题目内容
已知向量m=(2cosx,2sinx),n=(cosx,
cosx),设f(x)=m•n-1.
(I)求f(
)的值;
(Ⅱ)求函数f(x)的最小正周期单调递增区间.
| 3 |
(I)求f(
| π |
| 6 |
(Ⅱ)求函数f(x)的最小正周期单调递增区间.
(I)f(x)=2cos2x+2
sinxcosx-1=cos2x+
sin2x=2sin(2x+
),
∴f(
)=2sin(2×
+
)=2sin
=2;
(Ⅱ)∵ω=2,∴T=
=π,
∵2kπ-
≤2x+
≤2kπ+
,k∈Z,
∴kπ-
≤x≤kπ+
,k∈Z,
则函数f(x)的单调递增区间为[kπ-
,kπ+
],k∈Z.
| 3 |
| 3 |
| π |
| 6 |
∴f(
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 2 |
(Ⅱ)∵ω=2,∴T=
| 2π |
| 2 |
∵2kπ-
| π |
| 2 |
| π |
| 6 |
| π |
| 2 |
∴kπ-
| π |
| 3 |
| π |
| 6 |
则函数f(x)的单调递增区间为[kπ-
| π |
| 3 |
| π |
| 6 |
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