题目内容

(1)已知tanx=-2,求下列各式的值:①
cosx+sinxsinx-cosx
;②2sin2x-3cos2x.
(2)求值:sin(-1071°)sin99°+sin(-171°)sin(-261°)-2sin(-420°)+tan(-330°).
分析:(1)把已知tanx=-2代入 ①
cosx+sinx
sinx-cosx
=
1+tanx
tanx-1
,运算求得结果.把已知tanx=-2代入 ②2sin2x-3cos2x=
2sin2x-3cos2
cos2x+sin2x
=
2tan2x-3
1+tan2x
,运算求得结果.
(2)利用诱导公式把要求的式子化为sin9°cos9°-sin9°sin99°+2sin60°+tan30°,运算求得结果.
解答:解:(1)∵已知tanx=-2,∴①
cosx+sinx
sinx-cosx
=
1+tanx
tanx-1
=
-1
-3
=
1
3

②2sin2x-3cos2x=
2sin2x-3cos2
cos2x+sin2x
=
2tan2x-3
1+tan2x
=
5
5
=1.
(2)sin(-1071°)sin99°+sin(-171°)sin(-261°)-2sin(-420°)+tan(-330°)
=sin(-3×360°+9°)cos9°+sin(9°-180°)sin(-360°+99°)-2sin(-360°-60°)+tan(-360°+30°)
=sin9°cos9°-sin9°sin99°+2sin60°+tan30°=2sin60°+tan30°=
3
+
3
3
=
4
3
3
点评:本题主要考查同角三角函数的基本关系、诱导公式的应用,属于中档题.
练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网