若cos(π/4+x)=3/5,17/12π
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若cos(π/4+x)=3/5,17/12π
若cos(π/4+x)=3/5,17/12π
若cos(π/4+x)=3/5,17/12π
∵(17/12)π
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...
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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
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cos(π/4+x)=3/5,17π/12
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cos(π/4+x)=3/5,17π/12
cos(π/4+x)=3/5,17π/12
cos(π/4+x)=3/5,17/12π
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若cos(π/4+x)=3/5 17π/12
若cos(π/4+x)=3/5,17/12π