In ∆ABC, prove that a

Solution:

Consider

$a(\cos C-\cos B)$

$=k(\sin A \cos C-\sin A \cos B) \quad\left[\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\right]$

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

$=\frac{k}{2}[\sin (A+C)+\sin (A-C)-\sin (A+B)-\sin (A-B)]$

$=\frac{k}{2}[\sin (\pi-\mathrm{B})+\sin (\mathrm{A}-\mathrm{C})-\sin (\pi-\mathrm{C})-\sin (\mathrm{A}-\mathrm{B})] \quad(\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi)$

$=\frac{k}{2}[\sin B-\sin C+\sin (A-C)-\sin (A-B)]$

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

$=k \sin \left(\frac{B-C}{2}\right)\left[\cos \left(\frac{\pi}{2}-\frac{\mathrm{A}}{2}\right)+\cos \left\{\frac{2 \mathrm{~A}-(\pi-\mathrm{A})}{2}\right\}\right]$

$=k \sin \left(\frac{B-C}{2}\right)\left(\sin \frac{A}{2}+\sin \frac{3 A}{2}\right)$

$=k \sin \left(\frac{B-C}{2}\right)\left[2 \sin \left(\frac{\frac{A}{2}+\frac{3 A}{2}}{2}\right) \cos \left(\frac{\frac{3 A}{2}-\frac{A}{2}}{2}\right)\right]$

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

$=4 k \sin \left(\frac{B-C}{2}\right) \sin \frac{A}{2} \cos ^{2} \frac{A}{2}$

Now,

Consider

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

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

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

$=4 k \sin \left(\frac{B-C}{2}\right) \cos \left(\frac{\pi}{2}-\frac{A}{2}\right) \cos ^{2} \frac{A}{2}$

$=4 k \sin \left(\frac{B-C}{2}\right) \sin \frac{A}{2} \cos ^{2} \frac{A}{2} \quad \ldots(2)$

From (1) \& (2), we get

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

Hence proved.