Question

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

Answer 1

REF.Image

$\frac{a}{sinA}=\frac{b}{sinB}=\frac{C}{sinC}$ [Sin Formula]

= x(say)

x sin A = a

x sin B = b

x sin c = c

$A+B+C={180}^{\circ }$

$\Rightarrow \frac{B+C}{2}=90-\frac{A}{2}$

$\Rightarrow cos\left(\frac{B+C}{2}\right)=$

$cos(90-\frac{A}{2})$

$\Rightarrow cos\frac{(B+C)}{2}=sin\frac{A}{2}$

$\frac{b-c}{a}=\frac{x(sinB-sinC)}{xsinA}$

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

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