已知函数fx=√3cosπ 2-2x 2cosx
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fx=-√3cos2x-sin2x=-2sin(2x+π/3)所以最小正周期为πf'x=-4cos(2x+π/3),f'x>0时递增x在(π/12,π/3)上递增f'x=0,x=π/12.极小值f(π
f(x_=(cosx+sinx)(cosx-sinx)=cos²x-sin²x=cos2x所以T=2π/2=πf(α/2)=cosα=1/3sin²α+cos²
f(x)=1+cos2x+cos[2x+π/3]-1=cos2x+cos[2x+π/3]=2cos(2x+π/6)cos(π/6)=√3cos(2x+π/6)1)最小正周期T=2π/2=π单调增区间:
若cosα=3/5.α属于(3π/2,2π),sinα=-4/5把f(2α+π/3)代入fx=√2cos(x-π/12),化简原式=cos2α-sin2αcos2α-sin2α怎么化简的就不用我说了吧
f(x)=sin²x+√3sinxcosx+2cos²x,=√3sinxcosx+cos²x+1=√3/2sin2x+1/2(1+cos2x)+1=√3/2sin2x+1
f(x)=√3sin2x+cos2x=2(sin2x*√3/2+cos2x*1/2)=2(sin2xcosπ/6+cos2xsinπ/6)=2sin(2x+π/6)所以f(π/6)=2sin(2×π/
1、函数可化为f(x)=(√2/2)*sin[2wx+(π/4].===>(2π)/(2w)=π,===>w=1.2、不懂==
先化简f(x)=2根号3sinxcosx+2cos^2x-1=根号3sin2x+cos2x=2(根号3/2sin2x+1/2cos2x)=2sin(2x+π/6)则T=2π/ω=2π/2=πy=sin
fx=sin2x-根号3*(1+cos2x)+a+根号3=2sin(2x-60°)+aT=pi,增区间[k*pi-pi/6,k*pi+5pi/12],k属于Z 2.由题意得-5pi/6<
1.f(x)=√3sinxcosx-cos²x+1/2=(√3/2)(2sinxcosx)-(1/2)(2cos²x-1)二倍角公式:2sinxcosx=sin(2x),2cos&
f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=cos(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)]=cos(2x-π/3)+2sin(x-π/
f(x)=cos(2x-π/3)-cos2x=1/2cos2x+√3/2sin2x-cos2x=√3/2sin2x-1/2cos2x=sin(2x-π/6)最小正周期T=2π/2=π(2)0
已知函数fx=√3sinxcosx+(cos∧2)x+a(1)求fx的最小正周期及单调递减区间(2)若fx在区间[~π/6,π/3]上的最大值与最小值的和为3/2,求a的值.(1)解析:f(x)=√3
解1当2kπ-π/2≤2x+π/3≤2kπ+π/2,k属于Z时,y是增函数即2kπ-5π/6≤2x≤2kπ+π/6,k属于Z时,y是增函数即kπ-5π/12≤x≤kπ+π/12,k属于Z时,y是增函数
f(x)=√3sin2x+cos2x=2sin(2x+π/6)∴f(x0)=2sin(2x0+π/6)=6/5∴sin(2x0+π/6)=3/5∵x0∈[π/4,π/2]∴2x0+π/6∈[2π/3,
解f(x)=2cos^2x+2√3sinxcosx-1=√3sin2x+cos2x=2sin(2x+π/6)∴最小正周期为:2π/2=π再答:不懂追问再问:在三角形ABC中,角ABC所对的边分别是ab
(1)、f(x)=2cos²x-(sinx-cosx)²=2cos²x-(1-sin2x)=cos2x+sin2x运用一下化一公式得f(x)=√2sin(2x+π/4),
f(x)=2√3sinxcosx+2cos²x+m=√3sin2x+1+cos2x+m=2sin(2x+π/6)+m+1.0再问:在三角形ABC中角ABC所对的边长abc若F(A)=1,si