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