proszeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
Jakub: ( tg alfa+1))tg alfa−1)(1−sin alfa)(1+sin alfa)= (sin alfa− cos alfa)(sin alfa+cos
alfA)
8 mar 19:44
Jakub: prosze bardzo pomocy
chyba nikt nie moze mi pomoc czy jak?
8 mar 19:56
IchIch: (tgα+1)(tgα−1)(1−sinα)(1+sinα)=(sinα−cosα)(sinα+cosα)
(tgα
2−1)(1−sinα
2)=sinα
2−cosα
2
(tgα
2−1)cosα
2=sinα
2−cosα
2
tgα
2−1=tgα
2−1
8 mar 20:11
Goblin: (tgx+1)(tgx−1)=tg
2x−1
(1−sinx)(1+sinx)=1−sin
2x=cos
2x
(sinx−cosx)(sinx+cosx)=sin
2x−cos
2x=−cos2x=P
| sin2x | | cos2x | | sin2x−cos2x | | cos2x | |
tg2x−1= |
| − |
| = |
| (bo 1= |
| |
| cos2x | | cos2x | | cos2x | | cos2 | |
| sin2x−cos2x | |
(tg2x−1)*cos2x= |
| *cos2x}= sin2x−cos2x =−cos2x=L |
| cos2x | |
wniosek
Dla cosα≠0 (ze wzgledu na tangens ) ta tozsamosc jest prawdziwa
8 mar 20:18