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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.

cpiegore
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Manuel Algaba
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    If you write a cosine as $\cos A=\frac12(e^{iA}+e^{-iA})$ then $\cos A+\cos B$ will factor but alas $\cos A+\cos B+\cos C$ won't. – Angina Seng Mar 27 '18 at 01:09
  • In wikipedia the last entry in https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities --Product to sum box could answer your question with a small trick; matrix inversion. My trouble is I can not find a proof of it (looks nice and somewhat reasonable though) and their A&S reference does not have it. So I will treat it as an "internet rumor". If you want, and can explain how e_x comes from S I , I will demonstrate the answer; as if it were true. – rrogers Mar 28 '18 at 16:04
  • Here is the inverse problem: https://math.stackexchange.com/questions/2460563 I had almost reached that; almost mean I got started and finished 10% :) The inversion process, relating the product terms to summation terms shows that the simple extrapolation to over 2 angles is overdetermined and additional terms are needed. But.. that doesn't seem to rule out special cases or offsets in the product variables as far as I have seen. And, since this resembles the permutation group, subgroups as special cases; or not. – rrogers Mar 29 '18 at 12:12

3 Answers3

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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.

Manuel Algaba
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COMMENT (just to try to help you a bit).

The surface $z=\cos x+\cos y$ is complicated:

enter image description here

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 $O\left(\dfrac{x^8+y^8+z^8}{40320}\right)$.

Piquito
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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?

Manuel Algaba
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