题目内容
已知向量
=(sinx,0),
=(cosx,1),其中 0<x<
,求|
-
|的取值范围.
| a |
| b |
| 2π |
| 3 |
| 1 |
| 2 |
| a |
| ||
| 2 |
| b |
∵向量
=(sinx,0),
=(cosx,1),
∴|
-
|2=|(
cosx-
sinx,
)|2(2分)
=(
cosx-
sinx)2+
(3分)
=sin2(x-
)+
.(3分)
0<x<
,∴-
<x-
<
,(2分)
∴0≤sin2(C-
)<
,(2分)
得|
-
|∈[
,
).(2分)
| a |
| b |
∴|
| 1 |
| 2 |
| a |
| ||
| 2 |
| b |
| ||
| 2 |
| 1 |
| 2 |
| ||
| 2 |
=(
| ||
| 2 |
| 1 |
| 2 |
| 3 |
| 4 |
=sin2(x-
| π |
| 3 |
| 3 |
| 4 |
0<x<
| 2π |
| 3 |
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
∴0≤sin2(C-
| π |
| 3 |
| 3 |
| 4 |
得|
| 1 |
| 2 |
| a |
| ||
| 2 |
| b |
| ||
| 2 |
| ||
| 2 |
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