| 5 | π | |||
ctg x = | i x ∊ ( 0; | ) | ||
| 12 | 2 |
| 5 a | 5 | |||
cos x = | = | |||
| 13 a | 13 |
| 12 a | 12 | |||
sin x = | = | |||
| 13 a | 13 |
| 12 | ||
tg x = | ||
| 5 |
| 1 | ||
no to sin2x= | wylicz z tego sinx i x∊(0,π) | |
| 1+ctg2x |
| 1 | ||
I sprawdz sobie czy to jest prawda 1+ctg2x= | ||
| sin2x |
| 1 | ||
a także 1+tg2x= | ||
| cos2x |
| π | ||
x∊( | π) | |
| 2 |
| √5 | ||
cosx= √1−sin2x= − | ||
| 3 |
| sinx | ||
tgx= | podstaw i policz | |
| cosx |
| 1 | ||
ctg x = | to samo | |
| tgx |