已知函数f(x)=cos4x-2sinx-sin4x.     

原体应该是
函数f(x)=cos4x-2sinxcosx-sin4x 吧
cos4x指cosx的四次方吧!
f(x) = (cosx)^4 - 2sinx*cosx - (sinx)^4
= [(cosx)^2 + (sinx)^2] * [(cosx)^2 - (sinx)^2] - 2sinx*cosx
= cos(2x) + sin(2x)
= √2*sin(2x + π/4)
x属于R
所以sin(2x + π/4)属于-1,1,可等于
所以f(x)的最小值 为-√2
2）0＜x＜π÷2
2x + π/4属于（π/4,5π/4）
sin(2x + π/4)属于（-√2,1）
x范围应是0≤x≤π÷2
所以f(x)的最大=√2,最小=-2

f(x)=(cosx)^2-(sinx)^4-2sinx
=[(cosx)^2+（sinx)^2][(cosx)^2-(sinx)^2]-2sinx
=1-2(sinx)^2-2sinx
令sinx=t(-1<=t<=1)
f(t)=-2t^2-2t+1
=-2(t^2+t+1/4-1/4)+1
=-2(t+1/2...

1.f(x) = (cosx)^4 - 2sinx*cosx - (sinx)^4
= [(cosx)^2 + (sinx)^2] * [(cosx)^2 - (sinx)^2] - 2sinx*cosx
= cos(2x) + sin(2x)
= √2*[sin(2x + π/4)]
f(x)的最小值=-√2
2.当x=π/8 f(x)的最大=√2
当x=π/2 f(x)最小值 =-1