已知sinα=5/13,且α∈(π/2,π),求cos2α,sin2α及sinα/2的值.

已知sinα=5/13,且α∈(π/2,π)
所以cosα<0,即cosα=-根号(1-25/169)=-12/13
cos2α=(cosα)^2-(sinα)^2=119/169
sin2α=2sinαcosα=-120/169
[sin(α/2)]^2=(1-cosα)/2=25/26,α∈(π/2,π),α/2∈(π/4,π/2),sin(α/2)>0
所以sin(α/2)=5/根号26