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emezenoula
@emezenoula
January 2021
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Bonsoir à tous.j'aurai besoin d'aide sur l'exercice 15.merci d'avance
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Geijutsu
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Bonsoir,
1. On sait que :
cos(a)*cos(b)+sin(a)*sin(b) = cos(a-b)
cos(a)*cos(b)-sin(a)*sin(b) = cos(a+b)
Donc :
A = cos(π/12)*cos(5π/12)+sin(π/12)*sin(5π/12) = cos((5π/12)-(π/12)) = cos(4π/12) = cos(π/3) = 1/2
B = cos(π/12)*cos(5π/12)-sin(π/12)*sin(5π/12) = cos((5π/12)+(π/12)) = cos(6π/12) = cos(π/2) = 0
2. On a donc le système suivant :
Donc (E1)+(E2) ⇒ 2(cos(π/12)*cos(5π/12)) = 1/2 ⇒ cos(π/12)*cos(5π/12) = 1/4
Et (E1)-(E2) ⇒ 2(sin(π/12)*sin(5π/12)) = 1/2 ⇒ sin(π/12)*sin(5π/12) = 1/4
Donc sin(π/12)*sin(5π/12) = cos(π/12)*cos(5π/12) = 1/4
2 votes
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Geijutsu
(E1) et (E2) sont les noms que j'ai donné aux deux équations, au cas où tu te demandes ce que c'est
guestnosdevoir
Si tu veut l'exercice 14) ,je faire pour vous.
emezenoula
oui faites le .je serai ravi
guestnosdevoir
Envoyer le maintenant,car je peut pas le posté en commentaire car très long.
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bonsoir.besoin d'aide sur cet exercice ppur demain.merci d'avance
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emezenoula
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bonsoir quelqu'un peut m'aider sur cet exercice .
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Lista de comentários
Verified answer
Bonsoir,1. On sait que :
cos(a)*cos(b)+sin(a)*sin(b) = cos(a-b)
cos(a)*cos(b)-sin(a)*sin(b) = cos(a+b)
Donc :
A = cos(π/12)*cos(5π/12)+sin(π/12)*sin(5π/12) = cos((5π/12)-(π/12)) = cos(4π/12) = cos(π/3) = 1/2
B = cos(π/12)*cos(5π/12)-sin(π/12)*sin(5π/12) = cos((5π/12)+(π/12)) = cos(6π/12) = cos(π/2) = 0
2. On a donc le système suivant :
Donc (E1)+(E2) ⇒ 2(cos(π/12)*cos(5π/12)) = 1/2 ⇒ cos(π/12)*cos(5π/12) = 1/4
Et (E1)-(E2) ⇒ 2(sin(π/12)*sin(5π/12)) = 1/2 ⇒ sin(π/12)*sin(5π/12) = 1/4
Donc sin(π/12)*sin(5π/12) = cos(π/12)*cos(5π/12) = 1/4