∵α、β属于[-π/2,π/2],则α+β∈[-π,π],
sinαcosβ+sinβcosα=sin(α+β)=1
∴α+β=π/2
∴sinα+sinβ=sinα+sin(π/2-α)=sinα+cosα=√2sin(π/4+α)
∵α属于[-π/2,π/2]
∴π/4+α∈[-π/4,3π/4]
∴sin(π/4+α)∈[-√2/2,1]
∴√2sin(π/4+α)∈[-1,√2]
即sinα+sinβ的取值范围是[-1,√2]
∵α、β属于[-π/2,π/2],则α+β∈[-π,π],
sinαcosβ+sinβcosα=sin(α+β)=1
∴α+β=π/2
∴sinα+sinβ=sinα+sin(π/2-α)=sinα+cosα=√2sin(π/4+α)
∵α属于[-π/2,π/2]
∴π/4+α∈[-π/4,3π/4]
∴sin(π/4+α)∈[-√2/2,1]
∴√2sin(π/4+α)∈[-1,√2]
即sinα+sinβ的取值范围是[-1,√2]