Consider this equation
$$ e =\left(\cos \left({\frac{A - B}{2}}\right)\right)^{2} $$
Is there a solution for functions $f$ and $g$ such that
$$e = \frac{f(A)}{g(B)}$$
The initial application of difference of cosine formula will give us
$$e = \left({cos(\frac{A}{2})cos(\frac{B}{2}) + sin(\frac{A}{2})sin(\frac{B}{2})}\right)^2$$
But then on it becomes really challenging to separate the cos of A and B. Is this solvable? If yes how to solve it?
It's not possible, because if
$$\frac{f(A)}{g(B)} = \cos^2{\frac{A-B}{2}}$$
then you can put $A=B$, and we have $\cos(0) = 1$ so $f(A)/g(A) = 1$.
Therefore, $f(A) = g(A)\ \forall A \in R$, which means $f$ and $g$ are the same function and if $\frac{f(A)}{g(B)} = \cos^2{\frac{A-B}{2}}$ then, putting $B=0$ and substituting $g(0)$ with $f(0)$ you get
$$ f(A)/f(0) = \cos^2(A/2) $$
which means $f(A) = f(0) \times \cos^2(A/2)$ then, $$\frac{f(A)}{f(B)} = \frac{\cos^2(A/2)}{\cos^2(B/2)} = \cos^2((A-B)/2)$$ which is false.