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(sinC)^2=2sinAsinB
c^2=2ab
a^2+b^2-c^2=2abcosC
6abcosC-2ab=2abcosC
4cosC=2
cosC=1/2
C=π/3
2)2R=a/sinA=b/sinB=c/sinC=√3/√3/2=2 ,
b-a=2RsinB-2RsinA=2sin(2π/3-A)-2sinA
=2(√3/2cosA+1/2sinA)-2sinA
=√3cosA-sinA
=2(√3/2cosA-1/2sinA)
=2cos(A+π/6)
A∈(0,2π/3)
A+π/6∈(π/6,5π/6)
cos(A+π/6)在(π/6,5π/6)上值域为:(-1/2,√3/2)
2cos(A+π/6)在(π/6,5π/6)上值域为:(-1,√3)
b-a∈(-1,√3)
c^2=2ab
a^2+b^2-c^2=2abcosC
6abcosC-2ab=2abcosC
4cosC=2
cosC=1/2
C=π/3
2)2R=a/sinA=b/sinB=c/sinC=√3/√3/2=2 ,
b-a=2RsinB-2RsinA=2sin(2π/3-A)-2sinA
=2(√3/2cosA+1/2sinA)-2sinA
=√3cosA-sinA
=2(√3/2cosA-1/2sinA)
=2cos(A+π/6)
A∈(0,2π/3)
A+π/6∈(π/6,5π/6)
cos(A+π/6)在(π/6,5π/6)上值域为:(-1/2,√3/2)
2cos(A+π/6)在(π/6,5π/6)上值域为:(-1,√3)
b-a∈(-1,√3)
追问
但是答案是(-/3,/3)啊
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