三角形中一直c=b(1 2cosA),∠A=2∠B.求三角形形状
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sin(B+C/2)=sin[B+(π-A-B)/2])=sin[π/2+(B-A)/2]=cos{π/2-[π/2+(B-A)/2]}=cos[(A-B)/2)=4/5cos(A-B)=2cos
由正弦定理:b/2R=sinB,c/2R=sinC所以(b+c)/2c=[(2RsinB)+(2RsinC)]/[2(2RsinC)]=(sinB+sinC)/2sinC所以:cos^2(A/2)=(
cos^2A=cos^2(B+C)=1-sin^2(B+C)sin(B+C)=sinBcosC+sinCcosB所以cos^2A+cos^2B+cos^2C=cos^2B+cos^2C-(sin^2B
(4sinc)^2-4*6*cosc16-24cosc-16cosc^22-3cosc-2cosc^2(2cosc+1)(2-cosc)0=>2cosc+1cosc>=-1/2c再问:第一步什么意思?
直角再答:要过程吗再答:
直角三角形,解析:∵(cosB)^2-(cosC)^2=(sinA)^2∴1-(sinB)^2-1+(sinC)^2=(sinA)^2即(SinC)^2-(sinB)^2=(sinA)^2(c/2R)
三角形ABC中,已知COSA=3/5,COSB=12/13,则SINA=4/5,SINB=5/13COSC=COS(180-A-B)=-COS(A+B)=-(COSA*COSB-SINA*SINB)=
交叉相乘得sinBsinA+sinBsin(C-B)=cosBcos(C-B)拆开得sinBsinA+sinCsinBcosB-cosC(sinB)^2=(cosB)^2*cosC+sinCsinBc
cosC/cosB=(3a-c)/b用余弦定理:【(a^2+b^2-c^2)/2ab】/【(a^2+c^2-b^2)/2ac】=(3a-c)/b化简后得:2ac=3a^2+3c^2-3b^2(a^2+
用正弦定理换掉,sinAcosA+sinBcosB=SinCcosCsin2A+sin2B=sin2C和差化积,2sin(A+B)cos(A-B)=2sinCcosC即cos(A-B)=cosC=-c
B+C=180-ACOS(180-A)=-COSA诱导公式
交叉相乘得sinBsinA+sinBsin(C-B)=cosBcos(C-B)拆开得sinBsinA+sinCsinBcosB-cosC(sinB)^2=(cosB)^2*cosC+sinCsinBc
tanA=-3/4
2sinAcosB=sin(A+B)+sin(A-B)=sinC+sin(A-B)=sinC所以sin(A-B)=0所以A=B所以,△ABC是等腰三角形.完毕.
cos²B-cos²C=sin²Acos²B=1-sin²Bcos²C=1-sin²Csin²C-sin²B=
cos(A-C)+cosB=cos(A-C)-cos(A+C)=cosAcosC+sinAsinC-cosAcosC+sinAsinC=2sinAsinC=3/2即sinAsinC=3/4根据正弦定理
sin²A+1-sin²B-(1-sin²C)+sinAsinC=0正弦定理令a/sinA=b/sinB=c/sinC=1/ksinA=ka,sinB=kb,sinC=k
cos(A-B)cos(B-C)cos(C-A)=1,必有一项大于0cos(A-B)≤1cos(B-C)≤1cos(C-A)≤1得cos(A-B)=1cos(B-C)=1cos(C-A)=1A=B=C
A=B=C是等边三角形