【答案】
分析:本題考查的知識點是三角函數(shù)的定義,及倍角公式,由sinαcosβ+cosαsinβ=sin(α+β)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213102501556169/SYS201310232131025015561017_DA/0.png)
+cosαsinβ,sinαcosβ-cosαsinβ=sin(α-β)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213102501556169/SYS201310232131025015561017_DA/1.png)
-cosαsinβ,結合正弦函數(shù)的值域為[-1,1],解不等式組即可得到cosαsinβ的取值范圍.
解答:解:∵sinαcosβ+cosαsinβ=sin(α+β)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213102501556169/SYS201310232131025015561017_DA/2.png)
+cosαsinβ,
∴-1≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213102501556169/SYS201310232131025015561017_DA/3.png)
+cosαsinβ≤1
即-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213102501556169/SYS201310232131025015561017_DA/4.png)
≤cosαsinβ≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213102501556169/SYS201310232131025015561017_DA/5.png)
∵sinαcosβ-cosαsinβ=sin(α-β)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213102501556169/SYS201310232131025015561017_DA/6.png)
-cosαsinβ,
∴-1≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213102501556169/SYS201310232131025015561017_DA/7.png)
-cosαsinβ≤1
即-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213102501556169/SYS201310232131025015561017_DA/8.png)
≤cosαsinβ≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213102501556169/SYS201310232131025015561017_DA/9.png)
∴-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213102501556169/SYS201310232131025015561017_DA/10.png)
≤cosαsinβ≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213102501556169/SYS201310232131025015561017_DA/11.png)
∴cosαsinβ的取值范圍為[-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213102501556169/SYS201310232131025015561017_DA/12.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213102501556169/SYS201310232131025015561017_DA/13.png)
].
點評:觀察題目中已知與未知的量,并根據(jù)它們的關系選擇計算sinαcosβ+cosαsinβ=sin(α+β)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213102501556169/SYS201310232131025015561017_DA/14.png)
+cosαsinβ,sinαcosβ-cosαsinβ=sin(α-β)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213102501556169/SYS201310232131025015561017_DA/15.png)
-cosαsinβ,是解決本題的關鍵,要求大家熟練掌握三角函數(shù)的相關公式.