△ABC中,c=10 且cosA cosB=b a=4 3,证明∠C=90°
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![△ABC中,c=10 且cosA cosB=b a=4 3,证明∠C=90°](/uploads/image/f/935193-57-3.jpg?t=%E2%96%B3ABC%E4%B8%AD%2Cc%3D10+%E4%B8%94cosA+cosB%3Db+a%3D4+3%2C%E8%AF%81%E6%98%8E%E2%88%A0C%3D90%C2%B0)
然后呢.再问:题被吞了?!1.sin²B+C/2+cos²2A2.若b=2,ABC的面积S=3,求a再问:1.求sin²B+C/2+cos2A2.若b=2,三角形ABC的
由ccosB=bcosC可得bc=cosBcosC,由正弦定理知,bc=sinBsinC,∴sinBsinC=cosBcosC,化简得sinBcosC-cosBsinC=0,即sin(B-C)=0,∴
答:三角形ABC中:cosA=1/3A+B+C=180°所以:sin(B+C)=sinA=√(1-cos²A)=√[1-(1/3)²]=2√2/3所以:sin(B+C)=2√2/3
²-bc-2c²=0,=>(b-2c)(b+c)=0,=>b=2c或b=-c,b>0,c>0,=>b=2c,cosA=7/8,=>sinA=√(1-cos²A)=√15/
/>(1)sinC=sin(2A)=2sinAcosAsinC/sinA=2cosA=3/2a/sinA=c/sinCc/a=sinC/sinA=3/2c=3a/2a+c=10a+3a/2=10a=4
(1)sin2B+C2+cos2A=cos2A2+cos2A=1+cosA2+2cos2A−1=1+452+2×1625−1=5950(6分)(2)∵cosA=45∴sinA=35 &nbs
m*n=1*1*cos60=1/2=sinA*sinB-cosA*cosB=-cos(A+B)=cosCC=π/31/2absinc=S=2根号3ab=8c平方=a平方+b平方-2abcosC=a平方
ccosB=bcosC,正弦定理b/sinB=c/sinC,推出sinCcosB=sinBcosC,sinC/cosC=sinB/cosB,也就是tanC=tanB,B=CB=90-A/2sinB=s
(1)sin^(B+C)=[1-cos2(B+C)]/2=[1-cos2(π-A)]/2=[1-cos(-2A)]/2=[1-cos2A]/2原式=[1+cos2A]/2=cos^A=1/9;(2)c
(1)[sin(B+C)/2]^2=[1-cos(B+C)]/2=(1+cosA)/2=9/10,cos2A=2(cosA)^2-1=7/25,所以[sin(B+C)/2]^2+cos2A=59/50
1、cosA·(√3sinA-cosA)=√3sinAcosA-cos²A=√3/2sin2A-(1+cos2A)/2=√3/2sin2A-cos2A/2-1/2=sin(2A-π/6)-1
(Ⅰ)由题设及正弦定理知:cosAcosB=sinBsinA,得sin2A=sin2B∴2A=2B或2A+2B=π,即A=B或A+B=π2当A=B时,有sin(π-2A)=cosA,即sinA=12,
即cosA*cosA-cosA+1/4=0cosA=1/2A=60余弦定理b*b+c*c-bc=a*a---(1)3a*a=(b+c)(b+c)=b*b+c*c+2bc==>bc=2/3a*a带入(1
(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
根据正弦定理,a=2RsinA,b=2RsinB,c=2RsinC,代入已知的式子,整理有,2sinB*cosA=√3sin(A+C)=√3sinB,即cosA=√3/2,所以A=π/6设AC=2x,
(1)在△ABC中中,由cosA=45得sinA=1−(45)2=35则2sin(2A+π4)=2(sin2Acosπ4+cos2Asinπ4)=sin2A+cos2A=2sinAcosA+2cos2
sinA=tanB得sinAcosB=sinB又a/sinA=b/sinB得a/b=sinA/sinB,a=b(1+cosA)得a/b=1+cosA则sinA/sinB=1+cosAsinA=sinB
由正弦定理可得:a/sinA=b/sinB=c/sinC=2R代入(2c-b)cosA-acosb(2sinC-sinB)cosA=sinAcosB2sinCcosA=sin(A+B)=sinCcos
是求b,a的长度吧,c边有了啊.可以利用正弦定理和三角变换cosA/cosB=b/a由正弦定理cosA/cosB=sinB/sinAsinAcosA=sinBcosBsin2A=sin2B2A=2B或
cotA/2-tanA/2=1/tanA/2-tanA/2=[(tanA/2)^2-1]/tanA/2=-2{[1-(tanA/2)^2]/2tanA/2}=-2/tanA所以(cosA)^2*tan