116问答网 > 化简(2cos눀α-1)⼀2tan(π⼀4+α)sin눀(π⼀4-α)

化简(2cos눀α-1)⼀2tan(π⼀4+α)sin눀(π⼀4-α)

2025-06-21 09:05:58

推荐回答（2个）

回答1：

=(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
=

回答2：

(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