Question

If b+c=2a, then prove that that $cos\left(\frac{B-C}{2}\right)=2sin\frac{A}{2}$.

Answer 1

Given $b+c=2a$

$⟹2R\sin B+2R\sin C=4R\sin A$

$⟹\sin B+\sin C=2\sin A$

$⟹2\sin (\frac{B+C}{2})\cos (\frac{B-C}{2})=4\sin \frac{A}{2}\cos \frac{A}{2}$

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

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

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