a,b∈(3π/4,π),a+b∈(3π/2,2π),则sin(a+b)=-3/5,cos(a+b)=4/5b-π/4∈(π/2,3π/4),sin(b-π/4)=12/13,则cos(b-π/4)=-5/13cos(a+π/4)=cos[(a+b)-(b-π/4)]=cos(a+b)cos(b-π/4) + sin(a+b)sin(b-π/4) = 4/5 * (-5/13) + (-3/5) * 12/13 = -56/65