题目内容
已知向量
=(sin
,
cos
),
=(1,1),函数f(x)=
•
cos
.
(1)将f(x)写成Asin(ωx+φ)+B的形式,并求其图象的对称中心;
(2)如果△ABC的三边a、b、c满足b2=ac,且边b所对的角为x,试求x的取值范围及此时函数f(x)的值域.
| a |
| x |
| 3 |
| 3 |
| x |
| 3 |
| b |
| a |
| b |
| x |
| 3 |
(1)将f(x)写成Asin(ωx+φ)+B的形式,并求其图象的对称中心;
(2)如果△ABC的三边a、b、c满足b2=ac,且边b所对的角为x,试求x的取值范围及此时函数f(x)的值域.
(1)f(x)=sin
cos
+
cos2
=
sin
+
(1+cos
)
=
sin
+
cos
+
=sin(
+
)+
,
令sin(
+
)=0,即
+
=kπ(k∈Z),解得x=
π(k∈Z),
则对称中心为(
π,
)(k∈Z);
(2)∵b2=ac,
∴根据余弦定理得:cosx=
=
≥
=
,
∴
≤cosx<1,即0<x≤
,
∴
<
+
≤
,
∵|
-
|>|
-
|,
∴sin
<sin(
+
)≤1,
∴
<sin(
+
)+
≤1+
,
则x∈(0,
]时,函数f(x)的值域为(
,1+
].
| x |
| 3 |
| x |
| 3 |
| 3 |
| x |
| 3 |
=
| 1 |
| 2 |
| 2x |
| 3 |
| ||
| 2 |
| 2x |
| 3 |
=
| 1 |
| 2 |
| 2x |
| 3 |
| ||
| 2 |
| 2x |
| 3 |
| ||
| 2 |
=sin(
| 2x |
| 3 |
| π |
| 3 |
| ||
| 2 |
令sin(
| 2x |
| 3 |
| π |
| 3 |
| 2x |
| 3 |
| π |
| 3 |
| 3k-1 |
| 2 |
则对称中心为(
| 3k-1 |
| 2 |
| ||
| 2 |
(2)∵b2=ac,
∴根据余弦定理得:cosx=
| a2+c2-b2 |
| 2ac |
| a2+c2-ac |
| 2ac |
| 2ac-ac |
| 2ac |
| 1 |
| 2 |
∴
| 1 |
| 2 |
| π |
| 3 |
∴
| π |
| 3 |
| 2x |
| 3 |
| π |
| 3 |
| 5π |
| 9 |
∵|
| π |
| 3 |
| π |
| 2 |
| 5π |
| 9 |
| π |
| 2 |
∴sin
| π |
| 3 |
| 2x |
| 3 |
| π |
| 3 |
∴
| 3 |
| 2x |
| 3 |
| π |
| 3 |
| ||
| 2 |
| ||
| 2 |
则x∈(0,
| π |
| 3 |
| 3 |
| ||
| 2 |
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