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Prove $$\dfrac{\cos A - \cos B}{\sin A + \sin B} = \dfrac{\sin B - \sin A}{\cos A + \cos B}$$

I tried as shown below and am not sure how to do it. Your help is appreciated. Thanks.

Proving from left hand side:

$$\dfrac{\cos A}{\sin A + \sin B} - \dfrac{\cos B}{\sin A + \sin B}$$

$$=\dfrac{\dfrac{\cos A}{\sin A}} {\dfrac{\sin A+\sin B}{\sin A}} - \dfrac{\dfrac{\cos B}{\sin B}}{\dfrac{\sin A+\sin B}{\sin B}}$$

$$= \dfrac{\dfrac{1}{\tan A}}{1+\dfrac{\sin B}{\sin A}}- \dfrac{\dfrac{1}{\tan B}}{1+\dfrac{\sin A}{\sin B}}$$

Blue
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Joe
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  • Try to use $A=(A+B)/2 + (A-B)/2$, $B=(A+B)/2 - (A-B)/2$. – Ivan Kaznacheyeu Sep 25 '21 at 07:46
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    Simpler way is to use $c/d=f/g \Leftrightarrow cg=fd$ – Ivan Kaznacheyeu Sep 25 '21 at 07:47
  • Does your teacher or professor expect you to work exclusively on one side? If so, then @user 's answer can still work, but needs some significant modification. Most likely if the right hand side, = C/D , then multiplying left hand side by C/D * D/C will work. The D/C together with the preexisting left hand side should simplify into 1, leavind C/D. – nickalh Sep 25 '21 at 08:25

1 Answers1

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We have that (for $\sin A + \sin B \neq 0$ and $\cos A + \cos B \neq 0$)

$$\dfrac{\cos A - \cos B}{\sin A + \sin B} = \dfrac{\sin B - \sin A}{\cos A + \cos B} $$

$$\iff (\cos A - \cos B)(\cos A + \cos B)=(\sin B - \sin A)(\sin B + \sin A)$$

$$\iff \cos^2 A - \cos^2 B=\sin^2 B - \sin^2 A$$

$$\iff \cos^2 A + \sin^2 A= \cos^2 B + \sin^2 B$$

Edit

As an alternative, by sum to product identities we have

  • LHS

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

  • RHS

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

user
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    Is it as such: $cos^2A - cos^2B = sin^2B - sin^2A$ => $sin^2A+cos^2A = sin^2B + cos^2B=1$, hence proven. – Joe Sep 25 '21 at 08:20
  • @Joe Yes exactly, finally rearrange and use that $\cos^2 \theta + \sin^2 \theta =1$. – user Sep 25 '21 at 08:22
  • I thought we are going to show proving from left hand side to become right hand side, how come we can use the right hand side of the equation in our proving? – Joe Sep 25 '21 at 08:23
  • @Joe Of course it is another possible way to proceed using others identities but the way suggested is very effective and neat since uses just the fundamental identity. – user Sep 25 '21 at 08:26
  • @Joe That's a good question: yes, "given identity $\implies\ldots\implies \sin^2\theta+\cos^2\theta=1$" does not constitute a valid proof, but "given identity $\iff\ldots\iff\sin^2\theta+\cos^2\theta=1$" is totally valid. In the latter, the important direction is "$\impliedby$". (In the above comment, user is suggesting that after having worked it out on scratch paper, you can just finally present your working the reverse direction using only "$\implies$".) – ryang Sep 25 '21 at 08:29
  • @RyanG Indeed this is an important step, this is why I've added the condition $\sin A + \sin B \neq 0$ and $\cos A + \cos B \neq 0$. – user Sep 25 '21 at 08:30
  • @Joe Yes, the last step when going in the reverse direction requires the restriction spelt out in the parenthesis, which is no problem since that restriction is implicit in the original eqn. To be clear, if you're working starting from the given equation, you MUST explicitly display the "$\iff$" symbol on every line. – ryang Sep 25 '21 at 08:36
  • @Joe I've also added an alternative proof operating side by side. – user Sep 25 '21 at 08:40
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    @RyanG Yes I agree, let me know if the present form is not sufficiently clear. Thanks – user Sep 25 '21 at 08:41
  • Thank you everyone – Joe Sep 25 '21 at 09:44
  • @Joe Welcome to Mathematics StackExchange! The best way to say "thanks!" is to accept and/or upvote answers: it scores points, signals resolution, and prevents bumping. – ryang Sep 25 '21 at 16:16