【答案】
分析:(I)根據(jù)函數(shù)是一個(gè)齊次式,利用二倍角公式進(jìn)行化簡,再利用兩角和與差的正弦公式化簡成Asin(ωx+φ)+B的形式,最后解三角方程即可;
(II)根據(jù)(I)化簡得到的函數(shù)解析式可直接求出函數(shù)的最值,特別要注意定義域.
解答:解:(Ⅰ)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122730313656875/SYS201310251227303136568015_DA/0.png)
.
令f(x)=0,得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122730313656875/SYS201310251227303136568015_DA/1.png)
.
因?yàn)?img src="http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122730313656875/SYS201310251227303136568015_DA/2.png">,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122730313656875/SYS201310251227303136568015_DA/3.png)
.…(4分)
所以,當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122730313656875/SYS201310251227303136568015_DA/4.png)
,或
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122730313656875/SYS201310251227303136568015_DA/5.png)
時(shí),f(x)=0.
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122730313656875/SYS201310251227303136568015_DA/6.png)
或x=π時(shí),f(x)=0.
綜上,函數(shù)f(x)的零點(diǎn)為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122730313656875/SYS201310251227303136568015_DA/7.png)
或π.…(10分)
(Ⅱ)由(Ⅰ)可知,
當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122730313656875/SYS201310251227303136568015_DA/8.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122730313656875/SYS201310251227303136568015_DA/9.png)
時(shí),f(x)的最大值為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122730313656875/SYS201310251227303136568015_DA/10.png)
;
當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122730313656875/SYS201310251227303136568015_DA/11.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122730313656875/SYS201310251227303136568015_DA/12.png)
時(shí),f(x)的最小值為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122730313656875/SYS201310251227303136568015_DA/13.png)
.…(12分)
點(diǎn)評:本題主要考查了兩角和與差的正弦函數(shù),以及正弦函數(shù)的值域,同時(shí)考查了計(jì)算能力和轉(zhuǎn)化的能力,屬于中檔題.