题目内容
若sin(
-α)=-
,α∈(-
,
π),则cos2α=
.
| π |
| 6 |
| 1 |
| 3 |
| π |
| 3 |
| 2 |
| 3 |
7-4
| ||
| 18 |
7-4
| ||
| 18 |
分析:由α∈(-
,
π),sin(
-α)=-
,可求得cosα,再由二倍角的余弦即可求得cos2α.
| π |
| 3 |
| 2 |
| 3 |
| π |
| 6 |
| 1 |
| 3 |
解答:解:∵sin(
-α)=-
,
∴sin(α-
)=
,
∵α∈(-
,
π),
∴α-
∈(-
,
),
∴cos(α-
)=
,
∴cosα=cos[(α-
)+
]
=cos(α-
)cos
-sin(α-
)sin
=
×
-
×
=
.
∴cos2α=2cos2α-1
=2×
-1
=
.
故答案为:
.
| π |
| 6 |
| 1 |
| 3 |
∴sin(α-
| π |
| 6 |
| 1 |
| 3 |
∵α∈(-
| π |
| 3 |
| 2 |
| 3 |
∴α-
| π |
| 6 |
| π |
| 2 |
| π |
| 2 |
∴cos(α-
| π |
| 6 |
2
| ||
| 3 |
∴cosα=cos[(α-
| π |
| 6 |
| π |
| 6 |
=cos(α-
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
=
2
| ||
| 3 |
| ||
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
=
2
| ||
| 6 |
∴cos2α=2cos2α-1
=2×
24-4
| ||
| 36 |
=
7-4
| ||
| 18 |
故答案为:
7-4
| ||
| 18 |
点评:本题考查二倍角的余弦,考查“拼凑角”的技巧与运算能力,属于中档题.
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