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 $\cos(A+B)$?
Are trig identities commutative?
3
$\begingroup$
trigonometry
-
2When writing functions in $\LaTeX$, many of them get the right font if you write cosine as \cos etc. – 2012-11-15
3 Answers
5
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.
-
1@AA: No. I was saying that you can replace $A$ with any expression and $B$ with any expression (same or different) and the identity is still true. In fact the specific example I showed can be reduced to $\sin (-A)=-\sin (A)$ by taking $B=0$ – 2012-11-16
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*}
1
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 $\cos(1+x)\cos(q^2)-\sin(1+x)\sin(q^2)=\cos(1+x+q^2)$