a(cosc √3sinc)

若(sinA+sinB+sinC)/(cosA+cosB+cosC)=√3.据三角形恒等式:sinA+sinB+sinC=s/R,cosA+cosB+cosC=(R+r)/R.即s=√3*(R+r).

sinB+sinC=2sin[(B+C)/2]cos[(B-C)/2].cosB+cosC=2cos[(B+C)/2]cos[(B-C)/2].所以由条件可得：sin[(B+C)/2]=sinAcos

解题思路：由正弦定理及两角和的正弦公式可得sinAcosC+根号3sinAsinC=sinB+sinC=sin（A+C）+sinC=sinAcosC+sinCcosA+sinC，整理可求A解题过程：

sinA+sinB=-sinCcosA+cosB=-cosC两式分别左右平方,后相加得2+2(cosAcosB+sinAsinB)=1所以cosAcosB+sinAsinB=-1/2cos(A-B)=

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

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

sinC-√3/2+√3cosC-2sinCcosC=0(sinC-√3/2)-2cosC(sinC-√3/2)=0(sinC-√3/2)(cosC-1/2)=0∴sinC=√3/2或cosC=1/2

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

根据:cos(A+B)=cosAcosB-sinAsinBcos2A=2cos^2A-1=cos^2A-sin^2Acos(A+B)=-cosC所以：(sinA+sinB+sinC)^2=0.1)(c

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

xsina+ycosa=A(x/Asina+y/Acosb)=A(cosbsina+sinbcosa)=Asin(a+b),其中A=√(x^2+y^2),b=arctan(y/x)所以,这种题首先要除

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

（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

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