已知sin(α-β)=3/5,sin(α+β)=-3/5,且α-β∈(π/2,π),α+β∈(3π/2,2π),求cos2β,cos2α,sin2α的值
问题描述：
已知sin(α-β)=3/5,sin(α+β)=-3/5,且α-β∈(π/2,π),α+β∈(3π/2,2π),求cos2β,cos2α,sin2α的值
答
设:α+β=A,α-β=B,则2α=A+B,2β=A-BsinB=3/5===>cosB=-4/5,sinA=-3/5===>cosA=4/5cos2β=cos(A-B)=cosAcosB+sinAsinB=(4/5)*(-4/5)+(-3/5)*(3/5)=-1cos2α=cos(A+B)=cosAcosB-sinAsinB=(4/5)*(-4/5)-(-3/5)*(3/5)...