题目内容
设α为锐角,若cos(α+
)=
,则sin(α-
)= .
| π |
| 6 |
| 3 |
| 5 |
| π |
| 12 |
考点:两角和与差的正弦函数,两角和与差的余弦函数
专题:三角函数的求值
分析:根据题意求得sin(α+
)=
,再根据sin(α-
)=sin[(α+
)-
],再利用两角差的正弦公式计算求得结果.
| π |
| 6 |
| 4 |
| 5 |
| π |
| 12 |
| π |
| 6 |
| π |
| 4 |
解答:
解:∵α为锐角,cos(α+
)=
为正数,
∴α+
是锐角,sin(α+
)=
,
∴sin(α-
)=sin[(α+
)-
]
=sin(α+
)cos
-cos(α+
)sin
=
×
-
×
=
,
故答案为:
.
| π |
| 6 |
| 3 |
| 5 |
∴α+
| π |
| 6 |
| π |
| 6 |
| 4 |
| 5 |
∴sin(α-
| π |
| 12 |
| π |
| 6 |
| π |
| 4 |
=sin(α+
| π |
| 6 |
| π |
| 4 |
| π |
| 6 |
| π |
| 4 |
=
| 4 |
| 5 |
| ||
| 2 |
| 3 |
| 5 |
| ||
| 2 |
| ||
| 10 |
故答案为:
| ||
| 10 |
点评:本题着重考查了两角和与差的正弦公式,考查了三角函数中的恒等变换应用,属于中档题.
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