Question

asin(B-C)+bsin(C-A)+csin(a-B)=0

Answer 1

a=2R·sinA
a·sin(B-C)=2R·sinA·sin(B-C)=2R·1/2·[cos(A-B+C)-cos(A+B-C)]=R[cos(A-B+C)-cos(A+B-C)]
Facand permutari ciclice obtinem:
a·sin(B-C)+b·sin(C-A)+c·sin(A-B)=
=R[cos(A-B+C)-cos(A+B-C)+cos(B-C+A)-cos(B+C-A)+cos(C-A+B)-cos(C+A-B)]=0
Pe ultimul rand, evidentiati  cu acelasi format se reduc si obtinem 0.