题目内容
已知A (-sin
, sin
),B ( sin
, -2 cos
),C ( cos
, 0 )三点.
(1)求向量
和向量
的坐标;
(2)设f(x)=
•
,求f(x)的最小正周期;
(3)求f(x)的单调递减区间.
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
(1)求向量
| AC |
| BC |
(2)设f(x)=
| AC |
| BC |
(3)求f(x)的单调递减区间.
分析:(1)由已知点的坐标,由向量的坐标的定义可得;(2)可得f(x)=
•
=(cos
+sin
)•(cos
-sin
)+(-sin
)•2cos
,由三角函数的运算法则化简可得周期;(3)由2kπ≤x+
≤π+2kπ,k∈Z,解不等式可得单调递减区间.
| AC |
| BC |
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
| π |
| 4 |
解答:解:(1)∵A (-sin
, sin
),B ( sin
, -2 cos
),C ( cos
, 0 ),
∴
=(cos
+sin
,-sin
),
=(cos
-sin
,2cos
),
(2)由(1)知:f(x)=
•
=(cos
+sin
)•(cos
-sin
)+(-sin
)•2cos
=cos2
-sin2
-2sin
cos
=cosx-sinx
=
(cosx•
-sinx•
)=
cos(x+
),
∴f(x)的最小正周期T=2π
(3)由2kπ≤x+
≤π+2kπ,k∈Z,可得
-
+2kπ≤x≤
+2kπ,k∈Z.
∴f(x)的单调递减区间是[-
+2kπ,
+2kπ](k∈Z)
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
∴
| AC |
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
| BC |
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
(2)由(1)知:f(x)=
| AC |
| BC |
=(cos
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
=cos2
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
=
| 2 |
| ||
| 2 |
| ||
| 2 |
| 2 |
| π |
| 4 |
∴f(x)的最小正周期T=2π
(3)由2kπ≤x+
| π |
| 4 |
-
| π |
| 4 |
| 3π |
| 4 |
∴f(x)的单调递减区间是[-
| π |
| 4 |
| 3π |
| 4 |
点评:本题考查平面向量数量积的运算,涉及三角函数的周期性和单调性,属中档题.
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