{"id":37277,"date":"2019-10-29T15:14:19","date_gmt":"2019-10-29T14:14:19","guid":{"rendered":"https:\/\/groupe-reussite.fr\/?page_id=37277"},"modified":"2022-09-05T16:32:13","modified_gmt":"2022-09-05T15:32:13","slug":"cours-en-ligne-trigonometrie-maths-sup-exercices-corriges","status":"publish","type":"page","link":"https:\/\/groupe-reussite.fr\/ressources\/cours-en-ligne-trigonometrie-maths-sup-exercices-corriges\/","title":{"rendered":"Trigonom\u00e9trie en Maths Sup – Exercices corrig\u00e9s"},"content":{"rendered":"
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\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Cours en ligne Maths en Maths Sup<\/a><\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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Chapitres Maths en MPSI, PCSI, MP2I, PTSI<\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tR\u00e9currence<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSommes et produits<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tNombres complexes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tTrigonom\u00e9trie<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tNombres r\u00e9els<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tEnsembles et applications<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFonctions (intro)<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFonctions usuelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tPrimitives<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\t\u00c9quations diff\u00e9rentielles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSuites num\u00e9riques<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tLimites et continuit\u00e9<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tD\u00e9riv\u00e9es<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSyst\u00e8mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tPolyn\u00f4mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFractions rationnelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tArithm\u00e9tique et polyn\u00f4mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tArithm\u00e9tique et fractions rationnelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tAnalyse asymptotique<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tD\u00e9veloppements limit\u00e9s<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tD\u00e9nombrements<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tEspaces vectoriels<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tMatrices<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tInt\u00e9grations<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tEspaces pr\u00e9hilbertiens<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tEspaces euclidiens<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tS\u00e9ries num\u00e9riques<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tProbabilit\u00e9s<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tD\u00e9terminants<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tVariables al\u00e9atoires<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tCONTACTEZ-NOUS<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Cours en ligne Maths en Maths Sup<\/a><\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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Chapitres Maths en MPSI, PCSI, MP2I, PTSI<\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tR\u00e9currence<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSommes et produits<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tNombres complexes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tTrigonom\u00e9trie<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tNombres r\u00e9els<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tEnsembles et applications<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFonctions (intro)<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFonctions usuelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tPrimitives<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\t\u00c9quations diff\u00e9rentielles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSuites num\u00e9riques<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tLimites et continuit\u00e9<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tD\u00e9riv\u00e9es<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tSyst\u00e8mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tPolyn\u00f4mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFractions rationnelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tArithm\u00e9tique et polyn\u00f4mes<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tArithm\u00e9tique et fractions rationnelles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Exercices corrig\u00e9s de Trigonom\u00e9trie en Maths Sup<\/h1>\n<\/div>\n

R\u00e9sum\u00e9 de cours<\/a> Exercices et corrig\u00e9s<\/a><\/p>\n

Cours en ligne de Maths en Maths Sup<\/a><\/p>\n

Plan des exercices : Equation, In\u00e9quations, Lin\u00e9arisation<\/span><\/h2>\n
\n

1. Des calculs
\n2. Des \u00e9quations
\n3. Des in\u00e9quations
\n4. Syst\u00e8mes d’\u00e9quations<\/a>
\n5. Lin\u00e9arisation<\/p>\n

1. Des calculs de trigonom\u00e9trie en maths sup<\/h2>\n

Exercice 1<\/b>
\nTrouver \"a \in \, ] \pi/2\, , \, \pi [\" tel que \"\sin(a) = \displaystyle \frac {\sqrt{5} - 1} 4\". On utilisera \"\cos(4 \, a)\".
\nOn obtient \"a = \displaystyle \frac {\alpha\, \pi} {10}\"<\/p>\n

avec \"\alpha =\"<\/p>\n

Correction :<\/strong>\u00a0\"\ast\" \"\cos(2 \,a) = 1 - 2 \sin ^2(a)\"
\n\"\cos(2 \, a) = \displaystyle 1 - \frac 1 8 \left ( 6 - 2 \sqrt{5} \right )\"
\n\"\cos(2 \, a) = \displaystyle \frac 1 4 + \frac{\sqrt{5} } 4\".<\/p>\n

\"\ast\" \"\cos(4 \, a) = 2 \, \cos^2(2\, a) - 1\"
\n\"\cos(4 \, a) = \displaystyle \frac 1 8 \left (6 + 2 \sqrt{5} \right ) - 1\"
\n\"\cos(4 \, a) = \displaystyle \frac 1 4 \left ( - 1 + \sqrt{5} \right )\"<\/p>\n

\"\ast\" On a obtenu :
\n\"\cos(4 \, a) = \sin(a) =\displaystyle \cos\left ( \frac {\pi} 2 - a \right )\"
\ndonc \"\exists\, k \in \mathbb{Z}, \, 4\, a = \displaystyle \frac {\pi} 2 - a + 2 \, k \, \pi\"
\nou \"\exists \, p \in \mathbb{Z}, \, 4\, a = \displaystyle a - \frac {\pi} 2 + 2 \, p \, \pi\",
\nce qui donne
\n\"\; \; \exists\, k \in \mathbb{Z}, \, a = \displaystyle \frac {\pi} {10} + \frac {2 \, k \, \pi} 5\"
\nou \"\exists\, p \in \mathbb{Z}, \, a = \displaystyle - \frac {\pi} {6} + \frac {2 \, p \, \pi} 3\"<\/p>\n

\"\displaystyle - \frac {\pi} {6} + \frac {2 \, \pi} 3 = \frac {3 \, \pi} 6\" et\u00a0\"\displaystyle - \frac {\pi} {6} + \frac {4 \, \pi} 3 = \frac {7 \, \pi} 6\"
\nIl n’y a aucune valeur de \"p\" donnant une valeur de la forme \"a = \displaystyle - \frac {\pi} {6} + \frac {2 \, p \, \pi} 3\" dans \"]\pi/2 \, , \, \pi[\".<\/p>\n

La seule valeur de la forme \"\displaystyle \frac {\pi} {10} + \frac {2 \, k \, \pi} 5\" dans \"]\pi/2 \, , \, \pi[\" est obtenue pour \"k = 2\"
\nDonc \"a = \displaystyle \frac {9 \, \pi} {10}\".<\/p>\n

