化简(2cos²α-1)/2tan(π/4+α)sin²(π/4-α)
问题描述：
化简(2cos²α-1)/2tan(π/4+α)sin²(π/4-α)
答
=(2cos²α-1)/2tan(π/4+α)sin²[π/2-(π/4+α)]
=(2cos²α-1)/2tan(π/4+α)cos²(π/4+α)
=cos2α/2tan(π/4+α)cos²(π/4+α)
=cos2a/2sin(π/4+α)cos(π/4+α)
=cos2a/sin2(π/4+α)
=cos2a/sin(π/2+2α)
=cos2a/cos2a
=
答
(2cos²α-1)/2tan(π/4+α)sin²(π/4-α)
=(2cos²α-1)/2tan(π/4+α)sin²[π/2-(π/4+α)]
=(2cos²α-1)/2tan(π/4+α)cos²(π/4+α)
=cos2α/2tan(π/4+α)cos²(π/4+α)
=cos2a/2sin(π/4+α)cos(π/4+α)
=cos2a/sin2(π/4+α)
=cos2a/sin(π/2+2α)
=cos2a/cos2a
=1