116问答网 > 已知cos（π⼀4+x）=3⼀5,17π⼀12＜x小于7π⼀4，求sin2x+sin^2x⼀1-tanx

已知cos（π⼀4+x）=3⼀5,17π⼀12＜x小于7π⼀4，求sin2x+sin^2x⼀1-tanx

2024-11-17 22:18:10

推荐回答（1个）

回答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�0�5(x+π/4)]=-√[1-(3/5)�0�5]=-4/5
sin(2x)=-cos(2x+π/2)=-cos[2(x+π/4)]=1-2cos�0�5(x+π/4)=1-2�6�1(3/5)�0�5=7/25
[sin(2x)+2sin�0�5x]/(1-tanx)
=2(sinxcosx+sin�0�5x)/(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�6�17/25/(3/5)
=-28/75