cos(b-a)=cos[(b-a+60°)-60°]=cos(b-a+60°)cos60°+sin(b-a+60)sin60°cos(b-a+60°)=cos[(120°+b)-(60°+a)]=cos(120°+b)cos(60°+a)+sin(120°+b)sin(60°+a）------（*）已知cos(60°+a)=-3/5,则sin(60°+a)...