题目内容
已知向量
=(cos
,sin
),
=(cos
,-sin
),x∈[-
,
]
(1)求证:(
-
)⊥(
+
);
(2)|
+
|=
,求sin2x的值.
| a |
| 3x |
| 2 |
| 3x |
| 2 |
| b |
| x |
| 2 |
| x |
| 2 |
| π |
| 5 |
| 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(3分)
∴(
-
)•(
+
)=
2-
2=0(4分)
∴(
-
)⊥(
+
)(5分)
(2)∵|
+
|=
=
=
=
(8分)
又∵|
+
|=
∴cos2x=-
(10分)
∵x∈[-
,
]
∴2x∈[-
,π]
∴sin2x=
(12分)
| 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 |
又∵|
| a |
| b |
| 1 |
| 3 |
∴cos2x=-
| 17 |
| 18 |
∵x∈[-
| π |
| 5 |
| π |
| 2 |
∴2x∈[-
| 2π |
| 5 |
∴sin2x=
| ||
| 18 |
练习册系列答案
相关题目