Exercice 2<\/b>
\nSoient \"x\" et \"y\" dans \"]0 \, , \, \pi /2[\" v\u00e9rifiant \"\displaystyle \tan x = \frac 1 7\" et \"\tan y = 2\".
\na)\u00a0<\/b>En utilisant \"\tan(x + 2\, y)\", calculer \"x + 2 \, y\".<\/p>\n

Correction<\/strong> : \"\displaystyle \tan(2 \, y) = \frac {2 \, \tan(y)} {1 - \tan^2(y)} = - \frac 4 3\"
\n\"u = \tan(x + 2\, y) = \displaystyle \frac {\tan(x) + \tan(2 \, y) } {1 - \tan(x) \, \tan(2\, y)}\"
\n\"u = \displaystyle \frac {1/7 - 4/3} {1 + 1/7 . 4/3} = \frac { - 25} {25} = - 1\".
\nOn en d\u00e9duit qu’il existe \"k \in \mathbb{Z}\" tel que \"x + 2\, y = \displaystyle \frac {3 \, \pi} 4 + k \, \pi\".
\nOn remarque que \"0 < x + 2\, y < \displaystyle \frac {3 \pi } 2\" et \"\tan(x + 2 \, y) < 0\" donc \"\displaystyle \frac {\pi} 2 < x + 2\, y < \displaystyle \frac {3 \pi } 2\".
\n\"\displaystyle \frac {7\, \pi } 4\" est la seule valeur de la forme \"\displaystyle \frac {3 \, \pi} 4 + k \, \pi\" dans l’intervalle \"\displaystyle \left] \frac {\pi} 2 , \, \frac {3 \, \pi} 2 \right [\", donc \"x + 2 \, y = \displaystyle \frac {7\, \pi } 4\".<\/p>\n

b)<\/b>\u00a0Calculer \"\cos(2 \, y)\".
\nSoit \"a \in\displaystyle \left ]0 , \, \frac { \pi} 2 \right [\" tel que \"\cos(a) = \displaystyle \frac 3 5\". Exprimer \"x\" et \"y\" en fonction<\/a> de \"a\".<\/p>\n

Correction<\/strong> :\u00a0On a vu que \"\tan(2\, y) =\displaystyle - \frac 4 3 < 0\" avec \"2 \, y \in \; ]0 \, , \, \pi[\" donc \"2 \, y \in\; ]\pi/2\, , \, \pi[\".
\n\"\cos^2(2 \, y) = \displaystyle \frac 1 {1 + \tan^2(2 y) } = \frac 1 {1 + 16/9}\"
\n\"\cos^2(2 \, y) = \displaystyle \frac 9 {25 }\".
\nComme \"\cos(2 \, y) < 0\" car \"2 \, y \in\; ]\pi/2\, ,\, \pi[\".
\n\"\cos(2 \, y) = \displaystyle - \frac 3 {5 } = \cos (\pi - a)\"
\navec \"\pi - a \in \; ]\pi / 2\, ,\, \pi[\", donc \"2\, y = \pi - a\"
\nComme \"x + 2\, y = \displaystyle \frac { 3 \, \pi} 4\",\u00a0\"x = \displaystyle \frac { - \, \pi} 4 + a\".<\/p>\n

On a prouv\u00e9 que \"\displaystyle y = \frac {\pi - a} 2\" et \"\displaystyle \frac {- \pi} 4 + a\".<\/p>\n

Exercice 3<\/b>
\nSoit \"S = \displaystyle \frac 1 2 + \cos \frac {2 \pi } 5 + \cos \frac {4 \pi} 5\".
\nCalculer \"S \, \displaystyle \sin \frac \pi 5\".
\nEn d\u00e9duire la valeur de \"\displaystyle \cos \frac {2 \pi } 5\".<\/p>\n

Correction<\/strong> : On utilise \"\sin p \, \cos q = \displaystyle \frac 1 2 \left ( \sin \frac {p + q} 2 - \sin \frac {p - q} 2 \right )\"
\n\"2 \, S = \displaystyle \sin \frac \pi 5 + \left ( \sin \frac {3\pi} 5 - \sin \frac \pi 5 \right )\" \"\quad \quad \quad \quad \displaystyle + \, \left ( \sin \frac {5\pi} 5 - \sin \frac {3 \pi} 5 \right )\"
\ndonc \"S = 0\".<\/p>\n

Donc en posant \"x = \displaystyle \cos \frac {2 \pi } 5\", \"S = 0\" donne \"\displaystyle \frac 1 2 + x + 2 x ^2 - 1 = 0\" soit \"4 \, x^2 + 2 \,x - 1 = 0\".<\/p>\n

Cette \u00e9quation admet deux racines dont une seule est positive :
\non en d\u00e9duit que \"\displaystyle \cos \frac { 2\, \pi } 5 = \frac {\sqrt{5} - 1} 4\".<\/p>\n

Exercice 4<\/b>
\nCalculer \"\tan \displaystyle \frac {\pi} {12}\" puis \"\tan \displaystyle \frac {\pi} {24}\".<\/p>\n

Correction<\/strong> : On utilise la formule \"\tan(2\, x) = \displaystyle \frac {2 \, \tan(x)} {1 - \tan^2(x)}\" d’abord pour \"x = \displaystyle \frac {\pi} {12}\" et on pose \"t = \displaystyle \tan \frac {\pi} {12}\"
\n\"\displaystyle \tan \frac {\pi} 6 = \frac {2 \, t } {1 - t ^2} = \frac 1 {\sqrt{3}}\"
\nce qui donne \"1 - t ^2 = 2 \, \sqrt{3}\, t\"
\nsoit \"t ^2 + 2 \, \sqrt{3}\, t - 1 = 0\".
\ncette \u00e9quation a deux racines : \"2 - \sqrt{3}\" et \"- 2 - \sqrt{3} < 0\",
\ndonc \"\displaystyle \tan \frac {\pi} {12} = 2 - \sqrt{3}\".<\/p>\n

