题目内容
已知cos(α+
)=
,α∈(0,
),则cosα= .
| π |
| 4 |
| 1 |
| 3 |
| π |
| 2 |
考点:两角和与差的余弦函数
专题:三角函数的求值
分析:由同角三角函数的基本关系可得sin(α+
),而cosα=cos[(α+
)-
]=
cos(α+
)+
sin(α+
),代值化简即可.
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| ||
| 2 |
| π |
| 4 |
| ||
| 2 |
| π |
| 4 |
解答:
解:∵α∈(0,
),
∴α+
∈(
,
)
又∵cos(α+
)=
,
∴sin(α+
)=
=
,
∴cosα=cos[(α+
)-
]
=
cos(α+
)+
sin(α+
)
=
×
+
×
=
故答案为:
| π |
| 2 |
∴α+
| π |
| 4 |
| π |
| 4 |
| 3π |
| 4 |
又∵cos(α+
| π |
| 4 |
| 1 |
| 3 |
∴sin(α+
| π |
| 4 |
1-(
|
2
| ||
| 3 |
∴cosα=cos[(α+
| π |
| 4 |
| π |
| 4 |
=
| ||
| 2 |
| π |
| 4 |
| ||
| 2 |
| π |
| 4 |
=
| ||
| 2 |
| 1 |
| 3 |
| ||
| 2 |
2
| ||
| 3 |
4+
| ||
| 6 |
故答案为:
4+
| ||
| 6 |
点评:本题参考两角和与差的余弦公式,涉及同角三角函数的基本关系,属基础题.
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