Question

设函数f(x)=√2∕2cos[2x+π/4]+sin²x 1、求f(x)最小正周期

2设函数g(x)对任意x∈R,有g(x+π/2)=g(x),且当x∈[0,π/2]时,g(x)=1/2-f(x),求g(x)在区间
[-π,0]上的解析式

Answer 1

(1)
f(x=1/2cos2x-1/2sin2x+(1-cos2x)/2
=1/2cos2x-1/2sin2x+1/2-1/2cos2x=
=- 1/2sin2x+1/2
即f(x)= - 1/2sin2x+1/2
w=2 ;由周期公式得T=2π/2=π

(2)
当-π≤x≤ - π/2时
0≤x+π≤ π/2
g(x+π)=1/2-f(x+π)
=1/2-[ - 1/2sin2(x+π)+1/2]
=1/2sin2(x+π)
=1/2sin2x
g(x+π)=g(x+π/2)=g(x)
即g(x)=1/2sin2x

当-π/2<x≤ 0时
0<x+π/2≤ π/2
g(x+π)= 1/2-f(x+π)
=1/2-[]
=1/2-[ - 1/2sin2(x+π/2)+1/2]
=1/2sin2(x+π/2)
= - 1/2sin2x
综合可知：g(x)={1/2sin2x (-π≤x≤ - π/2)
g(x)={-1/2sin2x (-π2<x≤ 0)