| 1 | ||
Liczba p jest sinusem kąta ostrego β. Wykaż, że (tgβ)−2= | − 1 | |
| p2 |
| 1 | 1 | 1−sin2β | cos2β | ||||
−1= | −1= | = | =ctg2β | ||||
| p2 | sin2β | sin2β | sin2β |
| 1 | 1 | 1−sin2β | cos2β | ||||
−1= | −1= | = | =ctg2β | ||||
| p2 | sin2β | sin2β | sin2β |
Jakbym mogła to postawiłabym piwko
prawą stronę możesz zapisać tak: (p2+1)2
| 1 | |
=(p2+1)2 | |
| cos4α |
| 1 | |
=(p2+1)2 | |
| (cos2α)2 |
| 1 | |
=p2+1 | |
| cos2α |
| 1 | sin2α | ||
= | + 1 | ||
| cos2α | cos2α |
| 1 | sin2α | cos2α | |||
= | + | ||||
| cos2α | cos2α | cos2α |
| 1 | 1 | ||
= | CKD. | ||
| cos2α | cos2α |