# Proof of the identity: $c\sin \frac{A-B}{2} \equiv (a-b) \cos \frac{C}{2}$

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.

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

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

Here's a trigonograph:

(This space intentionally left blank.)

• 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?
– Naz
Aug 12, 2015 at 19:35
• oh, so that's the watermark.. What software do you use to plot those?
– Naz
Aug 12, 2015 at 19:38
• @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).
– Blue
Aug 12, 2015 at 20:27
• Awesome! Thanks! You should include this in your blog, not being biased, of course :)
– Naz
Aug 12, 2015 at 20:29
• @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.
– Blue
Aug 12, 2015 at 20:31