题目内容
已知函数f(x)=sin2ωx+
sinωxsin(ωx+
)(ω>0)的最小正周期为π.
(Ⅰ)求ω的值;
(Ⅱ)求函数f(x)在区间[0,
]上的取值范围.
| 3 |
| π |
| 2 |
(Ⅰ)求ω的值;
(Ⅱ)求函数f(x)在区间[0,
| 2π |
| 3 |
(Ⅰ)f(x)=
+
sin2ωx=
sin2ωx-
cos2ωx+
=sin(2ωx-
)+
.
∵函数f(x)的最小正周期为π,且ω>0,
∴
=π,解得ω=1.
(Ⅱ)由(Ⅰ)得f(x)=sin(2x-
)+
.
∵0≤x≤
,
∴-
≤2x-
≤
,
∴-
≤sin(2x-
)≤1.
∴0≤sin(2x-
)+
≤
,即f(x)的取值范围为[0,
].
| 1-cos2ωx |
| 2 |
| ||
| 2 |
| ||
| 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 |
(Ⅱ)由(Ⅰ)得f(x)=sin(2x-
| π |
| 6 |
| 1 |
| 2 |
∵0≤x≤
| 2π |
| 3 |
∴-
| π |
| 6 |
| π |
| 6 |
| 7π |
| 6 |
∴-
| 1 |
| 2 |
| π |
| 6 |
∴0≤sin(2x-
| π |
| 6 |
| 1 |
| 2 |
| 3 |
| 2 |
| 3 |
| 2 |
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