On r\u00e9utilise la m\u00eame m\u00e9thode en posant \"u = \tan \displaystyle \frac {\pi} {24}\".
\nOn obtient l’\u00e9quation \"2 - \sqrt{3} = \displaystyle \frac {2 \, u } {1 - u ^2}\"
\nsoit \"\left ( 2 - \sqrt{3} \right ) u ^2 + 2 \, u - 2 + \sqrt{3} = 0\"
\nadmet un discriminant \"\Delta = 16 ( 2 - \sqrt{3} )\"<\/p>\n

Une seule des racines est positive :
\n\"u = \displaystyle \frac {- 1 }{2 - \sqrt{3} } + \frac {2 \, \sqrt{2 - \sqrt3}} {2 - \sqrt{3} }\"
\npuis \"\displaystyle \frac { 1 }{2 - \sqrt{3} } = \frac {2 + \sqrt{3} } {4 - 3} = 2 + \sqrt{3}\"
\n\"u = -2 - \sqrt{3} + 2 \sqrt{2 + \sqrt{3}}\".
\n\"\tan\displaystyle \frac {\pi} {24} = -2 - \sqrt{3} + 2 \sqrt{2 + \sqrt{3}}\".<\/p>\n

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Les meilleurs professeurs particuliers<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Avis Google France <\/b>\u2605\u2605\u2605\u2605\u2605<\/b><\/span> 4,9 sur 5<\/b><\/p>\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t

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2. \u00c9quations<\/h2>\n

Exercice 1<\/b>
\nR\u00e9soudre \"\displaystyle \cos \left ( \frac {3 \, x} 4 \right ) = \sin \left ( x + \frac {3 \pi} 4 \right )\"<\/p>\n

Correction<\/strong> : \"\displaystyle \cos \left ( \frac {3 \, x} 4 \right ) = \cos \left ( \frac {\pi} 2 - x - \displaystyle \frac {3 \pi} 4 \right )\"
\nssi \"\displaystyle \cos \left ( \frac {3 \, x} 4 \right ) = \cos \left ( - \frac {\pi} 4 - x \right ) = \cos \left ( \frac {\pi} 4 + x \right )\"
\nssi \"\exists \, k \in \mathbb{Z} \, , \, \displaystyle \frac {3 \, x} 4 = \frac {\pi} 4 + x + 2 \, k \, \pi\"
\nou \"\exists \, p \in \mathbb{Z} \, , \, \displaystyle \frac {3 \, x} 4 = - \frac {\pi} 4 - x + 2 \, p \, \pi\"
\nssi \"\exists \, k \in \mathbb{Z} \, , \, \displaystyle \frac { x} 4 = - \frac {\pi} 4 - 2 \, k \, \pi\"
\nou \"\exists \, p \in \mathbb{Z} \, , \, \displaystyle \frac {7 \, x} 4 = - \frac {\pi} 4 + 2 \, p \, \pi\"
\nssi \"\exists \, k \in \mathbb{Z} \, , \, \displaystyle x = {\pi} - 8 \, k \, \pi\"
\nou \"\exists \, p \in \mathbb{Z} \, , \, \displaystyle x = - \frac {\pi} 7 + \frac {8 \,p \, \pi} 7\".<\/p>\n

Exercice 2<\/b>
\nEnsemble des r\u00e9els<\/a> \"m\" tels que l’\u00e9quation
\n\"m \cos(x) + \sqrt{1 - m ^2} \, \sin(x)\" \"\quad \quad \quad \quad \quad \quad \quad + \,3\, m - 4 \, m ^3 = 0\" .
\nait des solutions.
\nLes d\u00e9terminer.<\/p>\n

Correction<\/strong> : Pour que l’\u00e9quation ait un sens, il est n\u00e9cessaire que \"m \in [- 1\, , \, 1]\".
\nComme \"m ^2 +\left ( \sqrt{1 - m ^2} \right ) ^2 = 1\", il existe \"\varphi \in \mathbb{R}\" tel que
\n\"\quad \quad m = \cos(\varphi)\" et \"\sqrt{1 - m ^2} = \sin(\varphi)\".
\nL’\u00e9quation s’\u00e9crit alors
\n\"\quad \quad \cos(x - \varphi) = 4\, m ^3 - 3 \, m\".<\/p>\n

Elle admet des solution ssi \"- 1 \leq 4\, m ^3 - 3 \, m \leq 1\"
\nssi \"4\, m ^3 - 3 \, m + 1 \geq 0\"<\/p>\n

\n

\u00a0 et \"4\, m ^3 - 3 \, m - 1 \leq 0\".
\n\"\ast\" \"- 1\" est racine \u00e9vidente de \"4\, m ^3 - 3 \, m + 1\" et on peut \u00e9crire :
\n\"4\, m ^3 - 3 \, m + 1 = (m + 1)(4 \, m ^2 - 4 \, m + 1)\" \" 4\, m ^3 - 3 \, m + 1 =(m + 1)(2 \, m - 1)^2\"
\ndonc \"4\, m ^3 - 3 \, m + 1 \geq 0\" ssi \"m + 1 \geq 0\".<\/p>\n

\"\ast\" \"1\" est racine \u00e9vidente de \"4\, m ^3 - 3 \, m - 1\" et on peut \u00e9crire :
\n\"4\, m ^3 - 3 \, m - 1 = (m - 1)(4 \, m ^2 + 4 \, m + 1)\" \" 4\, m ^3 - 3 \, m - 1 =(m - 1)(2 \, m + 1)^2\"
\ndonc \"4\, m ^3 - 3 \, m - 1 \leq 0\" ssi \"m - 1 \leq 0\".<\/p>\n

L’\u00e9quation<\/a> a toujours des solutions lorsque \"m \in [ - 1 \, , \, 1]\".<\/p>\n

On transforme
\n\"4\, m ^3 - 3 \, m = 4 \, \cos^3(\varphi) - 3 \cos(\varphi)\"
\n\"\displaystyle = \frac 1 2 \left ( \textrm{e} ^{\textrm{i} \, \varphi} + \textrm{e} ^{ - \textrm{i} \, \varphi} \right ) ^3 - \frac 3 2 \left ( \textrm{e} ^{\textrm{i} \, \varphi} + \textrm{e} ^{\textrm{i} \, \varphi} \right )\"
\n\"\displaystyle = \frac 1 2 \left ( \textrm{e} ^{3\, \textrm{i} \, \varphi} + \textrm{e} ^{- 3 \, \textrm{i} \, \varphi} \right )\"
\n\"4\, m ^3 - 3 \, m = 4 \, \cos(3\, \varphi)\"<\/p>\n

