题目内容
已知向量
=(cosα,sinα),
=(cosβ,sinβ),|
-
|=
.
(1)求cos(α-β)的值;
(2)若0<α<
,AH⊥BE,且sinβ=-
,求sinα.
| a |
| b |
| a |
| b |
2
| ||
| 5 |
(1)求cos(α-β)的值;
(2)若0<α<
| π |
| 2 |
| 5 |
| 13 |
(1)∵
=(cosα,sinα),
=(cosβ,sinβ),
∴
-
=(cosα-cosβ,sinα-sinβ).
∵|
-
|=
,
∴
=
,即2-2cos(α-β)=
,
∴cos(α-β)=
.(7分)
(2)∵0<α<
, -
<β<0, ∴0<α-β<π,
∵cos(α-β)=
,∴sin(α-β)=
.
∵sinβ=-
,∴cosβ=
,
∴sinα=sin[(α-β)+β]
=sin(α-β)cosβ+cos(α-β)sinβ
=
•
+
•(-
)=
(14分)
| a |
| b |
∴
| a |
| b |
∵|
| a |
| b |
2
| ||
| 5 |
∴
| (cosα-cosβ)2+(sinα-sinβ)2 |
2
| ||
| 5 |
| 4 |
| 5 |
∴cos(α-β)=
| 3 |
| 5 |
(2)∵0<α<
| π |
| 2 |
| π |
| 2 |
∵cos(α-β)=
| 3 |
| 5 |
| 4 |
| 5 |
∵sinβ=-
| 5 |
| 13 |
| 12 |
| 13 |
∴sinα=sin[(α-β)+β]
=sin(α-β)cosβ+cos(α-β)sinβ
=
| 4 |
| 5 |
| 12 |
| 13 |
| 3 |
| 5 |
| 5 |
| 13 |
| 33 |
| 65 |
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