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
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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.
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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