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sinAsinBsinC成等比数列

sinB=sinAsinC,由正弦定理,b=ac,ac=3,余弦定理,b=a+c-2accosB2ac-2accosB ,cosB1/2,0sinB√3/2,S=1/2acsinB=3√3/4.

a,b,c成等差数列:2b=a+c正弦定理:b/sinB=a/sinA=c/sinC=2R2sinB=sinA+sinCsinB=(sinA+sinC)/2sinA,sinB,sinC成等比数列:sin^2B=sinA*sinC[(sinA+sinC)/2]^2=sinA*sinC(sinA-sinC)^2=0sinA=sinCA=B=C=60°等边三角形

由正弦定理知:sinA,SinB,sinC成等比数列即a,b,c,成等比,即b=ac,则CosB=(a+c-b)/2ac=(a+C-ac)/2ac≥1/2即B∈[0,兀/3] 得 B的最大值为兀/3.此时sinAsinc=3/4 即sinAsin(2兀/3-A)=√3/4sin2A-1/4Cos2A+1/4=1/2sin(2A-兀/6)+1/4=3/4,则sin(2A-兀/6)=1则A=兀/3,故为等边角形

sinA sinB sinC成等差数列 ,a b c也是等差数列不是sinA sinB sinC成等比数列 ,a b c也是等比数列成立sinA/a= sinB/b= sinC/c等比成立,等差不成了;

a,b,c成等差数列:2b=a+cb/sinB=a/sinA=c/sinC=2R2sinB=sinA+sinCsinB=(sinA+sinC)/2sinA,sinB,sinC成等比数列:sin^2B=sinA*sinC[(sinA+sinC)/2]^2=sinA*sinC(sinA-sinC)^2=0sinA=sinCA=B=C=60°所以是等边三角形

∵sinA,sinB,sinC成等比数列,∴sin2B=sinAsinC,根据正弦定理化简得:b2=ac,∴cosB=a2+c2b22ac=a2+c2ac2ac≥2acac2ac=12∵B∈(0,π)∴B∈(0,π3].故答案为:(0,π3].

因为:sinA,sinB,sinC成等比数列所以:2RsinA,2RsinB,2RsinC成等比数列即:a,b,c等比b^2=aca^2=c(a+c-b)=ac+c^2-bc=b^2+c^2-bccosA=(b^2+c^2-a^2)/2bc=1/2A=60度c/(b*sinB)=c/(b*b/2R)=2Rc/b^2=2R/a=1/(a/2R)=1/sinA=2根3/3

sinA、sinB、sinC依次成等比数列,则:sinB=sinA*sinC利用正弦定理,得:b=ac又:a+c≥2ac且:cosB=(a+c-b)/(2ac)=[a+c-ac]/(2ac)≥[2ac-ac]/(2ac)=1/2

a,b,c成等差数列:2b=a+cb/sinb=a/sina=c/sinc=2r2sinb=sina+sincsinb=(sina+sinc)/2sina,sinb,sinc成等比数列:sin^2b=sina*sinc[(sina+sinc)/2]^2=sina*sinc(sina-sinc)^2=0sina=sinca=b=c=60°等边三角形

成等比数列.根据正弦定理:a/sina=b/sinb=c/sinc=2r则有:a=2rsina,b=2rsinb,c=2rsinc因为a,b,c成等比数列,即:b^2=ac所以:(2rsinb)^2=(2rsina)(2rsinc)即:(sinb)^2=sina*sinc所以说成等比数列.

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