题目内容
若函数![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103103516864716325/SYS201311031035168647163014_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103103516864716325/SYS201311031035168647163014_ST/images1.png)
【答案】分析:由
T=π可求得ω,再由(-
)•ω+φ=0可求得φ,从而可得答案.
解答:解:由f(x)=sin(ωx+φ)的部分图象可知,
T=π,
∴T=4π,又T=
,
∴ω=
;
又(-
)×
+φ=0,
∴φ=
.
∴ω+ϕ=
+
.
故答案为:
+
.
点评:本题考查由y=Asin(ωx+φ)的部分图象确定其解析式,考查识图用图的能力,属于中档题.
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103103516864716325/SYS201311031035168647163014_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103103516864716325/SYS201311031035168647163014_DA/1.png)
解答:解:由f(x)=sin(ωx+φ)的部分图象可知,
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103103516864716325/SYS201311031035168647163014_DA/2.png)
∴T=4π,又T=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103103516864716325/SYS201311031035168647163014_DA/3.png)
∴ω=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103103516864716325/SYS201311031035168647163014_DA/4.png)
又(-
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103103516864716325/SYS201311031035168647163014_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103103516864716325/SYS201311031035168647163014_DA/6.png)
∴φ=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103103516864716325/SYS201311031035168647163014_DA/7.png)
∴ω+ϕ=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103103516864716325/SYS201311031035168647163014_DA/8.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103103516864716325/SYS201311031035168647163014_DA/9.png)
故答案为:
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103103516864716325/SYS201311031035168647163014_DA/10.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103103516864716325/SYS201311031035168647163014_DA/11.png)
点评:本题考查由y=Asin(ωx+φ)的部分图象确定其解析式,考查识图用图的能力,属于中档题.
![](http://thumb2018.1010pic.com/images/loading.gif)
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