Oblicz wartosc wyrazenia wiedzac, ze
sin²α+cos²=1 [jedynka trygonometrzyczna]
stąd
sin²α=1-cos²α
-------------------
√[cos²α-sin²α]=
=√[cos²α-(1-cos²α)]=
=√[cos²α-1+cos²α]=
=√[2cos²α-1]=
---
cos²α=12/13
=√[2 * 12/13 - 1]=
=√[24/13 - 13/13]=
=√[11/13]=
=√11/√13 * √13/√13=√143/13
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
sin²α+cos²=1 [jedynka trygonometrzyczna]
stąd
sin²α=1-cos²α
-------------------
√[cos²α-sin²α]=
=√[cos²α-(1-cos²α)]=
=√[cos²α-1+cos²α]=
=√[2cos²α-1]=
---
cos²α=12/13
---
=√[2 * 12/13 - 1]=
=√[24/13 - 13/13]=
=√[11/13]=
=√11/√13 * √13/√13=√143/13