已知函数f(x)=cos(2x-3/派)+2sin(x-4/派)sin(x+4/派)1.求f(x)的最小正周期和图像对称轴方程.2.求函数f(x)在区间[-12/派,2/派]上的值域.
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![已知函数f(x)=cos(2x-3/派)+2sin(x-4/派)sin(x+4/派)1.求f(x)的最小正周期和图像对称轴方程.2.求函数f(x)在区间[-12/派,2/派]上的值域.](/uploads/image/z/404473-49-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%EF%BC%89%3Dcos%EF%BC%882x-3%2F%E6%B4%BE%EF%BC%89%2B2sin%28x-4%2F%E6%B4%BE%EF%BC%89sin%28x%2B4%2F%E6%B4%BE%EF%BC%891.%E6%B1%82f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E5%92%8C%E5%9B%BE%E5%83%8F%E5%AF%B9%E7%A7%B0%E8%BD%B4%E6%96%B9%E7%A8%8B.2.%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%E5%8C%BA%E9%97%B4%5B-12%2F%E6%B4%BE%2C2%2F%E6%B4%BE%5D%E4%B8%8A%E7%9A%84%E5%80%BC%E5%9F%9F.)
已知函数f(x)=cos(2x-3/派)+2sin(x-4/派)sin(x+4/派)1.求f(x)的最小正周期和图像对称轴方程.2.求函数f(x)在区间[-12/派,2/派]上的值域.
已知函数f(x)=cos(2x-3/派)+2sin(x-4/派)sin(x+4/派)1.求f(x)的最小正周期和图像对称轴方程.
2.求函数f(x)在区间[-12/派,2/派]上的值域.
已知函数f(x)=cos(2x-3/派)+2sin(x-4/派)sin(x+4/派)1.求f(x)的最小正周期和图像对称轴方程.2.求函数f(x)在区间[-12/派,2/派]上的值域.
已知 f(x)=cos (2x-派/3)+2sin (x-派/4)sin (x+派/4)
=cos (2x-派/3)-2sin (x+派/4-派/2)sin (x+派/4)
=cos (2x-派/3)-2cos (x+派/4)sin (x+派/4)
=cos (2x-派/3)+sin (2x+派/2)
=cos (2x-派/3)-cos 2x
根据和差化积的公式,得
原式=2sin (2x-派/6)sin (派/6)
=sin (2x-派/6)
周期T=2派/2=派,对称轴:2x-派/6=派/2+k派 (k属于整数)
x=派/3+k派/2 (k属于整数)
因为函数 f(x)在-派/6到派/3为单调增函数
所以函数 f(x)在-派/12到派/2上的值域为:[-根号3/2,1]
已知 f(x)=cos (2x-派/3)+2sin (x-派/4)sin (x+派/4)
=cos (2x-派/3)-2sin (x+派/4-派/2)sin (x+派/4)
=cos (2x-派/3)-2cos (x+派/4)sin (x+派/4)
=cos (2x-派/3)+sin (2x+派/2)
=cos (2x-派/3)-cos 2x
根据和差化积的...
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已知 f(x)=cos (2x-派/3)+2sin (x-派/4)sin (x+派/4)
=cos (2x-派/3)-2sin (x+派/4-派/2)sin (x+派/4)
=cos (2x-派/3)-2cos (x+派/4)sin (x+派/4)
=cos (2x-派/3)+sin (2x+派/2)
=cos (2x-派/3)-cos 2x
根据和差化积的公式,得
原式=2sin (2x-派/6)sin (派/6)
=sin (2x-派/6)
周期T=2派/2=派,对称轴:2x-派/6=派/2+k派 (k属于整数)
x=派/3+k派/2 (k属于整数)
因为函数 f(x)在-派/6到派/3为单调增函数
所以函数 f(x)在-派/12到派/2上的值域为:[-根号3/2,1]就是这样了
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(1)f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
=cos(2x-π/3)+cos[(x-π/4)-(x+π/4)]-cos[(x-π/4)+(x+π/4)]
=cos(2x-π/3)-cos2x+cos(-π/2)
=cos(2x-π/3)-cos2x
=-2sin(2x-π/6)sin(-...
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(1)f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
=cos(2x-π/3)+cos[(x-π/4)-(x+π/4)]-cos[(x-π/4)+(x+π/4)]
=cos(2x-π/3)-cos2x+cos(-π/2)
=cos(2x-π/3)-cos2x
=-2sin(2x-π/6)sin(-π/6)
=sin(2x-π/6)
故最小正周期为π
图像的对称轴方程x=kπ/2+π/3
(2)由x∈[-π/12,π/2]得2x-π/6∈[-π/3,5π/6]
由π/2∈[-π/3,5π/6]
函数最大值为1
最小值为sin(-π/3)=-√3/2
故函数的值域是[-√3/2,1]
这题我们昨天考试卷上出了
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