解:由題意f(x)=sinxcosx+
=
sin2x+
cos2x
=sin(2x-
)
(1)T=
=π
(2)令2kπ+
≤2x-
≤2kπ+
,k∈z
解得kπ+
≤x≤kπ+
,k∈z
函數(shù)f(x)的單調(diào)減區(qū)間是[kπ+
,kπ+
]k∈z
(3)令2x-
=kπ+
,解得x=kπ+
,k∈z即為函數(shù)的對(duì)稱(chēng)軸方程;
可令2x-
=kπ,k∈z,解得x=
,對(duì)稱(chēng)中心的坐標(biāo)是(
,0),k∈z
分析:先用恒等變換公式對(duì)函數(shù)f(x)化簡(jiǎn)整理,易得f(x)=sinxcosx+
=sin(2x-
)
(1)求函數(shù)f(x)的最小正周期,用周期公式求解即可;
(2)求函數(shù)f(x)的單調(diào)減區(qū)間,利用正弦函數(shù)的性質(zhì),令2kπ+
≤2x-
≤2kπ+
,k∈z即可解出;
(3)求函數(shù)f(x)的對(duì)稱(chēng)軸方程,可令2x-
=kπ+
,k∈z求對(duì)稱(chēng)中心坐標(biāo)可令2x-
=kπ,k∈z
點(diǎn)評(píng):本題考查三角函數(shù)恒等變換應(yīng)用、三角函數(shù)的周期性的求法,函數(shù)的單調(diào)遞減區(qū)間等,解題關(guān)鍵是掌握住三角恒等變換公式,以及三角函數(shù)的性質(zhì).