f(x)=2根号2cosx(sinx-cosx) 根号2
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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)=m.n=2根号3sinxcosx+cos^2x-sin^2x=根号3sin2x+cos2x=2sin(2x+π/6)
f(x)=向量a乘向量b=2sinx*√3cosx+(√2cosx+1)(√2cosx-1)=√3sin2x+2(cosx)²-1=√3sin2x+cos2x=2sin(2x+π/6)∴T=
f(x)=2cosx(cosx+根号3sinx)=2cos^2x+2*根号3sinx*cosx=cos2x+根号3sin2x+1当cos2x=1/2sin2x=根号3/2有最大值3当cos2x=-1/
定义域是:1.sinx>=0,即有2kPai
f(x)=2cosx*sin(x+π/3)-√3sinx^2+sinx*cosx=2cosx*(sinxcosπ/3+cosxsinπ/3))-√3sinx^2+sinx*cosx=sinxcosx+
ac=b^2=a^2+c^2-2accosx==>ac(1+2cosx)=a^2+c^2≥2ac==>cosx≥1/2==>0<x≤π/3f(x)=sinx/3cosx/3+√3(cosx/3)^2=
√2sinx+√2cosx+sinxcosx设sinx+cosx=t,√2sinx+√2cosx+sinxcosx=√2t+(t^2-1)/2=t^2/2+√2t-1/2=(t+√2)^2/2-3/2
(1)f(x)=.2sinx3(sinx−cosx)sinx+cosxcosx.=sin2x+3cos2x=2sin(2x+π3)…(3分)所以函数f(x)的最小正周期为π…(3分)(2)
设y=√2sinx+cosx,求导得:y‘=√2cosx-sinx,当y‘=√2cosx-sinx>0时,y=√2sinx+cosx为增函数,tanx0,cosx√2时,为减函数;当sinx=√6/3
这个简单:f(x)=2cosx(sinxcos(pi/3)+cosxsin(pi/3))-根号33sin^2x+sinx*cosx=2sinxcosx+根号3cos2x=2sin(x+pi/3)所以:
f(x)=√2(sinx-cosx)=2*【sinxcos(-π/4)+cosxsin(-π/4)】=2sin(x-π/4)1)、f(x)的最小正周期为T=2π值域为【-2,2】2)、图像过点(α,6
f(x)=2√3sinxcosx-2cos²x+1=√3sin2x-(2cos²x-1)=√3sin2x-cos2x=2sin(2x-π/6)
1、f(x)=(√3/2)sin2x+(1/2)cos2x=sin(2x+π/6)周期T=2π/2=π当x∈【0,π/2】时,2x+π/6∈【π/6,7π/6】则sin(2x+π/6)∈【-1/2,1
y=2(sinx*1/2+cosx*√3/2)=2(sinxcosπ/3+cosxsinπ/3)=2sin(x+π/3)-π/6
因为0《2cosx《2所以-1《1-2cosx《1又因为1-2cosx》0所以0《1-2cosx《1所以0《cosX《2分之1所以60《X《90
f'(x)=2(√x)'sinx+2√x*(sinx)'+(cosx)'=sinx/√x+2√x*cosx-sinx
f(x)=1/2sin(2x/3)+√3(1+cos(2x/3))/2=1/2sin(2x/3)+√3/2cos(2x/3)+√3/2=sin(2x/3+π/3)+√3/2
由题意得sinx-cosx≥0√2sin(x-π/4)≥0得2Kπ+π/4≤x≤2Kπ+5/4π-x平方-2x+3>0x平方+2x-3<0(x-1)(x+3)<0-3<x<1所以取交集得[-3,-3π
(1)f(x)=2√3cosx/2sinx/2+2cos²x/2-1=√3sinx+cosx=2sin(x+π/6),最小正周期T=2π.(2)由f(α)=2得2sin(α+π/6)=2,∴