在三角形ABC中,已知a,b,c,分别是三个内角A,B,C的对边,2b–c/a=cosC/cosA
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答:
三角形ABC中:
(2b-c)/a=cosC/cosA
根据正弦定理:a/sinA=b/sinB=c/sinC=2R
所以:(2sinB-sinC)/sinA=cosC/cosA
所以:2sinBcosA-cosAsinC=sinAcosC
所以:2sinBcosA=cosAsinC+sinAcosC=sin(A+C)=sinB>0
所以:cosA=1/2
解得:A=60°
B+C=120°
y=√3sinB+sin(C-π/6)
=√3sin(2π/3-C)+sin(C-π/6)
=√3*[(√3/2)cosC+(1/2)sinC]+(√3/2)sinC-(1/2)cosC
=cosC+√3sinC
=2sin(C+π/6)
因为:0<C<2π/3)
所以:π/6<C+π/6<5π/6
所以:1/2<sin(C+π/6)<=1
所以:1<y<=2
三角形ABC中:
(2b-c)/a=cosC/cosA
根据正弦定理:a/sinA=b/sinB=c/sinC=2R
所以:(2sinB-sinC)/sinA=cosC/cosA
所以:2sinBcosA-cosAsinC=sinAcosC
所以:2sinBcosA=cosAsinC+sinAcosC=sin(A+C)=sinB>0
所以:cosA=1/2
解得:A=60°
B+C=120°
y=√3sinB+sin(C-π/6)
=√3sin(2π/3-C)+sin(C-π/6)
=√3*[(√3/2)cosC+(1/2)sinC]+(√3/2)sinC-(1/2)cosC
=cosC+√3sinC
=2sin(C+π/6)
因为:0<C<2π/3)
所以:π/6<C+π/6<5π/6
所以:1/2<sin(C+π/6)<=1
所以:1<y<=2
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