求函数f(x)=cos²x asinx α 1

∵f（x）＝cos（2x－π/3）＋（sinx）^2－（cosx）^2＝cos（2x－π/3）－cos2x＝2sin（2x－π/6）sin（π/6）＝sin（2x－π/6）.∴g（x）＝［sin（2x

先化简,f（x）=sin（2x-π/6）（1）最小正周期T=2π/2=π（2）x∈[0,π/4]2x-π/6∈[-π/6,π/3]∴f（x）单调递增最小值为f（0）=sin（-π/6）=-1/2最小值

是“函数f(x)=(1/2)cosx平方+((根号3)/2)sinxcosx-(1/4)”吧f（x)=(cos(2x)+1)/4+(（根号3)sin2x)/4-1/4=1/2(sin(2x+∏/6))

f(x)=（6cos^4x-5cos^2x+1）/（2cos^2x-1）=3cos^2x-1,2cos^2x-1≠0cos2x≠02x≠kπ+π/2,k∈Z解出x得定义域f(x)=3cos^2x-1=

解f(x)=cos^2x+sinxcosx=1/2(2sinxcosx)+1/2(2cos^2x-1)+1/2=1/2sin2x+1/2cos2x+1/2=√2/2sin(2x+π/4)+1/2当-π

f(x)=sin2x-2cos^2x=sin2x-cos2x-1=√2(sin2x-π/4)-1∴T=2π/2=π函数f(X)最小正周期是π

f(x)=cos(x^2+√x)f′(x)=-[sin(x^2+√x)][2x-1/(2√x)]

f(x)=sin²(x)+(√3)sin(x)cos(x)+2cos²(x)=3/2+√3/2sin2x+1/2cos2x=3/2+sin(2x+π/6)函数f(x)的最小正周期T

f(x)=2sin(x/4)cos(x/4)+(√3)cos(x/2)=sin(x/2)+(√3)cos(x/2)=2sin(x/2+π/3)T=4π最大值2最小值-2（2）令g(x)=f(x+π/3

f(x)=cos(x-π/3)-sin(π/2-x)=(1/2)cosx+(√3/2)sinx-cosx=(√3/2)sinx-(1/2)cosx=sin(x-π/6）,它的最小值=-1.

y=f(cos²x)y'=f'(cos²x)*(cos²x)'=f'(cos²x)*(-2sinxcosx)

设t=sinx,又cos2x=1-2(sinx)^2则f(t)=1-2(sinx)^2+1=2（1-t^2）所以f(cosx)=2（1-t^2）=2（1-(cosx)^2）=2(sinx)^2

令1＋cosx=x,则有x=arccos(x-1)cos^2x=(x-1)^2,即f(x)=(x-1)^2这个图会画了吧!

f(x)=cos^2x+1/2(2sinxcosx)    =cos^2x+1/2*sin^2x用辅助角公式f(x)=√5/2sin(2x+arctan2)&n

f(x)=sin(2x)+√2cos(x-π/4)=sin(2x)+√2[cosxcos(π/4)+sinxsin(π/4)]=sin(2x)+cosx+sinx=sin(2x)+√2sin(x+π/

f(x)=cosx+cos(x+π/2)=cosx-sinx=-(sinx-cosx)=-√2sin(x-π/4)f(x)的最大值√2最小值-√2

f（x）=sin^2x+2√3sinxcosx+3cos^2x=1+√3sin2x+2cos^2x-1+1=√3sin2x+cos2x+2=2(sin2x*√3/2+cos2x*1/2)+2=2sin

f(x)=sin2x-2sin^2x=sin2x-(1-cos2x)=sin2x+cos2x-1=√2(√2/2*sin2x+√2/2cos2x)-1(提取√2）=√2(sin2xcosπ/4+cos

f(x)=[cos(x)+sin(x)]sin(x)=cos(x)sin(x)+sin^2(x)=1/2sin2x+1/2(1-cos2x)=√2/2[cos(p/4)sin2x-sin(p/4)co

已知cosx=2cos²(x/2)-1所以,2cos²(x/2)=cosx+1则,f(x)=cos²x-cosx-1令cosx=t（-1≤t≤1）所以,f(t)=t