（2cos^2-1）/2tan(π/4-α)sin^2(π/4+α)
问题描述：
（2cos^2-1）/2tan(π/4-α)sin^2(π/4+α)
答
（2cos^2α-1）/2tan(π/4-α)sin^2(π/4+α)
=（1+cos2α-1）/[2tan(π/4-α)sin^2(π/4+α)]
=cos2x/[2tan(π/4-α)sin^2(π/4+α)]
2tan(π/4-α)sin^2(π/4+α)=[2(sin(π/2-2α))/(cos(π/2-2α)+1)]x[(1-cos(2α+π/2))/2]
=[2(cos2α/(sin2α+1)]*[(1+sin2α)/2]
=cos2x