α∈(0,π/2),cosα=3/5,
∴sinα=4/5,
β∈(-π/2,0),α+β∈(-π/2,π/2),sin(α+β)=-√2/4,
∴cos(α+β)=√14/4,
∴sinβ=sin(α+β-α)=sim(α+β)cosα-cos(α+β)sinα
=-(3√2+4√14)/20.
α∈(0,π/2),β∈(-π/2,0),则α+β∈(-π/2,π/2); cos(α+β)>0
由sin(α+β)=-√2/4得:cos(α+β)=√[1-(-√2/4)²]=√14/4;
cosα=3/5;α∈(0,π/2) 所以:sinα=4/5;
那么:sinβ=sin[(α+β)-α]=sin(α+β)cosα-cos(α+β)sinα
=(-√2/4)×(3/5)-(√14/4)×(4/5)=-(3√2+4√14)/20