sinA.sinB.sinC成等差数列,且C-A=π 3
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a,b,c成等差数列:2b=a+cb/sinB=a/sinA=c/sinC=2R2sinB=sinA+sinCsinB=(sinA+sinC)/2sinA,sinB,sinC成等比数列:sin^2B=
A=B=C=6时0最大,为3/2根号3证明:sinA+sinB+sinc=2sin[(A+B)/2]cos[(A-B)/2]+sinC>=2sin[(A+B)/2]+sinC=2sin(90-C/2)
∵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
题目应该是在锐角三角形中.诚如是,则解答如下:先证明sinA+sinB>1+cosC.由A、B是锐角得A-B0,所以sinA+sinB>1+cosC.所以sinA+sinB+sinC>1+cosC+s
1.假设a/sinA=b/sinB=c/sinC=2R那么sinA=a/2RsinB=b/2RsinC=c/2R因为(sinA)平方=(sinB)平方+sinC(sinB+sinC)所以(a/2R)^
120°利用前两个比例:5(sinB+sinC)=4(sinC+sinA)化简得到sinC=4sinA-5sinB利用后两个比例:6(sinC+sinA)=5(sinA+sinB)化简得到sinA=5
做出来啦!不过这题目有点小问题,只有锐角三角形时此题成立钝角三角形不等式反向若A=120,B=30,C=30直角三角形为等号设q=(A-B)/2sinA+sinB-cosA-cosB=2cos(C/2
锐角三角形,因为以直角三角形为界限sinA^2+sinB^2恰好等于1等于SinC^2=2,sinA^2+sinB^2+sinC^2的值若大于2则是钝角,小于2则是锐角.至于直角三角形sinA^2+s
已知sinA,sinB,sinC成等差数列则sinA+sinC=2sinB由正弦定理,化为边的形式得a+c=2bb=(a+c)/2由余弦定理cosB=(a²+c²-b²)
(b-a)(sinA+sinB)=bsinA,(b-a)/b=sinA/(sinA+sinB),(1)根据正弦定理,sinA/a=sinB/b,(sinA+sinB)/(a+b)=sinA/a,(等比
证明:设sinA/a=sinB/b=sinC/c=k,则sinA=ak,sinB=bk,sinC=ck,sinA/(sinB+sinC)+sinB/(sinA+sinC)+sinC(sinA+sinB
由sinA/a=sinB/b=sinC/c(其中a,b,c为角A,B,C对应的三条边)设sinA/a=sinB/b=sinC/c=k则a=sinA/k,b=sinB/k,c=sinC/k带入(sinB
把它变化为正玄定理(a+b+c)(a+b-c)=aba^2+b^2+2ab-c^2=ab(a^2+b^2-c^2)/ab=-1由余弦定理(a^2+b^2-c^2)/2ab=-1/2=cosCc=120
锐角三角形则A+B>90度所以A>90-B且A和90-B都是锐角sin再次范围内递增所以sinA>sin(90-B)即sinA>cosB同理sinB>cosCsinC>cosA三个加起来即可再问:很好
证:∵△ABC为锐角三角形,∴A+B>90°得A>90°-B∴sinA>sin(90°-B)=cosB,即sinA>cosB,同理可得sinB>cosC,sinC>cosA上面三式相加:sinA+si
因为m垂直n所以m×n=0(要加向量符号)即(sinB+sinC,sinA-sinB)×(sinB-sinC,sin(B+C))=0又sin(B+C)=sin(π-A)=sinA所以原式=[(sinB
∵acosA+bcosB=ccosC∴sinAcosA+sinBcosB=sinCcosC∴sin2A+sin2B=sin2C=sin(2π-2A-2B)=-sin(2A+2B)∴0=sin2A+si
a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=2RsinAsinB-2RsinAsinC+2RsinBsinC-2RsinBsinA+2RsinCsinA-2RsinC
sinC=[2sin((A+B)/2)cos((A-B)/2)]/2[cos((A+B)/2)cos((A-B)/2)](和差化积)=sin((A+B)/2)/cos((A+B))/2=tan((18
设△ABC的外接圆半径为R,由正弦定理可得,sinB=b2R,sinA=a2R,sinC=c2R,所以a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=a(b2R−c2R)