（1）g(x)=1/2sin(2x+2π/3)=g(x)=1/2sin[2(x+π/3)]
向左平移π/4個(gè)單位長度,再向上平移1/2個(gè)單位長度后
f(x)=1/2sin[2(x+π/3+π/4)]+1/2
=1/2sin[2(x+π/3)+π/2]+1/2
=1/2cos[2(x+π/3)]+1/2
=1/2[2cos^2(x+π/3)-1]+1/2
=cos^2(x+π/3)
∴ a=1 b=0
(2)φ(x)=g(x)-√3f(x)
=1/2sin[2(x+π/3)]-√3cos^2(x+π/3)
=1/2sin[2(x+π/3)]-√3[1/2cos[2(x+π/3)]+1/2]
=1/2sin[2(x+π/3)]-√3/2cos[2(x+π/3)]-√3/2
=sin[2(x+π/3)-π/6]-√3/2
=sin[2x+π/2]-√3/2
=cos2x-√3/2
所以 φ(x)的單調(diào)增區(qū)間由下式確定
-π+2kπ≤2x≤2kπ
即 -π/2+kπ≤x≤kπ (k=0,±1,±2,…)為所求的增區(qū)間