化简(2cos²α-1)/[2tan(π/4-α)sin²(π/4+α)]

题目

化简(2cos²α-1)/[2tan(π/4-α)sin²(π/4+α)]

答案

显然
tan(π/4-α)
=sin(π/4-α) / cos(π/4-α)
而cos(π/4-α)=sin[π/2 -(π/4-α)]=sin(π/4+α),
所以
2tan(π/4-α)sin²(π/4+α)
=2sin(π/4-α)sin²(π/4+α) / sin(π/4+α)
=2sin(π/4-α)sin(π/4+α)
=2sin(π/4-α)cos(π/4-α) 由公式sin2x=2sinx*cosx
=sin(π/2-2α)
=cos2α
而2cos²α-1=cos2α
故
原式=(2cos²α-1)/[2tan(π/4-α)sin²(π/4+α)]
=cos2α / cos2α
=1