Connexion

Connexion à votre compte

Identifiant
Mot de passe
Maintenir la connexion active sur ce site

Créer un compte

Pour valider ce formulaire, vous devez remplir tous les champs.
Nom
Identifiant
Mot de passe
Répétez le mot de passe
Adresse e-mail
Répétez l'adresse e-mail
Captcha
Vous êtes ici : AccueilCLASSESCorrection des exercices nombres complexes : Formules de Moivre et formules d’Euler
Terminale
C & E & D & TI
Mathématiques
Correction exercice
Bonjour ! Groupe telegram de camerecole, soumettrez-y toutes vos préoccupations. forum telegram

Correction exercice I

En utilisant les formules de Moivre, linéarisons les expressions suivantes :
a) \(A(x) = {\cos ^3}x\)
\(A(x) = \) \({\left( {\frac{{Z + \overline Z }}{2}} \right)^3}\) \( = \frac{1}{8}({Z^3} + \) \(3{Z^2}\overline Z + \) \(3Z{\overline Z ^2} + \) \({\overline Z ^3})\)
\({Z^n} + {\overline Z ^n}\) \( = 2\cos \left( {n\theta } \right)\)
\({Z^3} + {\overline Z ^3}\) \( = 2\cos \left( {3\theta } \right)\)
\({Z^3} + {\overline Z ^3}\) \( = 2\cos \left( {3\theta } \right)\)
\(Z + \overline Z = \) \(2\cos \left( \theta \right)\)
\(A(\theta) = \) \(\frac{1}{8}\left( {{Z^3} + {{\overline Z }^3}} \right) + \) \(\frac{3}{8}Z\overline Z \left( {Z + \overline Z } \right)\)
\(A(\theta) = \) \(\frac{1}{8}2\cos \left( {3\theta } \right)\) \( + \frac{3}{8}2\cos \left( \theta \right)\) \( = \frac{1}{4}\cos \left( {3\theta } \right)\) \( + \frac{3}{4}\cos \left( \theta \right)\)

\(A(x) = \) \(\frac{1}{4}\cos \left( {3x} \right)\) \( + \frac{3}{4}\cos \left( x \right)\)

b) \(B(x) = {\sin ^3}x\)
\(B(x) = \) \({\sin ^3}x = \) \({\left( {\frac{{Z + \overline Z }}{2}} \right)^3}\)
\({Z^n} - {\overline Z ^n}\) \( = 2i\sin \left( {n\theta } \right)\)
\({Z^n} \times {\overline Z ^n} = 1\)
\(B(x) = \) \({\left( {\frac{{Z - \overline Z }}{2}} \right)^3}\) \( = \frac{1}{{{{\left( {2i} \right)}^3}}}\) \((2({Z^3} - {\overline Z ^3})\) \( + 3(Z - \overline Z ))\)
\(B(x) = \) \(\frac{1}{{ - 8i}}(2i\sin 3x)\) \( + \frac{1}{{ - 8i}}(6i\sin x)\)

\(B(x) = \) \( - \frac{1}{4}\sin 3x\) \( - \frac{3}{4}\sin x\)

c) \(C(x) = {\cos ^5}\frac{x}{2}\)
\(C(x) = \) \(\frac{1}{{32}}\left( {{Z^5} + {{\overline Z }^5}} \right) + \) \(\frac{1}{{32}}5Z\overline Z \left( {{Z^3} + {{\overline Z }^3}} \right)\) \( + \frac{1}{{32}}10{Z^2}{\overline Z ^2}\) \(\left( {Z + \overline Z } \right)\)
\(C(x) = \) \(\frac{1}{{32}}2\cos 5\frac{x}{2}\) \( + \frac{1}{{32}}10\cos 3\frac{x}{2}\) \( + \frac{1}{{32}}20\cos \frac{x}{2}\)

\(C(x) = \) \(\frac{1}{{16}}\cos \frac{{5x}}{2} + \) \(\frac{5}{{16}}\cos \frac{{3x}}{2} + \) \(\frac{{10}}{{16}}\cos \frac{x}{2}\)

d) \(D(x) = \) \({\cos ^3}x{\sin ^3}x\)
\(D(x) = \) \({\left( {\frac{{Z + \overline Z }}{2}} \right)^3}\) \({\left( {\frac{{Z - \overline Z }}{{2i}}} \right)^3}\) \( = \frac{1}{{ - 64i}}\left( {{Z^6} - {{\overline Z }^6}} \right)\) \( - \frac{3}{{ - 64i}}{Z^2}{\overline Z ^2}\) \(\left( {{Z^2} - {{\overline Z }^2}} \right)\)
\(D(x) = \) \(\frac{1}{{ - 64i}}\left( {2i\sin 6x} \right)\) \( - \frac{3}{{ - 64i}}\left( {2i\sin 2x} \right)\)

