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已知函数F(x)=23sinxCosx%2sin2x+1(x∈R).(1...

f'(x)=1/2cos2x*(2x)]+cosx =cos2x+cosx 显然这是偶函数 cos2x+cosx =2cos²x-1+cosx =2(cosx+1/4)²-9/8 -1

y=sin2x+sinx+cosx+2 y=2sinxcosx+1+sinx+cosx+1 y=[2sinxcosx+sin^2(x)+cos^2(x)]+sinx+cosx+1 y=(sinx+cosx)^2+(sinx+cosx)+1 y=(sinx+cosx+1/2)^2+3/4 又因为:负根号2

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(...

(1)f(x)=32sin2x-12cos2x=sin(2x-π6),∵ω=2,∴f(x)的最小正周期T=2π2=π,令2x-π6=kπ+π2,得到x=kπ2+π3(k∈Z),则图象的对称轴方程为x=kπ2+π3(k∈Z);(2)要得到函数g(x)=sinx的图象,只需将函数f(x)的图象首先向左平移π12,得到y...

函数f(x)=2sinxcosx+23sin2x?3=sin2x+-3cos2x=2sin(2x-π3),y=f(x)的图象向左平移π6个单位,再向上平移1个单位,得到函数y=g(x)=2sin[2(x+π6)-π3]+1=2sin2x+1的图象,由题意可得,g(x)在[a,b]上至少含有1012个零点.令g(x)=0,得...

1、 f(x)=sinx(sinxcosπ/6+cosxsinπ/6) =√3/2*sin²x+1/2*sinxcosx =√3/2*(1-cos2x)/2+1/2*(sin2x)/2 =√3/4+1/2*(1/2*sin2x-√3/2*cos2x) =√3/4+(sin2xcosπ/3-cos2xsinπ/3) =√3/4+sin(2x-π/3) 所以T=2π/2=π 2、 0≤x≤π/2 -π/3≤2x-π/3≤2π/3 -√3...

f(x)=根号3sin2x+cos2x=2sin(2x+π/6) 1) T=2π/2=π x∈[0,π/2] 2x+π/6[π/6,7π/6] f(x)小=2sin(π/6)=1 f(x)大=2sin(π/2)=2 2) sin(2x0+π/6)=3/5 co2x0=cos(2x0+π/6-π/6)=(3-4根号3)/10 字限制

f(x)=sin^2x+√3sinxcosx+2cos^2x =cos^2x+√3/2*2sinxcosx+1 =1/2cos2x+√3/2sin2x+3/2 =sinx(2x+π/6)+3/2 T=2π/2=π 2x+π/6在[2kπ-π/2,2kπ+π/2]上单调递增 x在[kπ-5π/12,kπ+π/6]上单调递增 2)y=sin2x的图像经X轴向左平移π/12个单位得到y=sinx(2x+...

已知函数f(x)=-(√2)sin(2x+π/4)+6sinxcosx-2cos²x+1,x∈R; (1).求f(x)的最小正周期;(2).求f(x)在区间[0,π/2]上的最大值和最小值. 解:(1)。f(x)=-(sin2x+cos2x)+3sin2x+cos2x=2sin2x,故最小正周期T=π; (2)。在区间[0,π/2]上的最大...

F(x)=|cos2x+sin2x+Ax+B|=|2sin(2x+π4)+Ax+B|,(1)若F(x)是周期函数,F(x+π)=F(x),即|2sin(2x+π4)+Ax+B|=|2sin(2π+2x+π4)+Ax+Aπ+B|,可得A=0,B为任意实数;(2)∵0≤x≤3π2,∴π4≤2x+π4≤13π4,∴-1≤2sin(2x+π4)≤2,当A=0,B=-2...

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