用三角函数万能公式化简 y=(1-sinx)/(1+cosx),得y=1/2×(tan a/2-1)^2,
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/06/03 19:48:25
用三角函数万能公式化简 y=(1-sinx)/(1+cosx),得y=1/2×(tan a/2-1)^2,
就是用万能公式做的:
万能公式:sinx=2tan(x/2)/(1+tan(x/2)^2)
cosx=(1-tan(x/2)^2)/(1+tan(x/2)^2)
令:t=tan(x/2)
y=(1-2t/(1+t^2))/(1+(1-t^2)/(1+t^2))
=(t-1)^2/2
=(1/2)(tan(x/2)-1)^2
求导是一样的:
y=(1-sinx)/(1+cosx)
y'=(-cosx(1+cosx)+(1-sinx)sinx)/(1+cosx)^2
=(sinx-cosx-1)/(4cos(x/2)^4)
=(2sin(x/2)cos(x/2)-2cos(x/2)^2)/(4cos(x/2)^4)
=2cos(x/2)(sin(x/2)-cos(x/2))/(4cos(x/2)^4)
=(sin(x/2)-cos(x/2))/(2cos(x/2)^3)
y=(1/2)(tan(x/2)-1)^2
y'=(1/2)(2(tan(x/2)-1)sec(x/2)^2/2)
=(1/2)((tan(x/2)-1)/cos(x/2)^2
=(1/2)(sin(x/2)/cos(x/2)-1)/cos(x/2)^2
=(1/2)(sin(x/2)-cos(x/2))/cos(x/2)^3
万能公式:sinx=2tan(x/2)/(1+tan(x/2)^2)
cosx=(1-tan(x/2)^2)/(1+tan(x/2)^2)
令:t=tan(x/2)
y=(1-2t/(1+t^2))/(1+(1-t^2)/(1+t^2))
=(t-1)^2/2
=(1/2)(tan(x/2)-1)^2
求导是一样的:
y=(1-sinx)/(1+cosx)
y'=(-cosx(1+cosx)+(1-sinx)sinx)/(1+cosx)^2
=(sinx-cosx-1)/(4cos(x/2)^4)
=(2sin(x/2)cos(x/2)-2cos(x/2)^2)/(4cos(x/2)^4)
=2cos(x/2)(sin(x/2)-cos(x/2))/(4cos(x/2)^4)
=(sin(x/2)-cos(x/2))/(2cos(x/2)^3)
y=(1/2)(tan(x/2)-1)^2
y'=(1/2)(2(tan(x/2)-1)sec(x/2)^2/2)
=(1/2)((tan(x/2)-1)/cos(x/2)^2
=(1/2)(sin(x/2)/cos(x/2)-1)/cos(x/2)^2
=(1/2)(sin(x/2)-cos(x/2))/cos(x/2)^3
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