【答案】
分析:可利用函数图象的向量平移公式解决问题,设出平移向量
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/0.png)
=(a,b),得向量平移公式
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/1.png)
,代入平移后函数解析式得平移前函数解析式,与已知函数解析式比较即可求得a、b值
解答:解:设
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/2.png)
=(a,b),函数
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/3.png)
的图象上任意一点(x,y)沿向量
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/4.png)
平移后的对应点为(x′,y′)
则
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/5.png)
,
∵平移后得到函数g(x)=cos2x的图象,∴(x′,y′)满足函数g(x)=cos2x的解析式,
代入,得y+b=cos[2(x+a)]
化简,得,y=cos[2(x+a)]-b,即y=sin[
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/6.png)
+2(x+a)]-b=sin(2x+2a+
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/7.png)
)-b
∴原函数图象上的任意一点满足关系式y=sin(2x+2a+
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/8.png)
)-b
即原函数解析式为y=sin(2x+2a+
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/9.png)
)-b
又∵原函数为
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/10.png)
∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/11.png)
与y=sin(2x+2a+
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/12.png)
)-b为同一个函数.
∴2a+
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/13.png)
=-
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/14.png)
+2kπ(k∈Z),-b=1
解得,a=-
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/15.png)
+kπ(k∈Z),b=-1
∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/16.png)
可取
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024184639997800904/SYS201310241846399978009003_DA/17.png)
故选 D
点评:本题考查了三角函数的图象变换,函数图象的向量平移公式的运用,简单的三角变换公式的运用