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已知函数F(x)=3Cos2x+sinxCosx2,x∈R.(I)求...

(1)f(x)=3cos2x+2sinxcosx=3cos2x+sin2x=2sin(2x+π3),T=2π2=π,(2)∵x∈[0,π4],∴2x+π3∈[π3,5π6],∴12≤sin(2x+π3)≤1∴1≤f(x)≤2,即函数的值域为[1,2]

解答:解:(1)函数f(x)=3?cos2x2-4t?sinx2cosx2+2t2-6t=sin2x-2tsinx+2t2-6t+1 =(sinx-t)2+t2-6t+1.当t<-1时,g(t)=(-1-t)2+t2-6t+1=2t2-4t+2.当-1≤t≤1时,g(t)=t2-6t+1.当t>1时,g(t)=(1-t)2+t2-6t+1=2t2-8t+2.综上可得...

f(x)=√3sinxcosx+cos²x =(√3/2)·2sinxcosx+½(2cos²x-1)+½ =(√3/2)sin2x+½cos2x+½ =sin(2x+π/6) +½ 最小正周期T=2π/2=π

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)解:f(x)= sin x cos x + √3 cos ² x + √3 / 2 = (1 / 2)2 sin x cos x + √3 / 2 (2 cos ² x - 1) = (1 / 2)sin 2x + (√3 / 2)cos 2x = sin 2x cos π / 3 + sin π / 3 cos 2x = sin(2 x + π / 3) ∴ 最小正周期为:T ...

(Ⅰ)f(x)=2sinxcosx+2cos2x=sin2x+cos2x+1=2sin(2x+π4)+1,由2kπ-π2≤2x+π4≤2kπ+π2(k∈Z)得:kπ-3π8≤x≤kπ+π8,k∈Z,∴f(x)的单调递增区间是[kπ-3π8,kπ+π8](k∈Z);(Ⅱ)由题意得:g(x)=2sin(2x-π4)+1,由A(0,-1),得2sin(2α-π4...

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+...

题干多题,不能正常作答。

f(x)=12sin2x3+32(1+cos2x3)=12sin2x3+32cos2x3+32=sin(2x3+π3)+32,(Ⅰ)由sin(2x3+π3)=0即2x3+π3=kπ(k∈z)得x=3k?12π,k∈z,即对称中心的横坐标为3k?12π,k∈z;(Ⅱ)由已知b2=ac,cosx=a2+c2?b22ac=a2+c2?ac2ac≥2ac?ac2ac=12,∴

f(x)=√3sinxcosx-1/2cos2x =√3/2sin2x-1/2cos2x =sin(2x-π/6). (1)最小值:f(x)|min=-1, 此时2x-π/6=2kπ-π/2, 即x=kπ-π/6 (k为整数); 最小正周期:T=2π/2=π. (2)f(C)=1,则 sin(2C-π/6)=1,即C=π/3. R=c/(2sinC)=√3/(2·√3/2)=1 (正...

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