L’\u00e9quation est donc \u00e9quivalente \u00e0
\n\"\cos(x - \varphi) = \cos(3 \, \varphi)\"
\nssi \"\exists \, k \in \mathbb{Z}, \, x - \varphi =3 \, \varphi + 2 \, k \, \pi\"
\nou \"\exists \, p \in \mathbb{Z}, \, x - \varphi = - 3 \, \varphi + 2 \, p \, \pi\"
\nssi \"\exists \, k \in \mathbb{Z}, \, x =4 \, \varphi + 2 \, k \, \pi\"
\nou \"\exists \, p \in \mathbb{Z}, \, x = - 2 \, \varphi + 2 \, p \, \pi\".<\/p>\n<\/div>\n

Exercice 3<\/b>
\nR\u00e9soudre \"\cos^4 (x) + \sin ^4 (x) = \displaystyle \frac {6 + \sqrt{3} } 8\"<\/p>\n

Correction<\/strong> : \"\cos^4 (x) + \sin ^4 (x) =\" \" (\cos^2(x) + \sin ^2(x) ) ^2 - 2 \sin^2 (x) \, \cos^2(x)\"
\n\"= 1 - \displaystyle \frac 1 2 \sin ^2(2 \, x)\"
\n\"= 1 - \displaystyle \frac 1 4 \left ( 1 - \cos(4 \, x) \right )\"
\n\"= \displaystyle \frac 3 4 + \frac { \cos(4 \, x)} 4\"
\nl’\u00e9quation est \u00e9quivalente \u00e0
\n\"\displaystyle \cos(4 \, x) = \frac {\sqrt{3}} 2\"
\nssi \"\displaystyle \exists \, k \in \mathbb{Z}, \, 4 \, x = \frac {\pi } 6 + 2 \,k \, \pi\"
\nou \"\displaystyle \exists \, p \in \mathbb{Z}, \, 4 \, x = - \frac {\pi } 6 + 2 \,k \, \pi\"
\nssi \"\displaystyle \exists \, k \in \mathbb{Z}, x = \frac {\pi } {24} + \frac {k \, \pi} 2\"
\nou \"\displaystyle \exists \, p \in \mathbb{Z}, \, x = - \frac {\pi } {24} + \frac {p \, \pi} 2\".<\/p>\n

Exercice 4<\/b>
\nR\u00e9soudre \"\tan(2 x) - \tan(x) = \sin(2 x)\"<\/p>\n

Correction<\/strong> : \"\bullet\" Premi\u00e8re m\u00e9thode\u00a0<\/b>
\nOn pose \"t = \tan(x)\"
\n\"\displaystyle \frac {2\, t} {1 - t ^2} - t = \frac {2\, t} {1 + t ^2}\"
\nssi \"2\, t(1 + t ^2) - t( 1 - t ^4) = 2 \, t(1 - t ^2)\"
\nssi \"t (t ^4 + 4 \, t ^2 - 1) = 0\"<\/p>\n

L’\u00e9quation \"x ^2 + 4 \,x - 1 = 0\" admet une seule racine positive \"x_1 = \sqrt{5} - 2\"<\/p>\n

\"t (t ^4 + 4 t ^2 - 1) = 0\" ssi \"t = 0\" ou\"t = \sqrt{\sqrt{5} - 2 }\" ou \"t = - \sqrt{\sqrt{5} - 2 }\".
\nOn introduit \"\varphi\" tel que \"\tan(\varphi) = \sqrt{\sqrt{5} - 2 }\"
\nLes solutions sont les r\u00e9els \"k \, \pi\" , \"\varphi + k \, \pi\" et \"- \varphi + k \, \pi\" lorsque \"k\in \mathbb{Z }\"<\/p>\n

\"\bullet\"\u00a0Autre m\u00e9thode\u00a0<\/b>
\n\"\tan( 2 x) - \tan(x) =\" \"\displaystyle \frac {\sin(2 \, x) \, \cos(x) - \cos(2 \, x) \, \sin(x) } { \cos(2 \, x) \, \cos(x)}\"
\n\"\tan( 2 x) - \tan(x) = \displaystyle \frac {\sin(x) } { \cos(2 \, x) \, \cos(x)}\"
\net comme \"\sin(2 \, x) = 2\, \sin(x) \, \cos(x)\",\u00a0 l’\u00e9quation est \u00e9quivalente \u00e0
\n\"\quad \sin(x) = 0\" ou \"2 \, \cos^2(x) \, \cos(2 x) = 1\"
\n\"2 \, \cos^2(x) \, \cos(2 x) = (\cos(2\, x) + 1) \, \cos(2 \, x)\"
\nLa deuxi\u00e8me \u00e9quation s’\u00e9crit
\n\"\quad \cos^2 (2 \, x) + \cos(2 \, x) - 1 = 0\"
\nL’\u00e9quation \"u ^2 + u - 1 = 0\" admet une seule racine dans \"[- 1 \,, \, 1]\" : \" \displaystyle \frac {\sqrt{5} - 1} 2\"
\nOn note \"a\" tel que \"\cos(a) = \displaystyle \frac {\sqrt{5} - 1} 2\".<\/p>\n

L’ensemble<\/a> des solutions est form\u00e9 par les r\u00e9els
\n\"\ast\" \"k \, \pi\" o\u00f9 \"k\in \mathbb{Z}\"
\n\"\ast\" \"\displaystyle \frac a 2 + k\, \pi\" o\u00f9 \"k\in \mathbb{Z}\"
\n\"\ast\" \"\displaystyle - \frac a 2 + k \, \pi\" o\u00f9 \"k\in \mathbb{Z}\".<\/p>\n

On pourra choisir \"a = \textrm{Arccos} \displaystyle \frac {\sqrt{5} - 1} 2\".<\/p>\n

3. In\u00e9quations<\/h4>\n

Exercice 1<\/b>
\nR\u00e9soudre l’in\u00e9quation
\n\"\quad \displaystyle \sqrt{3} \, \cos ( x) + \sin(x) - \sqrt{2} \leq 0\".<\/p>\n

