Réponse :
Explications étape par étape
Exercice 3
1)
cos(2pi) = cos(2pi-2pi) = cos(0) = 1
cos(4pi) = cos(4pi-2pi) = cos(2pi) = 1
cos(6pi) = cos(6pi-2pi) = cos(4pi) = 1
2020pi = 1010 x 2pi donc cos(2020pi) = cos(2pi) = 1
2) cos(2pi x n) = 1 pour tout n dans Z
3) comme 3pi = pi + 2pi
cos(3pi) = cos(pi) = -1
de meme comme 5pi = pi + 4pi, 7pi = pi + 6pi, 1999pi = 1998pi + pi
cos(1999pi)=cos(7pi)=cos(5pi) = -1
4) cos((2n+1)pi) = cos(pi) = -1
5) sin(npi) = 0 car sin(pi)= sin(-pi) = 0
6) un = sin(npi)
u0 = u1 = u2 = u2020 = u2021 = u3000 = u3001 = 0
on peut conjecturer que la limite de la suite (un) est 0
