已知fx=sinx cosx 二分之根号三cos2x
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/02 14:36:29
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)=-√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√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)=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/
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<
f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2sinxcosx=cos(3x/2+x/2)-2sinxcosx=cos2x-sin2x=√2(√2/2*cos2x
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
1.g(x)+f(x)=x^(1/2)----(1).g(x)-f(x)=x^(-1/2)---(2).(1)+(2):2g(x)=x^(1/2)+x^(-1/2).g(x)=(1/2)[x^(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]
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。