| 1 | ||
sin(arctg5−arccos | ) | |
| 5 |
| π | π | π | ||||
arctg(5)=α, α∊(− | , | ) wtedy tg(arctg(5))=tgα⇔tgα=5, α∊(0, | ) | |||
| 2 | 2 | 2 |
| 1 | 1 | 1 | ||||
arccos( | )=β, β∊<0,π> wtedy cos(arccos( | )=cosβ⇔cosβ= | ||||
| 5 | 5 | 5 |
| 1 | ||
sin(α−β)=sinα*cosβ−sinβ*cosα=sinα* | −sinβ*cosα | |
| 5 |
| √24 | ||
sinβ=√1−125= | ||
| 5 |
| sinα | |
=5 ⇔sinα=5cosα | |
| cosα |
| 1 | 5 | |||
cosα= | i sin α= | ⇔ | ||
| √26 | √26 |
| 5 | 1 | √24 | 1 | |||||
sin(α−β)= | * | − | * | = | ||||
| √26 | 5 | 5 | √26 |
| √26 | 2√6 | |||
= | *(1− | ) | ||
| 26 | 5 |
