在三角形ABC中,角A,B,C所对边分别为a,b,c,且1+tanA/tanB=2c/b

1个回答

  • |m+n|取最小值√2.

    由1+tanA/tanB=2c/b得,

    tanB+tanA=2tanB*c/b,由正弦定理得c/b=sinC/sinB,故得

    tanB+tanA=2tanB*sinC/sinB=2sinC/cosB

    即tanB+tanA=2sinC/cosB

    sinB*cosA+sinA*cosB=2sinC*cosA

    sin(A+B)=2sinC*cosA,

    sinC=sin(180-A-B)=sin(A+B)=2sinC*cosA,sinC=2sinC*cosA,

    由sinC不等于零,故得cosA=1/2,A=30,

    m+n=(cosB,-1+2(cosC/2)^2)=(cosB,-cosC),B+C=150,C=150-B,

    |m+n|^2=(cosB)^2+(cosC)^2=(cosB)^2+(cos(150-B))^2

    =2+cos2B+cos(300-2B)=2+cos2B+cos300cos2B+sin300sin2B

    =2+cos2B+(1/2)cos2B-(√3/2)sin2B=2+(3cos2B-√3sin2B)/2

    =2+√3sin(D-2B),其中tanD=3/(2√3),

    当D-2B=0时,|m+n|^2取得最小值2,即|m+n|取得最小值√2,当D-2B=90时,|m+n|^2取得最大值2+√3,即|m+n|取得最大值√(2+√3).