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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 $\cos(A+B)$?

Ross Millikan
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A A
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3 Answers3

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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.

beauby
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    Not to mention $\cos(A+B) = \cos(B+A)$ because $A+B = B+A$. – EuYu Nov 15 '12 at 04:54
  • I meant the equation being commutative, but that is probably the wrong term. Thanks for your help though! – A A Nov 15 '12 at 04:55
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    @AA: But it also works if the operator doesn't commute in $\mathbb R$. We have $\sin (A-B)=\sin A \cos B - \cos A \sin B$ and it is fair to write $\sin (B-A)=\sin B \cos A - \cos B \sin A$ – Ross Millikan Nov 15 '12 at 05:05
  • @RossMillikan Are you implying sin(A-B)=sin(B-A)? – A A Nov 16 '12 at 00:54
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    @RossMillikan No he is not. He just said that you can "switch" (i.e. replace every occurence of $A$ with $B$ and reciprocally). – beauby Nov 16 '12 at 01:12
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    @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 $B=0$ – Ross Millikan Nov 16 '12 at 03:35
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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*}

Peter Grill
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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 $\cos(1+x)\cos(q^2)-\sin(1+x)\sin(q^2)=\cos(1+x+q^2)$

Ross Millikan
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