Correction<\/strong> : On cherche la forme trigonom\u00e9trique de \"\sqrt{3 } + \, \textrm {i} = \displaystyle 2 \left ( \frac {\sqrt{3} } 2 + \, \textrm{i } \, \frac 1 2 \right ) = 2 \, \textrm{e} ^{\textrm{i} \, \pi / 6}\".
\n\"\displaystyle \sqrt{3} \, \cos ( x) + \sin(x) =\" \"\quad \quad \quad \displaystyle 2 \left ( \cos \frac {\pi} 6 \, \cos(x) + \sin \frac {\pi} 6 \, \sin(x) \right )\"
\n\"\displaystyle \sqrt{3} \, \cos ( x) + \sin(x) = 2 \cos \left (x - \frac {\pi} 6 \right )\"<\/p>\n

On doit donc r\u00e9soudre :
\n\"\quad \displaystyle \cos \left (x - \frac {\pi} 6 \right ) \leq \frac {\sqrt{2}} 2\"
\nssi \"\displaystyle \cos \left (x - \frac {\pi} 6 \right ) \leq \cos \left ( \frac {\pi} 4 \right )\"
\nssi \"\displaystyle \exists \, k \in \mathbb{Z} \, , \, x - \frac {\pi} 6 \in \left [ \frac {\pi} 4 + 2 \, k\, \pi \, , \, \frac {5 \, \pi} 4 + 2 \, k\, \pi \right]\"
\nssi \"\displaystyle \exists \, k \in \mathbb{Z} \, , \, x \in \left [ \frac {5 \pi} {12} + 2 \, k\, \pi \, , \, \frac {17 \, \pi} {12} + 2 \, k\, \pi \right]\"<\/p>\n

Exercice 2<\/b>
\nR\u00e9soudre\u00a0\u00a0\"2 \, \sin^2(x) + 3 \sin(x) - 2 \leq 0\"<\/p>\n

Correction<\/strong> : Les racines de l’\u00e9quation \"\quad \quad 2 \, t ^2 + 3\, t - 2 = 0\"
\nsont \"- 2\" et \"1/2\"
\ndonc \"2 \, t ^2 + 3\, t - 2 = (2\, t - 1) (t + 2)\"
\net on doit r\u00e9soudre :
\n\"\quad (2 \, \sin(x) - 1) (\sin(x) + 2) \geq 0\"
\nssi \"2 \, \sin(x) - 1 \geq 0\"
\nssi \"\displaystyle \sin(x) \geq \frac 1 2\"
\nssi \"\displaystyle \exists \, k \in \mathbb{Z} \, , \, x \in \left [ \frac {2\, \pi} 3 + 2 \, k \, \pi \, , \, \frac {7\, \pi} 3 + 2 \, k \, \pi \right ]\"<\/p>\n

Exercice 3<\/b>
\nR\u00e9soudre si \"x \in [0 \, , \, \pi]\",
\n\"\sin(x) + \sin(2 \, x) + \sin(5 \, x)\" \"\quad \quad \quad \quad \quad \quad \quad \quad +\, \sin(6 \, x) \leq 0\".<\/p>\n

Correction<\/strong> : On note \"f(x) =\" \"\quad \sin(x) + \sin(2 \,x) + \sin(5 \,x) + \sin(6\, x)\"
\net en utilisant \"\sin p + \sin q = \displaystyle 2 \sin\left ( \frac {p + q} 2 \right ) \, \cos \left ( \frac {p - q} 2 \right )\" pour transformer \"\sin(x) + \sin(6\, x)\" et \"\sin(2 \, x) + \sin(3\, x)\",
\n\"\sin(x) + \sin(6\, x) = \displaystyle 2 \, \sin \left ( \frac {7\, x} 2 \right ) \, \cos \left ( \frac {5\, x} 2 \right )\"
\n\"\sin(2 \,x) + \sin(5 \,x) = \displaystyle 2 \, \sin \left ( \frac {7\, x} 2 \right ) \, \cos \left ( \frac {3\, x} 2 \right )\"<\/p>\n

donc \"f(x)\" s’\u00e9crit\u00a0 \"\quad \displaystyle 2 \, \sin \left ( \frac {7\, x} 2 \right ) \left ( \cos \left ( \frac {5\, x} 2 \right ) + \cos \left ( \frac {3\, x} 2 \right ) \right )\"
\nen utilisant ensuite
\n\"\cos p + \cos q = \displaystyle 2 \cos\left ( \frac {p + q} 2 \right ) \, \cos \left ( \frac {p - q} 2 \right )\"
\n\"f(x) = \displaystyle 4 \, \sin \left ( \frac {7\, x} 2 \right ) \, \cos ( 2 \, x)\, \cos \left ( \frac { x} 2 \right )\".<\/p>\n

On cherche les \"x\in [0 \, , \, \pi]\" qui v\u00e9rifient \"f(x) = 0\".
\n\"\ast\" \"\displaystyle \sin \left ( \frac {7\, x} 2 \right ) = 0\" ssi \"x = \displaystyle \frac {2\, k \, \pi} 7\" o\u00f9 \"k = 0 , 1, 2 , 3\".
\n\"\displaystyle \sin \left ( \frac {7\, x} 2 \right )\" s’annule et change de signe en ces 4 points et est strictement positif sur\u00a0\"\displaystyle \left ] 0 , \, \, \frac {2\, \pi} 7 \right [\".
\n\"\ast\" \"\cos(2 x) = 0\" ssi \"x = \displaystyle \frac {\pi} 4\" ou \"\displaystyle \frac{3\, \pi} 4\" .
\n\"\cos(2 \, x)\" s’annule et change de signe en ces 2 points et est strictement positif sur \"\displaystyle \left [ 0 , \, \, \frac {3\, \pi} 4 \right [\".
\n\"\ast\" \"\displaystyle \cos \left ( \frac { x} 2 \right ) = 0\" ssi \"x = \pi\".
\nce dernier facteur est strictement positif sur \"[0 \, , \, \pi [\".<\/p>\n

