f(x)=2sin(2x-π 3)的单调区间和对称轴

f(x)=[2sin(x+π/3)+sinx]cosx-根3sin^2x=[sinx+(√3）cosx+sinx]cosx-(√3）（sinx)^2=2sinxcosx+(√3）[(cosx)^2-(

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/

(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

2sin(x-π/4)sin(x+π/4)=cos(x-π/4-x-π/4)-cos(x-π/4+x+π/4)=-cos2xf(x)=cos(2x-π/3)-cos2x=cos(2x-π/6-π/6)

∵f（x）=[-sinx(-sinx)cos(π+x)]／[-2cosxsin(π-x)]=[sin²x(-cosx)]／(-2cosxsinx)=1/2sinx∴最小正周期T=2π∴函数图

f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1=sin2xcosπ/3+cos2xsinπ/3+sin2xcosπ/3-cos2xsinπ/3+cos2x=2sin2xc

f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx=2cosx(1/2sinx+√3/2cosx)-^3sin^2x+sinx*cosx=sin2x+√3cos2x=2si

f(x)=sin(π-x)cos(3π/2+x)+sin(π+x)sin(3π/2-x)=(sinx)(sinx)+(-sinx)(-cosx)=sinx(sinx+cosx)f'(x)=cosx(s

f(x)=cos(3x)*cos(2x)+sin(3x)*sin(2x)=cos(3x-2x)=cosxf'(x)=-sinx

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

cos2x+cos(x+π/3)+sin(x+π/6)+3sin^2x=cos2x+1/2*cosx-根号3/2*sina+根号3/2*sinx+1/2*cosx+3sin^2x=cos2x+cosx

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

f(x)=sin2(x+π)+根号3sin(x+π)sin(π-x)-1\2=sin2x－根号3sin²x-1/2=sin2x＋根号3/2cos2x－1=根号7/2sin（2x＋γ）－1co

经济将积极

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

hello!^-^令u=2x-π/6,则f（x）=sin(2x-π/6)=sinu=fu).因为-π/12≤x≤π/2,所以-π/3≤2x-π/6≤5π/6,即-π/3≤u≤5π/6.所以根据函数图像

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(x)=cos(2x-π/3)-(cos^2x-sin^2x)=cos(2x-π/3)-cos2x=2sin(2x-π/6)sinπ/6=sin(2x-π/6)因为y=sinx的单减区间为[π/2+

f(x)=cos(2x-π\3)+sin²x-cos²x=1/2cos2x+√3/2sin2x-cos2x=√3/2sin2x-1/2cos2x=-cos(2x+π\3)-1

（1）化解函数：√3∵f(x)=sin²wx+√3sinwxcoswx+2cos²wx=√3/2sin2wx+sin²wx+cos²wx+cos²wx