m=(1,sqrt(3)),n=(sin(π-A),sin(A-π/2))=(sinA,-cosA),m与n垂直,则：
m·n=(1,sqrt(3))·(sinA,-cosA)=sinA-sqrt(3)cosA=2sin(A-π/3)=0,因A是内角
故：0y=2sinB^2+cos(π/3-2B)=1-cos2B+cos2B/2+sqrt(3)sin2B/2=1+sqrt(3)sin2B/2-cos2B/2
=1+sin(2B-π/6)---------------------后面求什么?