已知函数fx 2根号3sin(π-x)sinx-(sinx-cosx)的平方

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＜

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

因为：函数f(x)=2cosx·sin（x+π/3)-√3(sinx)^2+sinx·cosx=2cosx·sin(x+∏/3)-√3(sinx)^2+sinx·cosx=2cosx·sinx·cos

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)+

(1)因为sin²x=（1-cos2x）/2,sin(2x-π/3)=1/2sin2x-根号3/2cos2x所以函数f(x)=2倍根号3sin²x-sin(2x-π/3)=-sin

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

y=2cosxsin(x+π/3)-根号3*(sin^2)x+sinxcosx,后两项先提出一个sinx,然后括号内部分用叠加原理,得到y=2cosxsin(x+π/3)+2sinxcos(x+π/3

经济将积极

这个简单：f(x)=2cosx(sinxcos(pi/3)+cosxsin(pi/3))-根号33sin^2x+sinx*cosx=2sinxcosx+根号3cos2x=2sin(x+pi/3)所以：

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)=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

已知函数f(x)=根号3sin(wx+φ)++(w>0,-π/2x=(π-2φ)/4=π/3==>φ=-π/6∴f(x)=√3sin(2x-π/6)(2)解析：设f(a/2)=√3/4,(π/6＜a＜

y=sin(2x-π/3)+根号3cos2x=sin2xcosπ/3-cos2xsinπ/3+2sinπ/3cos2x=sin2xcosπ/3+cos2xsinπ/3=sin(2x+π/3)1.振幅=

f(x)=sin(2x+π/3)(积化和差公式）所以最小正周期T=2π/2=π

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)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

（1）真数大于零,2kπ