Question

化简(2sinxcosx-2sin²x)(2sinxcosx+2sin²x)/sin4x 两种方法

Answer 1

(2sinxcosx-2sin²x)(2sinxcosx+2sin²x)/sin4x
=4sin²x(cosx-sinx)(cosx+sinx)/(2sin2xcos2x)
=4sin²x(cos²x-sin²x)/(4sinxcosxcos2x)
=sinxcos2x/(cosxcos2x)
=sinx/cosx
=tanx

二倍角公式：
sin2a=2sinacosa
cos2a=cos²a-sin²a

(2sinxcosx-2sin²x)(2sinxcosx+2sin²x)/sin4x
=[sin2x-(1-cos2x)](sin2x+1-cos2x)/sin4x
=[(sin²2x-(1-cos2x)²]/(2sinxcos2x)
=[1-cos²2x-1+2cos2x-cos²2x)/(2sin2xcos2x)
=2cos2x(1-cos2x)/(2sin2xcos2x)
=(1-cos2x)/sin2x
=tanx