已知-π/2
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已知-π/2已知-π/2已知-π/2∵sinx+cosx=1/5∴(sinx+cosx)^2=(sinx)^2+(cosx)^2+2sinxcosx=1+2sinxcosx=(1/5)^2=1/25∴
已知-π/2
已知-π/2
已知-π/2
∵sinx+cosx=1/5
∴(sinx+cosx)^2=(sinx)^2+(cosx)^2+2sinxcosx=1+2sinxcosx=(1/5)^2=1/25
∴2sinxcosx=-24/25
∴(cosx-sinx)^2=(cosx)^2+(sinx)^2-2sinxcosx=1+24/25=49/25
∵-π/2
∴cosx-sinx>0
∴cosx-sinx=7/5
∴[sin(2x)+2(sinx)^2]/(1-tanx)
=[2sinxcosx+2(sinx)^2]/(1-sinx/cosx)
=[2sinx(cosx)^2+2(sinx)^2(cosx)]/(cosx-sinx)
=2sinxcosx(sinx+cosx)/(cosx-sinx)
=(-24/25)×(1/5)/(7/5)
=-25/168