题目内容
设sin(
+θ)=
,则sin2θ等于( )
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824035919055302.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824035919071301.png)
A.-![]() | B.![]() | C.![]() | D.![]() |
A
方法一:利用公式:(sinθ+cosθ)2=1+
2sinθcosθ=1+sin2θ.
由sin(
+θ)=
,得
(sinθ+cosθ)=
,
化简得sinθ+cosθ=
.
两边平方得1+sin2θ=
.
从而sin2θ=-
,
方法二:变角利用二倍角余弦公式:cos2θ=1-2sin2θ.
sin2θ=-cos(
+2θ)
=-cos[2(
+θ)]
=2sin2(
+θ)-1=-
.
2sinθcosθ=1+sin2θ.
由sin(
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824035919149322.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824035919071301.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824035919180376.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824035919071301.png)
化简得sinθ+cosθ=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824035919102368.png)
两边平方得1+sin2θ=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824035919118306.png)
从而sin2θ=-
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824035919087309.png)
方法二:变角利用二倍角余弦公式:cos2θ=1-2sin2θ.
sin2θ=-cos(
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824035919274313.png)
=-cos[2(
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824035919149322.png)
=2sin2(
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824035919149322.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824035919087309.png)
![](http://thumb.zyjl.cn/images/loading.gif)
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