三角函数难题 3.已知函数f(x)=4cosxsin(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.4.已知函数f(x)=cos^2ωx+√3sinωxcosωx(ω>0)的最小正周期为π.(1)求f(2/3π)的值
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![三角函数难题 3.已知函数f(x)=4cosxsin(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.4.已知函数f(x)=cos^2ωx+√3sinωxcosωx(ω>0)的最小正周期为π.(1)求f(2/3π)的值](/uploads/image/z/7070646-30-6.jpg?t=%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E9%9A%BE%E9%A2%98+3.%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D4cosxsin%28x%2B%5B%CF%80%2F6%5D%29-1%281%29%E6%B1%82f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%EF%BC%9B%EF%BC%882%EF%BC%89%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.4.%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dcos%5E2%CF%89x%2B%E2%88%9A3sin%CF%89xcos%CF%89x%28%CF%89%3E0%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E4%B8%BA%CF%80.%EF%BC%881%EF%BC%89%E6%B1%82f%282%2F3%CF%80%29%E7%9A%84%E5%80%BC)
三角函数难题 3.已知函数f(x)=4cosxsin(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.4.已知函数f(x)=cos^2ωx+√3sinωxcosωx(ω>0)的最小正周期为π.(1)求f(2/3π)的值
三角函数难题
3.已知函数f(x)=4cosxsin(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.
4.已知函数f(x)=cos^2ωx+√3sinωxcosωx(ω>0)的最小正周期为π.
(1)求f(2/3π)的值;
(2)求函数f(x)的单调区间及其图像的对称轴方程
三角函数难题 3.已知函数f(x)=4cosxsin(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.4.已知函数f(x)=cos^2ωx+√3sinωxcosωx(ω>0)的最小正周期为π.(1)求f(2/3π)的值
3.
(1)
f(x)=4cosx(sinxcosπ/6+cosxsinπ/6)-1
=2√3sinxcosx+2cos²x-1
=√3sin2x+cos2x
=2(sin2x*√3/2+cos2x*1/2)
=2(sin2xcosπ/6+cos2xsinπ/6)
=2sin(2x+π/6)
∴f(x)的最小正周期:T=2π/2=π ,
(2)∵-π/6≤x≤π/4 ∴-π/6≤2x+π/6≤2π/3
∴2x+ π/6=π/2时,f(x)取得最大值 2
2x+ π/6=-π/6时,f(x)取得 最小值-1
4.
(1)f(x)=1/2(1+cos2ωx)+√3/2*sin2ωx
=1/2+√3/2*sin2ωx+1/2*cos2ωx
=1/2+sin(2ωx+π/6)
∵ f(x)的最小正周期为π,
∴T=2π/2ω=π,解得ω=1
∴ f(x)=sin(2x+π/6)+1/2
∴f(2π/3)=sin(4π/3+π/6)+1/2
=sin(3π/2)+1/2=1/2
(2)
由2kπ-π/2≤2x+π/6≤2kπ+π/2(k∈Z),
得:kπ-π/3≤x≤kπ+π/6,(k∈Z)
∴函数f(x)的单调增区间为
[kπ-π/3,kπ+π/6](k∈Z)
由 2kπ+π/2≤2x+π/6≤2kπ+3π/2(k∈Z),
得:kπ+π/6≤x≤kπ+2π/3(k∈Z)
函数f(x)的单调减区间为
[kπ+π/6,kπ+2π/3](k∈Z)
由2x+π/6=kπ+π/2(k∈Z)
得x=kπ/2+π/6(k∈Z)
∴f(x)图象的对称轴方程为
x=kπ/2+π/6(k∈Z)
3.
f(x)=4cosx(sinxcosπ/6+cosxsinπ/6)+1
=2√3sinxcosx+2cos²x-1+2
=√3sin2x+cos2x+2
=2(sin2x*√3/2+cos2x*1/2)+2
=2(sin2xcosπ/6+cos2xsinπ/6)+2
=2sin(2x+π/6)+2
则:T=2π/2=...
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3.
f(x)=4cosx(sinxcosπ/6+cosxsinπ/6)+1
=2√3sinxcosx+2cos²x-1+2
=√3sin2x+cos2x+2
=2(sin2x*√3/2+cos2x*1/2)+2
=2(sin2xcosπ/6+cos2xsinπ/6)+2
=2sin(2x+π/6)+2
则:T=2π/2=π ,因为-π/6≤2x+π/6≤2π/3
所以-1/2≤sin(2x+π/6)≤1 ,所以最大值:2×1+2=4,最小值:2×(-1/2)+2=1
4.
