题目内容
已知向量
=(cos
x,sin
x),
=(cos
,-sin
),x∈[0,
].
(1)求
•
及|
+
|;
(2)求函数f(x)=
•
-2|
+
|值域.
| a |
| 3 |
| 2 |
| 3 |
| 2 |
| b |
| x |
| 2 |
| x |
| 2 |
| π |
| 2 |
(1)求
| a |
| b |
| a |
| b |
(2)求函数f(x)=
| a |
| b |
| a |
| b |
(1)
•
=cos
x•cos
x-sin
x•sin
x=cos(
x+
x)=cos2x.
∵(
+
)2=(cos
x+cos
x)2+(sin
x-sin
x)2=2+2(cos
x•cos
x-sin
x•sin
x)
=2+2cos2x=2+2(2cos2x-1)=4cos2x
且x∈[0,
]
∴|
+
|=2cosx.
(2)由(1)知f(x)=
•
-2|
+
|=cos2x-4cosx
=2cos2x-4cosx-1=2(cosx-1)2-3
∵x∈[0,
]∴cosx∈[0,1]
∴函数f(x)=
•
-2|
+
|值域是[-3,-1].
| a |
| b |
| 3 |
| 2 |
| 1 |
| 2 |
| 3 |
| 2 |
| 1 |
| 2 |
| 3 |
| 2 |
| 1 |
| 2 |
∵(
| a |
| b |
| 3 |
| 2 |
| 1 |
| 2 |
| 3 |
| 2 |
| 1 |
| 2 |
| 3 |
| 2 |
| 1 |
| 2 |
| 3 |
| 2 |
| 1 |
| 2 |
=2+2cos2x=2+2(2cos2x-1)=4cos2x
且x∈[0,
| π |
| 2 |
∴|
| a |
| b |
(2)由(1)知f(x)=
| a |
| b |
| a |
| b |
=2cos2x-4cosx-1=2(cosx-1)2-3
∵x∈[0,
| π |
| 2 |
∴函数f(x)=
| a |
| b |
| a |
| b |
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