已知函数f(x)=根号3sin^2+sinxcosx-二分之根号3 x属于R

1个回答

  • f(x)=√3sin^2 x+sinxcosx-(√3/2)(x∈R)

    =√3*[(1-cos2x)/2]+(1/2)sin2x-(√3/2)

    =(√3/2)-(√3/2)cos2x+(1/2)sin2x-(√3/2)

    =(1/2)sin2x-(√3/2)cos2x

    =sin[2x-(π/3)]

    (1)f(π/4)=sin[(π/2)-(π/3)]=sin(π/6)=1/2

    (2)当x∈(0,π)时,2x∈(0,2π)

    2x-(π/3)∈(-π/3,5π/3)

    所以,f(x)的最大值为1

    (1)A<B,且f(A)=f(B)=1/2

    ===> sin[2A-(π/3)]=sin[2B-(π/3)]=1/2

    ===> 2A-(π/3)=π/6;2B-(π/3)=5π/6

    ===> 2A=π/2;2B=7π/6

    ===> A=π/4,B=7π/12

    所以,C=π-(π/4)-(7π/12)=π/6

    在△ABC中,BC=a,AB=c

    所以,由正弦定理有:BC/AB=a/c=sinA/sinC=√2.