已知函数f(x)=2√3sinxcosx-3sin²x-cos²x 3

1、由于函数g(x)=sin(2(x－a)+π/3)为偶函数,所以g(x)的图像关于y轴对称,即函数g(x)当x=0时取得最值,所以g(0)=±1,解得sin(π/3－2a)=±1,sin(2a－π/

f(x)＝2sin²[(π/4)－x]－√3cos2x＝1－cos[(π/2)－2x]－√3cos2x＝1－sin2x－√3cos2x＝－2sin(2x＋π/3)＋1∵－1≤sin(2x＋π

1.f（x）=2cosx*sin(x+π/3)-√3sin^2x+sinx*cosx=2cosx*sin(x+π/3)-2sinx*[(√3/2)sinx-(1/2)cosx]=2cosx*sin(x

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

5.（1）f(x)=(√3)cos2x+2sinxsin(x+π/2).=√3cos2x+2sinxcosx=sin2x+√3cis2x=2sin(2x+π/3).∴最小正周期T=2π/2=π,f(x

f(x)=-√3sin²x+sinxcosx=(1/2)*sin2x+(√3/2)*cos2x-√3/2=sin(2x+π/3)-√3/2所以函数f(x)的最小正周期是T=2π/2=π最大值

f(x)=3/2-3sin^2x-√3sinxcosx=3/2-3*(1-cos2x)/2-(√3/2)*sin2x=(3/2)*cos2x-(√3/2)*sin2x=√3*[(√3/2)*cos2x

f(x)=sin²(x)+(√3)sin(x)cos(x)+2cos²(x)=3/2+√3/2sin2x+1/2cos2x=3/2+sin(2x+π/6)函数f(x)的最小正周期T

答：f(x)=(cosx)^2-√3sinxcosx+2(sinx)^2-1/2f(x)=(1/2)*cos(2x)-(√3/2)sin2x+1-cos(2x)f(x)=-(√3/2)sin2x-(1

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

f(x)=-√3sin2x+1/2sin2x+√3/2=(-√3+1/2)sin2x+√3/2因为sin2x的最小正周期为2π/2=π因为-√3+1/2小于0所以单调递增区间为（π/4+kπ,3π/4

f(x)=sin(x/2)-√3[1-cos(x/2)]+√3=2[(1/2)sin(x/2)+(√3/2)cos(x/2)]=2sin(x/2+π/3)(1)g(x)=f(x+π/3)=2sin[(

fx=2cosx(0.5sinx+根号3/2cosx)-根号3sin*2x+sinxcosx=2sinxcosx+根号3（cos*2x-sin*2x）=sin2x+根号3cos2x=2sin（2x+派

(1)F(X)=√3sin2x+2sin²x=√3sin2x+1-cos2x=2(√3/2sin2x-1/2cos2x)+1=2sin(2x-π/6)+1F(x)的最小正周期T=2π/2=π

sin(2x+π/3)-√3sin²x+sinxcosx+√3/2=sin(2x+π/3)-2sinx【（√3/2）sinx-（1/2）cosx】+（√3/2）=sin(2x+π/3)-2s

f(x)=(√3sinωx+cosωx)*sin(-3π/2+ωx)=(√3sinωx+cosωx)*sin(π/2+ωx)=(√3sinωx+cosωx)*cosωx=(1/2)*(√3*2sinω

已知函数f(x)=sin(x/2)+(√3)cos(x/2),x∈R；(1)求f(x)的最小正周期,并求函数f(x)在x∈[-2π,2π]上的单调增区间；（2）函数f(x)=sinx(x∈R)的图像经

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

f(x)=[2(sinx*1/2+cosx*√3/2)+sinx]cosx-√3sin²x=(2sinx+√3cosx)cosx-√3sin²x=2sinxcosx+√3(cos&

f(x)=(cosx)^4-(sinX)^4+2√3sinxcosx=[(cosx)^2+(sinx)^2][(cosx)^2-(sinx)^2]+√3sin2x=(cosx)^2-(sinx)^2+