已知cos(四分之π+x)=五分之三,求1-tanx分之sin2x-2sin^2x的值
来源:学生作业帮 编辑:拍题作业网作业帮 分类:数学作业 时间:2024/06/11 00:28:17
已知cos(四分之π+x)=五分之三,求1-tanx分之sin2x-2sin^2x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-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
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
已知cos(四分之π+x)=五分之三,求1-tanx分之sin2x-2sin^2x的值
已知cos(4分之π+x)=5分之4,x属于(-2分之π,-4分之π),求(1+tanx)分之(sin2x-2*(sin
已知cos(π/4+x)=3\5,求(sin2x-2sin^x)/1-tanx的值
已知COS(π/4-X)=-4/5,求(SIN2X-2SIN^X)/(1+TANX)
已知cos(x+p/4)=4/5,求sin2x-2sin^2x/1-tanx的值
已知sin(x-45°)=四分之根号2,求sinxcosx,tanx+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cos(π/4+x)=3/5,x∈(3π/4,7π/4),求sin2x+2Sin^x /1-tanx的值
已知,cosx-sinx=五分之三倍根2,求sin2x-2sin^2x/1-tanx的值 麻烦写一下过程,O(∩_∩)O
已知函数fx=cosx-cos{x+π/2},x属于R.若fx等于四分之三,求sin2x的值
已知函数f(x)=sin2x+2sin(四分之派–x)cos(四分之派–x).
已知sin(x 4分之派)=负五分之三,则sin2x的值等于 怎么解决