| 3 |
| 1 |
| 2 |
| ||
| 2 |
| π |
| 3 |
所以函数f（x）的最小正周期是2π．
（2）由（1）得，f(x)＝2sin(x+
| π |
| 3 |
因为f(α−
| π |
| 3 |
| 6 |
| 5 |
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
| 6 |
| 5 |
即sinα＝
| 3 |
| 5 |
因为α∈(0，
| π |
| 2 |
| 1−sin2α |
| 4 |
| 5 |
所以f(2α−
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
=4sinαcosα
=4×
| 3 |
| 5 |
| 4 |
| 5 |
| 48 |
| 25 |
| 3 |
| π |
| 3 |
| 6 |
| 5 |
| π |
| 2 |
| π |
| 3 |
| 3 |
| 1 |
| 2 |
| ||
| 2 |
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
| 6 |
| 5 |
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
| 6 |
| 5 |
| 3 |
| 5 |
| π |
| 2 |
| 1−sin2α |
| 4 |
| 5 |
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
| 3 |
| 5 |
| 4 |
| 5 |
| 48 |
| 25 |