(1)f(x)=1/2(1+cos2ωx)+√3/2sin2ωx=1/2+sin(2ωx+π/6)
因为f(x)的最小正周期为π,所以2π/2ω=π,解得ω=1
所以f(x)=sin(2x+π/6)+1/2
所以f(π3/2)=0
(2)由2kπ-π/2≤2x+π/6≤2kπ+π/2(k∈Z),2kπ+π/2≤2x+π/6≤2kπ+3π/2(k∈Z),
可得kπ-π/3≤x≤kπ+π/6(k∈Z),kπ+π/6≤x≤kπ+2π/3(k∈Z)
所以,函数f(x)的单调增区间为(kπ-π/3,kπ+π/6)(k∈Z)
函数f(x)的单调减区间为(kπ+π/6,kπ+2π/3)(k∈Z)
由2x+π/6=kπ+π/2(k∈Z)得x=kπ/2+π/6(k∈Z)
所以f(x)图象的对称轴方程为x=kπ/2+π/6(k∈Z)
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3.
f(x)=4cosx(sinxcosπ/6+cosxsinπ/6)+1
=2√3sinxcosx+2cos²x-1+2
=√3sin2x+cos2x+2
=2(sin2x*√3/2+cos2x*1/2)+2
=2(sin2xcosπ/6+cos2xsinπ/6)+2
=2sin(2x+π/6)+2
则:T=2π/2=...
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3.
f(x)=4cosx(sinxcosπ/6+cosxsinπ/6)+1
=2√3sinxcosx+2cos²x-1+2
=√3sin2x+cos2x+2
=2(sin2x*√3/2+cos2x*1/2)+2
=2(sin2xcosπ/6+cos2xsinπ/6)+2
=2sin(2x+π/6)+2
则:T=2π/2=π ,因为-π/6≤2x+π/6≤2π/3
所以-1/2≤sin(2x+π/6)≤1 ,所以最大值:2×1+2=4,最小值:2×(-1/2)+2=1
4.
(1)f(x)=1/2(1+cos2ωx)+√3/2sin2ωx=1/2+sin(2ωx+π/6)
因为f(x)的最小正周期为π,所以2π/2ω=π,解得ω=1
所以f(x)=sin(2x+π/6)+1/2
所以f(π3/2)=0
(2)由2kπ-π/2≤2x+π/6≤2kπ+π/2(k∈Z),2kπ+π/2≤2x+π/6≤2kπ+3π/2(k∈Z),
可得kπ-π/3≤x≤kπ+π/6(k∈Z),kπ+π/6≤x≤kπ+2π/3(k∈Z)
所以,函数f(x)的单调增区间为(kπ-π/3,kπ+π/6)(k∈Z)
函数f(x)的单调减区间为(kπ+π/6,kπ+2π/3)(k∈Z)
由2x+π/6=kπ+π/2(k∈Z)得x=kπ/2+π/6(k∈Z)
所以f(x)图象的对称轴方程为x=kπ/2+π/6(k∈Z)
(1)
f(x)=4cosx(sinxcosπ/6+cosxsinπ/6)-1
=2√3sinxcosx+2cos²x-1
=√3sin2x+cos2x
=2(sin2x*√3/2+cos2x*1/2)
=2(sin2xcosπ/6+cos2xsinπ/6)
=2sin(2x+π/6)
∴f(x)的最小正周期:T=2π/2=π ,
(2)∵-π/6≤x≤π/4 ∴-π/6≤2x+π/6≤2π/3
∴2x+ π/6=π/2时,f(x)取得最大值 2
2x+ π/6=-π/6时,f(x)取得 最小值-1
4.
(1)f(x)=1/2(1+cos2ωx)+√3/2*sin2ωx
=1/2+√3/2*sin2ωx+1/2*cos2ωx
=1/2+sin(2ωx+π/6)
∵ f(x)的最小正周期为π,
∴T=2π/2ω=π,解得ω=1
∴ f(x)=sin(2x+π/6)+1/2
∴f(2π/3)=sin(4π/3+π/6)+1/2
=sin(3π/2)+1/2=1/2
(2)
由2kπ-π/2≤2x+π/6≤2kπ+π/2(k∈Z),
得:kπ-π/3≤x≤kπ+π/6, (k∈Z)
∴函数f(x)的单调增区间为
[kπ-π/3,kπ+π/6](k∈Z)
由 2kπ+π/2≤2x+π/6≤2kπ+3π/2(k∈Z),
得:kπ+π/6≤x≤kπ+2π/3(k∈Z)
函数f(x)的单调减区间为
[kπ+π/6,kπ+2π/3](k∈Z)
由2x+π/6=kπ+π/2(k∈Z)
得x=kπ/2+π/6(k∈Z)
∴f(x)图象的对称轴方程为
x=kπ/2+π/6(k∈Z)
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