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已知函数F(x)=23sinx?Cosx+2Cos2x,x∈R.(1)求...

(1)由f(x)=23sinxcosx+2cos2x-1,得f(x)=3(2sinxcosx)+(2cos2x-1)=3sin2x+cos2x=2sin(2x+π6),所以函数f(x)的最小正周期为π.(2)因为f(x)=2sin(2x+π6)在区间[0,π6]上为增函数,在区间[π6,π2]上为减函数,又f(0)=1,f(π...

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

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(...

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+...

(1)函数f(x)=2cos2x+23sinxcosx+1=cos2x+3sin2x=2sin(2x+π6)∵x∈[0,π],∴两零点关于x=π6或者x=2π3对称,∵f(x)=a有两异根,∴两根之和为π3或4π3;(2)根据对称性可知,函数y=f(x),x∈[π6,7π6]的图象与直线y=4围成图形的面积等于一矩形...

f(cosx)=f[sin(x+π/2)]=cos2(x+π/2)+1=cos(2x+π)+1=-cos2x+1

f(x)=cos2x+2sinxcosx+1 =cos2x+sin2x+1 =√2(√2/2*cos2x+√2/2*sin2x)+1 =√2cos(2x-π/4)+1 ∵-1≤cos(2x-π/4)≤1 ∴1-√2≤√2cos(2x-π/4)+1≤1+√2 则f(x)值域为:[1-√2,1+√2] 行家解答,质量保证

(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...

f(x)=(1-cos2x)/2+√3/2sin2x+1+cos2x =√3/2sin2x+1/2cos2x+3/2 =sin(2x+π/6)+3/2 的最小正周期 T=2π/2=π 2kπ-π/2

f(x)=(1-cos2x)/2+√3/2sin2x+cos2x+1 =1/2cos2x+√3/2sin2x+3/2 =sin(2x+π/6)+3/2 1. 最小正周期T=2π/2=π 单调增区间 2kπ-π/2

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