sinα=±√[1-(cosα)^2]=±12/13∵α∈（0,π/2）∴sinα=12/13令α+β=θ ,则cos(α+β)=cosθsinθ=±√[1-(cosθ)^2]=±3/5∵α,β∈（0,π/2）∴α+β∈（0,π）即：θ∈（0,π）∴sinθ=3/5即：sin(α+β)=3/5...