题目内容
已知sin(
+α)=
,α∈(0,
),求
的值.
| π |
| 4 |
| 12 |
| 13 |
| π |
| 4 |
cos(α-
| ||
| cos2α |
∵α∈(0,
),∴
+α∈(
,
)
∴cos(
+α)=
=
(3分)
∴sinα=sin[(
+α)-
]=sin(
+α)cos
-cos(
+α)sin
=
(6分)
∴cos2α=1-2sin2α=
(9分)
∵cos(α-
)=cos(
-α)=sin(
+α)=
(11分)
∴
=
(12分)
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| π |
| 2 |
∴cos(
| π |
| 4 |
1-sin2(
|
| 5 |
| 13 |
∴sinα=sin[(
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
7
| ||
| 26 |
∴cos2α=1-2sin2α=
| 120 |
| 169 |
∵cos(α-
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| 12 |
| 13 |
∴
cos(α-
| ||
| cos2α |
| 13 |
| 10 |
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