∵tan(π/4 +α)=(tanπ/4+tanα)/(1-tanπ/4*tanα)=(1+tanα)/(1-tanα)=-1/2 ∴tanα=-3 ∴sinα/cosα=tanα=-3,即sinα=-3cosα 代入sinα+cosα=1,解得sinα=9/10,cosα=1/10 又∵sinα/cosα=-3<0 ∴sinαcosα<0 ∴sinαcosα=-√sinαcosα=-3/10 ∴sin2α=2sinαcosα=-3/5 ∴原式=(-3/5-2×1/10)/(1-3)=2/5