题目内容
已知函数f(x)=sin(π-x)sin(
-x)+cos2x.
(Ⅰ)求函数f(x)的最小正周期;
(Ⅱ)当x∈[-
,
]时,求函数f(x)的单调区间.
| π |
| 2 |
(Ⅰ)求函数f(x)的最小正周期;
(Ⅱ)当x∈[-
| π |
| 8 |
| 3π |
| 8 |
(Ⅰ)f(x)=sinx•cosx+
cos2x+
=
sin2x+
cos2x+
=
sin(2x+
)+
∴函数f(x)的最小正周期T=
=π
(Ⅱ)当x∈[-
,
]时,2x+
∈[0,π]
∴当2x+
∈[0,
]即x∈[-
,
]时,函数f(x)单调递增
当2x+
∈[
,π]即x∈[
,
]时,函数f(x)单调递减
| 1 |
| 2 |
| 1 |
| 2 |
=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
=
| ||
| 2 |
| π |
| 4 |
| 1 |
| 2 |
∴函数f(x)的最小正周期T=
| 2π |
| 2 |
(Ⅱ)当x∈[-
| π |
| 8 |
| 3π |
| 8 |
| π |
| 4 |
∴当2x+
| π |
| 4 |
| π |
| 2 |
| π |
| 8 |
| π |
| 8 |
当2x+
| π |
| 4 |
| π |
| 2 |
| π |
| 8 |
| 3π |
| 8 |
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