题目内容
已知向量
=(cos
,sin
),
=(cos
,-sin
),x∈[-
,
]
(1)求证:(
-
)⊥(
+
);
(2)|
+
|=
,求cosx的值.
| a |
| 3x |
| 2 |
| 3x |
| 2 |
| b |
| x |
| 2 |
| x |
| 2 |
| π |
| 3 |
| π |
| 2 |
(1)求证:(
| a |
| b |
| a |
| b |
(2)|
| a |
| b |
| 1 |
| 3 |
(1)∵
=(cos
,sin
),
=(cos
,-sin
),
∴
2=cos2
+sin2
=1,
2=cos2
+sin2
=1,
∴(
-
)•(
+
)=
2-
2=0,
∴(
-
)⊥(
+
).
(2)∵|
+
|=
=
=
=
=
,
∴2+2cos2x=
,即cos2x=-
,
∴2cos2x-1=-
,
∴cos2x=
,
∵x∈[-
,
],
∴cosx=
.
| a |
| 3x |
| 2 |
| 3x |
| 2 |
| b |
| x |
| 2 |
| x |
| 2 |
∴
| a |
| 3x |
| 2 |
| 3x |
| 2 |
| b |
| x |
| 2 |
| x |
| 2 |
∴(
| a |
| b |
| a |
| b |
| a |
| b |
∴(
| a |
| b |
| a |
| b |
(2)∵|
| a |
| b |
(
|
|
=
1+2(cos
|
=
| 2+2cos2x |
=
| 1 |
| 3 |
∴2+2cos2x=
| 1 |
| 9 |
| 17 |
| 18 |
∴2cos2x-1=-
| 17 |
| 18 |
∴cos2x=
| 1 |
| 36 |
∵x∈[-
| π |
| 3 |
| π |
| 2 |
∴cosx=
| 1 |
| 6 |
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