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已知函数Fx)=根号3sinxCosx%Cosx^2%1\2求函数的最...

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(...

f(x)=√3sinxcosx-cos²x+1/2 =√3/2*sin2x-1/2(2cos²x-1) =√3/2*sin2x-1/2*cos2x =sin(2x-π/6) (1)、最小正周期:T=2π/2=π 设sin(2x-π/6)=±1 则2x-π/6=π/2+kπ,k∈Z ∴对称轴方程为:x=π/3+kπ/2,k∈Z (2)、∵在△ABC中 f(A/2)=sin(A-π/6)=1/...

解fx=sinx的平方+根号3sinxcosx=(1-cos2x)/2+√3/2sin2x=√3/2sin2x-cos2x/2+1/2=sin(2x-π/6)+1/2故函数的周期T=2π/2=π.

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]

1. f(x)=√3sinxcosx-cos²x+1/2=(√3/2)(2sinxcosx)-(1/2)(2cos²x-1) 二倍角公式:2sinxcosx=sin(2x),2cos²x-1=cos(2x),于是有: f(x)=(√3/2)sin(2x)-(1/2)cos(2x), cos30°=cos(π/6)=√3/2,sin30°=sin(π/6)=1/2,于是有: f(x)=sin(2...

f(x) = sin(π/2-x)sinx - √3cos²x = cosxsinx - √3cos²x = 1/2sin2x - √3/2cos2x - √3/2 = sin2xcosπ/3-cos2xsinπ/3 - √3/2 = sin(2x-π/3) - √3/2 最小正周期:2π/2 = π 最大值:1 - √3/2 = (2-√3)/2

1:∵函数f(x) =√3sinxcosx-(cosx)^2-1/2 =cosx(√3sinx-cosx)-1/2=2cosxsin(x-π/6)-1/2 =sin(2x-π/6)-sin(π/6)-1/2 =sin(2x-π/6)-1 ∴f(x)的最小值为-2,最小正周期为π 2。f(C)=sin(2c-π/6)-1=0 则sin(2c-π/6)=1 当2c-π/6=π/2时。f(c)=0求...

你好:f(x)=√3sinxcosx–cos²x+1/2 =√3sinxcosx–1/2(1-cos²x)+1/2 =√3/2sin2x+1/2cos²x-1/2+1/2 =sin(2x+π/6)T=2π/2=π f(x)=√3sinxcosx-cos²x+1/2=(√3/2)(2sinxcosx)-(1/2)(2cos²x-1) 二倍角公式:2sinxcosx=sin(2x),...

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