【答案】
分析:把已知的等式兩邊平方,利用二倍角的正弦函數(shù)公式即可求出sin2α的值,然后在把已知的等式提取
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/0.png)
,利用兩角和與差的正弦函數(shù)公式及特殊角的三角函數(shù)值化為一個(gè)角的正弦函數(shù),根據(jù)正弦的值,判斷得到α的范圍,進(jìn)而得到2α的范圍,利用同角三角函數(shù)間的基本關(guān)系由sin2α的值和2α的范圍即可求出cos2a的值.
解答:解:把sina+cosa=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/1.png)
,兩邊平方得:1+2sinαcosα=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/2.png)
,
即1+sin2α=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/3.png)
,解得sin2α=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/4.png)
,
又sina+cosa=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/5.png)
sin(α+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/6.png)
)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/7.png)
,解得:sin(α+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/8.png)
)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/9.png)
<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/10.png)
,
得到:0<α+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/11.png)
<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/12.png)
(舍去)或
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/13.png)
<α+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/14.png)
<π,
解得:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/15.png)
<α<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/16.png)
,所以2α∈(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/17.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/18.png)
),
則cos2α=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/19.png)
=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181802886483407/SYS201310241818028864834013_DA/20.png)
.
故答案為:-
點(diǎn)評(píng):此題考查學(xué)生靈活運(yùn)用同角三角函數(shù)間的基本關(guān)系及兩角和與差的正弦函數(shù)公式化簡(jiǎn)求值,靈活運(yùn)用二倍角的正弦函數(shù)公式化簡(jiǎn)求值,是一道中檔題.求出2α的范圍確定出cos2α的正負(fù)是解題的關(guān)鍵.