Question

Prove that $a\sin \left(B-C\right)+b\sin \left(C-A\right)+c\sin \left(A-B\right)=0$

Answer 1

Use $\sin$ formula

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

$\sin A=ak,\sin B=bk,\sin c=ck$

Also, $\sin (A-B)=\sin A..cosB-\cos A-\sin B$

$=ak\cos B-\cos A.bk$

$k(a\cos B-b\cos A)$

Similarity, $\sin (B-C)=k(b\cos C-c\cos B)$

$\sin (C-A)=k(c\cos A-a\cos c)$

$LHS=a\sin (B-C)+b\sin (C-A)+c\sin (A-B)$

$=ak(b\cos c-\cos B)+bk(c\cos A-a\cos C)+ck(a\cos B-b\cos A)$

$=k(bc\cos A-bc\cos A)+k(ac\cos B-ac\cos B)+(ab\cos -ab\cos c)$

$=0+0+0$

$=RHS$