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)對(duì)稱軸2x+π/4=kπ-->x=kπ/2-π/8 （k∈z） 追問(wèn)：f(x)的單調(diào)區(qū)間是? 回答：遞減區(qū)間2kπ≤2x+π/4≤2π+π--->kπi-π/8≤x≤kπ+3π/8遞增區(qū)間2kπ-π≤2x+π/4≤2kπ--->kπ-5π/8≤x≤kπ-π/8