将f(x)表示成cosx的多项式
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/19 06:28:30
将f(x)表示成cosx的多项式
设g(x)=sin(5x/2)-sin(x/2)
=sin(x/2+2x)-sin(x/2)
=[sin(x/2)cos2x+sin2xcos(x/2)]-sin(x/2)
=sin(x/2)[cos2x-1]+sin2xcos(x/2)
=2sin(x/2)(sinx)^2+2sinxcosxcos(x/2)
=2sinx[sin(x/2)sinx+cosxcos(x/2)]
因为cosx=cos-x
所以g(x)=2sinxcos(2x-x/2)=2sinxcos(3x/2)
f(x)=g(x)/[2sin(x/2)]
=2sinxcos(3x/2)/[2sin(x/2)]
=sinxcos(3x/2)/[sin(x/2)]
=2cos(x/2)cos(3x/2)
=2cos(x/2)cos(2x-x/2)
=2cos(x/2[cos2xcos(x/2)+sin2xsin(x/2)]
=2cos(x/2)cos2xcos(x/2)+2cos(x/2)sin2xsin(x/2)
=2cos2x(cos(x/2))^2+sin2xsinx
=cos2x(1+cosx)+2(sinx)^2cosx
=cos2x+cosx[cos2x+2(sinx)^2]
=cos2x+cosx[1-(sinx)^2+(sinx)^2)]
=cos2x+cosx
=2(cosx)^2+cosx-1
再问: =sin(x/2)[cos2x-1]+sin2xcos(x/2) =2sin(x/2)(sinx)^2+2sinxcosxcos(x/2) 这个步骤中是不是少了一个负号~
再答: 不好意思,你说下加在哪,我没看出来。。。
再问: cos2x不是等于1-2(sinx)²么~
再答: 嗯嗯,那里少了负号,多谢指正 =sin(x/2)[cos2x-1]+sin2xcos(x/2) =2sinx[-sin(x/2)(sinx)+cosxcos(x/2)] =2sinxcos(x+1/2x) =2sinxcos(3x/2)
=sin(x/2+2x)-sin(x/2)
=[sin(x/2)cos2x+sin2xcos(x/2)]-sin(x/2)
=sin(x/2)[cos2x-1]+sin2xcos(x/2)
=2sin(x/2)(sinx)^2+2sinxcosxcos(x/2)
=2sinx[sin(x/2)sinx+cosxcos(x/2)]
因为cosx=cos-x
所以g(x)=2sinxcos(2x-x/2)=2sinxcos(3x/2)
f(x)=g(x)/[2sin(x/2)]
=2sinxcos(3x/2)/[2sin(x/2)]
=sinxcos(3x/2)/[sin(x/2)]
=2cos(x/2)cos(3x/2)
=2cos(x/2)cos(2x-x/2)
=2cos(x/2[cos2xcos(x/2)+sin2xsin(x/2)]
=2cos(x/2)cos2xcos(x/2)+2cos(x/2)sin2xsin(x/2)
=2cos2x(cos(x/2))^2+sin2xsinx
=cos2x(1+cosx)+2(sinx)^2cosx
=cos2x+cosx[cos2x+2(sinx)^2]
=cos2x+cosx[1-(sinx)^2+(sinx)^2)]
=cos2x+cosx
=2(cosx)^2+cosx-1
再问: =sin(x/2)[cos2x-1]+sin2xcos(x/2) =2sin(x/2)(sinx)^2+2sinxcosxcos(x/2) 这个步骤中是不是少了一个负号~
再答: 不好意思,你说下加在哪,我没看出来。。。
再问: cos2x不是等于1-2(sinx)²么~
再答: 嗯嗯,那里少了负号,多谢指正 =sin(x/2)[cos2x-1]+sin2xcos(x/2) =2sinx[-sin(x/2)(sinx)+cosxcos(x/2)] =2sinxcos(x+1/2x) =2sinxcos(3x/2)
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