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 |