已知函数f(x)=cos(2x-pai/3)+2sin(x-pai/4).sin(x+pai/4)求函数在区间[-pai

2个回答

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

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

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

    =cos(2x-π/3)+sin(2x-π/2)

    =cos(2x-π/3)-cos2x

    =(1/2)cos2x+(√3/2)sin2x-cos2x

    =(√3/2)sin2x-(1/2)cos2x

    =sin(2x-π/6)

    ﹣π/12≤x≤π/12

    ﹣π/6≤2x≤π/6

    ﹣π/3≤2x-π/6≤0

    当2x-π/6=0时,y最大

    y(max)=0

    当2x-π/6= -π/3时,y最小

    y(min)= - √3/2