| 1−cos2(3π/5) | ||
Wykaż tożsamość | =tg2(π/5) | |
| (1−cos(3π/5))2 |
| (1−cosα)(1+cosα) | 1+cosα | |||
L= | = | , 1−cosα≠0 | ||
| (1−cosα)2 | 1−cosα |
| α | α | π | ||||
1+cosα=2cos2 | i 1−cosα= 2sin2 | i ctgα=tg( | −α) | |||
| 2 | 2 | 2 |
| α | π | α | ||||||||||||||
L= | = ctg2 | = tg2( | − | ) | |||||||||||||
| 2 | 2 | 2 |
| α | 3π | ||
= | |||
| 2 | 10 |
| π | 3π | π | ||||
to L= tg2( | − | )= tg2 | = P | |||
| 2 | 10 | 5 |
| (1−cosx)(1+cosx) | 1+cosx |
| |||||||||||||
L= | = | = | = | ||||||||||||
| (1−cosx)2 | 1−cosx |
|
| x | π | x | ||||
=ctg2 | =(tg( | − | ))2. | |||
| 2 | 2 | 2 |
| 3π | ||
Dla x= | jest | |
| 5 |
| π | x | π | 3π | π | ||||||
− | = | − | = | , | ||||||
| 2 | 2 | 2 | 10 | 5 |
| π | ||
L=(tg | )2=P, | |
| 5 |
