解:(1):①②?③④.
由①得ω×
![](http://thumb.1010pic.com/pic5/latex/257.png)
+∅=kπ+
![](http://thumb.1010pic.com/pic5/latex/73.png)
,k∈z. 由②得ω
![](http://thumb.1010pic.com/pic5/latex/196.png)
+∅=kπ,k∈z.
又∵ω>0,
![](http://thumb.1010pic.com/pic5/latex/53538.png)
,故有ω=2,∅=
![](http://thumb.1010pic.com/pic5/latex/196.png)
.
∴
![](http://thumb.1010pic.com/pic5/latex/6533.png)
,其周期为π.
令
![](http://thumb.1010pic.com/pic5/latex/23112.png)
,可得
![](http://thumb.1010pic.com/pic5/latex/23113.png)
.
故函数f(x)的增区间为[
![](http://thumb.1010pic.com/pic5/latex/53891.png)
],k∈z.
∵
![](http://thumb.1010pic.com/pic5/latex/53892.png)
,
∴f(x)在区间[
![](http://thumb.1010pic.com/pic5/latex/53893.png)
]上是增函数,
故可得 ①②?③④.
(2):还可①③?②④.
由③它的周期为π,可得ω=2,故 f(x)=sin(2x+∅).
由①得 2×
![](http://thumb.1010pic.com/pic5/latex/257.png)
+∅=kπ+
![](http://thumb.1010pic.com/pic5/latex/73.png)
,k∈z.再由
![](http://thumb.1010pic.com/pic5/latex/53538.png)
可得φ=
![](http://thumb.1010pic.com/pic5/latex/196.png)
,故函数f(x)=sin(2x+
![](http://thumb.1010pic.com/pic5/latex/196.png)
).
显然它的图象关于点(
![](http://thumb.1010pic.com/pic5/latex/196.png)
,0)对称,由(1)可得 f(x)在区间[
![](http://thumb.1010pic.com/pic5/latex/53893.png)
]上是增函数.
故可得 ①③?②④.
故答案为 (1):①②?③④; (2):①③?②④.
分析:(1)由①得ω×
![](http://thumb.1010pic.com/pic5/latex/257.png)
+∅=kπ+
![](http://thumb.1010pic.com/pic5/latex/73.png)
; 再由②得ω
![](http://thumb.1010pic.com/pic5/latex/196.png)
+∅=kπ,k∈z,以及ω、∅的范围,求得ω、∅的值,从而得函数解析式,从而求出周期和单调增区间,可得③④正确,故得①②?③④.
(2)由③可得ω=2,故 f(x)=sin(2x+∅),再由①得 2×
![](http://thumb.1010pic.com/pic5/latex/257.png)
+∅=kπ+
![](http://thumb.1010pic.com/pic5/latex/73.png)
,k∈z,结合∅的范围可得φ=
![](http://thumb.1010pic.com/pic5/latex/196.png)
,故函数f(x)=sin(2x+
![](http://thumb.1010pic.com/pic5/latex/196.png)
),由此推出②④成立.
点评:本题主要考查三角函数的周期性,单调性,对称性,以及学生构造命题拓展问题的能力,属中档题.