f(x)=(cos2/x)2+sin2/xcosx/2=1/2(1-cosx)+1/2sinx=1/2(sinx-cosx)+1/2=√2/4sin(x-π/4)+1/2
令 -π/2+2kπ≤x-π/4≤π/2+2kπ
得 -π/4+2kπ≤x≤3π/4+2kπ
即函数f(x)=a*b的单调递增区间为 [-π/4+2kπ,3π/4+2kπ] ,k∈Z