\(D(x) = \) \( - \frac{1}{{32}}\sin 6x + \) \(\frac{3}{{32}}\sin 2x\)

Exercice II

En utilisant la formule d'Euler, linéarisons les expressions suivantes :
a) \(A(x) = {\cos ^3}x\)
\(A(x) = \) \(\frac{1}{8}{\left( {{e^{ix}} + {e^{ - ix}}} \right)^3}\) \(\frac{1}{8}\left( {{e^{3ix}} + {e^{ - 3ix}}} \right)\) \( + \frac{3}{8}\left( {{e^{ix}} + {e^{ - ix}}} \right)\)

• \({e^{nix}} + {e^{ - nix}}\) \( = 2\cos nx\)
• \({e^{nix}} \times {e^{ - nix}}\) \( = 1\)

\(A(x) = \) \(\frac{1}{4}\left( {\cos 3x} \right)\) \( + \frac{3}{4}\left( {\cos x} \right)\)

b) \(B(x) = {\sin ^3}x\)
\(B(x) = \) \(\sin 3x = \) \({\left( {\frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}}} \right)^3}\) \( = \frac{1}{{ - 8i}}\) \( = \frac{1}{{ - 8i}}\) \( + \frac{1}{{ - 8i}}\) \(\left( {{e^{ix}} - {e^{ - ix}}} \right)\)

\(B(x) = - \) \(\frac{1}{4}\sin 3x - \) \(\frac{3}{4}\sin x\)

c) \(C(x) = {\cos ^5}\frac{x}{2}\)
d) \(D(x) = \) \({\cos ^3}x{\sin ^3}x\)

Correction exercice III

Linéarisons en utilisant :
a) Les formules d’Euler;
\({\cos ^2}x = \) \({\left( {\frac{{{e^{ix}} + {e^{ - ix}}}}{2}} \right)^2}\) \( = \frac{1}{4}({\left( {{e^{ix}}} \right)^2}\) \( + 2{e^{ix}}{e^{ - ix}}\) \( + {\left( {{e^{ - ix}}} \right)^2} = \) \(\frac{1}{4}\left( {{e^{i2x}} + {e^{ - 2ix}}} \right)\) \( + \frac{1}{2}\)
b) Les formules usuelles trigonométriques
\(\cos (x + x) = \) \(\cos (2x) = \) \({\cos ^2}x - {\sin ^2}x\) \( = {\cos ^2}x - \) \((1 - {\cos ^2}x)\) \( = 2{\cos ^2}x - 1\)
\({\cos ^2}x = \) \(\frac{1}{2}\cos 2x + \frac{1}{2}\)

Linéarisons en utilisant :
a) Les formules d’Euler;
\({\sin ^3}x\)
\({\sin ^3}x = \) \({\left( {\frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}}} \right)^3}\) \( = - \frac{1}{4}\) \(\left( {\frac{{{e^{i3x}} - {e^{ - i3x}}}}{{2i}}} \right)\) \( - \frac{3}{4}\) \(\left( {\frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}}} \right)\)
\({\sin ^3}x = \) \( - \frac{1}{4}\sin 3x\) \( - \frac{1}{4}\sin 3x\)

Correction exercice IV

Linéarisons \(2\left( {1 + {{\sin }^2}x} \right)\) \({\cos ^2}x\)
\(2\left( {1 + {{\sin }^2}x} \right)\) \({\cos ^2}x\) \( = 2\) \(\left( {1 + {{\left( {\frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}}} \right)}^2}} \right)\) \({\left( {\frac{{{e^{ix}} + {e^{ - ix}}}}{{2i}}} \right)^2}\) \( = \frac{1}{8}\) \(\left( {6 - {e^{2ix}} - {e^{ - 2ix}}} \right)\) \(\left( {2 + {e^{2ix}} + {e^{ - 2ix}}} \right)\)
\(2\left( {1 + {{\sin }^2}x} \right)\) \({\cos ^2}x = \) \(\frac{1}{4}(5 + 4\cos 2x\) \( - \cos 4x)\)