∴<+α<π,-<-β<0.
又已知sin(+α)=,cos(-β)=,∴cos(+α)=-,sin(-β)=-
∴cos(α+β)=sin[+(α+β)]=sin[(+α)-(-β)]
=sin(+α)cos(-β)-cos(+α)sin(-β)
=×-(-)×(-)=-.