已知f(x)=根号3sin^4x (sinx cosx)²-根号3cos^4x

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

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)=5cos²x+sin²x-4√3sinxcosx=4cos²x+cos²x+sin²x-2√3sin2x=2(cos2x+1)+1-2√3s

函数f(x)=2sin^2(x+π/4)-√3cos2x-1=-cos(2x+π/2)-√3cos2x=sin2x-√3cos2x=2sin(2x-π/3)1.当x属于R时,函数f(x)的最小正周期T

是四份之π还是π分之四?再问：四分之π再答：你的题给的有点特殊，所以只能按照我理解的给答案，如果有错可以追问f(x)=ab={√2sin(π/4+x)+1}x{√2sin(π/4+x)-1}-√3co

由题意可得:f(x)=sin(2x+π/3)-√3sin^2x+sinxcosx+√3/2=sin(2x+π/3)-√3(1/2-1/2cos2x)+1/2sin2x+√3/2=2sin(2x+π/3

函数f(x)=[2sin^2(x+π/4)-1]-√3cos2x+1=[-cos(2x+π/2)]-√3cos2x+1=sin2x-√3cos2x+1=2(1/2*sin2x-√3/2*cos2x)+

f(x)=2根号3sin(x/2+派/4)cos(x/2+派/4)-sin(x+派).=√3sin(x+π/2)+sinx=sinx+√3cosx=2(1/2sinx+√3/2cosx)=2sin(x

f(x)=cos^4x-2根号3sinxcosx-sin^4x=f(x)=(cos^2x-sin^2x)(cos^2x+sin^2x)-2根号3sinxcosx=cos2x-根号3sin2x=-2si

f(x)=2sin(x/4)cos(x/4)-2√3sin^2(x/4)+√3=sin（x/2）+√3cos（x/2）=2sin（x/2+π/3）最小正周期T=2π/（1/2）=4π最大值是2最小值是

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

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

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

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

(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)=2sin²(π/4+x)-√3cos2x=1-cos(π/2+2x)-√3cos2x=1+sin2x-√3cos2x=1+2sin(2x-π/3)π/4≤x≤π/2π/6≤2x-π