已知涵数f(x)=从事(2x-π /3)+2sin(x-π /4)sin(x+π /4)

1个回答

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

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

    =1/2cos2x+√3/2sin2x+2sin(x-π/4)cos(x-π/4)

    =1/2cos2x+√3/2sin2x+sin2(x-π/4)

    =1/2cos2x+√3/2sin2x+sin(2x-π/2)

    =1/2cos2x+√3/2sin2x-cos2x

    =√3/2-1/2cos2x

    =sin(2x-π/6)

    (1)

    最小正周期:T=2π/2=π

    对称轴:

    2x-π/6=π/2+kπ(k∈Z)

    2x=2π/3+kπ

    x=π/3+kπ/2

    (2)

    2x-π/6=-π/2+2kπ(k∈Z)

    2x=-π/3+2kπ

    x=-π/6+kπ

    2x-π/6=π/2+2kπ(k∈Z)

    2x=2π/3+2kπ

    x=π/3+kπ

    当x=-π/12时,f(x)=-√3/2

    当x=π/3是,f(x)=1

    f(x)在区间[-π /12,π /2]上的值域 为[-√3/2,1]