2cosc-2a-c
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根据余弦定理,得:2abcosC=a^2+b^2-c^22bccosA=b^2+c^2-a^2所以2b×cosA-c×cosA=(2b-c)×cosA=(b^2+c^2-a^2)(2b-c)/(2bc
(1).因为:cosB/cosC=-b/2a+c=-sinB/(2sinA+sinC)所以:2cosBsinA+cosBsinC=-sinBcosC就有:2cosBsinA+cosBsinC+sinB
cosC/cosB=(2a-c)/b=(2sinA-sinC)/sinBcoscsinB=2sinAcosB-sinCcosBcoscsinB+sinCcosB-2sinAcosB=0sin(B+C)
用正弦定理化等式右边为角,得到:cosB/cosC=-sinB/(2sinA+sinC),去分母后有cosB(2sinA+sinC)+sinBcosC=0,2cosBsinA+(cosBsinC+si
cosA=(b^2+c^2-a^2)/(2bc)s=a^2-(b-c)^2s=1/2bcsinA得到cosA=15/17sinA=8/17得到直角三角形cosC=0或cosC=8/17
利用三角函数的正弦定理做啊:a/(sina)=b/(sinb)=c/(sinc)=2R,其中R是三角形外接圆的半径就有:(a^2-b^2)=4R*R*(sin(a)*sin(a)-sin(b)*sin
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
m⊥n=>m.n=0(2cosc/2,-sinc).(cosc/2,2sinc)=02(cosc/2)^2-2(sinc)^2=0cosC+1-2(1-(cosC)^2)=02(cosC)^2+cos
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
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
(cosA-2cosC)/cosB=(2sinC-sinA)/sinBsinBcosA-2sinBcosC=2cosBsinC-cosBsinA2sinBcosC+2cosBsinC=sinBcosA
因为a/SinA=b/SinB=c/SinC由已知(根号2*SinA-SinC)CosB=SinBCosC即[根号2*Sin(B+C)-SinC]CosB=SinBCosC即有根号2*Sin(B+C)
√2/2A+C=2[180-(A+C)]=>A+C=1201/cosA+1/cosC=-√2/cosB=>(cosA+cosC)/cosAcosC=√2cos(A+C)带入A+C=120=>(cosC
由正弦定理a/sinA=b/sinB=c/sinC=2R得:a=2RsinA,b=2RsinB,c=2RsinC,将上式代入已知cosB/cosC=-(b/2a+c),得cosB/cosC=-sinB
由正弦定理可知sinA/a=sinB/b=sinC/v所以cosB/cosC=–b/2a+c可以化成cosB/cosC=-sinB/(2sinA+sinC)得到-sinBcosC=2sinAcosB+
因为:a/sinA=b/sinB=c/sinC=2R所以:a^2=4R^2*sinAb^2=4R^2*sinBc^2=4R^2*sinC所以:(a^2-b^2)/(cosA+cosB)=4R^2*(s
答:三角形ABC三边满足:(2b-c)/a=cosC/cosA根据正弦定理有:a/sinA=b/sinB=c/sinC=2R结合得:(2sinB-sinC)/sinA=cosC/cosA2sinBco
sinA=sin(B+C)=sinBcosC+cosBsinC=2sinBcosC故sinBcosC=cosBsinC,有sinBcosC-cosBsinC=0即sin(B-C)=0故B=C(这步可以