已知A=π/3,则B+C=2π/3
则,sin²B+sin²C=sin²B+sin²(2π/3-B)
=sin²B+[(√3/2)cosB+(1/2)sinB]²
=sin²B+(3/4)cos²B+(1/4)sin²B+(√3/2)sinBcosB
=(5/4)sin²B+(3/4)cos²B+(√3/4)sin2B
=(3/4)+(1/2)sin²B+(√3/4)sin2B
=(3/4)+(1/4)(1-cos2B)+(√3/4)sin2B
=1+(1/2)[sin2B*(√3/2)-cos2B*(1/2)]
=1+(1/2)sin[2B-(π/6)]
当且仅当B=π/3时有最大值=3/2