Is there any expression to calculate the sum of (at least) 3 cosines? I'm involved in a waves problem and I have to calculate $\cos(A)+\cos(B)+\cos(C)$, where $A$, $B$ and $C$ are independent angles. I want to find an expression similar to the sum-product identity: $$\cos(A)+\cos(B)=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) $$
I have tried the same path that leads you to the mentioned identity, but it is different. In the two cosines case it is easy because you lose two terms as they have opposite signs, however, in the three cosines case, you can't lose them as they are odd.
Any help is welcome!
Thank you for your time :)
EDIT:
Someone told me that you can use:
$$
\cos(A)+(\cos(B)+\cos(C))+(\cos(A)+\cos(B))+\cos(C)+\cos(B)+(\cos(A)+\cos(C))
$$
And use the sum-product identity in those sums in brackets in order to obtain the triple of the sum we want to obtain, so we obtain:
$$
\cos(A)+\cos(B)+\cos(C)=\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)+\cos\left(\frac{B+C}{2}\right)\cos\left(\frac{B-C}{2}\right)+\cos\left(\frac{A+C}{2}\right)\cos\left(\frac{A-C}{2}\right)
$$
While it is not a product, it is an interesting thing.
 A: A classmate found this.
If you write:
$$
[cosA+cosB]+cosC=[2cos(\frac{A+B}{2})cos(\frac{A-B}{2})]+cosC
$$
$$
cosA+[cosB+cosC]=...
$$
$$
cosB+[cosA+cosC]=...
$$
And you do the same as done in the first equation, but in the following ones, and them sum up, you obtain the following expression:
$$
cosA+cosB+cosC=cos(\frac{A+B}{2})cos(\frac{A-B}{2})+cos(\frac{B+C}{2})cos(\frac{B-C}{2})+cos(\frac{A+C}{2})cos(\frac{A-C}{2})
$$
It is not a unique product, but is the most near solution I got.
Thank you all. I like this place, it is a nest of creativity.
A: COMMENT (just to try to help you a bit).
The surface $z=\cos x+\cos y$ is complicated:

but $\cos z=\cos x+\cos y$ is even more complicated.
I believe that if there were a closed way of expressing your sum of three cosines, then it would be well known in school textbooks of trigonometry. In my opinion, what I would do is take into account that $$\cos x=\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k}}{(2k)!};\space\space x\in\mathbb R$$
so I would work on my waves problem with 
$$\cos x+\cos y+\cos z\approx3-\frac{x^2+y^2+z^2}{2}+\frac{x^4+y^4+z^4}{24}-\frac{x^6+y^6+z^6}{720}$$ with an little error of $O\left(\dfrac{x^8+y^8+z^8}{40320}\right)$.
A: Well, I was trying to sleep but an idea came to my mind xD
If you use the sum-product identity twice:
$$
cos(A)+cos(B)+cos(C)+cos(D)=2cos(\frac{A+B}{2})cos(\frac{A-B}{2})+2cos(\frac{C+D}{2})cos(\frac{C-D}{2}) 
$$
Now, if we want to have the sum of the first three cosines, we just have to do $ D= \frac \pi 2$, so the cosine of D will equal zero. Then:
$$
2cos(\frac{C+D}{2})cos(\frac{C-D}{2})=2\left(cos(\frac C 2)cos(\frac \pi 4)-sen(\frac C 2)sen(\frac \pi 4)\right)\left(cos(\frac C 2)cos(\frac \pi 4)+sen(\frac C 2)sen(\frac \pi 4)\right)=\frac{2}{\sqrt{2}}\left(cos(\frac C 2)-sen(\frac C 2)\right)\left(cos(\frac C 2)+sen(\frac C 2))\right)
$$
Where we used the cosine of a sum of angles. And if we use the equation for the cosine of a double angle:
$$
{\sqrt{2}}\left(cos^2(\frac C 2)-sen^2(\frac C 2)\right)={\sqrt{2}}cos(C)
$$
So the final result should be:
$$
cos(A)+cos(B)+cos(C)=2cos(\frac{A+B}{2})cos(\frac{A-B}{2})+{\sqrt{2}}cos(C)
$$
However, there is something very strange in this equation as we can use the sum-product identity again to remove cos(A)+cos(B) and we will obtain:
$$
cos(C)={\sqrt{2}}cos(C)
$$
What can only be true if cos(C)=0 and then $C=\frac \pi 2$
Why does this happen?
