Confused about identity for product of cosines into a sum of cosines

[swampwiz]

What I mean is the way that a product of cosines in which the angles increment the same amount is equal, with some extra terms, of the sum of the cosines.

It is discussed here:

https://math.stackexchange.com/ques...sines-be-rewritten-as-a-finite-sum-of-cosines

But I have no idea how this summation entity is applied.

j_{k + 1} ∈{ +1 , -1 }

I have been trying to search for a good explanation of this, but most of the time all I get is the stupid identity for the product of a pair of cosines.

 

[Office_Shredder]

I don't know what you mean by "the way this summation entity is applied", but this is just repeated application of the two term product rule. For example
$$\cos(a)\cos(b)\cos(c) = \frac{1}{2}\left(\cos(a+b)+\cos(a-b)\right)\cos(c)$$

Using the product rule on a and b. Then you can distribute the $\cos(c)$, and apply the product rule to that and $a+b$, and that with $a-b$ to get

$$\frac{1}{4}\left( \cos(a+b+c)+\cos(a+b-c)+\cos(a-b+c)+\cos(a-b-c)\right)$$

This is just summing every way to pick a sign for b and c, so you can rewrite this as
$$\frac{1}{4} \sum \cos(a\pm b \pm c)$$

Which I think is effectively what is being written in that post. If you had k terms, you would just make that 4 a $2^{k-1}$ and add more $\pm$ terms.

There are eight ways to pick a sign for each of a,b and c, and half of them show up here. The other half are exactly negative of one of the ones that we wrote down, and cosine is an even function, so you could just include them and divide by another factor of two to get
$$\frac{1}{8} \sum \cos( \pm a \pm b \pm c)$$

And similar for if you have more terms.
Hopefully this helps a bit.