题目内容
已知cos(α+
)=
,其中α∈(0,
),则sinα=( )
| π |
| 4 |
| 1 |
| 3 |
| π |
| 2 |
分析:由已知结合同角三角函数的基本关系可得sinα,而sinα=sin[(α+
)-
],由两角差的正弦公式计算可得.
| π |
| 4 |
| π |
| 4 |
解答:解:∵α∈(0,
),∴α+
∈(
,
),
又因为cos(α+
)=
,∴sin(α+
)=
=
,
故sinα=sin[(α+
)-
]=sin(α+
)cos
-cos(α+
)sin
=
×
-
×
=
故选A
| π |
| 2 |
| π |
| 4 |
| π |
| 4 |
| 3π |
| 4 |
又因为cos(α+
| π |
| 4 |
| 1 |
| 3 |
| π |
| 4 |
1-(
|
2
| ||
| 3 |
故sinα=sin[(α+
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
=
2
| ||
| 3 |
| ||
| 2 |
| 1 |
| 3 |
| ||
| 2 |
4-
| ||
| 6 |
故选A
点评:本题考查两角差的正弦公式,把α表示为[(α+
)-
]是解决问题的关键,属中档题.
| π |
| 4 |
| π |
| 4 |
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