On utilise le tableau de signes qui suit pour donner l’ensemble des solutions lorsque \"x \in [0 , \pi]\".
\nDans ce tableau, on fait appara\u00eetre les valeurs annulant \"f(x)\" et on \u00e9crit trois lignes r\u00e9sumant les signes des trois facteurs \u00e9tudi\u00e9s ci-dessus.
\nL’ensemble des solutions est d\u00e9fini par
\n\"\quad \displaystyle \left [0 , \frac {\pi} 4 \right ] \cup \left [\frac {2 \pi} 7 , \frac {4 \pi} 7 \right ] \cup \left [\frac {3 \pi} 4 , \frac {6 \pi} 7 \right ]\".<\/p>\n

\"tableau<\/p>\n

4. Syst\u00e8me d’\u00e9quations<\/h2>\n

Exercice 1<\/b>
\nSi \"a\" est un r\u00e9el donn\u00e9, r\u00e9soudre le syst\u00e8me \"\mathcal {S}\" :
\n\"\left \{\begin{matrix} \cos(a) + \cos(x + a ) + \cos(y + a )=0\\ \sin(a) + \sin(x+ a) + \sin(y+ a) =0\end{matrix} \right.\"<\/p>\n

Correction<\/strong> :\u00a0\"\bullet\" \"\mathcal {S}\" est \u00e9quivalent \u00e0 l’\u00e9quation obtenue en prenant la somme de la premi\u00e8re \u00e9quation et de \"\textrm{i}\" fois la deuxi\u00e8me soit \u00e0
\n\"\textrm{e} ^{\textrm{i} \, a} + \textrm{e} ^{\textrm{i} \, (x + a)}+ \textrm{e} ^{\textrm{i} \, (y + a)} = 0\"
\nsoit en multipliant par \"\textrm{e} ^{-\textrm{i} \, a}\" :
\n\"1 + \textrm{e} ^{\textrm{i} \, x}+ \textrm{e} ^{\textrm{i} \, y} = 0\"
\n\"\Leftrightarrow 1 + 2 \displaystyle \cos \left ( \frac {x - y} 2 \right ) \, \textrm{e} ^{\textrm{i} \, (x + y) / 2} = 0\"
\nen \u00e9galant les parties r\u00e9elles et imaginaires, on obtient le syst\u00e8me :
\n\"1 + \displaystyle 2 \, \cos \left ( \frac {x - y} 2 \right )\, \cos \left ( \frac {x + y} 2 \right ) = 0\"\u00a0 \u00a0(1)
\n\"\displaystyle 2 \cos \left ( \frac {x - y} 2 \right )\, \sin \left ( \frac {x + y} 2 \right ) = 0\"\u00a0 \u00a0(2)<\/p>\n

\"\bullet\" \"\displaystyle \cos \left ( \frac {x - y} 2 \right ) = 0\"\u00a0 est impossible car l’\u00e9quation (1) donnerait \"1 = 0\"
\ndonc (2) est \u00e9quivalent \u00e0
\n\"\quad \displaystyle \sin \left ( \frac {x + y} 2 \right ) = 0\"
\nssi \"\exists \, k \in \mathbb{Z}, \, x + y = 2\, k \, \pi\".<\/p>\n

\"\bullet\" Alors (1) s’\u00e9crit
\n\"\quad 1 + \displaystyle 2 \,\cos \left ( \frac {x - y} 2 \right )\, \cos ( k \, \pi) = 0\" (1)
\nssi \"\displaystyle \cos \left ( \frac {x - y} 2 \right ) = \frac {(-1) ^{k + 1} } 2\"<\/p>\n

\"\ast\" cas \"k = 2 \, q\"
\n\"\displaystyle \cos \left ( \frac {x - y} 2 \right ) = \frac { - 1 } 2\"
\nssi \"\exists \, p \in \mathbb {Z}, \, \displaystyle \frac {x - y} 2 = \frac {2 \pi } 3 + 2 \, p \, \pi\"\u00a0 \u00a0(a)
\nou \"\exists \, p \in \mathbb {Z}, \, \displaystyle \frac {x - y} 2 = \frac {-2 \pi } 3 + 2\, p \, \pi\"\u00a0 \u00a0(b)
\navec \"\displaystyle \frac {x + y} 2 = 2 \, q\, \pi\"\u00a0 \u00a0 (c)<\/p>\n

par somme et diff\u00e9rence de (c) et (a), on obtient :
\n\"x = \displaystyle \frac {2 \pi } 3 + 2 (p + q) \pi\"
\net \"y = \displaystyle \frac {-2 \pi } 3 + 2 (q - p) \pi\"
\npuis avec (c) et (b) :
\n\"x = \displaystyle \frac {-2 \pi } 3 + 2 (p + q) \pi\"
\net \"y = \displaystyle \frac {2 \pi } 3 + 2 (q - p) \pi\"<\/p>\n

\"\ast\" cas \"k = 2 \,q + 1\"
\n\"\displaystyle \cos \left ( \frac {x - y} 2 \right ) = \frac { 1 } 2\"
\nssi \"\exists\, p \in \mathbb {Z} , \, \displaystyle \frac {x - y} 2 = \frac {\pi } 3 + 2 \,p \,\pi\" (a)
\nou \"\exists \, p \in \mathbb {Z} , \, \displaystyle \frac {x - y} 2 = \frac {-\pi } 3 + 2 \,p\, \pi\" (b)
\navec \"\displaystyle \frac {x + y} 2 = (2 \,q + 1) \pi\" (c)<\/p>\n

Par somme et diff\u00e9rence de (c) et (a), on obtient :\u00a0\"x = \displaystyle \frac {\pi } 3 + (2 p + 2 q + 1) \pi\"
\net \"y = \displaystyle \frac {-\pi } 3 + (2 q + 1 - 2p) \pi\"
\nsoit
\n\"x = \displaystyle \frac {4 \pi } 3 + 2 (p + q) \pi\"
\net \"y = \displaystyle \frac {2 \pi } 3 + 2 (q - p) \pi\"<\/p>\n

puis avec (c) et (b),\u00a0\"x = \displaystyle \frac {- \pi } 3 + (2 p + 2 q + 1 ) \pi\"
\net \"y = \displaystyle \frac {\pi } 3 + (2 q + 1 - 2p) \pi\".
\nsoit
\n\"x = \displaystyle \frac {2 \pi } 3 + 2 (p + q ) \pi\"
\net \"y = \displaystyle \frac {4 \pi } 3 + 2( q - p) \pi\".<\/p>\n

