle\sum_{j_{2}=0}^1...\displaystyle\sum_{j_{m}=0}^1(\displaystyle\prod_{k=0}^m\gamma_{j_{k}k})]$

Apologizes for the explosion of indices.