题目内容
设α∈(
,
),β∈(0,
),cos(α-
)=
,sin(
+β)=
,则sin (α+β)=
.
| π |
| 4 |
| 3π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| 3 |
| 5 |
| 3π |
| 4 |
| 5 |
| 13 |
| 56 |
| 65 |
| 56 |
| 65 |
分析:先根据角的范围和同角三角函数的关系求出sin(α-
)与cos(
+β)的值,然后利用诱导公式得到sin (α+β)=-cos(α+β+
)=-cos[(α-
)+(
+β)],最后利用两角和的余弦函数可求出所求.
| π |
| 4 |
| 3π |
| 4 |
| π |
| 2 |
| π |
| 4 |
| 3π |
| 4 |
解答:解:∵α∈(
,
),β∈(0,
),cos(α-
)=
,sin(
+β)=
,
∴sin(α-
)=
,cos(
+β)=-
∴sin (α+β)=-cos(α+β+
)=-cos[(α-
)+(
+β)]
=sin(α-
)sin(
+β)-cos(α-
)cos(
+β)
=
×
-
×(-
)
=
故答案为:
| π |
| 4 |
| 3π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| 3 |
| 5 |
| 3π |
| 4 |
| 5 |
| 13 |
∴sin(α-
| π |
| 4 |
| 4 |
| 5 |
| 3π |
| 4 |
| 12 |
| 13 |
∴sin (α+β)=-cos(α+β+
| π |
| 2 |
| π |
| 4 |
| 3π |
| 4 |
=sin(α-
| π |
| 4 |
| 3π |
| 4 |
| π |
| 4 |
| 3π |
| 4 |
=
| 4 |
| 5 |
| 5 |
| 13 |
| 3 |
| 5 |
| 12 |
| 13 |
=
| 56 |
| 65 |
故答案为:
| 56 |
| 65 |
点评:本题主要考查了两角和与差的正弦函数余弦函数,以及同角三角函数,配角是解题的关键,属于中档题.
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