化简(2cos²α-1)/[2tan(π/4-α)sin²(π/4+α)]

来源：学生作业帮编辑：大师作文网作业帮分类：数学作业时间：2024/11/12 07:30:42

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

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