在三角形ABC中,A,B,C的对边分别为a,b,c.且acosC,bcosB,ccosA.成等差数列.求(1)B的值.(2)求2sin^2A+c...在三角形ABC中,A,B,C的对边分别为a,b,c.且acosC,bcosB,ccosA.成等差数列.求(1)B的值.(2)求2sin^2A+cos(A-C)的范围
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![在三角形ABC中,A,B,C的对边分别为a,b,c.且acosC,bcosB,ccosA.成等差数列.求(1)B的值.(2)求2sin^2A+c...在三角形ABC中,A,B,C的对边分别为a,b,c.且acosC,bcosB,ccosA.成等差数列.求(1)B的值.(2)求2sin^2A+cos(A-C)的范围](/uploads/image/z/2620542-30-2.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2CA%2CB%2CC%E7%9A%84%E5%AF%B9%E8%BE%B9%E5%88%86%E5%88%AB%E4%B8%BAa%2Cb%2Cc.%E4%B8%94acosC%2CbcosB%2CccosA.%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97.%E6%B1%82%281%29B%E7%9A%84%E5%80%BC.%282%29%E6%B1%822sin%5E2A%2Bc...%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2CA%2CB%2CC%E7%9A%84%E5%AF%B9%E8%BE%B9%E5%88%86%E5%88%AB%E4%B8%BAa%2Cb%2Cc.%E4%B8%94acosC%2CbcosB%2CccosA.%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97.%E6%B1%82%281%29B%E7%9A%84%E5%80%BC.%282%29%E6%B1%822sin%5E2A%2Bcos%28A-C%29%E7%9A%84%E8%8C%83%E5%9B%B4)
在三角形ABC中,A,B,C的对边分别为a,b,c.且acosC,bcosB,ccosA.成等差数列.求(1)B的值.(2)求2sin^2A+c...在三角形ABC中,A,B,C的对边分别为a,b,c.且acosC,bcosB,ccosA.成等差数列.求(1)B的值.(2)求2sin^2A+cos(A-C)的范围
在三角形ABC中,A,B,C的对边分别为a,b,c.且acosC,bcosB,ccosA.成等差数列.求(1)B的值.(2)求2sin^2A+c...
在三角形ABC中,A,B,C的对边分别为a,b,c.且acosC,bcosB,ccosA.成等差数列.求(1)B的值.(2)求2sin^2A+cos(A-C)的范围
在三角形ABC中,A,B,C的对边分别为a,b,c.且acosC,bcosB,ccosA.成等差数列.求(1)B的值.(2)求2sin^2A+c...在三角形ABC中,A,B,C的对边分别为a,b,c.且acosC,bcosB,ccosA.成等差数列.求(1)B的值.(2)求2sin^2A+cos(A-C)的范围
acosC,bcosB,ccosA成等差数列
2bcosB=acosC+ccosA.
根据正弦定理得
a/sinA=b/sinB=c/sinC=k
a=ksinA,b=ksinB,c=ksinC
代入上式得
2ksinBcosB=ksinAcosC+ksinCcosA
sin2B=sin(A+C)=sinB
B=60°
2sin^2A+cos(A-C)
=2sin^2A+cos[A-(120-A)]
=2sin^2A+cos(2A-120)
=1-cos2A+cos(2A-120)
=1+2sin(2A-60)sin60
=1+√3sin(2A-60)
由于0
acosC+ccosA=2bcosB
∴sinAcosC+sinCcosA=2sinBcosB
∴sin(A+C)=2sinBcosB
即sinB=2sinBcosB
∵sinB≠0,∴cosB=1/2
∴B=120°
2sin²A+cos(A-C)
=1-cos2A+cos(2A-120°)
=1-cos2A-1/2cos...
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acosC+ccosA=2bcosB
∴sinAcosC+sinCcosA=2sinBcosB
∴sin(A+C)=2sinBcosB
即sinB=2sinBcosB
∵sinB≠0,∴cosB=1/2
∴B=120°
2sin²A+cos(A-C)
=1-cos2A+cos(2A-120°)
=1-cos2A-1/2cos2A+√3/2sin2A
=√3/2sin2A-3/2cos2A+1
=√3sin(2A-π/3)+1
∵0∴-π/3<2A-π/3<π/3
∴所求范围是(1/2,3/2)
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(1) 2bcosB=acosC+ccosA
由正弦定理得2sinBcosB=sinAcosC+sinCcosA
∴2sinBcosB=sin(A+C)
∴2sinBcosB=sinB
∴cosB=1/2
∴B=60度
(2)
2sin²A+cos(A-C)
=1-cos2A+cos(2A-120)
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(1) 2bcosB=acosC+ccosA
由正弦定理得2sinBcosB=sinAcosC+sinCcosA
∴2sinBcosB=sin(A+C)
∴2sinBcosB=sinB
∴cosB=1/2
∴B=60度
(2)
2sin²A+cos(A-C)
=1-cos2A+cos(2A-120)
=1-2sin(2A-60)sin(-60)
=1+2sin60sin(2A-60)
∵B=60度
∴A∈(0,120)
∴2A-60∈(-60,180)
∴sin(2A-60)∈(-sin60,1]
(将sin60的值代入)(A=75时,原式为1)
∴原式的范围是(-1/2,1+根号3〕
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