| 5 | 3 | 3 | ||||
Oblicz sin(2x + | π), jeśli ctgx = | i x ∊ (π; | π) | |||
| 4 | 2 | 2 |
| 5 | π | ||
π= π+ | |||
| 4 | 4 |
| √2 | ||
zatem sin(2x+5π/4)= −sin(2x+π/4) =−(sin2x*cos(π/4)+cos2x*sin(π/4))= − | (sin2x+cos2x) | |
| 2 |
| 2tgx | 1−tg2x | |||
sin2x= | i cos2x= | |||
| 1+tg2x | 1+tg2x |
| 5^2 | 52 |
| 2^{10} | 210 |
| a_2 | a2 |
| a_{25} | a25 |
| p{2} | √2 |
| p{81} | √81 |
| Kliknij po więcej przykładów | |
|---|---|
| Twój nick | |