向量a=(1+cosA,sinA)=(2cos^2A/2,2sinA/2cosA/2)=2cosA/2(cosA/2,sinA/2)
向量b=(1-cosB,sinB)=(2sin^2B/2,2sinB/2cosB/2)=2sinB/2(sinB/2,cosA/2)
∴|向量a|=2cosA/2,|向量b|=2sinB/2
而向量a·向量c=(1+cosA,sinA)·(1,0)=1+cosA
向量b·向量c=(1-cosB,sinB)·(1,0)=1-cosB
∴cosC=向量a·向量c/(|向量a|·|向量c|)=(1+cosA)/(2cosA/2×1)=cosA/2
∵0<A<π
∴0<A/2<π/2
∴C=A/2
cosD=向量b·向量c/(|向量b|·|向量c|)=(1-cosB)/(2sinB/2×1)=sinB/2
∵π<B<2π
∴π/2<B/2<π
∴cosD=sinB/2=cos(B/2-π/2)
∴D=B/2-π/2
∴A/2-(B/2-π/2)=π/6
即:A-B=-2π/3