若cos(π/4+x)=3/5,17/12π<x<7/4π,求(sin2x+2sinx)/(1-tanx)的值
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/08/27 02:04:06
若cos(π/4+x)=3/5,17/12π<x<7/4π,求(sin2x+2sinx)/(1-tanx)的值
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-cos²(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
cos(x-π/4)=cos[(x+π/4)-π/2]=sin(x+π/4)=-√[1-cos²(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
已知cos(π/4+x)=3/5,求[sin2x-2(sinx)^2]/(1-tanx)
已知cos(π/4+x)=3/5,17π/12<x<7π/4,求sin2x+2sin平方x/(1-tanx)的值
已知tan(x+5π/4)=2 (1)求tanx的值(2)求sinx+cosx/sinx-cosx +sin2x+cos
已知cos(π/4+x)=3\5,求(sin2x-2sin^x)/1-tanx的值
已知cos(x+45°)=0.6 求(1)sin2x (2)若17π/12<x<7π/4 则(sinx +sin2x·t
已知cos(π4+x)=35,17π12<x<7π4,求sin2x+2sin2x1-tanx的值.
已知cos(π/4+x)=-3/5,3/4π<x<5π/4,求(sin2x+2sin²x)/(1-tanx)的
sinx+cosx=1/5,0<x<π,求sinx、cosx、tanx,求sin2x-cos2x的值
已知cos(π/4+x)=3/5,x∈(3π/4,7π/4),求sin2x+2Sin^x /1-tanx的值
已知cos(π/4+x)=3/5,5/4π<x<7π/4,求(sin2x+2sin²x)/(1-tanx)
已知COS(π/4-X)=-4/5,求(SIN2X-2SIN^X)/(1+TANX)
已知cos{П/4 x}=3/5,17П/12∠7П/4,求sin2x 2sin2x/1-tanx的值