1)求函数的最小正周期以及对称轴方程(2)求函数在区间[-pai/12,pai/2]上的值域

2个回答

  • f(x)=cos(2x-π/3)+2sin(x-π/4)*sin(x+π/4)

    =cos(2x-π/3)+2sin(x-π/4)cos(π/2-x-π/4)

    =cos2xcosπ/3+sin2xsinπ/3 +2sin(x-π/4)cos(x-π/4)

    =(1/2)*cos2x+(√3/2)sin2x+sin(2x-π/2)

    =(1/2)*cos2x+(√3/2)sin2x-sin(π/2-2x)

    =(1/2)*cos2x+(√3/2)sin2x-cos2x

    =(√3/2)sin2x-(1/2)cos2x

    =cos(π/6)sin2x-sin(π/6)cos2x

    =sin(2x-π/6)

    ∴最小正周期为:2π/2=π

    因为y=sinx的对称轴为:kπ+π/2,k∈Z

    ∴2x-π/6=kπ+π/2

    x=kπ/2 +π/3即为f(x)=sin(2x-π/6)的对称轴.

    2)

    x∈[-π/12,π/12]

    2x∈[-π/6,π/6]

    2x-π/6∈[-π/3,0]

    ∴sin(2x-π/6)∈[-√3/2,0]

    函数在区间[-π/12,π/12]上的值域为:[-√3/2,0]