化简(2cos²α-1)/[2tan(π/4-α)sin²(π/4+α)]
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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

原式=(2cos²α-1)/[2tan(π/4-α)sin²(π/4+α)]

=cos2α / cos2α

=1

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