题目内容
已知函数f(x)=sin2ωx+
sinωxsin(ωx+
)(ω>0)的最小正周期为π
(1)求f(x);
(2)当x∈[-
,
]时,求函数f(x)的值域.
| 3 |
| π |
| 2 |
(1)求f(x);
(2)当x∈[-
| π |
| 12 |
| π |
| 2 |
(1)f(x)=
+
sinωxcosωx=
sin2ωx-
cos2ωx+
=sin(2ωx-
)+
.
∵函数f(x)的最小正周期为π,且ω>0,
∴
=π,解得ω=1.
∴f(x)=sin(2x-
)+
.
(2)∵x∈[-
,
],∴2x-
∈[-
,
].
根据正弦函数的图象可得:
当2x-
=
,即x=
时,g(x)=sin(2x-
)取最大值1
当2x-
=-
,即x=-
时g(x)=sin(2x-
)取最小值-
.
∴
-
≤sin(2x-
)+
≤
,即f(x)的值域为[
,
].
| 1-cos2ωx |
| 2 |
| 3 |
| ||
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| π |
| 6 |
| 1 |
| 2 |
∵函数f(x)的最小正周期为π,且ω>0,
∴
| 2π |
| 2ω |
∴f(x)=sin(2x-
| π |
| 6 |
| 1 |
| 2 |
(2)∵x∈[-
| π |
| 12 |
| π |
| 2 |
| π |
| 6 |
| π |
| 3 |
| 5π |
| 6 |
根据正弦函数的图象可得:
当2x-
| π |
| 6 |
| π |
| 2 |
| π |
| 3 |
| π |
| 6 |
当2x-
| π |
| 6 |
| π |
| 3 |
| π |
| 12 |
| π |
| 6 |
| ||
| 2 |
∴
| 1 |
| 2 |
| ||
| 2 |
| π |
| 6 |
| 1 |
| 2 |
| 3 |
| 2 |
1-
| ||
| 2 |
| 3 |
| 2 |
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