In any triangle `A B C ,` prove that: `\ asin(B-C)+b s in(C-A)+csin(A-B)=0`

Explanation

`frac{sin A}{c}=frac{sin B}{c}=frac{sin C}{c}=k`

`sin A=ak, sin B=bk, sin C=ck`

`sin(A-B)=sin AcosB-cosA-sin B`

`=akcosB-cosAbk=>k(acosB-bcosA)`

`sin(B-C)=k(bcosC-c cosB)`

`sin(C-A)=k(c cosA-acosC)`

`=a sin(B-C)+bk(c cos A-a cosC)+ck(a cos B-b cos A)`

`=k(bc cos A-bc cosA)+k(ac cos B-ac cos B)+(ab cosC-ab cos C)`
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