Question

In $△ABC$ prove that $\sin \left(\frac{B-C}{2}\right)=\left(\frac{b-c}{a}\right)\cos \frac{A}{2}$

Answer 1

To prove

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

$R.H.S.=\left(\frac{b-c}{a}\right)\cos \frac{A}{2}$

$\left(\frac{k\sin B-k\sin C}{k\sin A}\right)\cos \frac{A}{2}$ $[\text{using sine rule}:\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k]$

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

$=\frac{2\left(\frac{B+C}{2}\right)\sin \left(\frac{B-C}{2}\right)}{2\sin \frac{A}{2}\cos \frac{A}{2}}$ $[\text{using C \& D rule}\sin C-\sin D=2\cos \frac{C+D}{2}\sin \frac{C-D}{2}]$

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

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