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函数y=3Cos( k 4 x+ π 3 ...

y=3cos(2x+A)的图像关于点(4π/3,0)中心对称 那么当x=4π/3时,y=0 ∴2×4π/3+A=kπ+π/2,k∈Z ∴A=kπ+π/2-8π/3=kπ-13π/6,k∈Z 当k=2时,A=-π/6, |A|取得最小π/6

f(k)=cos(3πx/4+π/4)+cos(πx/3+π/6) 解: cos(3πk/4+π/4)的最小正周期: 2π/(3π/4)=8/3 cos(πx/3+π/6)的最小正周期: 2π/(π/3)=6 ∵8/3和6的广义最小公倍数是24 ∴f(k)=cos(3πx/4+π/4)+cos(πx/3+π/6)的最小正周期是24

解:y=3cos(2x-pai/4)的单调递增区间 这个是复合函数,符合函数求单调递增区间则使用换元法。 令t=2x-pai/4 y=3cost 先求定义域: t(x)的定义域为R. 然后y(t)的定义域t:R t:R y(t)的定义域是t(x)的值域: t(x)是一次函数。 当一次函数的值域为R...

解: (1) tanx有意义,x≠kπ+ π/2,(k∈Z) 函数定义域为{x|x≠kπ+ π/2,k∈Z} f(x)=4tanxsin(π/2 -x)cos(x- π/3) -√3 =4tanxcosxcos(x-π/3)-√3 =4sinx[cosxcos(π/3)+sinxsin(π/3)] -√3 =4sinx[(1/2)cosx+(√3/2)sinx] -√3 =2sinxcosx+2√3sin²x-√...

f(x)=4sinxcos(x-π/3)-√3 =2×2sinxcos(x-π/3)-√3 =2[sin(x+x-π/3)+sin(x-x+π/3)]-√3 =2sin(2x-π/3)+2sinπ/3-√3 =2sin(2x-π/3) 故函数的周期T=π,令2x-π/3=kπ,k属于Z, 故函数的零点为x=kπ/2+π/6,k属于Z 2x属于【π/24,3π/4】 则2x属于【π...

f(x)=cosx(sinxcosπ/3+cosxsinπ/3)-√3cos²x+√3/4 =(1/2)sinxcosx+(√3/2)cos²x-√3cos²x+√3/4 =(1/2)sinxcosx-(√3/2)cos²x+√3/4 =(1/4)sin2x-(√3/4)cos2x =(1/2)sin(2x-π/3) 所以T=π x∈[-π/4,π/4]时, 则2x-π/3∈[-5π/6,π/6] ...

π约等于3.14; 4π就等于12.56; k=13

∵y=cos(x+π4)的值域为[-1,1],所以y=2-3cos(x+π4)的最大值为5,此时cos(x+π4)=-1,∴x+π4=2kπ+π,k∈Z,∴x=2kπ+3π4,(k∈Z).故答案为:5; 2kπ+3π4,(k∈Z)

cos(x+π/3)=sin(x+π/3+π/2)=sin(x+5π/6) 首先,将y=sin(x/2+π/3)的图像进行伸缩变换,纵坐标不变,横坐标变为原来的1/2,得到y=sin(x+π/3); 然后进行平移变换,将图像向左平移π/2个单位,得y=sin(x+π/2+π/3)=sin(x+5π/6)2401

f(x)=2sin²(π/4+x)-根号3cos2x =1-cos(π/2+2x)-√3cos2x =sin2x-√3cos2x+1 =2sin(2x-π/3)+1 ∵ x∈[π/4,π/2] ∴ 2x-π/3∈[π/6,2π/3] ∴ sin(2x-π/3)∈[1/2,1] ∴ 2x-π/3=π/6时,f(x)有最小值2 2x-π/3=π/2时,f(x)有最大值3

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