题目内容
已知向量
=(
,-sin
),
=(sin
x,2sin
),函数f(x)=
•
.
(1)求f(x)的对称轴方程;
(2)若0<α<
且f(α)=
,求f(α+
)的值.
| a |
| 3 |
| x |
| 3 |
| b |
| 2 |
| 3 |
| x |
| 3 |
| a |
| b |
(1)求f(x)的对称轴方程;
(2)若0<α<
| π |
| 2 |
| 3 |
| 5 |
| 3π |
| 8 |
分析:(1)由题意可得函数f(x)的解析式,由整体法可得对称轴;
(2)由(1)可得sin(
+
)=
,进而可得cos(
+
),而f(α+
)=2sin(
+
+
)-1=2sin[(
+
)+
]-1,由两角和与差的公式可得答案.
(2)由(1)可得sin(
| 2α |
| 3 |
| π |
| 6 |
| 4 |
| 5 |
| 2α |
| 3 |
| π |
| 6 |
| 3π |
| 8 |
| 2α |
| 3 |
| π |
| 4 |
| π |
| 6 |
| 2α |
| 3 |
| π |
| 6 |
| π |
| 4 |
解答:解:(1)由题意可得:函数f(x)=
•
=
sin
-2sin2
=
sin
+cos
-1=2sin(
+
)-1,
由
+
=kπ+
,k∈Z可得x=
kπ+
.
故f(x)的对称轴方程为:x=
kπ+
,k∈Z
(2)由(1)知:2sin(
+
)-1=
,解得sin(
+
)=
,
结合0<α<
可得cos(
+
)=
.
而f(α+
)=2sin(
+
+
)-1=2sin[(
+
)+
]-1
=2sin(
+
)cos
+2cos(
+
)sin
-1
=2×
×
+2×
×
-1
=
-1
| a |
| b |
| 3 |
| 2x |
| 3 |
| x |
| 3 |
=
| 3 |
| 2x |
| 3 |
| 2x |
| 3 |
| 2x |
| 3 |
| π |
| 6 |
由
| 2x |
| 3 |
| π |
| 6 |
| π |
| 2 |
| 3 |
| 2 |
| π |
| 2 |
故f(x)的对称轴方程为:x=
| 3 |
| 2 |
| π |
| 2 |
(2)由(1)知:2sin(
| 2α |
| 3 |
| π |
| 6 |
| 3 |
| 5 |
| 2α |
| 3 |
| π |
| 6 |
| 4 |
| 5 |
结合0<α<
| π |
| 2 |
| 2α |
| 3 |
| π |
| 6 |
| 3 |
| 5 |
而f(α+
| 3π |
| 8 |
| 2α |
| 3 |
| π |
| 4 |
| π |
| 6 |
| 2α |
| 3 |
| π |
| 6 |
| π |
| 4 |
=2sin(
| 2α |
| 3 |
| π |
| 6 |
| π |
| 4 |
| 2α |
| 3 |
| π |
| 6 |
| π |
| 4 |
=2×
| 4 |
| 5 |
| ||
| 2 |
| 3 |
| 5 |
| ||
| 2 |
=
7
| ||
| 5 |
点评:本题为三角函数和向量的数量积的结合,两角和与差的三角函数公式是解决问题的关键,属中档题.
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