已知函数f(x)=4cosx*sin(x+π/6)-11.求f(x)的最小正周期2.求f(x)在区间[-π/6,π/4]上的最大值和最小值
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![已知函数f(x)=4cosx*sin(x+π/6)-11.求f(x)的最小正周期2.求f(x)在区间[-π/6,π/4]上的最大值和最小值](/uploads/image/z/9319281-33-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D4cosx%2Asin%EF%BC%88x%2B%CF%80%2F6%29-11.%E6%B1%82f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F2.%E6%B1%82f%28x%29%E5%9C%A8%E5%8C%BA%E9%97%B4%5B-%CF%80%2F6%2C%CF%80%2F4%5D%E4%B8%8A%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC)
已知函数f(x)=4cosx*sin(x+π/6)-11.求f(x)的最小正周期2.求f(x)在区间[-π/6,π/4]上的最大值和最小值
已知函数f(x)=4cosx*sin(x+π/6)-1
1.求f(x)的最小正周期
2.求f(x)在区间[-π/6,π/4]上的最大值和最小值
已知函数f(x)=4cosx*sin(x+π/6)-11.求f(x)的最小正周期2.求f(x)在区间[-π/6,π/4]上的最大值和最小值
f(x)=4cosx*sin(x+π/6)-1
=4cosx(sinx*√3/2+cosx*1/2)-1
=2√3sinxcosx+2(cosx)^2-1
=√3sin2x+cos2x
=2sin(2x+π/6)
1.求f(x)的最小正周期 T =π
2.x∈[-π/6,π/4]
则2x+π/6∈[-π/6,2π/3]
最大值:y=2sin(π/2)=2
最小值:y=2sin(-π/6)=-1
用积化和差就行:2*sin(a)*cos(b)=sin(a+b)+sin(a-b)
f(x)=4sin(x+π/6)*cos(x)-1=2*(sin(2x+π/6)+2sin(π/6)-1=2*sin(2x+π/6)
所以T=π;当-π/6<=x<=π/4时,-π/6<=2x+π/6<=2π/3,
所以,f(x)_max=2*sin(π/2)=2,f(x)_min=2*sin(-π/6)=-1。
f(x)=4cosx*sin(x+π/6)-1
=4cosx(sinx*√3/2+cosx*1/2)-1
=2√3sinxcosx+2(cosx)^2-1
=√3sin2x+cos2x
=2sin(2x+π/6)
①:w=2,T=2π/w=π
②:令t=2x+π/6,因为x∈[-π/6,π/4],推出t∈[-π...
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f(x)=4cosx*sin(x+π/6)-1
=4cosx(sinx*√3/2+cosx*1/2)-1
=2√3sinxcosx+2(cosx)^2-1
=√3sin2x+cos2x
=2sin(2x+π/6)
①:w=2,T=2π/w=π
②:令t=2x+π/6,因为x∈[-π/6,π/4],推出t∈[-π/6,2π/3], 问题转化为求2sint在[-π/6,2π/3]的最大值和最小值,一画图,显然得到max=2,min=-1。这种问题,用这种方法是绝对不会错的。首先感谢楼上的人,我就不化简了,直接借用楼上化简的式子,但是关键部分的处理,楼上没有把这个思路写出来,等于就做了一个化简,就得到了答案。
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