题目内容
已知函数f(x)=sin(2x-
)-2cos(x-
)cos(x+
)+1,x∈R
(1)求函数f(x)的最小正周期:
(2)求函数f(x)在区间[0,
]上的值域.
| π |
| 6 |
| π |
| 4 |
| π |
| 4 |
(1)求函数f(x)的最小正周期:
(2)求函数f(x)在区间[0,
| π |
| 2 |
函数f(x)=sin(2x-
)-2cos(x-
)cos(x+
)+1
=
sin2x-
cos2x-2(
cosx+
sinx)(
cosx-
sinx)+1
=
sin2x-
cos2x-cos2x +1
=
sin(2x-
) +1.
(1)f(x)的最小正周期为:π;
(2)∵x∈[0,
],∴-
≤2x-
≤
∴-
≤sin(2x-
) ≤1,
所以
sin(2x-
) +1∈[-
,
+1];
故函数的值域为:[-
,
+1]
| π |
| 6 |
| π |
| 4 |
| π |
| 4 |
=
| ||
| 2 |
| 1 |
| 2 |
| ||
| 2 |
| ||
| 2 |
| ||
| 2 |
| ||
| 2 |
=
| ||
| 2 |
| 1 |
| 2 |
=
| 3 |
| π |
| 3 |
(1)f(x)的最小正周期为:π;
(2)∵x∈[0,
| π |
| 2 |
| π |
| 3 |
| π |
| 3 |
| 2π |
| 3 |
∴-
| ||
| 2 |
| π |
| 3 |
所以
| 3 |
| π |
| 3 |
| 1 |
| 2 |
| 3 |
故函数的值域为:[-
| 1 |
| 2 |
| 3 |
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