题目内容
已知向量
=(cosx,sinx),
=(sinx,cosx),且x∈[0,
],
(1)求
•
的取值范围;
(2)求证|
+
|=2sin(x+
);
(3)求函数f(x)=
•
-
|
+
|的取值范围.
| a |
| b |
| π |
| 2 |
(1)求
| a |
| b |
(2)求证|
| a |
| b |
| π |
| 4 |
(3)求函数f(x)=
| a |
| b |
| 2 |
| a |
| b |
(1)∵
•
=sinx•cosx+sinx•cosx=2sinx•cosx=sin2x (2′)
∵x∈[0,
],
∴2x∈[0,π]
∴
•
∈[0,1](4′)
(2)证明:∵
+
=(cos+sinx,sinx+cosx)
∴|
+
|=
(6')
=
=2|sin(x+
)|
∵x∈[0,
],
∴x+
∈[
,
],
∴sin(x+
)>0,
∴2|sin(x+
)|=2sin(x+
),
∴|
+
|=2sin(x+
).(8')
(3)∵x∈[0,
],
∴x+
∈[
,
]
∴f(x)=
•
-
|
+
|
=sin2x-2
sin(x+
)
=2sinxcosx-2(sinx+cosx)(9')
解法1:令t=sinx+cosx
∴sinx•cosx=
(1≤t≤
)
∴y=t2-1-2t(10')
=(t-1)2-2
∴y∈[-2,1-2
](12')
解法2:f(x)=sin2x-2
sin(x+
)(9')
=-cos[2(x+
)]-2
sin(x+
)
=2sin2(x+
)-2
sin(x+
)-1(10')
∵
≤sin(x+
)≤1
∴f(x)∈[-2,1-2
](12')
| a |
| b |
∵x∈[0,
| π |
| 2 |
∴2x∈[0,π]
∴
| a |
| b |
(2)证明:∵
| a |
| b |
∴|
| a |
| b |
| 2(cosx+sinx)2 |
=
2[
|
| π |
| 4 |
∵x∈[0,
| π |
| 2 |
∴x+
| π |
| 4 |
| π |
| 4 |
| 3π |
| 4 |
∴sin(x+
| π |
| 4 |
∴2|sin(x+
| π |
| 4 |
| π |
| 4 |
∴|
| a |
| b |
| π |
| 4 |
(3)∵x∈[0,
| π |
| 2 |
∴x+
| π |
| 4 |
| π |
| 4 |
| 3π |
| 4 |
∴f(x)=
| a |
| b |
| 2 |
| a |
| b |
=sin2x-2
| 2 |
| π |
| 4 |
=2sinxcosx-2(sinx+cosx)(9')
解法1:令t=sinx+cosx
∴sinx•cosx=
| t2-1 |
| 2 |
| 2 |
∴y=t2-1-2t(10')
=(t-1)2-2
∴y∈[-2,1-2
| 2 |
解法2:f(x)=sin2x-2
| 2 |
| π |
| 4 |
=-cos[2(x+
| π |
| 4 |
| 2 |
| π |
| 4 |
=2sin2(x+
| π |
| 4 |
| 2 |
| π |
| 4 |
∵
| ||
| 2 |
| π |
| 4 |
∴f(x)∈[-2,1-2
| 2 |
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