a(bcosB-ccosC)=(b²-c²)cosA
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![a(bcosB-ccosC)=(b²-c²)cosA](/uploads/image/f/430931-11-1.jpg?t=a%28bcosB-ccosC%29%3D%28b%C2%B2-c%C2%B2%29cosA)
直角三角形a/sinA=b/sinB=c/sinC=ta=tsinAb=tsinBc=tsinCacosA+bcosB=ccosCtsinAcosA+tsinBcosB=tsinCcosCsin2A+
∵acosA+bcosB=ccosC∴sinAcosA+sinBcosB=sinCcosC∴sin2A+sin2B=sin2C=sin(2π-2A-2B)=-sin(2A+2B)∴0=sin2A+si
用cosA=(b^2+c^2-a^2)/2bc把所有的余弦角全还成边再化简合并同类项(a2-b2)2=c2c2又a>0,b>0,c>0两边同时开方得出a2-b2=c2得出a2=b2+c2所以ABC为直
由正弦定理可知a=2rsinAb=2rsinBc=2rsinC代入acosA+bcosB=ccosC,得sinAcosA+sinBcosB=sinCcosCsin2A+sin2B=2sinCcosC即
令k=a/sinA=b/sinB=c/sinC所以a=ksinAb=ksinBc=ksinC代入acosA+bcosB=ccosC,并约去ksinAcosA+sinBcosB=sinCcosCsin2
正弦定理,得:sinAcosA+sinBcosB=sinCcosC,即:sin2A+sin2B=2sinCcosC,就是2sin(A+B)cos(A-B)=2sinCcosC,则2sinCcos(A-
根据正弦定理有,sinA(sinBcosB-sinCcosC)=(sinB*sinB-sinC*sinC)cosAsinA(sin2B-sin2C)=(cos2C-cos2B)cosAsinAcos(
∵bcosB+ccosC=acosA,由正弦定理得:sinBcosB+sinCcosC=sinAcosA,即sin2B+sin2C=2sinAcosA,∴2sin(B+C)cos(B-C)=2sinA
∵a=2bcosC,由正弦定理可得,2sinBcosC=sinA=sin(B+C)=sinBcosC+cosBsinC,∴sinBcosC-cosBsinC=0,即sin(B-C)=0,∴B-C=0,
∵bcosB+ccosC=acosA∴sinAcosA=sinBcosB+sinCcosC∴sin2A=sin2B+sin2C∴sin2A=2sin(B+C)cos(B-C)∴2sinAcosA-2s
∵acosA+bcosB=ccosC∴sinAcosA+sinBcosB=sinCcosC∴sin2A+sin2B=sin2C=sin(2π-2A-2B)=-sin(2A+2B)∴0=sin2A+si
三角形ABC形状是等边三角形.(a^3+b^3-c^3)/(a+b-c)=c^2,a^3+b^3-c^3=c^2(a+b-c),a^3+b^3=(a+b)*c^2,有a^2+b^2-c^2=ab,co
a(bCOSB-cCOSC)=(b^2-c^2)COSA,而,cosA=(b^2+c^c-a^2)/2bc,cosB=(a^2+c^2-b^2)/2ac,cosC=(a^2+b^2-c^2)/2ab,
用cosA=(b^2+c^2-a^2)/2bc把所有的余弦角全还成边再化简合并同类项(a²-b²)²=c²c²又a>0,b>0,c>0两边同时开方得出
cosA=(b平方+c平方-a平方)/2bc,同理可得cosb和cosc所以acosA+bcosB=ccosC可转化为(b平方+c平方-a平方)/2bc+(a平方+c平方-b平方)/2ac=(a平方+
cosA=(b^2+c^2-a^2)/2bccosB=(a^2+c^2-b^2)/2accosC=(a^2+b^2-c^2)/2abacosA+bcosB=ccosCa(b^2+c^2-a^2)/2b
∵acosA+bcosB=ccosC∴sinAcosA+sinBcosB=sinCcosC∴sin2A+sin2B=sin2C=sin(2π-2A-2B)=-sin(2A+2B)∴0=sin2A+si
将cosA=(b^2+c^2-a^2)/(2bc)cosB=(a^2+c^2-b^2)/(2ac),cosC=(a^2+b^2-c^2)/(2ab)代入得到:a[b*(a^2+c^2-b^2)/(2a
【解法1】由已知得acosA+bcosB=ccosCcosA=(b^2+c^2-a^2)/2bccosB=(a^2+c^2-b^2)/2accosC=(a^2+b^2-c^2)/2abacosA+bc
acosA+bcosB=ccosC,a*(b^2+c^2-a^2)/2bc+b*(a^2+c^2-b^2)/2ac=c*(a^2+b^2-c^2)/2ab,方程式各项同时乘以2abc,得到a^4+b^