已知b =ac 且cosb=3 4 求tana

(1)1/tanA+1/tanC=(tanA+tanC)/(tanA*tanC)=cosA*cosC(tanA+tanC)/(sinA*sinC)=(sinAcosC+sinCcosA)/(sinA*

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

∵cosB=3/4∴sinB=±√7/4又∵0

由b^2=ac知道a,b.c成等比数列,则SinASinBSinC也成等比数列Cos(A-C)+CosB=Cos(A-C)-Cos(A+C)展开得2SinASinC=3/2得出2倍SinB的平方=3/

∵A、B为锐角,∴sinA=√[1-(cosA)^2]=√{1-[√(1/10)]^2}=√(9/10)sinB=√[1-(cosB)^2]=√{1-[√(1/5)]^2}=√(4/5)sin(A+B

=(cosb,sinb),b+c=(2cosb,0),所以,c=(cosb,-sinb)ab=sinacosb+cosasinb=sin(a+b)=1/2ac=sinacosb-cosasinb=si

(2)因为向量BA*向量BC=3/2所以acCOSB=3/2,又cosB=3/4所以ac=2①在三角形ABC中,由余弦定理得,cosB=a^+c^-b^/2ac=3/4,又b^2=ac所以a^+c^-

cosB=cosC,∠B=∠C3b=2√3asinB,用正弦定理,两边消去2R,3sinB=2√3sinAsinBsinA=√3/2,A=60°,120°A=60,B=C=60°A=120,B=C=3

cosa=4/5a为锐角,所以sina>0,sina=√(1-cos²a)=3/5cos(a+b)=-16/65,a、b为锐角,所以a+b0即sin(a+b)=√【1-cos²（a

a,b为锐角,则sin(a+b)=5/13,sinb=3/5cosa=cos(a+b-b)=cos(a+b)·cosb+sin(a+b)·sinb=-12/13·cosb+5/13·3/5=4/5,故

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cosA=√(1-sinA^2)=4/5tanA=sinA/cosA=3/4tan(A-B)=(tanA-tanB)/(1+tanAtanB)=(3/4-tanB)/(1+3/4tanB)=-1/33

因为A,B都是锐角,则有sinA=5/13,sin(A+B)=3/5cosB=cos(A+B-A)=cos(A+B)cosA+sin(A+B)sinA=4/5*12/13+3/5*5/13=63/65

cosA=1/17,sinA=12根号2/17cos(A+B)=cosAcosB-sinAsinB=1/17cosB-12根号2/17sinB=-47/51根据cos平方B+sin平方B=1解方程得c

原式=sin²a+2sinasinb+sin²b+cos²a+2cosacosb+cos²b=(sin²a+cos²a)+2(cosacos

显然sinb>0sin²b+cos²b=1所以sinb=15/170

我想问题应该是+1/tanC吧?如果是这样,那马b^2=a^2+c^2-2ac*cosB=acyiwen1/tanA+1/tanC=cosA\sinA+cosC\sinC=(sinAcosC+cosA

由已知S=0.5ab又有S=0.5*AB*AC*sinA固0.5*AB*AC*sinA=0.5*AB*AC*cosA从而tanA=1固A=45°又有cosC=-cos（A+B)=sinA*sinB-c

siny=sinb-sinacosy=cosa+cosb分别平方,相加sin²y+cos²y=sin²a+cos²a+sin²b+cos²b