ABC内角ABC已知2cosc(acosb bcosa)=c
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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