【答案】
分析:(1)把角放在銳角三角形中,使一些運(yùn)算簡(jiǎn)單起來(lái),本題主要考查兩角和與差的正弦公式,根據(jù)分解后的結(jié)構(gòu)特點(diǎn),解方程組,做比得到結(jié)論.
(2)同角的三角函數(shù)之間的關(guān)系,換元解方程在直角三角形中,用定義求的結(jié)果
解答:(I)證明:∵sin(A+B)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810016_DA/0.png)
,sin(A-B)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810016_DA/1.png)
,
∴sinAcosB+cosAsinB=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810016_DA/2.png)
,sinAcosB-cosAsinB=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810016_DA/3.png)
,
∴sinAcosB=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810016_DA/4.png)
,cosAsinB=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810016_DA/5.png)
,
∴tanA=2tanB.
(2)解:∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810016_DA/6.png)
<A+B<π,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810016_DA/7.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810016_DA/8.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810016_DA/9.png)
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810016_DA/10.png)
,將tanA=2tanB代入上式并整理得2tan
2B-4tanB-1=0
解得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810016_DA/11.png)
,因?yàn)锽為銳角,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810016_DA/12.png)
,∴tanA=2tanB=2+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810016_DA/13.png)
.
設(shè)AB上的高為CD,則AB=AD+DB=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810016_DA/14.png)
,由AB=3得CD=2+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810016_DA/15.png)
故AB邊上的高為2+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810016_DA/16.png)
.
點(diǎn)評(píng):以銳角三角形為載體,應(yīng)用同角三角函數(shù)之間的關(guān)系,應(yīng)用兩角和與差的正弦公式,求解過(guò)程中應(yīng)用代數(shù)方法解題,構(gòu)造直角三角形用銳角三角函數(shù)解決問(wèn)題,這種問(wèn)題做起來(lái)有一定難度.