【答案】
分析:(1)由條件求得
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/0.png)
≤tanθ≤1,再根據(jù)0≤θ≤π,從而求出θ的取值范圍.
(2)利用兩角和差的正弦公式、二倍角公式花簡函數(shù)f(θ)的解析式為2sin(2θ-
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/1.png)
)+2,根據(jù)
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/2.png)
≤θ≤
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/3.png)
,求得2θ-
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/4.png)
的范圍,從而求得sin(2θ-
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/5.png)
)的范圍,從而求出f(θ)的最大值和最小值.
解答:解:(1)因?yàn)?img src="http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/6.png">,
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/7.png)
與
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/8.png)
的夾角為θ,所以,
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/9.png)
.
S=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/10.png)
=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/11.png)
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/12.png)
. (3分)
又
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/13.png)
,所以,
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/14.png)
≤
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/15.png)
•tanθ≤
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/16.png)
,即
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/17.png)
≤tanθ≤1,
又0≤θ≤π,所以,
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/18.png)
≤θ≤
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/19.png)
. (6分)
(2)函數(shù)
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/20.png)
=2sin
2θ+
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/21.png)
sin2θ+1
=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/22.png)
sin2θ-cos2θ+2=2sin(2θ-
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/23.png)
)+2,----(9分)
因?yàn)?img src="http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/24.png">≤θ≤
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/25.png)
,所以
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/26.png)
≤2θ-
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/27.png)
≤
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/28.png)
,(10分)
從而當(dāng) θ=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/29.png)
時(shí),f(θ)取得最小值為3,
當(dāng) θ=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/30.png)
時(shí),f(θ)取得最大值為
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112203662892288/SYS201312021122036628922016_DA/31.png)
.---------(12分)
點(diǎn)評(píng):本題主要考查兩個(gè)向量的數(shù)量積的運(yùn)算,正弦函數(shù)的定義域和值域,兩角和差的正弦公式、二倍角公式的應(yīng)用,屬于中檔題.