Question

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

Answer 1

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

L.H.S $=a(\cos C-\cos B)$

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

$a=k\sin A,b=k\sin B,c=k\sin C$

Now L.H.S

$=k\sin A(\cos C-\cos B)$

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

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

$=4k\times \sin \left[\frac{\pi -(B+C)}{2}\right]\times \cos \frac{A}{2}\times \cos \frac{A}{2}\times \sin \left(\frac{B-C}{a}\right)$.

$=4k\times {\cos }^2\frac{A}{2}\times \cos \frac{B+C}{2}\times \sin \frac{B-C}{2}$

$=2k\times {\cos }^2\frac{A}{2}\times (2\cos \frac{B+C}{2}.\sin \frac{B-C}{2})$

$=2k\times {\cos }^2\frac{A}{2}\times (\sin B-\sin C)$

$=2k(\frac{b}{k}-\frac{c}{k}){\cos }^2\frac{A}{2}$.

$=2k\frac{(b-c)}{k}{\cos }^2\frac{A}{2}$

$=2(b-c){\cos }^2\frac{A}{2}$

=R.H.S