Question

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

Answer 1

$\begin{array}{c}a\cos \left(\frac{b-c}{2}\right)=\left(b+c\right)\sin \frac{A}{2}\\ In\Delta ABC\\ here\frac{a}{b+c}=\frac{2R\sin A}{R\sin B+2R\sin c}\\ =\frac{\sin A}{\sin B+\sin C}\\ =\frac{2\sin A/2\cos A/2}{2\sin \left(\frac{b+c}{2}\right)\cos \left(\frac{B-c}{2}\right)}\\ =\frac{\sin A/2\cos A/2}{\sin \left(\frac{180-A}{2}\right)}\cos \left(\frac{b-c}{2}\right)\left[A+B+C={180}^0\right]\\ =\frac{\sin A/2\cos A/2}{\cos A/2\cos \left(\frac{B-c}{2}\right)}\\ \frac{a}{b+c}=\frac{\sin A/2}{\cos \left(\frac{B-c}{2}\right)}\\ =a\cos \left(\frac{B-c}{2}\right)=\left(b+c\right)\sin A/2proved.\end{array}$