已知函数f(x)=2cos2x sin.sinx-4cosx求f(兀 3)的

cos2x≠02x≠kπ+π/2,k∈Z∴f(x)定义域为{x|x≠kπ/2+π/4,k∈Z}∵对于x∈{x|x≠kπ/2+π/4,k∈Z},都有-x∈{x|x≠kπ/2+π/4,k∈Z}关于原点对称

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

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

1.sinx≠0,∴x≠kπ.∴f（x）的定义域为{x|x≠kπ,k∈Z}2.f(x)=(sin2x-cos2x+1)/(2sinx)=(2sixcosx+2sin²x)/(2sinx)=c

只要cos2x≠0即可,所以定义域为｛x｜x∈R,且x≠k∏+∏/2,k∈Z｝f（x）=1/2（tan2x+1/cos2x+1）tan2x和cos2x在第一象限时的值域为（0,+∞）tan2x和cos

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

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

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

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[

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+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]

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/

（Ⅰ）f(x)＝sinx•cosx+12cos2x+12=12sin2x+12cos2x+12=22sin(2x+π4)+12∴函数f（x）的最小正周期T＝2π2＝π（Ⅱ）当x∈[−π8，3π8]时，

令t=sinx,则cos2x=1-2t²f(x)=2(1-2t²)+t²=2-3t²因为|t|再问：可以加你qq吗再答：不常挂Q再问：哦！谢谢再问：设函数f（x

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

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

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

sqrt表示根号下；首先将原函数化成正弦型函数f(x)=sin2x+cos2x+2=(sqrt2)*(sin2x*(sqrt2)/2+cos2x*(sqrt2)/2)+2=(sqrt2)*sin(2x

f(x)=2(1－2sin²x)＋sin²x=2－3sin²x1、f(π/3)=2－3/4=5/4；2、f(x)的最大值是2,最小值是-1

f(x)=1-2x^2