zlh198771
花朵
共回答了11个问题采纳率:100% 举报
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
1年前
追问
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zlh198771
因7π/12<x<7π/4,所以5π/6<π/4+x<2π 因为cos(π/4+x)=3/5 >0 所以tan(π/4+x)=-4/5 sin2x=-cos(2x+π/2)=1-2[cos(π/4+x)]^2=7/25 原式=sin2x(1+tanx)/(1-tanx)=sin2x×tan(π/4+x) =(7/25)×(-4/5)=-28/125