ABC内角ABC已知2cosc(acosb bcosa)=c

sinBcosC=2sinAcosB-cosBsinCsin(B+C)=2sinAcosBsinA=2sinAcosBcosB=1/2B=60°49=a²+c²-2accos60°

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

再问：利用这个该怎么求出sinC/sinA再答：

已知三角形ABC的三个内角,满足A+B=2B,设x=cos(A-C)/2,f(x)=cosB(1/cosA+1/cosC)已知三角形ABC的三个内角,满足A+B=2B,设x=cos(A-C)/2,f(

A+C=2B3B=A+C+B=π∴B=π/31/cosA+1/cosC=-√2/(1/2)=-2√2cosA+cosC=-2√2cosAcosC2cos[(A+C)/2]cos[(A-C)/2]=-√

(cosA-2cosC)/cosB=(2c-a)/b正弦定理:a/sinA=b/sinB=c/sinC=2R∴(2c-a)/b=(4RsinC-2RsinA)/2RsinB=(2sinC-sinA)/

因A+B+C=π,又A+C=2B得B=π/31/cosA+1/cosC=-2√2=>(cosA+cosC)=-2√2cosAcosC=>2cos(A-C)/2cos(A+C)/2=-√2[cos(A+

1/cosA+1/cosC=-2√2(cosC+cosA)/cosAcosC=-2√2即：cosA+cosC=-2√2(cosAcosC)利用和差化积,积化和差公式,可得：2cos[(A+C)/2]c

第1小题(1)先用正弦定理化边为角；(2cosA-cosC)/cosB=(sinC-2sinA)/sinB(2)化分式为整式,并移项；2(cosAsinB+sinAcosB)=sinCcosB+cos

三个内角成等差数列所以B=60°cosC=根号6/3sin^2C+cos^2C=1sinC=根号3/3用正弦定理b/sinB=c/sinC可得c=根号2

1（cosA-2cosC）/cosB=（2c-a）/b（cosA-2cosC）/cosB=（2sinC-sinA）/sinBsinBcosA-2sinBcosC=2sinCcosB-sinAcosBs

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

3b=2asinB√3b/sinB=a2√3/3由正弦定理a/sinA=b/sinBsinA=3/(2√3)=√3/2A=60°或A=120°cosB=cosCB=CB=C=60°或B=C=30°

如A,B,C成等差,显然B=π/3sinA-sinC+√2/2cos(A-C)=√2/2这个方程用构造一元二次方程来解.由和差化积公式,易得:①sinA-sinC=2cos[(A+C)/2]sin[(

cosC=(a^2+b^2-c^2)/2abcosB=(a^2+c^2-b^2)/2accosA=(c^2+b^2-a^2)/2bc代入a／cosA=b+c／cosB+cosC化简得a^2=b^2+c

A=B-dC=B+dA+B+C=3B=180B=602^(1/2)*(sinA-cosC)+cos(A-C)=12^(1/2)*[sin(B-d)-cos(B+d)]+cos(2d)=12^(1/2)

(1)因为（cosA-2cosC）÷cosB=（2c-a）÷b根据正弦定理（cosA-2cosC）÷cosB=（sinA-2sinC）÷sinB因为cosB=-cos（A+C）sinB=sin（A+C

由已知得：(sinB+cosB)sinC+（sinB-cosB)cosC=-1/5即-cos(B+C)+sin(B+C)=-1/5即cosA+sinA=-1/5联立cosA^2+sin^2=1得sin