| π |
| 2 |
| π |
| 2 |
∴α−β∈(0,π),α+β∈(
| π |
| 2 |
| 3π |
| 2 |
∴sin(α−β)=
| 1−cos2(α−β) |
| 3 |
| 5 |
cos(α+β)=−
| 1−sin2(α+β) |
| 12 |
| 13 |
∴cos2α=cos[(α-β)+(α+β)]
=cos(α-β)cos(α+β)-sin(α-β)sin(α+β)
=
| 4 |
| 5 |
| 12 |
| 13 |
| 3 |
| 5 |
| 5 |
| 13 |
=−
| 33 |
| 65 |
| π |
| 2 |
| π |
| 2 |
| 4 |
| 5 |
| 5 |
| 13 |
| π |
| 2 |
| π |
| 2 |
| π |
| 2 |
| 3π |
| 2 |
| 1−cos2(α−β) |
| 3 |
| 5 |
| 1−sin2(α+β) |
| 12 |
| 13 |
| 4 |
| 5 |
| 12 |
| 13 |
| 3 |
| 5 |
| 5 |
| 13 |
| 33 |
| 65 |