已知函数f(x)=cos(2x-π3)+2sin(x-π4)sin(x+π4).(1)求函数f(x)的最小正周期和图象的对称

已知函数f(x)=cos(2x-π3)+2sin(x-π4)sin(x+π4).(1)求函数f(x)的最小正周期和图象的对称中心;(2)求函数f(x)在[?π3,π2]上的单减区间.

(1)f(x)=cos(2x?
π
3
)+2sin(x?
π
4
 )cos(x?
π
4
)

=cos(2x?
π
3
)+sin(2x?
π
2
)

=
3
2
sin2x?
1
2
cos2x
=sin(2x-
π
6

2x?
π
6
=kπ
x=
2
+
π
12
,故f(x)的最小正周期为π,对称中心(
2
+
π
12
,0)
,k∈Z         (6分)
(2)由2kπ+
π
2
≤2x?
π
6
≤2kπ+
2
得:kπ+
π
3
≤x≤kπ+ 
6

故f(x)在[?
π
3
π
2
]
上的单减区间[?
π
3
,?
π
6
],[
π
3
π
2
]
                                   (12分)
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