f(x)=2cos²x-1+2sinxcosx+2=cos2x+sin2x+2=√2sin(2x+π/4)+2（1）最小正周期T=2π/2=π（2）x∈【0,π/2】则：2x+π/4∈【π/4,5π/4】所以,sin(2x+π/4)∈【-√2/2,1】所以,f(x)∈【1,2+√2】即f(x)的最大值...�鷳��f(x)=2cos²x-1+2sinxcosx+2 =cos2x+sin2x+2����ô�õ��ģ�倍角公式：cos2x=2cos²x-1，sin2x=2sinxcosx