| 4 | ||
oblicz cos(α−60), ∊(90,180) sinα= | ||
| 5 |
| √3 | ||
cos60cosβ+sin60sinβ= −12cosβ+ | sinβ= i teraz moge uznac ze sinβ to cosα ? jak to | |
| 2 |
| 3 | ||
α∊(90o,180o) −− || ćw. to cosα= − | ||
| 5 |
| 3 | 1 | 4 | √3 | |||||
cos(α−60)= cosα*cos60o+sinα*sin60o = − | * | + | * | = | ||||
| 5 | 2 | 5 | 2 |
| −3+4√3 | ||
= | ||
| 10 |
| 5^2 | 52 |
| 2^{10} | 210 |
| a_2 | a2 |
| a_{25} | a25 |
| p{2} | √2 |
| p{81} | √81 |
| Kliknij po więcej przykładów | |
|---|---|
| Twój nick | |