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How would I prove the following trig identity

$$\frac{ \cos (A+B)}{ \cos A-\cos B}=-\cot \frac{A-B}{2} \cot \frac{A+B}{2}$$

My work thus far has been. ((2cos(1/2)(A+B)cos(1/2)(A-B))/((-2sin(1/2)(A+B)sin(1/2)(A-B)

-cot(1/2)(A+B)cot(1/2)(A-B)

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$\frac{cosA+cosB}{cosA-coSB}=\frac{2cos\frac{A-B}{2}cos\frac{A+B}{2}}{-2sin\frac‌​{A-B}{2}sin\frac{A+B}{2}}=-cot\frac{A-B}{2}cot\frac{A+B}{2}$, dividing the numerator & the denominator by $sin\frac{A-B}{2}sin\frac{A+B}{2}$, right? –  lab bhattacharjee Aug 4 '12 at 16:06
    
I'm probably being stupid here, but if this were true, wouldn't this be $0$. –  SiliconCelery Aug 4 '12 at 16:08
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Just a minor nit-pick: you solve equations, rather than prove them. Trigonometric things that you prove are identities rather than equations. –  Old John Aug 4 '12 at 16:26
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@Old: It's not such a minor nitpick in my view. Misuse of "prove" is one of the major English errors on this site, and I think a certain amount of insistence on correct use of terminology will be required if we don't want the bad examples to become so abundant that people start learning wrong mathematical English from this site. –  joriki Aug 4 '12 at 16:31
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I agree fully! I was just being polite, being something of a a newbie here. –  Old John Aug 4 '12 at 16:34

2 Answers 2

First, note that $\dfrac{\cos(A + B)}{\cos A - \cos B} \neq -\cot \left( \dfrac{A-B}{2} \right) \cot \left(\dfrac{A+B}{2} \right)$

In particular, if we use something like $A = \pi/6, B = 2\pi/6$, then the left is $0$ as $\cos(\pi/2) = 0$ and the right side is a product of two nonzero things.

I suspect instead that you would like to prove:

$$\dfrac{\cos A + \cos B}{\cos A - \cos B} = -\cot \left( \dfrac{A-B}{2} \right) \cot \left(\dfrac{A+B}{2} \right)$$

HINTS

And this looks to me like an exercise in the sum-to-product and product-to-sum trigonometric identities (wiki reference). In fact, if you just apply these identities to the top and the bottom, you'll get the result.

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A couple of hints: use the $\cot$ identity on the RHS to put in terms of $\sin$ and $\cos$, use a product to sum identity, and pay attention to odd and even functions. This should simplify it to an easier expression.

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I downvoted because the written identity is false. So this is hinting at the wrong direction. –  mixedmath Aug 4 '12 at 16:54
    
@mixedmath This only simplifies the expression. Then it is clear that the identity is false. I never said that this proves the identity true. –  Lance Helsten Aug 4 '12 at 16:58
    
I'm still not okay with this. If your intention is to prove something false, why not do it in the easy way and just choose two values? It's shorter than applying identities, especially when it can be difficult to say when two trig expressions are the same. I still think this misleads the OP. –  mixedmath Aug 4 '12 at 17:07
    
To you and me it is immediately obvious, and we can come up with explicit values to disprove the identity. To a new student this is not always the case: so using identities to simplify to an expression where it is obviously false is still a good exercise. You and I have different methods of teaching is all. –  Lance Helsten Aug 4 '12 at 17:13

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