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求函数y=2sin3x%1的最小正周期,最值,并求出y取得最...

f(x) = sin(π/2-x)sinx - √3cos²x = cosxsinx - √3cos²x = 1/2sin2x - √3/2cos2x - √3/2 = sin2xcosπ/3-cos2xsinπ/3 - √3/2 = sin(2x-π/3) - √3/2 最小正周期:2π/2 = π 最大值:1 - √3/2 = (2-√3)/2

fx=1/2sin2x-√3cos²x =1/2sin2x+√3/2cos2x-√3/2 =sin2xcosπ/3+cos2xsinπ/3-√3/2 =sin(2x+π/3)-√3/2 最小正周期=2π/2=π 最小值=-1-√3/2

sinx, 最小正周期为2π; 同理,sin3x,最小正周期为:3x=2π,x=2π/3 望采纳

(1)f(x)=3sin2x+cos2x=2(sin2xcosπ6+cos2xsinπ6)=2sin(2x+π6)∴T=2π2=π,当2x+π6=2kπ+π2,k∈Z,即x=π6+kπ,k∈Z时,函数取得最大值2.当2x+π6=2kπ-π2,即x=kπ-π3,k∈Z时,函数取得最小值-2.(2)当2kπ-π2≤2x+π6≤2kπ+π6,k∈Z时,即kπ-π3≤x...

解: (1) f(x)=cos²x-√3sinxcosx+½ =½[1+cos(2x)]-(√3/2)sin(2x)+½ =½cos(2x)-(√3/2)sin(2x)+1 =cos(2x+π/3)+1 最小正周期T=2π/2=π cos(2x+π/3)=1时,f(x)取得最大值f(x)max=1+1=2 cos(2x+π/3)=-1时,f(x)取得最小值f(...

解:据题意知  y=2sin(1/2x+60°)       所以 ymax=2   ymin=-2   T=2π/(1/2)=4π

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-√...

(Ⅰ)f(x)=3sinωxcosωx+1+cos2ωx2+1=32sin 2ωx+12cos 2ωx+32=sin(2ωx+π6)+32.∵ω>0,∴T=2πω=π,∴ω=1.故f(x)=sin(2x+π6)+32.令2kπ?π2≤2x+π6≤2kπ+π2,解得:kπ?π3≤x≤kπ+π6.f(x)的单调递增区间为[kπ?π3,kπ+π6](k∈Z)(Ⅱ)∵0≤x≤π2,...

答: (1) f(x)=(√3/2)sinwx-sin²(wx/2)+1/2 =(√3/2)sinwx+(1/2)coswx =sin(wx+π/6) f(x)最小正周期T=2π/w=π,w=2 所以:f(x)=sin(2x+π/6) f(x)单调递增区间满足:2kπ-π/2

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