| 1 |
| 5 |
| 1 |
| 5 |
∵sin2α+cos2α=1，
∴25sin2α-5sin α-12=0．
∵α是三角形的内角，∴
|
∴tanα=-
| 4 |
| 3 |
（2）
| 1 |
| cos2α−sin2α |
| sin2α+cos2α |
| cos2α−sin2α |
=
| ||
|
=
| tan2α+1 |
| 1−tan2α |
∵tanα=-
| 4 |
| 3 |
∴
| 1 |
| cos2α−sin2α |
| tan2α+1 |
| 1−tan2α |
| 25 |
| 7 |
| 1 |
| 5 |
| 1 |
| cos2α−sin2α |
| 1 |
| 5 |
| 1 |
| 5 |
|
| 4 |
| 3 |
| 1 |
| cos2α−sin2α |
| sin2α+cos2α |
| cos2α−sin2α |
| ||
|
| tan2α+1 |
| 1−tan2α |
| 4 |
| 3 |
| 1 |
| cos2α−sin2α |
| tan2α+1 |
| 1−tan2α |
| 25 |
| 7 |