已知α ∈(π/2,π),sin α =3 / 5
则cosα =√(1-sin²α )=-4/5
所以tanα=sinα/cosα=(3/5)/(-4/5)=-3/4
故tan(α+π/4)=[tanα+tan(π/4)]/[1-tanα*tam(π/4)]
=(-3/4+1)/(1+3/4)=1/7
tan(2α+π/4)=[tan(α+π/4)+tanα]/[1-tan(α+π/4)tanα]
=(1/7-3/4)/[1-(1/7)*(-3/4)]
=(4-21)/(28+3)
=-17/31