0°<α+β<180°cos(α+β)=-5/13,cos(α-β)=3/5sin(α+β)=√[1-(-5/13)^2=12/13sin(α-β)=±√[1-(3/5)^2=±4/5cos(2β)=cos[(α+β)-(α-β)]=cos(α+β)cos(α-β)-sin(α+β)sin(α-β)=-5/13*3/5-12/13*(±4/5)∴cos(2β)=-63/65,或cos(2β)=33/65
