已知函数f(x)=1/2sin2xsinψ+cos²xcosψ-1/2sin(π/2+ψ)(0《 ψ 》π),齐图象过点(π/6,1/2)(1)求ψ的值(2)将函数y=f(x)的图象上各点的横坐标缩短到原来的1/2,纵坐标不变,得到函数y=g(x)的图象,
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![已知函数f(x)=1/2sin2xsinψ+cos²xcosψ-1/2sin(π/2+ψ)(0《 ψ 》π),齐图象过点(π/6,1/2)(1)求ψ的值(2)将函数y=f(x)的图象上各点的横坐标缩短到原来的1/2,纵坐标不变,得到函数y=g(x)的图象,](/uploads/image/z/1618399-55-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D1%2F2sin2xsin%CF%88%2Bcos%26%23178%3Bxcos%CF%88-1%2F2sin%28%CF%80%2F2%2B%CF%88%EF%BC%89%EF%BC%880%E3%80%8A+%CF%88+%E3%80%8B%CF%80%EF%BC%89%2C%E9%BD%90%E5%9B%BE%E8%B1%A1%E8%BF%87%E7%82%B9%EF%BC%88%CF%80%2F6%2C1%2F2%EF%BC%89%EF%BC%881%EF%BC%89%E6%B1%82%CF%88%E7%9A%84%E5%80%BC%EF%BC%882%EF%BC%89%E5%B0%86%E5%87%BD%E6%95%B0y%3Df%28x%29%E7%9A%84%E5%9B%BE%E8%B1%A1%E4%B8%8A%E5%90%84%E7%82%B9%E7%9A%84%E6%A8%AA%E5%9D%90%E6%A0%87%E7%BC%A9%E7%9F%AD%E5%88%B0%E5%8E%9F%E6%9D%A5%E7%9A%841%2F2%2C%E7%BA%B5%E5%9D%90%E6%A0%87%E4%B8%8D%E5%8F%98%2C%E5%BE%97%E5%88%B0%E5%87%BD%E6%95%B0y%3Dg%28x%29%E7%9A%84%E5%9B%BE%E8%B1%A1%2C)
已知函数f(x)=1/2sin2xsinψ+cos²xcosψ-1/2sin(π/2+ψ)(0《 ψ 》π),齐图象过点(π/6,1/2)(1)求ψ的值(2)将函数y=f(x)的图象上各点的横坐标缩短到原来的1/2,纵坐标不变,得到函数y=g(x)的图象,
已知函数f(x)=1/2sin2xsinψ+cos²xcosψ-1/2sin(π/2+ψ)(0《 ψ 》π),齐图象过点(π/6,1/2)
(1)求ψ的值
(2)将函数y=f(x)的图象上各点的横坐标缩短到原来的1/2,纵坐标不变,得到函数y=g(x)的图象,求函数g(x)在区间【0,π/4】上的最大和最小值
已知函数f(x)=1/2sin2xsinψ+cos²xcosψ-1/2sin(π/2+ψ)(0《 ψ 》π),齐图象过点(π/6,1/2)(1)求ψ的值(2)将函数y=f(x)的图象上各点的横坐标缩短到原来的1/2,纵坐标不变,得到函数y=g(x)的图象,
(1)
f(x)=1/2sin2xsinψ+cos²xcosψ-1/2sin(π/2+ψ)
=1/2*sin2x*sinψ+cos²xcosψ-1/2*cosψ
=1/2*(sin2x*sinψ+cos2x*cosψ)=1/2*cos(2x-ψ)
由 f(π/6)=1/2*cos(2*π/6-ψ)=1/2
得 2*π/6-ψ=0 ==> ψ=π/3
(2)
令 x=2t
则 y=g(t)=1/2*cos(4t-π/3)
若 0 ≤ t ≤ π/4
则 -π/3 ≤ 4t-π/3 ≤ 2π/3
因为cos在区间 [-π/3 ,0]是增函数,在 [0,2π/3]上是减函数,所以
y=1/2*cos(4t-π/3) 在[0,π/4]的最大值 fmax=1/2*cos(0)=1/2
最小值 fmin=1/2*cos(2π/3)=-1/2*1/2= -1/4