已知sin(α+π/4)=1/2.α∈(π/2,π),求sin2α,cos2α,tan2α

sin(α+π/4)=1/2则cos(α+π/4)=-[√1-(1/2)^2]=-(√3)/2α∈(π/2,π)且α+π/4∈(π/2,π)则2α∈(π,3π/2)sin(2α+π/2)=2sin(α+π/4)cos(α+π/4)=-(√3)/2cos2α=-(√3)/2sin2α=-1/2tan2α=(√3)/3...