已知:w>0,a=(2sinwx+coswx,2sinwx-coswx),b=(sinwx,coswx),f(x)=a*

1个回答

  • 1

    f(x)=a·b=(2sin(wx)+cos(wx),2sin(wx)-cos(wx))·(sin(wx),cos(wx))

    =(2sin(wx)+cos(wx))sin(wx)+(2sin(wx)-cos(wx))cos(wx)

    =2sin(wx)^2+sin(wx)cos(wx)+2sin(wx)cos(wx)-cos(wx)^2

    =3sin(2wx)/2+(1-cos(2wx))-(1+cos(2wx))/2

    =3sin(2wx)/2-3cos(2wx)/2+1/2

    =(3√2/2)sin(2wx-π/4)+1/2

    f(x)图像相邻的两条对称轴的距离为π/2

    即f(x)的最小正周期:T=π

    即:2π/(2w)=π

    即:w=1

    即:f(x)=(3√2/2)sin(2x-π/4)+1/2

    2

    增区间:2x-π/4∈[2kπ-π/2,2kπ+π/2]

    即:x∈[kπ-π/8,kπ+3π/8],k∈Z

    减区间:2x-π/4∈[2kπ+π/2,2kπ+3π/2]

    即:x∈[kπ+3π/8,kπ+7π/8],k∈Z

    x∈[0,π/2],故当k=0时,增区间:x∈[0,3π/8]

    减区间:x∈[3π/8,π/2]

    x∈[0,π/2],2x-π/4∈[-π/4,3π/4]

    sin(2x-π/4)∈[-√2/2,1]

    故:(3√2/2)sin(2x-π/4)+1/2∈[-1,(3√2+1)/2]

    即:fmax=(3√2+1)/2

    fmin=-1