In any triangle ABC, prove the following:

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

Then,

Consider the RHS of the equation $\sin \left(\frac{B-C}{2}\right)=\frac{b-c}{a} \cos \frac{A}{2}$

$\mathrm{RHS}=\frac{b-c}{a} \cos \frac{A}{2}$

$=\frac{k(\sin B-\sin C)}{k \sin A} \cos \left(\frac{\pi-(\mathrm{B}+\mathrm{C})}{2}\right) \quad(\because A+B+C=\pi)$

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

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

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

$=\frac{\sin \left(\frac{B-C}{2}\right) \sin (\pi-\mathrm{A})}{\sin A}$

$=\frac{\sin A \sin \left(\frac{B-C}{2}\right)}{\sin A}$

$=\sin \left(\frac{B-C}{2}\right)=$ LHS

Hence proved.