a*cosC 根号3*a*sinC--b-c=0

（1）根号3cosa-sina/根号3cosa+sina=（√3-tana)/(√3+tana)分子分母同时除以cosa=(√3-3)/(√3+3)=(√3-3)²/(√3-3)(√3+3)

因为sin(a+π/2)=cosa,所以sin(2π/3+a)=sin[π/2+(π/6+a)]=cos(π/6+a)由sin(π/6+a)=（根号3）/3>0,得cos(π/6+a)=(±根号6)/

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]

cosB=1/2B=60sinB=根号3/2sinA=2/3cos^2A=(1-(2/3)^2)cosA=根号5/3orCosA=-根号5/3cosC=cos(180-A-B)=-cos(A+B)=s

cosA+cosB+cosC=2cos[(A+B)/2]cos[(A-B)/2]+cosC=2cos[(π-C)/2]cos[(A-B)/2]+cosC=2sin(c/2)cos[(A-B)/2]+1

根据正弦定理a/sinA=b/sinB=c/sinC,原式可变形为:bsin2A+bcos2A=跟号3a,即b=根号3a.将上面的结果带入余弦定理“cosC=(a^2+b^2-c^2)/(2·a·b)

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)

(1)因为cosc=3/4所以sinc=（根号7）/4再用正弦定理c/sinc=a/sina代入数（根号2）/（根号7）/4=1/sina解得sina=（根号14）/8(2)根据余弦定理b=2=1+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

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

A+B=π-Ccos(A+B)=cos(π-C)=-cosCcos[(A+B)/2]=cos[(π-C)/2]=cos(π/2-C/2)=sin(C/2)sin(3A+3B)=sin3*(A+B)=s

2sin^2C=3cosC2(1-cos^2C)=3cosC2cos^2C+3cosC-2=0(cosC+2)(2cosC-1)=0cosC=1/2角C=60°＝＝＝＝＝＝＝＝＝＝＝＝＝＝因为角C＝6

右边2sin(π/6+A)=2sin(π/6)cosA+2sinAcos(π/6)=cosA+根号3sin(A)=左式.得证#

2a=√3c,a=√3/2cc^2=a^2+b^2-2abcosC3/4c^2+b^2-√3/2*√3/2cb-c^2=0b^2-3/4cb+c^2/4=0(b-c)(b+c/4)=0得b=c,另一个

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

tanC=sinC/cosC=(sinA+sinB)/(cosA+cosB),交叉相乘得sin(C-A)=sin(B-C),故C-A=B-C或C-A+B-C=π,显然C-A=B-C成立,2C=A+B,

利用余弦定理逆定理可得cosA=(根号3)/2A=30'2sinBcosC-sin(B-C)拆开可化为sinBcosC+cosBsinC=sin(B+C)=sinA=1/2

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.cosC=√3/4sinC=√13/42a=√3c正弦定理2sinA=√3sinCsinA=√39/8cosA=5/8sinB=sin(A+C)=sinAcosC+cosAsinC=√39/8*√

A=45度,B+C=135度;Cos（B+C）=cosBcosC-sinBsinC=cos135度；sinA=cosA=代人角度很容易求得;已知cosB=,则sinB=也易求;有cosCsinC的分别