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Dannii1
@Dannii1
April 2019
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Besoin d'aide svp c urgent
On propose de résoudre l'équation cosx + sinx = -1 dans l'intervalle [ 0 ; 2\pi ]
1_ Montrer que (cosx + sinx +1)2 = 2(1 + cosx)(1+sinx)
2_ En déduire la résolution de l'équation proposée.
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maudmarine
Verified answer
1)
Montrer que (cosx + sinx +1)2 = 2(1 + cosx)(1+sinx)
(cosx + sinx +1)² = 2(1 + cosx)(1+sinx)
(1 + cosx + sinx)² = (1 + cosx + sinx) (1 +cosx + sinx)
= ( (1 + cosx) + sinx ) (cosx + (1 + sinx) )
= (1 + cosx) cosx + (1 + cosx) (1 + sinx) + sinx cosx + sinx (1 + sinx)
2)
En déduire la résolution de l'équation proposée
(1 + cosx) (1 + sinx) = (1 + cosx) cosx + sinx cosx + sinx. (1+sinx)
On a :
(1 + cosx) cosx + sinx cosx + sinx (1 + sinx) = cosx + (cosx)² + sinx cosx + sinx + (sinx)²
= cosx +1 + sinx (1 + cosx)
= (cosx + 1) (sinx + 1)
d'où on a le resultat
(1+cosx +sinx)²=2(cosx+1)(sinx+1)
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Dannii1
merci beaucoup
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Verified answer
1) Montrer que (cosx + sinx +1)2 = 2(1 + cosx)(1+sinx)(cosx + sinx +1)² = 2(1 + cosx)(1+sinx)(1 + cosx + sinx)² = (1 + cosx + sinx) (1 +cosx + sinx)
= ( (1 + cosx) + sinx ) (cosx + (1 + sinx) )
= (1 + cosx) cosx + (1 + cosx) (1 + sinx) + sinx cosx + sinx (1 + sinx)
2) En déduire la résolution de l'équation proposée
(1 + cosx) (1 + sinx) = (1 + cosx) cosx + sinx cosx + sinx. (1+sinx)
On a :
(1 + cosx) cosx + sinx cosx + sinx (1 + sinx) = cosx + (cosx)² + sinx cosx + sinx + (sinx)²
= cosx +1 + sinx (1 + cosx)
= (cosx + 1) (sinx + 1)
d'où on a le resultat
(1+cosx +sinx)²=2(cosx+1)(sinx+1)