已知F(SINX)=COS2X-1,则F(cos15°)?

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

f(x)=0.5(1-cos2x)+0.5sin2x+cos2x=0.5+0.5(sin2x+cos2x)=0.5+0.5根号2sin(2x+π/4)f(x)的最小值是（1-根号2)/2

5.（1）f(x)=(√3)cos2x+2sinxsin(x+π/2).=√3cos2x+2sinxcosx=sin2x+√3cis2x=2sin(2x+π/3).∴最小正周期T=2π/2=π,f(x

f(x)=cos2x+sinx(sinx+cosx)=cos2x+sinxsinx+sinxcosx=cos2x+1/2sin2x+(1-cos2x)/2=1/2cos2x+1/2sin2x+1/2=

F(cosx)=cos2x=2cosx^2-1设y=cosx,则F(y)=2y^2-1F(sinx)=2sinx^2-1=-(1-2sinx^2)=-cos2x新年快乐!

f(x)=2cos2x+sin²x-4cosx=2(2cos²x-1)+(1-cos²x)-4cosx=4cos²x-2+1-cos²x-4cosx=

f(x+π/6)=a×b=cos2xsinx+sin2xcosx=sin3xf(x)=sin[3(x-π/6)]=sin(3x-π/2)因此f(x)的最小正周期＝2π/3函数取最大值时3x-π/2=2

1)sinx≠0==>x≠kπ2)原式化简得f(x)=2(cosx+sinx)a是锐角所以cosa>0所以cosa=3/5所以f(a)=14/5

（Ⅰ）由题意y＝f(x)＝sin2x+sinx•cosx+cos2x＝1−cos2x2+12sin2x+cos2x（2分）=12(cos2x+sin2x)+12＝22(22cos2x+22sin2x)

解（1）∵f(x)=23sinx•cosx+cos2x-1，∴f(x)=2sin(2x+π6)-1．（2分）解2kπ-π2≤2x+π6≤2kπ+π2，k∈Z，得kπ-π3≤x≤kπ+π6，k∈Z．∴函

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[

存在,f(x)连续可导故要使f(x)为常数,则f'(x)=0恒成立而f'(x)=k(sinx)^(k-1)sin(k-1)x-k(cosx)^(k-1)sin(k-1)x+2k(cos2x)^(k-1

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

f（cosx）=cos2x=2cosx*cosx-1故f（sinx）=2sinx*sinx-1=-cos2x

（1）f（x）=（sinx+cosx）2-2cos2x=1+2sinxcosx-2cos2x（2分）=sin2x-cos2x（3分）=2(22sin2x−22cos2x)（4分）=2sin(2x−π4

f（sinx）=1-cos2x=2(sinx)^2,既有f(x)=2x^2,故f(cosx)=2*(cosx)^2=1+cos2x.

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)=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