题目内容
已知函数![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_ST/0.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_ST/1.png)
(Ⅱ)设△ABC的三边a,b,c满足b2=ac且边b所对的角θ的取值集合为A,当x∈A时,求f(x)的值域.
【答案】分析:(Ⅰ)利用两角和与差的余弦、正弦函数以及二倍角公式公式,化简函数f(x)=sin(ωx+
)+sin(ωx-
)-2cos2![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/2.png)
为:f(x)=
,然后利用在
处取得最大值,求出最小正整数ω的值.
(Ⅱ)设△ABC的三边a,b,c满足b2=ac,利用余弦定理、基本不等式求出a=c,推出θ的范围,利用三角函数的有界性,求f(x)的值域.
解答:解:(Ⅰ)
=![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/6.png)
由题意得
,k∈Z,得ω=6k+2,k∈Z
当k=0时,最小正整数ω的值为2,故ω=2.
(Ⅱ)因b2=ac且b2=a2+c2-2accosθ
则
当且仅当
,a=c时,等号成立
则
,又因θ∈(0,π),则
,即![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/12.png)
由①知:![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/13.png)
因
,则
,
-2<f(x)≤1,
故函数f(x)的值域为:(-2,1].
点评:本题是基础题,考查两角和与差的正弦函数、余弦函数以及二倍角的应用,函数的性质,最值的求法,处理相关的多个问题时,前一问的解答是后边解答的依据,考查学生的细心程度,计算能力.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/0.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/1.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/2.png)
为:f(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/3.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/4.png)
(Ⅱ)设△ABC的三边a,b,c满足b2=ac,利用余弦定理、基本不等式求出a=c,推出θ的范围,利用三角函数的有界性,求f(x)的值域.
解答:解:(Ⅰ)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/5.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/6.png)
由题意得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/7.png)
当k=0时,最小正整数ω的值为2,故ω=2.
(Ⅱ)因b2=ac且b2=a2+c2-2accosθ
则
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/8.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/9.png)
则
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/10.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/11.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/12.png)
由①知:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/13.png)
因
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/14.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/15.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213113608850901/SYS201310232131136088509014_DA/16.png)
故函数f(x)的值域为:(-2,1].
点评:本题是基础题,考查两角和与差的正弦函数、余弦函数以及二倍角的应用,函数的性质,最值的求法,处理相关的多个问题时,前一问的解答是后边解答的依据,考查学生的细心程度,计算能力.
![](http://thumb.zyjl.cn/images/loading.gif)
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