Question

在△abc中,角a,b,c所对的边分别为a,b,c,已知c=根号3,角C=π/3(1)若2sin2A+sin(A+B

Answer 1

2sin2A+sin(A-B)=sinC
2sin2A+sin(A-B)=sin(A+B)
4sinAcosA+sinAcosB-cosAsinB=sinAcosB+cosAsinB
4sinAcosA-2cosAsinB=0
cosA(2sinA-sinB)=0
cosA[2sinA-sin(2π/3-B)]=0
cosA[2sinA-√3/2cosB-1/2sinB]=0
cosA(3/2sinA-√3/2cosB)=0
√3cosA(√3/2sinB-1/2cosB)=0
√3cosAsin(B-π/6)=0
cosA=0 或sin(B-π/6)=0
A=π/2 或 B=π/6
2)2R=a/sinA=b/sinB=c/sinC=√3/(√3/2)=2
a=2RsinA=2sinA ,b=2RsinB=2sinB ,
a+b+c=√3+2sinA +2sinB
= √3+2sinA +2sin(2π/3-A)
= √3+2sinA +2*√3/2cosA+2*1/2sinA
=√3+2(3/2sinA+√3/2cosA)
=√3+2√3(√3/2sinA+1/2cosA)
=√3+2√3sin(A+π/6)
0<A<2π/3
π/6<A+π/6<5π/6
sin(A+π/6)在π/6<A+π/6<5π/6上值域为:(1/2,1]
√3+2√3sin(A+π/6)在π/6<A+π/6<5π/6上值域为:(2√3,3√3]
所以,a+b+c范围是:(2√3,3√3]