若f（x）=Sin^2ax-sinaxcosax（a>0）的最小正周期为兀/2，（1）求a的值（2）

f(x)=sin²ax-sinaxcosax =(1-cos2ax)/2-1/2sin2ax =1/2-1/2*(cos2ax+sin2ax) =1/2-√2/2*(√2/2sin2ax+√2/2cos2ax) =1/2-√2/2*sin(2ax+π/4) (1) 2π/(2a)=π/2 a=2 (2) f(x)=1/