f(x)=cox(2x-π/3)+2sin(x-π/4)sin(x+π/4)
=cox(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)]
=cox(2x-π/3)+2sin(x-π/4)cos(x-π/4)
=cox(2x-π/3)+sin(2x-π/2)
=cox2x*cosπ/3+sin2x*sinπ/3-cos2x
=1/2*cos2x+sin2x*sinπ/3-cos2x
=sin2x*sinπ/3-1/2cos2x
=sin2x*sinπ/3-cosπ/3cos2x
=-cos(2x+π/3)
=cos(2x+4π/3)
1. 最小正周期 2π/2=π；
令2x+4π/3=0,則x=-2π/3；
對(duì)稱(chēng)軸 x=-2π/3+kπ,k為自然數(shù)；
2.令-π/12≤-2π/3+kπ≤π/2,得7/12≤ k ≤ 7/6,k為自然數(shù),k取1,此時(shí)x=π/3；
函數(shù)f(x)在區(qū)間[-π/12,π/2]上的最大值,最小值在x=-π/12,x=π/2,x=π/3之間產(chǎn)生；
f(-π/12)=cos(7π/6)=-√3/2；
f(π/2)=cos(π+4π/3)=1/2；
f(π/3)=cos(2π/3+4π/3)=1；
函數(shù)f(x)的區(qū)間[-π/12,π/2]上的值域[-√3/2,1]；可以解釋一下-cos(2x+π/3)怎么化到cos(2x+4π/3)嗎？