| √3 | ||
Wykaż,że tg20o*tg40o*cos60o*tg80o= | ||
| 2 |
| 1 | ||
cos60o= | , tg60o= √3 | |
| 2 |
| 3−tg2α | ||
tg(3α)= tgα* | ||
| 1−3tg2α |
| tgx−tgy | tgx+tgy | |||
tg(x−y)= | i tg(x+y)= | |||
| 1+tgx*tgy | 1−tgx*tgy |
| √3−tg20o | ||
tg40o=tg(60o−20o) = | ||
| 1+√3*tg20o |
| √3+tg20o | ||
i tg80o=tg(60o+20o) = | ||
| 1−√3*tg20o |
| 1 | 3−tg220o | 1 | 1 | √3 | ||||||
L= | *tg20o* | = | *tg3*20o= | *√3= | =P | |||||
| 2 | 1−3*tg220o | 2 | 2 | 2 |