f(x)=(1/2)(cosx)^2-sinxcosx-(1/2)(sinx)^2==(1/2)[(cosx)^2-(sinx)^2]-(1/2)2sinxcosx=(1/2)(cos2x-sin2x)=(√2/2)[cos2xcos(π/4)-sin2xsin(π/4)]=(√2/2)cos(2x+π/4)1)周期T=2π/2=π2)对称轴2x+π/4=kπ-->x=kπ/2-π/8 （k∈z） 追问：f(x)的单调区间是? 回答：递减区间2kπ≤2x+π/4≤2π+π--->kπi-π/8≤x≤kπ+3π/8递增区间2kπ-π≤2x+π/4≤2kπ--->kπ-5π/8≤x≤kπ-π/8