fx=sinx cosx²
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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)=[(cosx)^2-(sinx)^2]+√3sin2x=cos2x+√3sin2x=2sin(2x+π/6),最小正周期T=π,由-π/2+2kπ≤2x+π/6≤π/2+2kπ,k∈Z解得:
f(x)=-√3sin²x+sinxcosx=√3/2cos2x+1/2sin2x-1/2=sin(2x+π/3)+1/2T=2π/2=πf(π/6)=sin(π/3+π/3)+1/2=(1
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根号3sinxcosx+cos²x-sin²xf(x)=根号2(2sinxcosx)+(cos²x-sin²x)f(x)=根号3sin2x+cos2
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
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<
解题思路:数列递推运算,由递推公式知道第一项求其他项解题过程:由得答案D最终答案:由得
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)=cos²x+sinxcosx=(cos2x+1)/2+1/2sin2x=(1/2cos2x+1/2sin2x)+1/2=√2/2*(√2/2cos2x+√2/2sin2x)+1/2
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。