| 4 | ||
cos ( arcsin ( | ) ) = | |
| 5 |
| 3 | ||
sin ( 2arccos ( | ) ) = | |
| 5 |
| 3 | π | |||
arccos | =α i α∊(0, | ) popatrz na wykres | ||
| 5 | 2 |
| 3 | 4 | |||
cosα= | i sinα>0 , sinα=√1−(3/5)2= | |||
| 5 | 5 |
| 3 | 3 | 4 | ||||
sin(arccos( | ))=2* | * | ||||
| 5 | 5 | 5 |
| 4 | ||
cos(arcsin | )=? | |
| 5 |
| 4 | π | π | ||||
arcsin | =α, α∊<− | , | > | |||
| 5 | 2 | 2 |
| 4 | π | π | 3 | |||||
sinα= | i cos α>0 dla α∊<− | , | >⇔cosα=√1−(4/5)2= | |||||
| 5 | 2 | 2 | 5 |
| 4 | 3 | |||
cos(arcsin | )=cosα= | |||
| 5 | 5 |