A coincidence or not? Hello it's related to  Boutin's Identity :
First remark :
I was working on this : show this inequality $\left(\sum_{i=1}^{n}x_{i}+n\right)^n\ge \left(\prod_{i=1}^{n}x_{i}\right)\left(\sum_{i=1}^{n}\frac{1}{x_{i}}+n\right)^n$
Wich is equivalent  after substitution to :
$$
\left(\dfrac{-\sum_{i=1}^{n}\sin(x_{i})^2+2n}{\sum_{i=1}^{n}\tan(x_i)^2+2n}\right)^n\le\prod_{i=1}^{n}\cos(x_i)^2 .
$$
So my idea was to use in the RHS the following identity :
$$2\cos(a)\cos(b)=\cos(a+b)+\cos(a-b)$$
For $n=3$ we get :
$$4\cos(a)\cos(b)\cos(c)=\big(\cos(a+b)+\cos(a-b)\big)\cos(c)$$
Since:
$$2(\cos(a+b)+\cos(a-b))\cos(c)=\cos(a+b+c)+\cos(a+b-c)+\cos(a+c-b)+\cos(a-b-c)$$
So we have :
$$4\cos(a)\cos(b)\cos(c)=\cos(a+b+c)+\cos(a+b-c)+\cos(a-b+c)+\cos(a-b-c)$$
So there exists an analogy between Boutin's identity and the decomposition of \cosinus product .
So the natural question is : It's a coincidence or not ? 

Second question :
What's  the particular functional equation behind this I mean if we take an expression of the form :
$$\sum_{2^{n-1}}\pm f(\pm x_1\pm x_2\pm\cdots\pm x_n)$$
What $f(x)$ should check as a condition to have :
$$\sum_{2^{n-1}}\pm f(\pm x_1\pm x_2\pm\cdots\pm x_n)=\alpha f(x_1)\cdots f(x_n)$$
Where $\alpha$ is a constant.
Example :
For cosinus it's $2f(a)f(b)=f(a+b)+f(a-b)$
Thanks a lot
 A: 
I think it's a very general problem and by the way very difficult...But it's a challenge !

Hi. :-) It was a challenge to my memory which I failed: few days before you asked the question I saw a paper considered a similar equation. Trying to find it, I looked more than a half of a thousand pdf files. At last, when I googled for trigonometric Identities and functional equations I found a paper [Kan] and then I was able to findt also in a few year old gif file. :-) Unfortunately, I can read only the first page of this article, so it is not very useful for me. 
But I also found a book [Eft], Section 7 of which is devoted to equations for trigonometric functions. For your question is relevant 
D’Alembert-Poisson I Equation, considered in Subsection 7.2 (see also Subsection 7.4). 

Find all continuous functions $f :\Bbb R\to \Bbb R$ satisfying for all $x,y\in\Bbb R$ the equality 

$$f (x + y) + f (x − y) = 2 f (x) f (y).$$  
These remarks allow us to answer your question. Indeed,
Taking into account the examples which you have presented, I guess that you question can be formulated more precisely, asking about only one equation for given $n$ and $\alpha$. Namely,
(1) $\sum f(x_1\pm x_2\pm\cdots\pm x_n)=\alpha f(x_1)\cdots f(x_n),$
where all signs before functions and a variable $x_1$  are positive, but and all other variables come with all $2^{n-1}$ possible combinations of signs. For instance, for $n=3$  we have the equation 
$$f(x_1+x_2+x_3)+ f(x_1+x_2-x_3)+ f(x_1-x_2+x_3)+ f(x_1-x_2+x_3)=\alpha f(x_1)f(x_2)f(x_3).$$
But it is not so complicated as it looks. Indeed,
For $n=1$ we have a boring equation $f(x_1) =\alpha f(x_1)$ or $ f(x_1)(\alpha-1)=0$, whose solution is the zero function for $\alpha\ne 1$ and any function for $\alpha=1$. 
So further we assume that $n>1$. If $f(0)=0$ or $\alpha=0$ then putting $x_2=\dots x_n=0$, we obtain $2^{n-1}f(x_1)=0$, that is $f$ is the zero function. So  further we assume also that $f(0)\ne 0$ and $\alpha\ne 0$. 
If $n=2$ then multiplying the function $f$ by $\alpha/2$ we see that without loss of generality we may assume that $\alpha=2$.
In this case a solution for continuous $f$  was considered in [Eft], a more general case, maybe, is solved in [Kan] (at  JSTOR is written that the registered users may read the paper online for free). 
We can reduce the case $n\ge 3$ to the previous. By putting $x_3=\dots x_n=0$, we obtain that the function $f$ satisfies Equality 1 with the constant $\alpha\left(\frac {f(0)}2\right)^{n-2}$ at the right hand side. Moreover, putting all $x_i$ equal to zero, we obtain that $2^{n-1}=\alpha f(0)^{n-1}$, that is  $\alpha\left(\frac {f(0)}2\right)^{n-2}=\frac 2{f(0)}$.
PS. Lou van den Dries in [vdDri] proved that all  valid  identities  in  terms  of  variables,  real  constants,  the arithmetic  operations  of  addition  and  multiplication,  and  the  trigonometric operations  of  sine  and  cosine  are  consequences  of  a  few  familiar identities  and  numerical  facts.  More precisely, consider  the  familiar  identities:
1) $\cos(x  +  y)  =  \cos x \cdot\cos y -  \sin x\cdot\sin y$
2) $\sin(x  +  y)  =  \cos x \cdot\sin y + \sin x\cdot\cos y$
3) $\cos(-x)  =  \cos x,  \sin(-x)  =  -  \sin x.$
From  these  identities  and  numerical  facts  like  $\cos 0  =  1$  we  can  derive  other identities  like  $\cos^2 x +  \sin^2  x = 1$.  It is showed that all  valid  identities  formulated  in  terms  of  variables,  individual  real  numbers,  the functions  $\cos$  and  $\sin$,  and  the  ring  operations  $+$,  $-$,  $\cdot$  can  be  derived  from  the identities 1-3  above,  plus  the  identities  defining  commutative  rings  with  unit $1$,  plus  the  "true  numerical  facts",  these  being  the  valid  identities  not  containing variables.  Terms  like  $\sin(\cos x)$ are allowed. The  addition  law  for  the  cosine  and  the  symmetry  law  for  the  cosine  ($\cos(-x)  =\cos x$)  are  derivable  from  the  other  identities,  but  that  one cannot  also  leave  out  $\sin(-x)  =  -  \sin x$.
References
[vdDri] Lou van den Dries A completeness  theorem for  trigonometric  identities and  various  results  on  exponential  functions,  Proc. Amer. Math. Soc. 35 (1972), *96**:2 (1986), 345-352.
[Eft] Costas Efthimiou, Introduction to functional equations: theory and problem-solving strategies for mathematical competitions and beyond,
[Kan] Pl. Kannappan Trigonometric Identities and Functional Equations, The Mathematical Gazette 88:512 (2004), 249-257. 
