| 1 | ||
wiedzac ze sinα−cosα=− | oblicz sin2α | |
| 6 |
| 1 | ||
sinα−cosα=− | /2 | |
| 6 |
| 1 | ||
sin2α−2sinαcosα+cos2α= | ||
| 36 |
| 1 | ||
−sin2α=−1+ | ||
| 36 |
| 35 | ||
sin2α= | ||
| 36 |
| 1 | ||
sin(α) − cos(α) = − | ||
| 6 |
| 5 | ||
sin2(α) + cos2(α) + sin(α) − cos(α) = | ||
| 6 |
| 5 | ||
sin2(α) − sin(2α) + cos2(α) + sin(α) − cos(α) + sin(2α) = | ||
| 6 |
| 5 | ||
[sin(α) − cos(α)]2 + sin(α) − cos(α) + sin(2α) = | ||
| 6 |
| 1 | 1 | 5 | |||
− | + sin(2α) = | ||||
| 36 | 6 | 6 |
| 35 | ||
sin(2α) = | ||
| 36 |