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 |