| √3 | ||
kąt α jest ostry i sin α = | ||
| 2 |
| √3 | ||
podstaw za sinα | i policz cosα | |
| 2 |
| √3 | |
+cos2α=1 | |
| 2 |
| √3 | ||
+1=cos2α | ||
| 2 |
| 2+√3 | |
| 2 |
| √3 | 3 | |||
czyli ( | )2= | |||
| 2 | 4 |
| 3 | |
+cos2α=1 | |
| 4 |
| 3 | ||
cos2α=1− | ||
| 4 |
| 3 | ||
sin2α−3cos2α=podstawiasz = | −3*... | |
| 4 |
| √3 | ||
sinα= | ||
| 2 |
| 3 | ||
sin2α= | ||
| 4 |
| 3 | 1 | |||
cos2α=1−sin2α=1− | = | |||
| 4 | 4 |
| 3 | 1 | ||
−3* | =0 | ||
| 4 | 4 |