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已知函数F(x)=2根号3sinXCosX+2Cos^2X%1(X属于R)

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)=2根号3sinxcosx+2cos^2x-1 =√3 sin2x+cos2x =2sin(2x+π/6) f(A)=2sin(2A+π/6)=2/3 sin(2A+π/6)=1/3 0

解: (1) f(x)=2√3sinxcosx+2cos²x-1 =√3sin(2x)+cos(2x) =2[(√3/2)sin(2x)+(1/2)cos(2x)] =2sin(2x+π/6) 最小正周期T=2π/2=π (2) f(x0)=6/5 2sin(2x0+π/6)=6/5 sin(2x0+π/6)=3/5 x0∈[π/4,π/2] 2π/3≤2x0+π/6≤7π/6 cos(2x0+π/6)

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(...

fx=1+cos2x/2 +√3/2sin2x +1 =(√3/2sin2x+1/2cos2x)+3/2 =sin(2x+π/6)+3/2 T=2π/w=π fx max=1+3/2=2.5 fx min=-1+3/2=0.5 2kπ-π/2≤2x+π/6≤2kπ+π/2,k∈Z kπ-π/3≤x≤kπ+π/6,k∈Z 函数的单调递增区间[kπ-π/3,kπ+π/6],k∈Z

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+...

f(x)=√3sinxcosx-cos²x-1/2 =(√3/2)sin2x-(1+cos2x)/2-1/2 =(√3/2)sin2x-(1/2)cos2x-1 =sin(2x-π/6)-1. 早小正周期T=2π/2=π. sin(2x-π/6)=-1, 即x=kπ-π/6时, 最小值f(x)|min=-2。

解: f(x)=2√3sinxcosx+1 -2sin²x =√3sin(2x)+cos(2x) =2[(√3/2)sin(2x)+(1/2)cos(2x)] =2sin(2x+π/6) -1≤sin(2x+π/6)≤1 -2≤2sin(2x+π/6)≤2 -2≤f(x)≤2 函数的值域为[-2,2]

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