在三角形abc中,aCos(c 2) cCos(α 2)=1.5b

用正弦定理sinA/a=sinB/b=sinC/c,得出(b-c)/[acos(60+C)]=(sinB-sinC)/[sinAcos(60+C)]=[2cos(B+C)/2*sin(B-C)/2]/

根据正弦定律得知a/sinA=b/sinB=c/sinC=kb-c=2acos(60°+C)ksinB-ksinC=2ksinAcos(60°+C)sinB-sinC=2sinAcos(60°+C)s

根据正弦定律得知a/sinA=b/sinB=c/sinC=kb-c=2acos(60°+C)ksinB-ksinC=2ksinAcos(60°+C)sinB-sinC=2sinAcos(60°+C)s

第二问答案在图中,点击图片查看大图

∵cos2C2=1+cosC2，cos2A2=1+cosA2∴由acos2C2+cos2A2=3b2，得a•1+cosC2+c•1+cosA2＝3b2…（4分）由正弦定理，得sinA（1+cosC）+

由正弦定理asinA＝bsinB＝csinC＝2R，得：sinB-sinC=2sinA•cos（60°+C），…（2 分）∵A+B+C=π，故有：sin(A+C)−sinC＝sinAcosC

a*cosc/2=a*(1/2(2cos^2c/2-1)+1/2)=a*(1/2cosc+1/2)=1/2a*cosc+1/2a.另一部分同理,整理后左边是：1/2a*cosc+1/2a+1/2c*c

acos^2C/2+ccos^2A/2=3/2b,a(1+cosC)+c(1+cosA)=3b,sinA(1+cosC)+sinC(1+cosA)=3sinB,sinA+sinC+sinAcosC+s

/>老师说的没错,o(∩_∩)o...哈哈!写到“sinB-sinC=sinAcosC-√3sinAsinC”的时候,因为sinB=sin（A+C）=sinAcosC+cosAsinC所以cosAsi

acos^2C/2+ccos^2A/2=3b/2a*(cosC+1)/2+c*(cosA+1)/2=3b/2acosC+a+ccosA+c=3bacosC+a+ccosA+c=2b+b,a/sinA=

明白了,是偶看错了刚才.A=2π/3因为b-c=2acos(π/3+C)所以sinB-sinC=2sinA(1/2cosC-√3/2sinC)所以sinB-sinC=sinAcosC-√3sinAsi

cos²(c/2)=(1+cosC)/2cos²（A/2)=(1+cosA)/2就有（a+c)/2+1/2(acosC+ccosA）=3b/2再用余弦定理把cos转化就出来了.

正弦定理知等价于证sinacosa+sinbcosb+sinccosc=2sinasinbsin（a+b）=2sin^2asinbcosb+2sin^2bsinacosa移项用二倍角公式等价于cos2

由正弦定理知,a/sinA=b/sinB=c/sinC∵acosA+bcosB=ccosC∴sinAcosA+sinBcosB=sinCcosC∴sin2A+sin2B=sin2C=sin(2π-2A

acos^2C/2+ccos^2A/2=3b/2a*(cosC+1)/2+c*(cosA+1)/2=3b/2acosC+a+ccosA+c=3bacosC+a+ccosA+c=2b+b,a/sinA=

acos^2(C/2)+ccos^2(A/2)=3/2b1/2a(2cos^2(C/2)-1)+1/2a+1/2c(2cos^2(A/2)-1)+1/2c=3/2b1/2acosC+1/2ccosA+

利用二倍角公式2cos平方二分之c=cosc+1,cos二分之a的平方=cosa+1代入原式,用余弦定理展开cosA,cosC即可

a[2cos²(C/2)]+c[2cos²(A/2)]=3b--->a(1+cosC)+c(1+cosA)=3b--->a(a²+b²-c²+2ab)