f(x)=2cos²x 2根号三sinxcosx-1

(1)f(x)=sinxcosx+√3cos²X-√3/2=sin2x/2+√3cos2x/2+√3/2-√3/2=sin(2x+π/3).(2)f(x)的最小正周期为π,值域是[-1,1]

函数f(x)=sin(2x+φ)+根号三cos(2x+θ)=2sin（2x+θ+π/3）的定义域为：R则f(0)=0所以：0=sin（θ+π/3）,=>θ+π/3=2kπ,k∈Z即θ=2kπ-π/3所

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,

1、f(x)=1/2*sin(2x/3)+√3*(1+cos2x/3)/2=1/2*sin(2x/3)+√3/2*cos(2x/3)+√3/2=sin(2x/3)cosπ/3+cos(2x/3)sin

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

(1)f(x)=2sin(x+θ/2).cos(x+θ/2)+2倍根号三cos²(x+θ/2)-根号三=sin(2x+θ)+√3[cos(2x+θ)+1]-√3=sin(2x+θ)+√3[c

由倍角公式：f(x)=sin2x-√3cos2x由辅助角公式：f(x)=2sin(2x-π/3)所以,最小正周期T=2π/2=π最大值为2

f(x)=6cos²x＋√3sin2x=3cos2x+3+√3sin2x=2√3(√3/2sin2x+1/2cos2x)+3=2√3sin(2x+π/6)+3f(a)=2√3sin(2a+π

f(x)=√3/2sin(x/2)+1/2cos(x/2)=sin(x/2+π/6)sinx的增区间是(2kπ-π/2,2kπ+π/2)所以这里有2kπ-π/2

f(x)=根号3cos^x+sinxcosx-根号3/2=根号3*(1+cos2x)/2+sin2x/2-根号3/2所以f(派/8)=根号3*(1+cos派/4)/2+sin(派/4)/2-根号3/2

f(x)=(12)sin2x＋√3cos²x=(1/2)sin2x＋(√3/2)[1＋cos2x]=(1/2)sin2x＋(√3/2)cos2x＋(√3/2)=sin(2x＋π/3)＋(√3

原式=sin2x/2+cosx^2根号3=1/2sin2x+根号3/2+根号3/2*cos2x=sin（2x+π/3)+根号3/2因为T=2π/ww=2所以T=π所以最小正周期为π

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

f(x)=1/2sin2wx+sqrt(3)/2(2cos^2wx-1)=1/2sin2wx+sqrt(3)/2cos2wx=sin(2wx+pi/3)周期为pi/2（因为对称轴距离Pi/4）w=2g

f(x)=cos（X+π/6）-cos（X-π/6）+根号三*cosx=cosxcosπ/6-sinxsinπ/6-(cosxcosπ/6+sinxsinπ/6)+√3cosx=-sinx+√3cos

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

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/

f（x）=√3sin2x+2cos²x+1=√3sin2x+1+cos2x+1=√3sin2x+cos2x+2tana=√3/2然后用万能公式sin2a=2tana/(1+tana^2)=2

f(x)=(sin²x+cos²x)+2cos²x+2√3sinxcosx-2=1+1+cos2x+√3sin2x-2=√3sin2x+cos2x=2sin(2x+π/6