| 4 | ||
arctg | =α | |
| 3 |
| 1 | ||
cos | α=? | |
| 2 |
| 4 | 4 | |||
arctg | =α⇔tgα= | |||
| 3 | 3 |
| sinα | 4 | ||
= | |||
| cosα | 3 |
| 4 | ||
cosα= | sinα | |
| 3 |
| 4 | ||
z jedynki trygonometrycznej: sin2α+( | sinα)2=1 | |
| 3 |
| 4 | 3 | |||
sinα= | ; cosα= | |||
| 5 | 5 |
| α | ||
cosα=2cos2 | −1 [ze wzoru na cos2α=2cos2α−1] | |
| 2 |
| 3 | α | 8 | α | ||||
=2cos2 | −1 ⇔ | =2cos2 | |||||
| 5 | 2 | 5 | 2 |
| α | 2 | 1 | 4 | 2 | ||||||
cos | = | ⇔cos( | arctg | )= | ||||||
| 2 | √5 | 2 | 3 | √5 |