△ABC中 cosC+(cosA-根号3sinA)cosB=0
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/07 02:11:04
证明:利用正弦定理a/(sina)=b/(sinb)=c/(sinc)=2R,就有:a^2=4R^2sin^2Ab^2=4R^2sin^2Bc^2=4r^2sin^2C(a^2-b^2)=4R^2(s
∵0∴0∴cos(C/2)>sin(C/2).又∵0∴-π∴-π/2∴cos((A-B)/2)>0,∴sin(A)+sin(B)=2sin((A+B)/2)cos((A-B)/2)=2sin((π-C
1+cosA+cosB+cosC-(sinA+sinB+sinC)=2[cos(A/2)]^2+2cos(B+C)/2*cos(B-C)/2-2[sin(A/2)*cos(A/2)+sin((B+C)
证明:∵A+B+C=180º.∴A=180º-(B+C).∴sinA=sin[180º-(B+C)]=sin(B+C)=sinBcosC+cosBsinC.即有sinA=
因a^2=b^2+c^2-2bc*cosA,c^2=a^2+b^2-2ab*cosC所以cosA=(b^2+c^2-a^2)/(2bc)cosC=(a^2+b^2-c^2)/(2ab)(√3*b-c)
证明:由三角形正弦定理得a/sinA=b/sinB=c/sinC所以a/b=sinA/sinB=cosA/cosB得sinAcosB-cosAsinB=0所以sin(A-B)=0所以A-B=π*n(n
三角和差公式:(cosA+cosB)=2*cos[(A+B)/2]*cos[(A-B)/2](cosA-cosB)=-2*sin[(A+B)/2]*sin[(A-B)/2]倍角公式:cosC=cos(
∵sinA:sinB:sinC=4:5:6根据正弦定理可得:a:b:c=4:5:6,不妨设a=4k,b=5k,c=6k(k>0)cosA=b2+c2−a22bc=25k2+36k2−16k22×30k
D如果是锐角每个角的正弦余弦都会是正的直径则会等于0只有是钝角时,会出现负值
在△ABC中,sinA(2sinC-sinA)=cosA(2cosC+cosA)⇔2sinA•sinC-sin2A=2cosA•cosC+cos2A⇔2sinA•sinC-2cosA•cosC=cos
证明:∵△ABC是锐角三角形,A+B>π2,∴π2>A>π2−B>0∴sinA>sin(π2−B),即sinA>cosB;同理sinB>cosC;sinC>cosA,∴sinA+sinB+sinC>c
a/b=sinA/sinB=cosA/cosBtgA=tgBA=B,同理B=C所以A=B=C为等边三角形.
(1)A=60度2R=a/sinA=b/sinB=c/sinC代入原式化简得4sinBcosA-2sinCcosA=2sinAcosC即4sinBcosA=2sinCcosA+2sinAcosC=2s
纠正一下题目:应该是tanC=(sinA+sinB)/(cosA+cosB)因为tanC=(sinA+sinB)/(cosA+cosB),sinC/cosC=(sinA+sinB)/(cosA+cos
cosA+cosB+cosC=2cos[(A+B)/2]*cos[(A-B)/2]+cosC≤2cos[(A+B)/2]]+cosC≤2sin(C/2)+cosC=-2sin(C/2)^2+2sin(
1.将a,b,c用sinA,sinB,sinC替换即√3*sinB*cosA-sinC*cosA=sinA*cosC移项:√3*sinB*cosA=sinC*cosA+sinA*cosC将右边合并:√
sinA=√(1-cos^2A)=√[1-(3/5)^2]=4/5.sinB=√(1-cos^2B)=√[1-(5/13)^2]=12/13.在△ABC中,C=180-(A+B).cosC=cos[1
三角形内角在0和180之间所以sinA>0,sinB>0sin²A+cos²A=1所以sinA=4/5同理sinB=12/13cosC=cos[180-(A+B)]=-cos(A+
cosC=cos[π-(A+B)]=-cos(A+B)=-[cosAcosB-sinAsinB]=-[15/17*9/41-8/17*20/41]=-25/697
因为a/CosA/2=b/CosB/2=c/CosC/2所以a/CosA=b/CosB=c/CosC根据正弦定理a=2R*SinAb=2R*SinBc=2R*sinC得SinA/cosA=SinB/C