fx=( sinx cosx)2 cos2x在0,二分之派最大最小值
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f(x)=2cos²x+2√3sinxcosx=1+cos(2x)+√3sin(2x)=2[(√3/2)sin(2x)+(1/2)cos(2x)]+1=2sin(2x+π/6)+1当sin(
f(x)=a(cos²x+sinxcosx)+b=a(cos²x-1/2+sinxcosx+1/2)+b=a(cos2x/2+sin2x/2)+b=a根号下2sin(2x+π/4)
f(x)=√3sin²x+sinxcosx=√3[(1-cos2x)/2]+1/2sin2x=1/2sin2x-√3/2cos2x+√3/2=sin(2x-π/3)+√3/2∵x∈[π/2,
f(x)=2√3sinxcosx+2sin^2x-1=√3sin2x-cos2x=2sin(2x-π/6)最小正周期T=π,单调递增区间:2kπ-π/2
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/
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=2cos^2x+2根号3sinxcosx-1=2cos^2x-1+2根号3sinxcosx根据倍角公式,sin2α=2sinαcosαcos2α=2cos^2(α)-1fx=cos2x+根号3s
1.T=πfx=2cosxsin(x+π/3)-√3sin^2x+sinxcosx=cosxsinx+√3cos^2x-√3sin^2x+sinxcosx=2sinxcosx+√3cos2x=sin2
f(x)=(2sinxcosx)/2=(sin2x)/2周期为2π/2=π最小值为-1/2,sin2x=-1时取得
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&
答:y=f(x)=2√3sinxcosx-2sin²x=√3sin2x+cos2x-1=2*[(√3/2)sin2x+(1/2)cos2x]-1=2sin(2x+π/6)-1y=f(x)关于
f(x)=2sinxcosx-(2cos²x-1)=sin2x-cos2x=√2sin(2x-π/4)所以值域是[-√2,√2]
解:原式=√3sin2x+cos2x+1=2(√3/2sin2x+1/2cos2x+1=2cos(2x-pai/3)+1.
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
函数fx=2根号3sinxcosx+1-2sinX=根号3sin2x+cos2x=2sin(2x+30度),fx的值域就是【-2,2】
f(x)=sin2x+cos2x=√2sin(2x+π/4)最小正周期T=2π/2=π最大值为√2再问:题目都不一样再答:哪不一样?2sinxcosx可化为sin2x呀。