Trigs is not my strongest apparently...

I need to prove $c\sin \frac{A-B}{2} = (a-b) \cos \frac{C}{2}$ for a general triangle $ABC$.

Here is what I do, or rather, here is how I fail at proving it:

$\cos \frac{C}{2} \equiv \sin \frac{A+B}{2}$, so $\displaystyle{\frac{\sin \frac{A-B}{2}}{\sin \frac{A+B}{2}} \equiv \displaystyle{\frac{a-b}{c}}}$.

This implies: $\displaystyle{\frac{\tan \frac{A}{2}-\tan \frac{B}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}} \equiv \frac{a-b}{c}}$.

Now, imagine graphing an angle bisector from angle $A$ and then from angle $B$, the point where they intersect, let's call it $K$. From that point drop a perpendicular on $AB$, let's call that point $L$. Hence, $\tan \frac{A}{2} = \frac{KL}{AL}$ and $\tan \frac{B}{2} = \frac{KL}{LB}$. Plugging those in, gives us: $$\frac{LB-AL}{c} \equiv \frac{a-b}{c}$$ And now I have no clue how to show that $LB-AL = a-b$.

If you could let me know how to show that and/or you know a better way of proving the identity, please share.


2 Answers 2


Using sine law, $$\dfrac{a-b}c=\dfrac{\sin A-\sin B}{\sin C}$$

Using Prosthaphaeresis & Double Angle Formula,

$$\dfrac{\sin A-\sin B}{\sin C}=\dfrac{2\sin\dfrac{A-B}2\cos\dfrac{A+B}2}{2\sin\dfrac C2\cos\dfrac C2}$$

Now $\dfrac{A+B}2=\dfrac\pi2-\dfrac C2\implies\cos\dfrac{A+B}2=?$

Finally, if $\sin\dfrac C2=0,\dfrac C2=n\pi\iff C=2n\pi$ where $n$ is any integer

But $0<C<\pi\implies\sin\dfrac C2\ne0$

  • $\begingroup$ $\cos \frac{C}{2} \equiv \sin \frac{A+B}{2}$ and $\sin \frac{C}{2} \equiv \cos \frac{A+B}{2}$. Thanks! I still wonder how I could have proceeded with my original proof. $\endgroup$
    – Naz
    Aug 12, 2015 at 17:26

Here's a trigonograph:

enter image description here

(This space intentionally left blank.)

  • $\begingroup$ hmm, i tried to follow the link to see whether I can plot something similar to the above, but cannot seem to find where exactly I am able to do it. As I understand it, it is your website? $\endgroup$
    – Naz
    Aug 12, 2015 at 19:35
  • $\begingroup$ oh, so that's the watermark.. What software do you use to plot those? $\endgroup$
    – Naz
    Aug 12, 2015 at 19:38
  • $\begingroup$ @isquared-KeepitReal: [trigonography.com](trigonography.com) is indeed my website (currently directing to a section of my math blog), where I intend to compile my growing list of trigonographs. As for the software, I use [GeoGebra](GeoGebra.org). $\endgroup$
    – Blue
    Aug 12, 2015 at 20:27
  • $\begingroup$ Awesome! Thanks! You should include this in your blog, not being biased, of course :) $\endgroup$
    – Naz
    Aug 12, 2015 at 20:29
  • 1
    $\begingroup$ @isquared-KeepitReal: I tend to get distracted between posting something here and re-posting it to my blog. :) Eventually, I hope to get everything to match. $\endgroup$
    – Blue
    Aug 12, 2015 at 20:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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