题目内容
设平面向量
=(cos2
,
sinx),
=(2,1),函数f(x)=
•
.
(Ⅰ)当x∈[-
,
]时,求函数f(x)的取值范围;
(Ⅱ)当f(α)=
,且-
<α<
时,求sin(2α+
)的值.
| m |
| x |
| 2 |
| 3 |
| n |
| m |
| n |
(Ⅰ)当x∈[-
| π |
| 3 |
| π |
| 2 |
(Ⅱ)当f(α)=
| 13 |
| 5 |
| 2π |
| 3 |
| π |
| 6 |
| π |
| 3 |
解析:(Ⅰ)∵
=(cos2
,
sinx),
=(2,1),
∴f(x)=(cos2
,
sinx)•(2,1)=2cos2
+
sinx
=cosx+
sinx+1=2sin(x+
)+1.
当x∈[-
,
]时,x+
∈[-
,
],
则-
≤sin(x+
)≤1,0≤2sin(x+
)+1≤3,
∴f(x)的取值范围是[0,3];
(Ⅱ)由f(α)=2sin(α+
)+1=
,得sin(α+
)=
,
∵-
<α<
,
∴-
<α+
<
,得cos(α+
)=
,
∴sin(2α+
)=sin[2(α+
)]=2sin(α+
)cos(α+
)=2×
×
=
.
| m |
| x |
| 2 |
| 3 |
| n |
∴f(x)=(cos2
| x |
| 2 |
| 3 |
| x |
| 2 |
| 3 |
=cosx+
| 3 |
| π |
| 6 |
当x∈[-
| π |
| 3 |
| π |
| 2 |
| π |
| 6 |
| π |
| 6 |
| 2π |
| 3 |
则-
| 1 |
| 2 |
| π |
| 6 |
| π |
| 6 |
∴f(x)的取值范围是[0,3];
(Ⅱ)由f(α)=2sin(α+
| π |
| 6 |
| 13 |
| 5 |
| π |
| 6 |
| 4 |
| 5 |
∵-
| 2π |
| 3 |
| π |
| 6 |
∴-
| π |
| 2 |
| π |
| 6 |
| π |
| 3 |
| π |
| 6 |
| 3 |
| 5 |
∴sin(2α+
| π |
| 3 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| 4 |
| 5 |
| 3 |
| 5 |
| 24 |
| 25 |
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