√3sinc- cosB＝cos(A-C)

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).

sinA=√(1-cos²A)=√[1-(3/5)²]=4/5.sinB=√(1-cos²B)=√[1-(5/13)²]=12/13.在△ABC中,C=180-

为(sinc)^2=(sina+sinb)^2=(sina)^2+2sina*sinb+(sinb)^2,同理(cosc)^2=(cosa)^2+2cosa*cosb+(cosb)^2,所以相加得1=

因为(sinc)^2=(sina+sinb)^2=(sina)^2+2sina*sinb+(sinb)^2,同理(cosc)^2=(cosa)^2+2cosa*cosb+(cosb)^2,所以相加得1

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)=

由正弦定理知a:b:c=2:3:4设a=2kb=3kc=4k由余弦定理cosA=(b²+c²-a²)/(2bc)=(9k²+16k²-4k²

因为(sinc)^2=(sina+sinb)^2=(sina)^2+2sina*sinb+(sinb)^2,同理(cosc)^2=(cosa)^2+2cosa*cosb+(cosb)^2,所以相加得=

sinA+sinB=sinC,cosA+cosB=cosC(sinA)^2+(sinB)^2+2sinAsinB=(sinC)^2(cosA)^2+(cosB)^2+2cosAcosB=(cosC)^

sina+sinb=-sinc;cosa+cosb=-cosc;两式平方再相加,化简得cosa*cosb+sina*sinb=-1/2;∴cos(a-b)=cosa*cosb+sina*sinb=-1

cos(B-C)=cosBcosC+sinBsinc又sinB+sinc=-sinAcosB+cosC=-cosA所以同时平方sinB^2+sinc^2+2sinBsinc=sinA^2cosB^2+

1、sinA(sinB＋√3cosB)=√3sinC=√3sin(A＋B),展开,sinAsinB=√3cosAsinB,tanA=√3,A=60°；2、a=3,A=60°,b＋c=(a/sinA)s

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

(1)已知sinA+sinB+sinC=0,cosA+cosB+cosC=0.求cos(B-C)的值.sinA=-(sinB+sinC)cosA=-(cosB+cosC)sinA^2+cosA^2=1

(sinA+sinB+sinC)²=sin²A+sin²B+sin²C+2(sinAsinB+sinAsinC+sinBsinC)(cosA+cosB+cosC

这题目抄丢了吧再问：。。。

三角形ABC中,若COSA＋COSB＝SINC,则三角形ABC是直角三角形证明如下cosA+cosB=sinC=sin[π-(A+B)]∴cosA+cosB=sin(A+B)∴cosA+cosB=si

cos(B-C)=cosBcosC+sinBsincsinB+sinc=-sinAcosB+cosC=-cosA所以同时平方sinB^2+sinc^2+2sinBsinc=sinA^2cosB^2+c

c=cos(sinc)>=cos(1)>=sin(1)>=sin(cosb)=

(sina)^2=(sinb+sinc)^2=(sinb)^2+2sinb*sinc+(sinc)^2,(cosa)^2=(cosb+cosc)^2=(cosb)^2+2cosb*cosc+(cosc

cos2A2=1+cosA2=1−cos(B+C)2=sinBcosC∴cosBcosC-sinBsinC=1-2sinBcosC∴cos（B-C）=1∴B-C=0，即B=C∴三角形为等腰三角形．