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解:(1)該函數(shù)的周期T= |
13π |
|
3 |
|
- |
π |
|
3 |
|
=4π, |
∴ω= |
2π |
|
T |
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=,又A=3, 所給圖象是曲線y=3sin |
沿x軸向右平移 |
π |
|
3 |
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而得到的, 故所求函數(shù)解析式是 |
y=3sin(x- |
π |
|
3 |
|
) |
=3sin(x- |
π |
|
6 |
|
) |
(2)設(shè)(x,y)為y=3sin(x- |
π |
|
6 |
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)上的任意一點(diǎn), 該點(diǎn)關(guān)于直線x=2π |
的對(duì)稱點(diǎn)應(yīng)為(4π-x,y)故與y=3sin(x- |
π |
|
6 |
|
) |
關(guān)于直線x=2π對(duì)稱的函數(shù)解析式是
y=3sin( |
4π-x |
|
2 |
|
- |
π |
|
6 |
|
)=-3sin(+ |
π |
|
6 |
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) |
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