已知f(x)={cos2x,x〈0. 兀x,x≥0.则f[-f(1 6)]=

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已知f(x)={cos2x,x〈0. 兀x,x≥0.则f[-f(1 6)]=
已知函数f(x)=2cos2x+sin²x

①原式=f(x)=2cos2x+sinx^2=2cos2x+1-cos2x/2=3/2cos2x+1/2故f(π/3)=3/2*cos2π/3+1/2=-3/4+1/2=-1/4②依f(x)=3/2c

已知函数f(x)=sin2x-cos2x+1

f(x)=sin2x-cos2x+1=√2*(√2/2*sin2x-√2/2*cos2x)+1=√2sin(2x-π/4)+1最小正周期为:T=2π/2=π∵-1≤sin(2x-π/4)≤1∴1-√2

已知函数f(x)=根号3sinxcosx+cos2x+1

f(x)=(√3)sinxcosx+cos2x+1f(x)=(√3)(2sinxcosx)/2+cos2x+1f(x)=(√3/2)sin2x+cos2x+1f(x)=(√7/2)[(√3/2)(2/

已知函数f(x)=2根号3sinxcosx-cos2x

f(x)=2√3sinxcosx-cos2x=√3sin2x-cos2x=2(sin2x*√3/2-cos2x*1/2)=2sin(2x-π/6)x=π/12;函数f(x)的图象可以由函数y(x)=2

已知函数f(x)=(sin2x+cos2x+1)/2cosx,

分式有意义,cosx≠0f(x)=[sin(2x)+cos(2x)+1]/(2cosx)=(2sinxcosx+cos²x-sin²x+cos²x+sin²x)

已知函数f(x)=cos2x+sinx(sinx+cosx)

字数限制f(x)=cos2x+(1-cos2x)/2+sin2x/2=(cos2x+sin2x)/2+1/2=cos(2x+π/4)/根号2+1/2其最小正周期为π,最大值为:(1+根号2)/2x在[

已知f(x)=(sin2x-cos2x+1)/(1+cotx)

f(x)=1-cos2x2)∵sin(x+π/4)=3/5∴cos2(x+π/4)=1-2sin²(x+π/4)=7/25∵cos2(x+π/4)=cos(2x+π/2)=-sin2x∴-s

已知函数f(x)=(sinx+cosx)平方+cos2x

一f(x)=sin"x+cos"x+2sinxcosx+cos2x=1+sin2x+cos2x=_/2sin(2x+派/4)+1所以T=2派/2=派"指平方“_/2”指根号2二因为X属于[0派/2]是

已知函数f(x)=根号3sin2x+cos2x+2

已知函数f(x)=根号3sin2x+cos2x+21求f(x)的最大值及f(x)取得最大值时自变量x集合f(x)=根号3sin2x+cos2x+2=2[(根号3/2)sin2x+(1/2)cos2x]

已知函数f(x)=cos2x/[sin(π/4-x)]

cos2x=sin(π/2-2x)=2sin(π/4-x)cos(π/4-x)cos2x/[sin(π/4-x)]=2sin(π/4-x)cos(π/4-x)/[sin(π/4-x)]=2cos(π/

已知函数f(x)=3sin2x+cos2x.

(1)f(x)=3sin2x+cos2x=2(sin2xcosπ6+cos2xsinπ6)=2sin(2x+π6)∴T=2π2=π,当2x+π6=2kπ+π2,k∈Z,即x=π6+kπ,k∈Z时,函数

已知函数f(x)=sin2x+cos2x-1

f(x)=√2sin(2x+π/4)-1(1)最小正周期π;最大值√2(2)2kπ-π/2≤2x+π/4≤2kπ+π/22kπ-3π/4≤2x≤2kπ+π/42kπ-3π/8≤x≤2kπ+π/8

已知f(x)=2sinxcosx+cos2x(x属于R)

f(x)=2sinxcosx+cos2x=sin2x+cos2x=√2sin(2x+π/4)2x+π/4=π/2+2kπ时f(x)有最大值f(x)=√2x=π/8+kπ2x+π/4=3π/2+2kπ时

已知f(x)=sin2x+根号3cos2x

f(x)=sin2x+√3cos2x=2(1/2sin2x+√3/2cos2x)=2sin(2x+π/3)f(x)的最大值=2,最小值=-2f(x)的单调区间:∵2kπ-π/2≤2x+π/3≤2kπ+

已知函数f(x)=2cos2x+sinx的平方.

f(x)=2cos2x+sinx=2-4*(sinX)^2+(sinX)^2=2-3*(sinX)^2f(π/3)=-3*(9/4)+2=-1/4f(x)的最大值2最小值-1

已知函数f(x)=2sinxcosx+cos2x

f(x)=sin2x+cos2x=√2sin(2x+π/4)f(π/4)=√2sin(2*π/4+π/4)=√2*√2/2=10

已知函数f(x)=2sinxcosx+cos2x(x

f(x)=sin2x+cos2x=√2sin(2x+π/4)所以T=2π/2=π最大值=√2f(θ+π/8)=√2sin(2θ+π/4+π/4)=√2cos2θ=√2/3cos2θ=1/3θ锐角则si

已知函数f(x)=根号3 sinxcosx+cos2x+1

f(x)=√3sinxcosx+cos2x+1=(√3/2)sin2x+cos2x+1=[(√7)/2][(√3/√7)sin2x+(2/√7)cos2x]+1=[(√7)/2]sin(2x+α)+1

已知f(sin-1)=cos2x+2,求f(x)

f(sinx-1)=cos2x+2cos2x=1-2sin^2xf(sinx-1)=3-2sin^2x=-2(sinx-3/4)^2+7.5f(x)=-2(x+1/4)^2+7.5