【答案】
分析:首先利用诱导公式求出sinα=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101231806901374336/SYS201311012318069013743009_DA/0.png)
,由角的正弦值为正,判断角在第一和第二象限,又已知α为第二象限角,余弦值一定小于零,从而求出余弦值,用二倍角公式得到2α的正弦值.
解答:解:sin(π+α)=-sinα=-
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101231806901374336/SYS201311012318069013743009_DA/1.png)
∴sinα=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101231806901374336/SYS201311012318069013743009_DA/2.png)
∵sinα=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101231806901374336/SYS201311012318069013743009_DA/3.png)
,
∴α是第二象限角,
∴cosα<0,
∴cosα=-
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101231806901374336/SYS201311012318069013743009_DA/4.png)
,
∴sin2θ=2sinθcosθ=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101231806901374336/SYS201311012318069013743009_DA/5.png)
故答案为:
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101231806901374336/SYS201311012318069013743009_DA/6.png)
.
点评:已知一个角的某个三角函数式的值,求这个角的其他三角函数式的值,一般需用三个基本关系式及其变式,通过恒等变形或解方程求解,熟记二倍角的正弦、余弦、正切公式是解题的关键.