Are trig identities commutative?

Question

If I have $\cos(B)\cos(A)-\sin(A)\sin(B)$, can I write that as $\cos(A)\cos(B)-\sin(A)\sin(B)$? And then combine it as $\cos(A+B)$?

Answer 1

It is not really about the trig identities themselves being commutative here (the meaning of which is not clear). Since $\mathbb{R}$ is commutative, $\cos a \cos b = \cos b \cos a$, so your exemple works.

Answer 2

The key here is that both $\cos(A)$, and $\cos(B)$ are just numbers, and when multiplying two numbers the order that they are written is not important. So \begin{align*} \cos(A) \cos(B) = \cos(B) \cos(A) \end{align*}

Answer 3

The variables in the identity are "dummy variables" that can stand for anything. Yes, you can interchange $A$ and $B$. You can also define (for this use) $A=1+x, B=q^2$ and conclude that $\cos(1+x)\cos(q^2)-\sin(1+x)\sin(q^2)=\cos(1+x+q^2)$