| x | x | |||
2(cos(2x)+ctg(2x))=ctg | −tg | |||
| 2 | 2 |
| 1 | ||
1cosec(x)= | to wtedy będzie (dobrze ?) | |
| sin(x) |
| 1 | ||
cosec(2x)= | ||
| sin(2x) |
| cos(2x) | ||
ctg(2x)= | ||
| sin(2x) |
| 1 | cos(2x) | |||
L= 2(cosec(2x)+ctg(2x))= 2[ | + | ] | ||
| sin(2x) | sin(2x) |
| 1+cos(2x) | 2*2cos2(x) | |||
L= 2[ | ]= | =2ctg(x) | ||
| sin(2x) | 2sin(x)*cos(x) |
| ctg2(x/2)−1 | x | x | ||||
2ctg(x)= 2* | = ctg | −tg | ||||
| 2ctg(x/2) | 2 | 2 |