已知向量m=(√3sinx/4,1),n=(cosx/4,cos2x/4)记函数f(x)=m*n

f(x)=m*n
=√3sin(x/4)cos(x/4)+ cos²(x/4)
=(√3/2)sin(x/2) +( cos(x/2) +1) /2
=((√3/2)sin(x/2)+(1/2)cosx/2) +1/2
=cos(π/3 -x/2) + 1/2
1)∵m*n=1
∴f(x)=cos(π/3-x/2)+1/2=1
cos(π/3 -x/2) = 1/2
cos(2π/3 -x) =2cos²(π/3-x/2)-1=1/2
则cos(x+π/3)=-cos(π-(2π/3-x))=-cos(2π/3 -x) = -1/2
2) 当π/3 -x/2=2kπ,即x=2π/3-4kπ时,f(x)取最大值,fmax=1+1/2=3/2
3) 当π/3 -x/2∈(2kπ-π,2kπ)时,f(x)递增
则f(x)增区间(8π/3-4kπ,2π/3-4kπ) （k∈Z）