Q:

Does a Bijective Commutative transformation on a vector of angles exist?

I have a problem where I have two vectors a and b  representing a list of angles.
I need to find a transformation T where T(a,b) = T(b,a), where T has a distance metric to compare two transformations, and that is inversible : I should be able to retrieve a and b (swapped is ok) from T(a,b)
For example: 
a+b, cos(a+b), sin(a)+sin(b), cos(a)*cos(b) or any combinations of these qualify
a.b doesn't qualify because I don't have a metric to compare square angles.
a-b doesn't qualify because it is not commutative
Do you think this is possible?
What I have tried that didn't work:
T(a,b) = {cos(a)*cos(b), cos(a)+cos(b), sin(a)*sin(b), sin(a)+sin(b)}
And solve the system for cos(a) cos(b) sin(a) sin(b). However because it is a quadratic equation, I get two solutions and my two arrays a and b are not consistent anymore, I got two new vectors that have either values from a or from b depending when the determinant of the equations reaches zero.
This was just one idea of transformation that was symmetric so it would satisfy my conditions, but I couldn't get back my original vectors.
Thank you!

A:

Notation convention: the $k$th element of an array, such as $a$, will be denoted with a subscript, as in $a_k$.

... I get two solutions and my two arrays a and b are not consistent anymore, I got two new vectors that have either values from a or from b depending ...

If you're working with functions of the form $c_k=f(a_k,b_k)$ for some symmetric $f$, this is unavoidable. We can swap $a_1$ and $b_1$, leave all of the other $a_k$ and $b_k$ alone, and the $c_k$ will still be the same. Repeat with $d_k=g(a_k,b_k)$ and the same thing happens; adding more functions will never solve the problem.
So then, we need functions that cross over, and use more than one $k$. Here's an idea:
\begin{align*}c_1 = a_1+b_1\quad d_1=\cos(a_1-b_1) &\\
c_2 = a_2+b_2\quad d_2=\cos(a_2-b_2) &\quad e_2=\cos(a_1+a_2-b_1-b_2)\\
c_3 = a_3+b_3\quad d_3=\cos(a_3-b_3) &\quad e_3=\cos(a_1+a_2+a_3-b_1-b_2-b_3)\end{align*}
and so on. By adding in the third vector there, we know how the angle differences combine, and can tell the difference between the likes of $((x,y),(u,v))$ and $((x,v),(y,u))$ since $|x+y-u-v|\neq |x+v-u-y|$ in general.

