Question

In $\Delta ABC$, prove that: $(b+c)\cos \left(\frac{B+C}{2}\right)=a\cos \left(\frac{B-C}{2}\right)$

Answer 1

We have, $\frac{a}{b+c}$$=\frac{2R\sin A}{2R\sin B+2R\sin C}$ {Using Sine rule]

$=\frac{\sin A}{\sin B+\sin C}$

$=\frac{2\sin \frac{A}{2}.\cos \frac{A}{2}}{2\sin \frac{B+C}{2}.\cos \frac{B-C}{2}}$

$=\frac{\sin \frac{\pi -(B+C)}{2}.\cos \frac{\pi -(B+C)}{2}}{\sin \frac{B+C}{2}.\cos \frac{B-C}{2}}$ [Since $A+B+C=\pi$]

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

So, $(b+c)\cos \left(\frac{B+C}{2}\right)=a\cos \left(\frac{B-C}{2}\right)$