三角形ABC中acosB-bsinB=C,sinA+sinB取值范围

三角形ABC中acosB-bsinB=C,sinA+sinB取值范围

acosB-bsinB=c
根据正弦定理
a=2RsinA,b=2RsinB,c=2RsinC
∴sinAcosB-sinB*sinB=sinC
又sinC=sin(A+B)=sinAcosB+cosAsinB
∴sinAcosB-sinB*sinB=sinAcosB+cosAsinB
∴-sinB*sinB=cosAsinB
∴sinB=-cosA
=-sin(π/2-A)=sin(A-π/2)
A+B=2A-π/2<π
∴2A<3π/2
∴A<3π/3
∵B=A-π/2>0,A>π/2
∴π/2那么sinA+sinB
=sinA-cosA
=√2sin(A-π/4)
π/4∴√2/2∴1<√2sin(A-π/4)<√2
即sinA+sinB的取值范围是(1,√2)