Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\cos(A+B)$?

share|improve this question
2  
When writing functions in $\LaTeX$, many of them get the right font if you write cosine as \cos etc. –  Ross Millikan Nov 15 '12 at 5:02

3 Answers 3

up vote 5 down vote accepted

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.

share|improve this answer
1  
Not to mention $\cos(A+B) = \cos(B+A)$ because $A+B = B+A$. –  EuYu Nov 15 '12 at 4:54
    
I meant the equation being commutative, but that is probably the wrong term. Thanks for your help though! –  A A Nov 15 '12 at 4:55
1  
@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 5:05
1  
@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 1:12
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 $B=0$ –  Ross Millikan Nov 16 '12 at 3:35

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*}

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.