题目内容
设α∈(0,
),若cos(α+
)=
,则sin(2α+
)的值为 .
| π |
| 2 |
| π |
| 6 |
| 4 |
| 5 |
| π |
| 12 |
考点:两角和与差的正弦函数,同角三角函数间的基本关系,两角和与差的余弦函数
专题:三角函数的求值
分析:由同角三角函数基本关系可得sin(α+
),由二倍角公式可得sin2(α+
)和cos2(α+
),而sin(2α+
)=sin[2(α+
)-
]=
sin2(α+
)-
cos2(α+
),代值计算可得.
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 12 |
| π |
| 6 |
| π |
| 4 |
| ||
| 2 |
| π |
| 6 |
| ||
| 2 |
| π |
| 6 |
解答:
解:∵α∈(0,
),cos(α+
)=
,
∴sin(α+
)=
=
,
∴sin2(α+
)=2sin(α+
)cos(α+
)=
,
cos2(α+
)=cos2(α+
)-sin2(α+
)=
,
∴sin(2α+
)=sin[2(α+
)-
]
=
sin2(α+
)-
cos2(α+
)
=
×
-
×
=
.
故答案为:
.
| π |
| 2 |
| π |
| 6 |
| 4 |
| 5 |
∴sin(α+
| π |
| 6 |
1-cos2(α+
|
| 3 |
| 5 |
∴sin2(α+
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| 24 |
| 25 |
cos2(α+
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| 7 |
| 25 |
∴sin(2α+
| π |
| 12 |
| π |
| 6 |
| π |
| 4 |
=
| ||
| 2 |
| π |
| 6 |
| ||
| 2 |
| π |
| 6 |
=
| ||
| 2 |
| 24 |
| 25 |
| ||
| 2 |
| 7 |
| 25 |
17
| ||
| 50 |
故答案为:
17
| ||
| 50 |
点评:本题考查两角和与差的三角函数,涉及二倍角公式,属中档题.
练习册系列答案
相关题目
已知m、n是两条不同的直线,α、β、γ是三个不同的平面,则下列命题正确的是( )
| A、若α⊥γ,α⊥β,则γ∥β |
| B、若m∥n,m?α,n?β,则α∥β |
| C、若m∥n,m∥α,则n∥α |
| D、若 m∥α,m?β,α∩β=n,则m∥n |