a=1cosA-根号3cosC

B=π/3,A+C=2π/3,①cosA+cosC=2cos[(A+C)/2]cos{(A-C)/2]=cos[(A-C)/2]>0,∴1+√2sin[(A-C)/2]=0,∴sin[(A-C)/2]

∵m//n∴﹙√3b-c﹚×cosA－a×cosC＝0,根据余弦定理得,﹙√3b-c﹚×﹙b²＋c²－a²﹚／2bc＝a×﹙a²＋b²－c²

因a^2=b^2+c^2-2bc*cosA,c^2=a^2+b^2-2ab*cosC所以cosA=（b^2+c^2-a^2）/(2bc)cosC=(a^2+b^2-c^2)/(2ab)(√3*b-c)

诱导公式cosA=cos(π-B-C)=-cos(B+C)故cos(B+C)=-1/3两角和公式cos(B+C)=cosBcosC-SinBsinC=-1/3代入下式(cosB+cosC)^2=cos

利用积化和差和和差化积B=60,A+C=1201/cosA+1/cosC=-2√2(cosA+cosC)/(cosAcosC)=-2√2cosA+cosC=2cos[(A+C)/2]cos[(A-C)

A+C=2B3B=A+C+B=π∴B=π/31/cosA+1/cosC=-√2/(1/2)=-2√2cosA+cosC=-2√2cosAcosC2cos[(A+C)/2]cos[(A-C)/2]=-√

因为:A+B+C=180,A+C=2B,3B=180,B=60度1/cosA+1/cosC=-√2/cosB,(cosC+cosA)/cosA*cosC=-√2/cosB,利用公式:cosA+cosC

因A+B+C=π,又A+C=2B得B=π/31/cosA+1/cosC=-2√2=>(cosA+cosC)=-2√2cosAcosC=>2cos(A-C)/2cos(A+C)/2=-√2[cos(A+

题目应该是这样的：(2b-[根号3]c)cosA)=[根号3]acosC求角A.利用正弦定理,有：(2sinB－√3sinC)×cosA=√3sinAcosC,展开后得到：2sinBcosA=√3si

1/cosA+1/cosC=-2√2(cosC+cosA)/cosAcosC=-2√2即：cosA+cosC=-2√2(cosAcosC)利用和差化积,积化和差公式,可得：2cos[(A+C)/2]c

∵ABC依次成等差数列∴2B=A+C∴3B=180°∴B=60°∴A+C=120°1/cosA+1/cosC=-√2/cosB∴(cosA+cosC)/cosAcosC=√2cos(A+C)带入A+C

（根号3）/3(“根号3”b-c)*cosA=a*cosC(根号3sinb-sinc）cosa=sina*cosC根号3*sinb*cosa=cosc*sina+sinc*cosa根号3*sinb*c

∵a＋c＝2b∴sinA＋sinc＝2sinB即sinA＋sinC＝2sin(A＋C)由和差化积、二倍角公式得：2sin[(A+C)/2]×cos[(A-C)/2]＝4sin[(A+C)/2]×cos

∵cosA=2/3.且0º＜A＜180º∴sinA=√(1-cos²A)=√5/3（1）∵sinB=√5cosC,A+C=π-B∴sin(A+C)=√5cosC∴sinA

1∵cosA=2/3,∴sinA=√(1-cos²A)=√5/3∵sinB=√5cosCsinB=sin(A+C)=sinAcosC+cosAsinC∴sinAcosC+cosAsinC=√

正弦定理学过吧!就是a/sinA=b/sinB=c/sinC=2R（R是△ABC外接圆半径）.这题用正弦定理代换一下就能够得到（√3sinB-sinC)cosA=sinA*cosC即√3sinB*co

1.将a,b,c用sinA,sinB,sinC替换即√3*sinB*cosA-sinC*cosA=sinA*cosC移项：√3*sinB*cosA=sinC*cosA+sinA*cosC将右边合并：√

由cosA=2/3.结合0º＜A＜180º可得:sinA=(√5)/3.cosA=2/3.[[[[1]]]]结合sinB=(√5)cosC及sinA=(√5)/3可得sinAsin

A+C=2B,B=60度(1/cosA)+(1/cosC)=-(根号2/cosB)=-2√2(cosA+cosC)/(cosAcosC)2cos[(A+C)/2]*cos[(A-C)/2]=-√2[c