【答案】
分析:(1)法一:直接利用兩角差的余弦函數(shù)展開,再用方程兩邊平方,求sin2β的值;
法二:利用sin2β=cos(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/0.png)
-2β),二倍角公式,直接求出sin2β的值;
(2)通過題意求出sin(β-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/1.png)
)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/2.png)
,cos(α+β)=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/3.png)
,根據(jù)cos(α+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/4.png)
)=cos[(α+β)-(β-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/5.png)
)],展開代入數(shù)據(jù),即可求cos(α+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/6.png)
)的值.
解答:解:(1)法一:∵cos(β-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/7.png)
)=cos
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/8.png)
cosβ+sin
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/9.png)
sinβ
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/10.png)
cosβ+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/11.png)
sinβ=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/12.png)
.
∴cosβ+sinβ=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/13.png)
.
∴1+sin2β=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/14.png)
,∴sin2β=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/15.png)
.
法二:sin2β=cos(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/16.png)
-2β)
=2cos
2(β-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/17.png)
)-1=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/18.png)
.
(2)∵0<α<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/19.png)
<β<π,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/20.png)
<β-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/21.png)
<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/22.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/23.png)
<α+β<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/24.png)
.
∴sin(β-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/25.png)
)>0,cos(α+β)<0.
∵cos(β-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/26.png)
)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/27.png)
,sin(α+β)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/28.png)
,
∴sin(β-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/29.png)
)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/30.png)
,cos(α+β)=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/31.png)
.
∴cos(α+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/32.png)
)=cos[(α+β)-(β-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/33.png)
)]
=cos(α+β)cos(β-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/34.png)
)+sin(α+β)sin(β-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/35.png)
)
=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/36.png)
×
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/37.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/38.png)
×
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/39.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137836308/SYS201310232145271378363016_DA/40.png)
.
點評:本題是基礎(chǔ)題,考查三角函數(shù)的化簡與求值,角的變換技巧在三角函數(shù)化簡求值中應(yīng)用比較普遍,不僅體現(xiàn)一個人的解題能力,同時體現(xiàn)數(shù)學(xué)素養(yǎng)的高低,可以說是智慧與能力的展現(xiàn)題目.