Question

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

Answer 1

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

追问

是7π，怎么17π也对

追答

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