题目内容：

判断△ABC的形状:(a²+b²)sin(A-B)=(a²-b²)sin(A+B)

优质解答

a/sinA=b/sinB=k(k不为零),即a=ksinA,b=ksinB.
(a²+b²)sin(A-B)=(a²-b²)sin(A+B)
b^2(sin(A-B)+sin(A+B))=a^2sin(A+B)-sin(A-B))
b^2sinAcosB=a^2sinBcosA
b^2×a/k×cosB=a^2×b/k×cosA
bcosB=acosA
ksinBcosB=ksinAcosA
sin(A-B)=0
A=B.
等腰三角形