已知函数f(x)=4sinxcos(x+π/3)+根号3 (1)若x∈R,求f(x)的最小正周期和单调
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解决时间 2021-04-06 03:38
- 提问者网友:记得曾经
- 2021-04-05 15:51
已知函数f(x)=4sinxcos(x+π/3)+根号3 (1)若x∈R,求f(x)的最小正周期和单调
最佳答案
- 五星知识达人网友:舍身薄凉客
- 2021-04-05 16:51
f(x)=4sinxcos(x+π/3)+根号3
=4sinx(1/2cosx-√3/2sinx)+√3
=2sinxcosx-2√3sin²x+√3
=sin2x-2√3sin²x+√3(sin²x+cos²x)
=sin2x+√3cos²x-√3sin²x
=sin2x+√3(cos²x-sin²x)
=sin2x+√3con2x
=2sin(2x+π/3)
(1)最小正周期T=2π/2=π
-π/2+2Kπ≤2x+π/3≤π/2+2Kπ
解得x∈[-5π/12+kπ,π/12+kπ],k∈{0,1,2。。。}
(2)x∈[-π/3,π/4] 则2x+π/3∈[-π/3,5π/6]
由图像知在2x+π/3=-π/3时取得最小值=-√3
2x+π/3=π/2取得最大值=2
=4sinx(1/2cosx-√3/2sinx)+√3
=2sinxcosx-2√3sin²x+√3
=sin2x-2√3sin²x+√3(sin²x+cos²x)
=sin2x+√3cos²x-√3sin²x
=sin2x+√3(cos²x-sin²x)
=sin2x+√3con2x
=2sin(2x+π/3)
(1)最小正周期T=2π/2=π
-π/2+2Kπ≤2x+π/3≤π/2+2Kπ
解得x∈[-5π/12+kπ,π/12+kπ],k∈{0,1,2。。。}
(2)x∈[-π/3,π/4] 则2x+π/3∈[-π/3,5π/6]
由图像知在2x+π/3=-π/3时取得最小值=-√3
2x+π/3=π/2取得最大值=2
全部回答
- 1楼网友:几近狂妄
- 2021-04-05 18:12
f(x)=4sinxcos(x+π/3)+√3=4sinx(1/2cosx-√3sinx)+√3=2sinxcosx-2√3sin²x+√3
=sin2x+√3﹙1-2sin²x)=sin2x+√3cos2x=2(1/2sin2X+√3/2cos2X)
=2(cosπ/3sin2x+sinπ/3cos2x)
=2sin(2x+π/3)
即f(x)=2sin(2x+π/3)
主要是化简吧,后面自己算算
=sin2x+√3﹙1-2sin²x)=sin2x+√3cos2x=2(1/2sin2X+√3/2cos2X)
=2(cosπ/3sin2x+sinπ/3cos2x)
=2sin(2x+π/3)
即f(x)=2sin(2x+π/3)
主要是化简吧,后面自己算算
- 2楼网友:佘樂
- 2021-04-05 17:45
已知函数f(x)=4sinxcos(x+π/3)+√3 ;(1)若x∈R,求f(x)的最小正周期和单调递增区间;
(2)若x∈[-π/3,π/4]求函数f(x)的最大值和最小值。
解:(1) f(x)=4sinxcos(x+π/3)+√3=2[sin(-π/3)+sin(2x+π/3)]+√3=2sin(2x+π/3)
故最小正周期T=2π/2=π;单增区间:由-π/2+2kπ≦2x+π/3≦π/2+2kπ,
得-5π/6+2kπ≦2x≦π/6+2kπ,故单增区间为-5π/12+kπ≦x≦π/12+kπ,k∈Z.
(2).在区间[-π/3,π/4]内,当x=π/12时,f(π/12)=2sin(π/6+π/3)=2sin(π/2)=2是其最大值;
当x=-π/3时,f(-π/3)=2sin(-2π/3+π/30=2sin(-π/3)=-2sin(π/3)=-√3是其最小值。
(2)若x∈[-π/3,π/4]求函数f(x)的最大值和最小值。
解:(1) f(x)=4sinxcos(x+π/3)+√3=2[sin(-π/3)+sin(2x+π/3)]+√3=2sin(2x+π/3)
故最小正周期T=2π/2=π;单增区间:由-π/2+2kπ≦2x+π/3≦π/2+2kπ,
得-5π/6+2kπ≦2x≦π/6+2kπ,故单增区间为-5π/12+kπ≦x≦π/12+kπ,k∈Z.
(2).在区间[-π/3,π/4]内,当x=π/12时,f(π/12)=2sin(π/6+π/3)=2sin(π/2)=2是其最大值;
当x=-π/3时,f(-π/3)=2sin(-2π/3+π/30=2sin(-π/3)=-2sin(π/3)=-√3是其最小值。
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