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设函数F(x)=2根号3sinxCosx+2Cos^2x+m(1)求函数F...

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(...

解: f(x)=sin²x-cos²x-2√3sinxcosx =-(cos²x-sin²x)-√3(2sinxcosx) =-cos2x-√3sin2x =-2[(√3/2)sin2x+(1/2)cos2x] =-2sin(2x+π/6) (1) f(2x/3)=-2sin[2(2x/3)+ π/6]=-2sin(4x/3 +π/6) (2) 最小正周期T=2π/2=π 2kπ+ π/2≤2x+...

f(x)=2√3sinxcosx+2cos^2x =√3*sin2x+cos2x+1 =2(√3/2*sin2x+1/2*cos2x)+1 =2sin(2x+π/6)+1 f(x)的最小正周期T=2π/2=π f(x)min=-2+1=-1,f(x)max=2+1=3 f(x)值域为[-1,3]

f(x)=cos2x+根号3sin2x=2sin(2x+π/2) 所以周期为π 对称轴2x+π/2=π/2+kπ(k是整数) 即x=kπ/2 k是整数 单调区间 -π/2+2kπ

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...

题干多题,不能正常作答。

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+...

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