mamy wyznaczyć: ctgα, sinα, cosα
| 1 | ||
ctgα= | = .......... dokończ
| |
| tgα |
| 1 | 1 | |||
tg2α+1 = | .. to 42 +1= | |||
| cos2α | cos2α |
| 1 | √17 | √17 | ||||
to cos2α= | więc cosα= | lub cosα= − | ||||
| 17 | 17 | 17 |
| sinα | ||
wiesz ,że: tgα= | ||
| cosα |
| sinα | ||
= 4 => sinα= 4*cosα= ........ podstaw dane w dwu przypadkach i | ||
| cosα |
| √17 | 4√17 | |||
Odp: ctgα= 14 , cosα= | , ... sinα= | |||
| 17 | 17 |
| √17 | 4√17 | |||
lub ctgα = 14 , cosα= − | , .... sinα= − | |||
| 17 | 17 |