| 1 | 1 | 4 | ||||
wykaz ze jesli α jest kątem ostrym i ( | − | ) = | , | |||
| 1−cosα | 1+cosα | sinα |
| 1 | ||
to tgα = | ||
| 2 |
| 1 | 1 | 2 | ||||
( | − | ) = | << no i co z tego ? | |||
| 1−cosα | 1+cosα | sin2 |
| (1+cosx)−(1−cosx) | 4 | ||
= | |||
| (1−cosx)(1+cosx) | sinx |
| 2cosx | 4 | ||
= | |||
| 1−cos2x | sinx |
| 2cosx | 4 | ||
= | /*sinx | ||
| sin2x | sinx |
| 2cosx | |
= 4 | |
| sinx |
| 1 | 1 | |||
tgx = | = | |||
| ctgx | 2 |
| 2cosx | 4 | |||
dlaczego | = | |||
| sin2x | sinx |
| 1 | ||
chyba nie zrozumialem zadania, mialem obliczyc nawias i wykazac ze tg x = | tak? | |
| 2 |