f(x)=(cosx)^4-2sinxcosx-(sinx)^4
=[(cosx)^2+(sinx)^2][(cosx)^2-(sinx)^2]-2sinxcosx
=(cosx)^2-(sinx)^2-2sinxcosx
=cos(2x)-sin(2x)
= √2*[cos(2x)*√2/2-sin(2x)*√2/2]
= √2*[cos(2x)cos(π/4)-sin(2x)sin(π/4)]
= √2cos(2x+π/4) .
（2）因?yàn)?0??????cos???? = ??2cos(2x+??/4) ??????y=Asin(wx+??)??????????????????????????????????????????? -??2sin(2x-??/4) ???????????????????????????????絥????????????ū???????=-??2sin??2x-??/4)??????????????t=2x-??/4??0??t???/2???/8??x??3??/8why??求出來(lái)是 f(x)= -√2sin（2x-π/4) ，要求 f(x) 的單調(diào)遞減區(qū)間，就要求 sin（2x-π/4) 的遞增區(qū)間（因?yàn)榍懊嫦禂?shù)是負(fù)數(shù)），令 t=2x-π/4 ，則 -π/2<=t<=π/2 ，（這點(diǎn)你弄錯(cuò)了）因此 -π/2<=2x-π/4<=π/2 ，解得 -π/8<=x<=3π/8 。