已知α,β属于(3TT/4,TT),sin(α+β)=-3/5,sin(β-TT/4)=12/13,则cos(α+TT/4)=

∵α,β属于(3π/4,π)
∴3π/2<α+β<2π,π/2<β-π/4<3π/4
∴cos(α+β)>0,cos(β-π/4)<0
∵sin(α+β)=-3/5,sin(β-π/4)=12/13
∴cos(α+β)=√[`1-sin²(α+β)]=4/5
cos(β-π/4)=-√[1-sin²(β-π/4)]=-5/13
故cos(α+π/4)=cos[(α+β)-(β-π/4)]
=cos(α+β)cos(β-π/4)+sin(α+β)sin(β-π/4)
=(4/5)(-5/13)+(-3/5)(12/13)
=-20/65-36/65
=-56/65