CosC=(a^2 b^2-c^2) 2ab 怎样在EXCEL中表达
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/29 03:36:31
1(cosA-2cosC)/cosB=(2c-a)/b根据正弦定理(cosA-2cosC)/cosB=(2sinC-sinA)/sinB∴sinBcosA-2cosCsinB=2sinCcosB-si
(1)由正弦定理可得:a/sinA=b/sinB=c/sinC那么:(cosA-2cosC)/cosB=(2c-a)/b可化为:(cosA-2cosC)/cosB=(2sinC-sinA)/sinB即
(1).因为:cosB/cosC=-b/2a+c=-sinB/(2sinA+sinC)所以:2cosBsinA+cosBsinC=-sinBcosC就有:2cosBsinA+cosBsinC+sinB
证明:∵A+B+C=180.∴A=180-(B+C).∴sinA=sin[180-(B+C)]=sin(B+C)=sinBcosC+cosBsinC.即有sinA=sinBcosC+cosBsinC.
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
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
cosB=(a²+c²-b²)/2accosC=(a²+b²-c²)/2ab代入cosB/cosC=-b/2a+c得:2ab(a²
(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)
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
cosC/cosB=(2sinA-sinC)/sinBsinBcosC=2sinAcosB-cosBsinCsinBcosC+cosBsinC=2sinAcosB∴sin(B+C)=sinA=2sin
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+
解,向量m⊥向量n∴m*n=0∴b*(cosA-2cosC)+(a-2c)*cosB=0利用正弦定理,b=sinB*2Rc=sinC*2R∴sinB*(cosA-2cosC)+(sinA-2sinC)
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