cos[2(α - β/2)]=cos(2α - β)
=2cos²(α - β/2) - 1
=2•(√5/5)² - 1
=2•(1/5) - 1=-3/5
∴sin(2α - β)=±√1 - cos²(2α-β)
=±4/5
∵0<α<π/2,则0<2α<π
∴2α - β∈(0,3π/2)
∵cos(2α - β)<0
∴sin(2α - β)=4/5
∴cos2α=cos[(2α-β)+β]
=cos(2α-β)cosβ - sin(2α-β)sinβ
=(-3/5)•(4/5) - (4/5)•(-3/5)
=-12/25 + 12/25=0
∵0<2α<π
∴2α=π/2,则α=π/4