∵α,β∈[-π/2,π/2] α+β<0,若sinα=1/3
∴α=arcsin1/3
-π/2<β<-arcsin1/3
∴sin(-π/2)<sinβ<sin(-arcsinβ1/3)
即 -1<sinβ<-1/3
有 -1<1-a<-1/3
→-2<-a<-4/3
→4/3<a<2
即实数a的取值范围是(-4/3,2)
∵α,β∈[-π/2,π/2] α+β<0,若sinα=1/3
∴α=arcsin1/3
-π/2<β<-arcsin1/3
∴sin(-π/2)<sinβ<sin(-arcsinβ1/3)
即 -1<sinβ<-1/3
有 -1<1-a<-1/3
→-2<-a<-4/3
→4/3<a<2
即实数a的取值范围是(-4/3,2)