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Junie
@Junie
April 2019
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Bonsoir. Le devoir c'est d'établir la relation suivante
sin a×sin(b-c)+sin b×(c-a) +sin c × sin (a-b)=0
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sin a×sin(b-c)+sin b×sin(c-a) +sin c × sin (a-b)
Formulaire :
sin(p)sin(q) = 1/2(cos(p - q) - cos(p + q))
donc on déduit que :
sin(a)sin(b-c)
=
1/2 (cos(a-b+c)-cos(a+b-c))
sin(b)sin(c-a)=
1/2 (cos(a+b-c)-cos(a-b-c))
sin(c)sin(a-b)=
1/2 (cos(a-b-c)-cos(a-b+c))
ainsi
sin a×sin(b-c)+sin b×sin(c-a) +sin c × sin (a-b)
=1/2(
cos(a-b+c)-cos(a+b-c)+cos(a+b-c)-cos(a-b-c)+
cos(a-b-c)-cos(a-b+c))
=0
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Lista de comentários
Formulaire :
sin(p)sin(q) = 1/2(cos(p - q) - cos(p + q))
donc on déduit que :
sin(a)sin(b-c)=1/2 (cos(a-b+c)-cos(a+b-c))
sin(b)sin(c-a)=1/2 (cos(a+b-c)-cos(a-b-c))
sin(c)sin(a-b)=1/2 (cos(a-b-c)-cos(a-b+c))
ainsi
sin a×sin(b-c)+sin b×sin(c-a) +sin c × sin (a-b)
=1/2(cos(a-b+c)-cos(a+b-c)+cos(a+b-c)-cos(a-b-c)+cos(a-b-c)-cos(a-b+c))
=0