已知sin(α-β)=3/5,sin(α+β)=-3/5,且α-β∈(π/2,π),α+β∈(3π/2,2π),求cos

设:α+β=A,α-β=B,则2α=A+B,2β=A-B
sinB=3/5===>cosB=-4/5,sinA=-3/5===>cosA=4/5
cos2β=cos(A-B)=cosAcosB+sinAsinB=(4/5)*(-4/5)+(-3/5)*(3/5)=-1
cos2α=cos(A+B)=cosAcosB-sinAsinB=(4/5)*(-4/5)-(-3/5)*(3/5)=-7/25
sin2α=sin(A+B)=sinAcosB+sinBcosA=(-3/5)*(-4/5)+(3/5)*(4/5)=24/25