Question:

In ∆ABC, prove that:

$a \sin \frac{A}{2} \sin \left(\frac{B-C}{2}\right)+b \sin \frac{B}{2} \sin \left(\frac{C-A}{2}\right)+c \sin \frac{C}{2} \sin \left(\frac{A-B}{2}\right)=0$

Solution:

Consider

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

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

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

$(\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi)$

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

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

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

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

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

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

$=k(0)$

$=0$

Hence proved.

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