Les couples solutions sont les couples
\n\"\displaystyle \left ( \frac {2 \pi } 3 + 2 (p + q) \pi\, , \, \frac {- 2 \pi } 3 + 2 (q - p) \pi \right )\"
\n\"\displaystyle \left ( \frac {- 2 \pi } 3 + 2 (p + q) \pi\, , \, \frac {2 \pi } 3 + 2 (q - p) \pi \right )\"
\n\"\displaystyle \left ( \frac {4 \pi } 3 + 2 (p + q) \pi\, , \, \frac { 2 \pi } 3 + 2 (q - p) \pi \right )\"
\n\"\displaystyle \left ( \frac {2 \pi } 3 + 2 (p + q) \pi\, , \, \frac {4 \pi } 3 + 2 (q - p) \pi \right )\"
\no\u00f9 \"(p, q) \in \mathbb{Z}^2\" .<\/p>\n

⚠️ N’oubliez pas de faire la synth\u00e8se des diff\u00e9rents cas !<\/p>\n

⚠️ Il ne faut surtout pas raisonner<\/a> avec les modulos dans un probl\u00e8me faisant intervenir des sommes<\/a> ou des diff\u00e9rences d’\u00e9quations valables modulo \"\pi\" ou \"2 \pi\".
\nPar exemple, en notant \"\alpha = \displaystyle \frac {2\, \pi} 3\", si l’on \u00e9crivait le premier groupe de solutions sous la forme :
\n\"\quad \quad x \equiv \alpha \; \; [2\, \pi]\" et \"y \equiv - \alpha \; \; [2\, \pi]\",
\nle couple \"(\alpha + 2 \pi\, , \, \alpha )\" serait solution alors qu’il est impossible de l’\u00e9crire sous la forme \"\quad \quad (\alpha + 2 (p + q) \pi\, ,\, \alpha + 2(q - p) \pi )\" puisque \"p + q = 1\" et \"q - p = 0\" n’admet pas de solution enti\u00e8re.<\/p>\n

Exercice 2<\/b>
\nR\u00e9soudre si \"a\" est un r\u00e9el donn\u00e9 le syst\u00e8me \"\mathcal {S}\" :
\n\"\left \{\begin{matrix} \cos(x ) + \cos(y )&=&1 - \sin(a) \\ \sin(x) - \sin(y) &=&\cos(a) \end{matrix} \right.\"<\/p>\n

\n

Correction<\/strong> : \"\mathcal {S}\" est \u00e9quivalent \u00e0 l’\u00e9quation obtenue en prenant la somme de la premi\u00e8re \u00e9quation et de \"\textrm{i}\" fois la deuxi\u00e8me soit \u00e0
\n\"\textrm{e} ^{\textrm{i} \, x }+ \textrm{e} ^{ -\textrm{i} \, y } = 1 + \textrm{i} \, \textrm{e} ^{\textrm{i} \, a }\"
\n\"\textrm{e} ^{\textrm{i} \, x }+ \textrm{e} ^{ -\textrm{i} \, y } = 1 + \, \textrm{e} ^{\textrm{i} \,( a+ \pi/2) }\"
\nssi
\n\"\displaystyle 2 \cos \left ( \frac {x + y} 2 \right ) \,\textrm{e} ^{\textrm{i} \, (x - y)/2 } =\" \"\displaystyle \quad \quad \quad \quad 2 \cos \left ( \frac {2 a + \pi} 4 \right ) \, \textrm{e} ^{\textrm{i} \, (a /2 + \pi / 4) }\"<\/p>\n

 <\/p>\n

On distingue alors deux cas
\n\"\bullet\" Cas 1
\n\"\displaystyle \cos \left ( \frac {x + y} 2 \right ) = \cos \left ( \frac {2 a + \pi} 4 \right ) \quad (1)\"
\net \"\displaystyle\textrm{e} ^{\textrm{i} \, (x - y)/2 } = \textrm{e} ^{\textrm{i} \, (a /2 + \pi / 4) } \quad (2)\"
\n\"(2)\" donne
\n\"\exists \, k \in \mathbb{Z},\, \displaystyle \frac {x - y} 2 = \frac {2 a + \pi} 4 + 2\, k \, \pi\"
\net (1) donne l’existence de \"p \in \mathbb{Z}\" tel que
\n(1 a) \"\displaystyle \frac {x + y} 2 = \frac {2 \,a + \pi} 4 + 2 \, p\, \pi\"
\nou
\n(1 b) \"\displaystyle \frac {x + y} 2 = - \frac {2 \,a + \pi} 4 + 2\, p\, \pi\"<\/p>\n

\"\ast\" en formant la somme et la diff\u00e9rence de (1a) et (2), on obtient :
\n\"x =\displaystyle a + \frac {\pi} 2 + 2(k + p) \pi\"
\n\"y = 2(p - k) \pi\"<\/p>\n

\"\ast\" en formant la somme et la diff\u00e9rence de (1b) et (2), on obtient :
\n\"x = 2(p + k) \pi\"
\n\"y =\displaystyle - a - \frac {\pi} 2 + 2(p - k) \pi\"<\/p>\n

\"\bullet\" Cas 2
\n\"\displaystyle \cos \left ( \frac {x + y} 2 \right ) = - \cos \left ( \frac {2 a + \pi} 4 \right ) \quad (3)\"
\net \"\displaystyle\textrm{e} ^{\textrm{i} \, (x - y)/2 } = - \textrm{e} ^{\textrm{i} \, (a /2 + \pi / 4) } \quad (4)\"<\/p>\n

\"(4)\" donne \"\exists \, k \in \mathbb{Z},\, \displaystyle \frac {x - y} 2 = \frac {2 a + \pi} 4 + \pi + 2\, k \,\pi\"
\net (3) donne l’existence de \"p \in \mathbb{Z}\" tel que
\n(3 a) \"\displaystyle \frac {x + y} 2 = \pi - \frac {2 a + \pi} 4 + 2 \, p\, \pi\"
\nou
\n(3 b) \"\displaystyle \frac {x + y} 2 = \pi + \frac {2 a + \pi} 4 + 2\, p \, \pi\"<\/p>\n

