f(x)=sin(2x π 3)-2cos²x
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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)=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
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
f(x)=[2(sinx*1/2+cosx*√3/2)+sinx]cosx-√3sin²x=(2sinx+√3cosx)cosx-√3sin²x=2sinxcosx+√3(cos&