题目内容
已知函数f(x)=sin(2x+
)+sin(2x-
)+2cos2x-1,x∈R.
(1)求函数f(x)的最小正周期;
(2)求函数f(x)在区间[-
,
]上的最大值和最小值.
| π |
| 3 |
| π |
| 3 |
(1)求函数f(x)的最小正周期;
(2)求函数f(x)在区间[-
| π |
| 4 |
| π |
| 4 |
(1)∵f(x)=sin2x•cos
+cos2x•sin
+sin2x•cos
-cos2x•sin
+cos2x
=sin2x+cos2x
=
sin(2x+
),
∴函数f(x)的最小正周期T=
=π.
(2)∵函数f(x)在区间[-
,
]上是增函数,在区间[
,
]上是减函数,
又f(-
)=-1,f(
)=
,f(
)=1,
∴函数f(x)在区间[-
,
]上的最大值为
,最小值为-1.
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
=sin2x+cos2x
=
| 2 |
| π |
| 4 |
∴函数f(x)的最小正周期T=
| 2π |
| 2 |
(2)∵函数f(x)在区间[-
| π |
| 4 |
| π |
| 8 |
| π |
| 8 |
| π |
| 4 |
又f(-
| π |
| 4 |
| π |
| 8 |
| 2 |
| π |
| 4 |
∴函数f(x)在区间[-
| π |
| 4 |
| π |
| 4 |
| 2 |
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