√3sinC-cosB=cos(A-C)

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

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+

展开合并得：cos(B-0.5A)=cos(C-0.5A)B-0.5A=C-0.5A或B-0.5A=0.5A-CB=C或B+C=A=90°三角形ABC的形状是等腰三角形或直角三角形

第一问算对,就不用详解了吧,设sinA=7m,sinB=5m,sinC=3m就可以算出cosA,cosB,cosC.第二问应该错了cosA=-1/2cosB=11/14cosC=13/14sinA=±

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

证:∵△ABC为锐角三角形,∴A+B>90°得A>90°-B∴sinA>sin(90°-B)=cosB,即sinA>cosB,同理可得sinB>cosC,sinC>cosA上面三式相加:sinA+si

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

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

得cosC=3/5再算出cos(B+C),cosA=cos(180-(B+C))继续代公式

考虑函数f(x)=x-cos(x),∵f'(x)=1+sin(x)≥0,∴f(x)单调递增.∵a=cos(a),∴f(a)=0.∵b∈(0,π/2),∴cos(b)>0,∴b=sin(cos(b))0

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

在△ABC中,C=180-(A+B).cosC=cos[180-A+B)]=-cos(A+B)=-(cosAcosB-sinAsinB).=sinAsinB-cosAcosB.=(4/5)*(12/1

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