正弦解三角形:在△ABC中,求证:a sin(B-C)+b sin(C-A)+c sin(A-B)=0

由正弦定理得a=2RsinA,b=2RsinB,c=2RsinC,故有
asin(B-C)+bsin(C-A)+csin(A-B)
=2R(sinAsin(B-C)+sinBsin(C-A)+sinCsin(A-B))
=2R(sinA(sinBcosC-cosBsinC)+sinB(sinCcosA-cosCsinA)+sinC(sinAcosB-cosAsinB))
=2R(sinAsinBcosC-sinAcosBsinC+sinBsinCcosA-sinBcosCsinA+sinCsinAcosB-sinCcosAsinB)=0

由正弦定理,知:
a/sinA=b/sinB=c/sinC=2R
故欲证
asin(B-C)+bsin(C-A)+csin(A-B)=0
即证sinAsin(B-C)+sinBsin(C-A)+sinCsin(A-B)=0(积化和差)
<==>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
<==>[cos(A+B-C)-cos(A+B-C)]+[cos(B+C-A)-cos(B+C-A)]+[cos(A+C-B)-cos(A+C-B)]=0
而末式显然成立.
故原等式成立.