已知cos(x+π/4)=3/5且17π/12<x<7π/4,求(sin2x+2sin^2x)/(1-tanx)的值过程,thanks

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已知cos(x+π/4)=3/5且17π/12<x<7π/4,求(sin2x+2sin^2x)/(1-tanx)的值过程,thanks
已知cos(x+π/4)=3/5且17π/12<x<7π/4,求(sin2x+2sin^2x)/
(1-tanx)的值
过程,thanks

已知cos(x+π/4)=3/5且17π/12<x<7π/4,求(sin2x+2sin^2x)/(1-tanx)的值过程,thanks
17π/12<x<7π/4,得5π/3<x+π/4<2π
cos(x-π/4)=cos[(x+π/4)-π/2]=sin(x+π/4)=-√[1-sin²(x+π/4)]=-√[1-(3/5)²]=-4/5
sin(2x)=-cos(2x+π/2)=-cos[2(x+π/4)]=1-2cos²(x+π/4)=1-2•(3/5)²=7/25
[sin(2x)+2sin²x]/(1-tanx)
=2(sinxcosx+sin²x)/(1-sinx/cosx)
=2(cosx+sinx)/(1/sinx-1/cosx)
=2(cosx+sinx)sinxcosx/(cosx-sinx)
=cos(x-π/4)sin(2x)/cos(x+π/4)
=-4/5•7/25/(3/5)
=-28/75

由cos(π/4+x)=3/5,得cos2(π/4+x)=2*cos(π/4+x)的平方-1=-7/25.而cos2(π/4+x)利用诱导公式=-sin2x.即sin2x=7/25,而sin2x=2sinxcosx=7/25,再联立Sin平方x+cos平方x=1,分别求出sinx和cosx(注意cosx有两个取值),自然tanx也可以求出,最后代进去进行运算就可以了