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