已知函数f(x)=cos(2x-pi/3)+2sin(x-pi)*sin(x+pi/4)

1个回答

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

    =cos(2x-π/3)-√2sinx(sinx+cosx)

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

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

    =√[(4+√2-√6)/2]sin(2x+θ)-√2/2;θ=arctan(2+√2+√3+√6)

    函数f(x)的最小正周期T=2∏/2=∏;

    对称轴方程2x+θ=(k+1/2)∏,x=(2k∏-2θ+1)/4

    函数f(x)在一个周期的区间内,sin(2x+θ)∈[-1,1]

    值域 f(x)∈[-√[(4+√2-√6)/2]-√2/2,√[(4+√2-√6)/2]-√2/2]