使函数f(x)=3sin(2x 5Ø)
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1,f(x)=sin²x+√3sinxcosx+2cos²x=1-cos²x+√3/2sin2x+2cos²x=cos²x+√3/2sin2x+1=(
f(x)=[2sin(x+π/3)+sinx]cosx-根3sin^2x=[sinx+(√3)cosx+sinx]cosx-(√3)(sinx)^2=2sinxcosx+(√3)[(cosx)^2-(
∵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
对f(x)作图,你可以很容易的发现,在0到π范围内,是sin(3x)的叠加,在π到2π范围内,sin(3x)为负,|sin(3x)|是正,它们抵消掉了,所以这个范围内f(x)是等于0的.后面的以此类推
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
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)=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
y=sinx^2+根3sinxcosx+2cosx^2=-1/2(1-2sinx^2)+1/2根3*2sinxcosx+2cosx^2-1+3/2=-1/2cos2x+二分之根3倍sin2x+cos2
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
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)=cos(x-π/3)-sin(π/2-x)=(1/2)cosx+(√3/2)sinx-cosx=(√3/2)sinx-(1/2)cosx=sin(x-π/6),它的最小值=-1.
(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
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
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)=[2sin(x+π/3)+sinx]cosx-√3sin²x=[sinx+√3cosx+sinx]cosx-√3sin²x=2sinxcosx+√3cos²x-