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

Question

已知cos（π/4+x）=3/5,17π/12＜x小于7π/4,求sin2x+sin^2x/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-sin��(x+π/4)]=-√[1-(3/5)��]=-4/5
sin(2x)=-cos(2x+π/2)=-cos[2(x+π/4)]=1-2cos��(x+π/4)=1-2��(3/5)��=7/25
[sin(2x)+2sin��x]/(1-tanx)
=2(sinxcosx+sin��x)/(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