题目内容
已知sin(α+
)+sinα=-
,则cos(α+
)=
.
| π |
| 3 |
4
| ||
| 5 |
| 2π |
| 3 |
| 4 |
| 5 |
| 4 |
| 5 |
分析:利用两角和的正弦函数与辅助角公式将已知转化为
sin(α+
)=-
,从而可求得sin(α+
)=-
,再利用诱导公式可求得cos(α+
)的值.
| 3 |
| π |
| 6 |
4
| ||
| 5 |
| π |
| 6 |
| 4 |
| 5 |
| 2π |
| 3 |
解答:解:∵sin(α+
)+sinα
=sinαcos
+cosαsin
+sinα
=
sinα+
cosα+sinα
=
sinα+
cosα
=
(
sinα+
cosα)
=
sin(α+
)
=-
,
∴sin(α+
)=-
.
又α+
=(α+
)+
,
∴cos(α+
)=cos[(α+
)+
]=-sin(α+
)=
.
故答案为:
.
| π |
| 3 |
=sinαcos
| π |
| 3 |
| π |
| 3 |
=
| 1 |
| 2 |
| ||
| 2 |
=
| 3 |
| 2 |
| ||
| 2 |
=
| 3 |
| ||
| 2 |
| 1 |
| 2 |
=
| 3 |
| π |
| 6 |
=-
4
| ||
| 5 |
∴sin(α+
| π |
| 6 |
| 4 |
| 5 |
又α+
| 2π |
| 3 |
| π |
| 6 |
| π |
| 2 |
∴cos(α+
| 2π |
| 3 |
| π |
| 6 |
| π |
| 2 |
| π |
| 6 |
| 4 |
| 5 |
故答案为:
| 4 |
| 5 |
点评:本题考查两角和的正弦与余弦及辅助角公式,求得sin(α+
)=-
是关键,考查观察分析与运算能力,属于中档题.
| π |
| 6 |
| 4 |
| 5 |
练习册系列答案
相关题目
已知sin(α+
)+sinα=-
,-
<α<0,则cos(α+
)等于( )
| π |
| 3 |
4
| ||
| 5 |
| π |
| 2 |
| 2π |
| 3 |
A、-
| ||
B、-
| ||
C、
| ||
D、
|