设函数f(x)=√2 2cos
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f(x)=sin2x+2√3cos²x=sin2x+√3(2cos²x-1)+√3=sin2x+√3cos2x+√3=2(sin2xcos60º+cos2xsin60
∵f(x)=cos(2x-π/3)+(sinx)^2-(cosx)^2=cos(2x-π/3)-cos2x=2sin(2x-π/6)sin(π/6)=sin(2x-π/6).∴g(x)=[sin(2x
(1)f(x)=cos(x+2π/3)+2cos²(x/2)=-(cosx)/2-(√3sinx)/2+1+cosx=1-[(√3sinx)/2-(cosx)/2]=1-[sin(x-π/6
1)f(x)=1+cos(2x+π/3)-(1+cos2x)/2=1/2-sin2x根号3/2最小值1/2-根号3/2最小正周期π2)c带入得sinC=根号3/2C=π/3A=π-B-C=2π/3-a
f(x)=sin²x+sin2x+3cos²x=1+2cos²x+sin2x=sin2x+cos2x周期=π再问:要过程,还有第三题的图像再答:(1)f(x)=sin
求导得:f′(x)=-4sinxcosx+23cos2x=-2sin2x+23cos2x=4sin(π3-2x),令f′(x)=0,得到x=π6,∵f(0)=2+a,f(π2)=a,f(π6)=3+a
f(f(f(x)))=f(f(arcsin(cos(x))))=f(arcsin(cos(arcsin(cos(x)))))=arcsin(cos(arcsin(cos(arcsin(cos(x)))
f(x)=cos(√3x+∮),则f’(x)=-√3sin(√3x+∮),f(x)+f’(x)=cos(√3x+∮)-√3sin(√3x+∮)=2[1/2cos(√3x+∮)-√3/2sin(√3x+
f(x)+f'(x)=cos(√3x+θ)-√3sin(√3x+θ)f(0)+f'(0)=cos(θ)-√3sin(θ)=0tan(θ)=√3θ=pi/3
(1)解析:∵函数f(x)=cos(wx+f)(w>0,-π/2<f<0)的最小正周期为π∴w=2π/π=2,f(x)=cos(2x+f)∵f(π/4)=√3/2f(π/4)=cos
f(x)=sinxcosx-√3cos(π+x)cosx(x∈R)可化为f(x)=(sin2x)/2+√3((cos2x)/2+1))=(sin2x)/2+(√3cos2x)/2+√3=cosπ/3s
f(x)=sinxcosx-√3COS(π+x)cosx=sinxcosx+√3cos²x=1/2sin2x+√3/2cos2x+√3/2=sin(2x+π/3)+√3/2按向量b=﹙π/4
f(x)=cos(√3x+A)f'(x)=-√3sin(√3x+A)f'(x)+f(x)=cos(√3x+A)-√3sin(√3x+A)-(f'(-x)+f(-x))=-cos(-√3x+A)+√3s
f'(x)=sinθx²+√3cosθxf'(1)=sinθ+√3cosθ=(1/2)×sin(θ+π/6)θ∈[0,5π/12],则θ+π/6∈[π/6,7π/12].∴θ+π/6=π/2
1.f(x)=1+cos2x-√3sin2x-1=2(1/2cos2x-√3/2sin2x)=2sin(π/6-2x).当sin(π/6-2x)=-1,即π/6-2x=2kπ-π/2,x=kπ-5π/
f(x)=cos(√3X+m)f'(x)=-sin(√3X+m)*(√3)f(X)+f'(X)=cos(√3X+m)-√3sin(√3X+m)f(X)+f'(X)为奇函数所以f(0)+f'(0)=0所
证明:f(x)=sinx-cosx+x+a求导:f'(x)=cosx+sinx+1=√2sin(x+π/4)+10
f'(x)=-sin(√3x+a)*(√3x+a)'=-√3sin(√3x+a)g(x)=f(x)+f'(x)=cos(√3x+a)-√3sin(√3x+a)=-[√3sin(√3x+a)-cos(√
(1)f(x)=sinwxcoswx+coswxcoswx=1/2sin2wx+1/2cos2wx+1/2=√(根号)2/2sin(2wx+π/4)+1/2因为f(x)的周期为π,所以w=1f(x)=