化简(2sinxcosx-2sin²x)(2sinxcosx+2sin²x)/sin4x 两种方法
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/09/02 21:40:13
化简(2sinxcosx-2sin²x)(2sinxcosx+2sin²x)/sin4x 两种方法
(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
=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
sin²X-sinxcosx+2cos²x=
sin²x+sinxcosx+2cos²x=
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