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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)=√3sin2x+cos2x=2sin(2x+π/6) ∴f(x0)=2sin(2x0+π/6)=6/5 ∴sin(2x0+π/6)=3/5 ∵x0∈[π/4,π/2] ∴2x0+π/6∈[2π/3,7π/6] ∴cos(2x0+π/6)=﹣4/5 ∴cos2x0=cos[(2x0+π/6)-π/6]=cos(2x0+π/6)cos(π/6)-sin(2x0+π/6)sin(π/6) =(﹣4/...

f(x)=2√3sinxcosx+2cos^2x-1=√3sin2x+cos2x=2sin(2x+π/6) 由π/2+2kπ≤2x+π/6≤3π/2+2kπ得:π/6+kπ≤x≤2π/3+kπ,因此单调递减区间为 [π/6+kπ,2π/3+kπ] f(x)=2sin(2x+π/6)的最大值为2,此时2x+π/6=π/2+2kπ,即x=π/6+kπ,所以自变量x的集合为 {x|x=π/6...

f(x)=2倍根号3sinxcosx+2cos^2x-1 =√3sin2x+cos2x =2sin(2x+π/6) 2sin(2x0+π/6)=6/5 sin(2x0+π/6)=3/5 cos(2x0+π/6)=-4/5 cos2xo=cos[(2x0+π/6)-π/6] =cos(2x0+π/6)cosπ/6+sin(2x0+π/6)sinπ/6 =-4/5*√3/2+3/5*1/2 =(3-4√3)/10

见图片吧

(1) f(x)=2√3sinxcosx+2cos²x-1=√3sin2x+cos2x=2(√3/2sin2x+1/2cos2x)=2sin(2x+π/6) 函数的最小正周期T=π (2) f(x)=2sin(2x+π/6)=6/5 sin(2x+π/6)=3/5 x∈[π/4,π/2], cos(2x+π/6)=-4/5 cos2x=cos(2x+π/6-π/6)=cos(2x+π/6)cosπ/6...

f(x)=2根号3sinxcosx+2cos^2x-1=根号3*sin2x+cos2x=2(cos派/6*sin2x+sin派/6*cos2x)=2sin(2x+派/6) g(x)=|f(x)|=|2sin(2x+派/6)| 由g(x)的图像可以明显看出,单调减区间是[(k+1/6)派 , (k+5/12)派],k是整数 f(A)=2sin(2A+派/6)=2/3 因为0

f(x)=√3sin2x+2cos2x-1 =√7sin(2x+φ) -1 (其中cosφ=√3/√7;sinφ=4/√7) T=2π/2=π 二. 0≤x≤π/2 φ≤x+φ≤π/2+φ 函数y=sint 在【φ,π/2+φ】上先增后减,所以函数sin(2x+φ)的最大值为 1 f(x)(MAX)=√7-1 最小值是两个端点中的一个或两个, sint的左端点si...

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: (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(...

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