题目内容
设向量
=(1+cosα,sinα),
=(1-cosβ,sinβ),
=(1,0),其中α∈(0,π),β∈(π,2π),
与
的夹角为θ1,
与
的夹角为θ2,且θ1-θ2=
,求sin
的值.
| a |
| b |
| c |
| a |
| c |
| b |
| c |
| π |
| 6 |
| α-β |
| 2 |
由题意可得
=2cos
•(cos
,sin
),
同理
=2sin
•(sin
,cos
).
又α∈(0,π),β∈(π,2π),
∴0<
<
,
<
<π.
∴|
|=2cos
,|
|=2sin
…4′
∴cosθ1=
=
=2cos
,
cosθ2=
=2sin
=cos(
-
).…8′
∵
、
-
∈(0,
),∴θ1=
,θ2=
-
.
∴
=θ1-θ2=
+
,即
=-
,
∴sin(
)=-
.…12′.
| a |
| α |
| 2 |
| α |
| 2 |
| α |
| 2 |
同理
| b |
| β |
| 2 |
| β |
| 2 |
| β |
| 2 |
又α∈(0,π),β∈(π,2π),
∴0<
| α |
| 2 |
| π |
| 2 |
| π |
| 2 |
| β |
| 2 |
∴|
| a |
| α |
| 2 |
| b |
| β |
| 2 |
∴cosθ1=
| ||||
|
|
2cos2
| ||
2cos
|
| α |
| 2 |
cosθ2=
| ||||
|
|
| β |
| 2 |
| β |
| 2 |
| π |
| 2 |
∵
| α |
| 2 |
| β |
| 2 |
| π |
| 2 |
| π |
| 2 |
| α |
| 2 |
| β |
| 2 |
| π |
| 2 |
∴
| π |
| 6 |
| π |
| 2 |
| α-β |
| 2 |
| α-β |
| 2 |
| π |
| 3 |
∴sin(
| α-β |
| 2 |
| 1 |
| 2 |
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