2) Cos(4n) = 1
Sin(4n) = 0
4) Cos(5n/2) = 0
Sin (5n/2) = 1
6) (2k+1)n , k - Z
пусть k = 0, то:
пусть k = 1 , то:
(0+1)n= n
(2+1)n = 2n+n= 3n.
cos (n) = 0
cos (3n) = 0
sin (n) = 1
sin (3n) = 1 ; Откуда:
cos ((2k+1)n) = 0
sin ((2k+1)n) = 1
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
2) Cos(4n) = 1
Sin(4n) = 0
4) Cos(5n/2) = 0
Sin (5n/2) = 1
6) (2k+1)n , k - Z
пусть k = 0, то:
пусть k = 1 , то:
(0+1)n= n
(2+1)n = 2n+n= 3n.
cos (n) = 0
cos (3n) = 0
sin (n) = 1
sin (3n) = 1 ; Откуда:
cos ((2k+1)n) = 0
sin ((2k+1)n) = 1