已知函数f(x)=根号2cos(x-π 12),x属于R,求f(-π 6)的值

1,f(x)=sin²x+√3sinxcosx+2cos²x＝1－cos²x＋√3/2sin2x+2cos²x＝cos²x＋√3/2sin2x+1=(

已知:函数f(x)=2sinxcosx+2√3cos²x-√3求:(1)单调增区间和最小正周期；(2)当x∈[-π/4,π/4]时求最值.f(x)=2sinxcosx+2√3cos²

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

f(x)=cos^2x-sin^2x+2(根号3)sinxcosx+1=cos2x+(根号3)sin2x+1=2{(1/2)cos2x+[(根号3)/2]sin2x}+1=2sin(2x+派/6)+1

f（x）=1+cos2x+根号3sin2x+a=2sin（2x+π/6）+a+11、若f（x）max=2,则sin（2x+π/6）=1,即2+a+1=2,得a=-12、正弦的单调减区间在第二和第三象限

(1)f(x)=2√3sinxcosx+2cos^2-t.=√3sin2x+1+cos2x-t.=2[(√3/2sin2x+(1/2)cos2x]+1-t.=2sin(2x+π/6)+1-t.令f(x

答：f(x)=(1/2)*(cosx)^2+(√3/2)sinxcosx+1=(1/2)*(cos2x+1)/2+(√3/4)sin2x+1=(1/2)[sin2xcosπ/6+cos2xsinπ/6

f(x)=√3(sin^2x-cos^2x)-2sinxcosx=-√3cos2x-sin2x=-2sin(2x+π/3)1.求最小正周期T=π2.设x∈[-π/3,π/3],求函数的值域和单调区间-

(1)f(x)=2√3sinxcosx+2cos^2-t.=√3sin2x+1+cos2x-t.=2[(√3/2sin2x+(1/2)cos2x]+1-t.=2sin(2x+π/6)+1-t.令f(x

f(x)=sin²x+2√3sinxcosx-cos²x=2√3sinxcosx-(cos²x-sin²x)=√3sin(2x)-cos(2x)=2sin(2x

y=sinx^2+根3sinxcosx+2cosx^2=-1/2(1-2sinx^2)+1/2根3*2sinxcosx+2cosx^2-1+3/2=-1/2cos2x+二分之根3倍sin2x+cos2

f(x)=sin2x-2√3(cosx)^2+√3=sin2x-√3(1+cos2x)+√3=sin2x-√3cos2x=2sin(2x-π/3)π/4=再问：π/6=

f(x)=sinxcosx+√3(cosx)^2-√3/2=(1/2)sin2x+(√3/2)cos2x=sin2xcosπ/3+cos2xsinπ/3=sin(2x+π/3)1.0

解题过程如下：

f(x)=cos2x+(√3)sin2x=2cos(2x-π/3)故Tmin=2π/2=π单增区间：由-π+2kπ≤2x-π/3≤2kπ,-2π/3+2kπ≤2x≤2kπ+π/3,得-π/3+kπ≤x

（1）f(x)=√3sin(2x+φ)-cos(2x+φ）=2[√3/2*sin(2x+φ)-1/2*cos(2x+φ）]=2sin(2x+φ-π/6)因为是偶函数∴函数f(x)在x=0处取最大值或最

f（x）=√3cos²x+sinxcosx-√3/2=√3(cos2x+1)/2+sin2x/2-√3/2=√3/2cos2x+√3/2+1/2sin2x-√3/2=1/2sin2x+√3/

(1)T=2π/2=π增区间：2kπ-π≤2x-π/4≤2kπ2kπ-3π/4≤2x≤2kπ+π/4kπ-3π/8≤x≤kπ+π/8所以增区间为[kπ-3π/8,kπ+π/8]k∈Z(2)x∈[-π/

2cos^2-1=cos2xcos^2=(1+cos2x)/2f(x)=sinxcosx-(根号3)cos^2+(根号3)/2=sin2x/2-根号3*(cos2x+1)/2=sin2x/2-根号3*

f(x)=sin(2x+α)+根号3cos(2x+α)=2sin(2x+α+π/3)∵f(x)图像过（π/12,1)∴f(π/12)=2sin(π/6+π/3+α)=2sin(π/2+α)=2cosα