(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
(2cos^2-1)/2tan(π/4-α)sin^2(π/4+α)
(2cos^2-1)/2tan(π/4-α)sin^2(π/4+α)
数学人气:254 ℃时间:2019-12-24 10:10:25
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