已知函数f[x]=根号3sin2x sinxcosx

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

此题如果去化归的话,可能会比较复杂所以用分析法（1）在定义域内,sin平方x的周期显然是π,根号3sinxcosx=根号1.5*sin2x,所以周期也是π；综上,函数f(x)的最小周期为π（2）(si

1f(x)=√3sinπx+cosπx=2（（√3/2）sinπx+（1/2）cosπx）=2sin（πx+π/3）∴最小正周期T=2π/w=2π/π=2值域f（x）∈[-2,2]2-π/2+2kπ＜

f(x)=2cosxsin(x+60°)-√3sin²x+sinxcosx=2cosx(sinxcos60º+cosxsin60º)-√3sin²x+sinxc

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)=√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-

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)=2根号3sin方x+sin2x+根号3＝根号3（2sin方x＋1）＋sin2x＝根号3（1－cos2x＋1）＋sin2x＝2根号3－根号3cos2x＋sin2x＝2sin（2x－60度）＋2

f(x)=2√3sin²x-sin(2x-π/3)=√3-√3cos2x-1/2sin2x+√3/2cos2x=√3-(1/2sin2x+√3/2cos2x)=√3-sin(2x+π/3)T

经济将积极

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

f(x)=sin^2x+2√3sinxcosx+sin(x+π/4)sin(x-π/4)=(1-cos2x)/2+√3sin2x+(1/2)2sin(x-π/4)cos(x-π/4)=2-2cos2x

（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)=(1-cos2x)/2+(√3/2)sin2x=(√3/2)sin2x-(1/2)cos2x+1/2=sin(2x-π/6)+1/2最小正周期T=2π/2=π递增区间：-π/2+2kπ

f(25π/6)=f(π/6）3sin²x+sinxcosx=3sin²x+0.5sin2xf（x)=--根号下3sin²x+0.5sin2x然后根据函数的单调性就可求出

1：(sinwx)^2+√3sinwxsin(wx+π\2)=(sinwx)^2+√3sinwxcoswx=2[(sinwx)^2+(√3\2)sin2wx]\2=[2(sinwx)^2+√3sin2

(1)已知函数f(x)=-(√3)sin²x+sinxcosx,求f(25π/6).f(x)=-(√3)sin²x+sinxcosx=(√3/2)(cos2x-1)+(1/2)si

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

f(x)=√3sin2x+(1-cos2x)-1=√3sin2x-cos2x=2sin(2x-π/6)最小正周期T=2π/2=π单调增区间：2kπ-π/2=