skoro alfa jest ostry, to ze wzorow redukcyjnych
tg (90−alfa) = ctg alfa
| 2 | ||
sin | , więc albo rysujesz sobie trojkat, albo z jedynki trygonometrycznej | |
| 3 |
| cos alfa | ||
Lub ctg alfa= | ||
| sin alfa |
| √5 | ||
cos α = | ||
| 3 |
| √5 | ||
ctg α = | ||
| 2 |
| √5 | √5 | 5 | ||||
cos α * tg(90 − α) = | * | = | ||||
| 3 | 2 | 6 |
| 2 | ||
α ∊ ( 0o ; 90o ) i sin α = | ||
| 3 |
| √5 | √5 | 5 | ||||
cos α*tg ( 90o − α) = cos α *ctg α = | * | = | ||||
| 3 | 2 | 6 |
| 4 | 5 | √5 | ||||
cos2α = 1 − sin2α = 1 − | = | ⇒ cos α = | ||||
| 9 | 9 | 3 |
| cos α | √5 | 2 | √5 | 3 | √5 | |||||||
ctg α = | = | : | = | * | = | |||||||
| sin α | 3 | 3 | 3 | 2 | 2 |