Question

In any triangle ABC, prove the following:

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

Answer 1

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

Then,
Consider the RHS of the equation $\sin \left(\frac{B-C}{2}\right)=\frac{b-c}{a}\cos \frac{A}{2}$
$$\begin{gathered} \mathrm{RHS}=\frac{b-c}{a}\cos \frac{A}{2} \\ =\frac{k\left(\sin B-\sin C\right)}{k\sin A}\cos \left(\frac{\mathrm{\pi }-\left(\mathrm{B}+\mathrm{C}\right)}{2}\right)\left(\because A+B+C=\mathrm{\pi }\right) \\ =\frac{2\sin \left({\displaystyle \frac{B-C}{2}}\right)\cos \left({\displaystyle \frac{B+C}{2}}\right)}{\sin A} \\ =\frac{\sin \left({\displaystyle \frac{B-C}{2}}\right)2\cos \left({\displaystyle \frac{B+C}{2}}\right)}{\sin A}\sin \left(\frac{B+C}{2}\right) \\ =\frac{\sin \left({\displaystyle \frac{B-C}{2}}\right)\sin \left(B+C\right)}{\sin A} \\ =\frac{\sin \left({\displaystyle \frac{B-C}{2}}\right)\sin \left(\mathrm{\pi }-\mathrm{A}\right)}{\sin A} \\ =\frac{\sin A\sin \left({\displaystyle \frac{B-C}{2}}\right)}{\sin A} \\ =\sin \left(\frac{B-C}{2}\right)=\mathrm{LHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$