a(sinC+根号3cosC)＝根号3b

(sinA+sinB+sinC)/(cosA+cosB+cosC)=√3sinA+sinB+sinC=√3cosA+√3cosB+√3cosC(sinA-√3cosA)+(sinB-√3cosB)+(

因为a/sinA=b/sinB=c/sinC所以-b/（2a+c）=-sinB/（2sinA+sinC）再问：麻烦写一下中间转化过程和约掉的东西。。3Q再答：a=ksinAb=ksinBc=ksinC

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

应该是这样的吧:(根号3sinB-cosB)(根号3sinC-cosC)=4cosBcosC.3sinBsinC-根号3sinBcosC-根号3cosBsinC+cosBcosC=4cosBcosC3

m⊥n=>m.n=0(2cos(C/2),-sinC).(cos(C/2),2sinC)=02(cosC/2)^2-2(sinC)^2=0(2(cosC/2)^2-1)-2(sinC)^2+1=0co

由正弦定理,sinAcosC+√3sinAsinC-sinB=sinC,sinAcosC+√3sinAsinC-sin(A+C)=sinC,sinAcosC+√3sinAsinC-sinAcosC-c

2sinA(cosC/2)^2+2sinC（cos(A/2)）^2=3sinBsinA(1+cosC)+sinC(1+cosA)=sinA+sinC+sin(A+C)=sinA+sinC+sinB=3

cosC=1-2(sinC/2)^2=1-2*5/8=-1/4

,{sin(A-B)+sinC)/{cos(A-B)+cosC}=,{sin(A-B)+sin(A+B))/{cos(A-B)-cos(A+B)}=2sinAcosB/2sinAsinB=cosB/s

1.sinC+cosC化成半角,2sinc/2cosc/2+1-2sinc/2sinc/2原式化为cosC/2-sinC/2=0两边平方,得到1-sinC=0即sinC=12.条件不足,看看题是否写错

A+B=180-C所以原式=cos(180-C)=-cosC选B再问：为什么A+B=180-C呢？再答：三角形内角和=？度

2cosc/2的是这啊~这不是直接约了啊成为COSC如果题这是这样我再给你说

(1).利用cosB+cosC=2(cos(B+C)/2)(cos(B-C)/2)和sinB+sinC=2(sin(B+C)/2)(cos(B-C)/2),由2sinA(cosB+cosC)=3(si

sinC-cosC=√2sin(C-π/4)=√2cos(3π/4-C)=-√2cos(π/4+C)显然C>=π/4,否则cosC>=√2/2,sinC+sinB=√3+cosC>=2,不可能成立因此

1向量点乘公式（X1,Y1）点乘(X2,Y2)=X1X2+Y1Y2故cos^2C-sin^2B-sinbsinc=cos^2A然后,你这没有问题啊?我猜是三角,接下来的可能变形是首先全变sin这是能做

MN=0则M垂直N,(sinC*1+cosC*(-√3)=0,sinC*1/2-√3/2*cosC=0,sin(C-60)=0,C=60度.A+B=180-C,(A+B)/2=90-(C/2).sin

（1）sinC+cosC=1-sinC/2,移项得sinC-sinC/2=1-cosC由二倍角公式得2sinC/2cosC/2-sinC/2=2(sinC/2)^2因为sinC/2≠0,所以两边消去s

sin(C/2)=√10/4由：[sin(C/2)]^2+[cos(C/2)]^2=1,解得[cos(C/2)]^2=3/8则：cosC=[cos(C/2)]^2-[sin(C/2)]^2=3/8-（

2边平方.化解,用余眩定理

由sinc+cosc=2sina平方可得1+2sinc*cosc=4sin^2a因sinc*cosc=sin^2b所以1+2sin^2b=4sin^2a2-4sin^2a=1-2sin^2b2cos2