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

f(x)
=sin²ax-sinaxcosax
=-1/2(2cos²ax-1)-1/2·2sinaxcosax-1/2
= -1/2cos2ax-1/2sin2ax-1/2
=-1/√2cos(2ax-π/4)-1/2
(1)最小正周期为π/2,由T=2π/ω得π/2=2π/2a,a=2
(2)f(x)=-1/√2cos(4x-π/4)-1/2,当4x-π/4∈[(2k-1)π,2kπ]时(第三四象限）,f(x)单调递减
即x∈[（2k-1)π/4+π/16,kπ/2+π/16],k∈z