| α | 4 | |||
1) Wiedząc że sin | = | i α∊ 90−180 stopni oblicz: | ||
| 2 | 5 |
| α | α | |||
sin2 | +cos2 | =1⇔ | ||
| 2 | 2 |
| 16 | α | ||
+cos2 | =1 | ||
| 25 | 2 |
| α | 9 | |||
cos2 | = | |||
| 2 | 25 |
| α | 3 | |||
cos | = | |||
| 2 | 5 |
| α | α | |||
2*sin | *cos | =sinα [ wzór: sin2x=2sinx *cosx] | ||
| 2 | 2 |
| 4 | 3 | |||
2* | * | =sinα | ||
| 5 | 5 |
| 24 | ||
sinα= | (kąt α∊ drugiej ćwiartki, to sinα>0) | |
| 25 |
| α | 4 | α | ||||
1/ sin | = | , cos | =............ | |||
| 2 | 5 | 2 |
| α | α | |||
sinα=2sin | *cos | =... | ||
| 2 | 2 |
| α | ||
cosα= 1−2sin2 | =......... | |
| 2 |
| sinα | ||
tgα= | =.......... | |
| cosα |
| 1 | ||
ctgα= | =.......... | |
| tgα |
tg2α= −3 <0 to 2α€ II ćw v IVćw
| 3√10 | 3√10 | |||
sin2α= | v sin2α= − | |||
| 10 | 10 |
| √10 | √10 | |||
cos2α= − | v cos2α= | |||
| 10 | 10 |