\"\ast\" en formant la somme et la diff\u00e9rence de (3a) et (4), on obtient
\n\"x = 2(p + k +1 ) \pi\"
\n\"y =\displaystyle a + \frac {\pi} 2 + 2(p - k) \pi\"<\/p>\n

\"\ast\" en formant la somme et la diff\u00e9rence de (3b) et (4), on obtient
\n\"x =\displaystyle a +\frac {\pi} 2 + 2(p + k +1 ) \pi\"
\n\"y = 2(p - k) \pi\"<\/p>\n

Les couples solutions sont les couples
\n\"\displaystyle \left ( \frac {\pi } 2 + a+ 2 (k + p) \pi\, , \, 2 (p - k) \pi \right )\"
\n\"\displaystyle \left ( 2 (p + k) \pi\, , \, - \frac { \pi } 2 - a + 2 (p - k) \pi \right )\"
\n\"\displaystyle \left ( 2 (p + k + 1) \pi\, , \, \frac { \pi } 2 + a + 2 (p - k) \pi \right )\"
\n\"\displaystyle \left ( \frac {\pi } 2 + a + 2 (p + k + 1) \pi\, , \, 2 (p - k) \pi \right )\"
\no\u00f9 \"(p, k) \in \mathbb{Z}^2\" .<\/p>\n<\/div>\n

⚠️ Les remarques faites \u00e0 la fin de l’exercice 1 restent valables ici.<\/p>\n

Exercice 3<\/b>
\nR\u00e9soudre le syst\u00e8me :
\n\"\mathcal{S} \; \; \left \{ \begin{matrix} x + y &=& \pi/3 \\ \sin x + \sin y &=& 1/2 \end{matrix} \right.\".<\/p>\n

Correction<\/strong> : \"\bullet\" On utilise \"\sin x + \sin y = 2\, \displaystyle \sin \frac {x + y} 2 \, \cos \frac {x- y} 2\"
\n\"\mathcal{S} \; \; \left \{ \begin{matrix} x + y &=& \pi/3 \\ \cos ((x -y)/2)& =& 1/2 \end{matrix} \right.\"<\/p>\n

\"\bullet\" \"\displaystyle \cos \frac {x- y} 2 = \frac 1 {2}\"
\n\"\displaystyle \Leftrightarrow \exists \, k \in \mathbb{Z} , \frac {x - y} 2 = \frac {\pm \, \pi} 3 + 2 \, k \, \pi\"<\/p>\n

\"\bullet\" Le syst\u00e8me
\n\"\mathcal{S} \; \; \left \{ \begin{matrix} x + y &= &\pi/3 \\ x - y &=& 2 \,\pi/3 + 4 \, k\, \pi \end{matrix} \right.\"
\ndonne \"\displaystyle x = \frac {\pi } 2 +2\, k \, \pi\" et \"\displaystyle y =- \frac {\pi} 6 - 2\, k \, \pi\"<\/p>\n

\"\bullet\" Le syst\u00e8me
\n\"\mathcal{S} \; \; \left \{ \begin{matrix} x + y &=& \pi/3 \\ x - y &=& - 2 \, \pi/3 + 4\, k \, \pi \end{matrix} \right.\"
\ndonne \"\displaystyle x = - \frac {\pi} 6 + 2\, k \, \pi\" et \"\displaystyle y = \frac {\pi } 2 - 2\, k \, \pi\"<\/p>\n

\"\bullet\" Les couples solutions sont \"\quad \quad \quad \displaystyle \left ( \frac {\pi } 2 +2\, k \, \pi \, , \, - \frac {\pi}6 - 2\, k \, \pi \right )\"
\n\"\quad \quad \textrm{ et} \displaystyle \left ( - \frac {\pi} 6 +2\, k \, \pi \, , \, \frac {\pi } 2 - 2\, k \, \pi \right )\"
\no\u00f9 \"k \in \mathbb{Z}\"<\/p>\n

⚠️ Il ne faut surtout pas raisonner avec les modulos dans un tel probl\u00e8me car on perdrait la condition \"x + y = \pi /3\".<\/p>\n

👍 La m\u00e9thode utilis\u00e9e ici est plus simple que l’utilisation de la premi\u00e8re \u00e9quation pour \u00e9crire la deuxi\u00e8me \u00e9quation sous la forme \"\sin(x) + \sin(\pi/3 - y) = 1/2\".<\/p>\n

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COURS PARTICULIERS EN LIGNE<\/h3>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Nous avons s\u00e9lectionn\u00e9 pour vous les meilleurs professeurs particuliers.<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\n

 <\/p>\n

5. Lin\u00e9arisation<\/h2>\n

Question 1<\/b>.
\nLin\u00e9arisation de \"\cos^n t\"\u00a0 lorsque \"n \in \mathbb{N }^*\" et \"t \in \mathbb{R}\".<\/p>\n

\n

Correction<\/strong> : On utilise les formules d’Euler :
\n\"\cos^n t = \displaystyle \left ( \frac {\textrm{e} ^{\textrm{i} \, t} +\textrm{e} ^{-\textrm{i} \, t} } 2 \; \; \right ) ^n\"
\npuis le bin\u00f4me de Newton :
\n\"\cos^n t = \displaystyle \frac 1 {2 ^{n} } \sum _{k = 0} ^n \binom {n} k (\textrm{e} ^{\textrm{i} \, t})^k \,( \textrm{e} ^{-\textrm{i} \, t}) ^{n - k}\"
\n\"\cos^n t = \displaystyle \frac 1 {2 ^{n} } \sum _{k = 0} ^n \binom {n} k \textrm{e} ^{\textrm{i} \, k \, t}\, \textrm{e} ^{\textrm{i} \,(k-n) \, t}\"
\n\"\cos^n t = \displaystyle \frac 1 {2 ^{n} } \sum _{k = 0} ^n \binom {n} k \textrm{e} ^{\textrm{i} \,(2 \, k - n) \, t}\"
\net en prenant la partie r\u00e9elle
\n\"\cos^n t = \displaystyle \frac 1 {2 ^{n} } \; \sum _{k = 0} ^n \binom {n} k \cos\left ( {(2 \,k - n) \, t} \right )\"<\/p>\n

 <\/p>\n

👍 Le principe est bien s\u00fbr le m\u00eame pour obtenir par exemple \"\cos^3 t = \displaystyle \frac 1 4 \left ( \cos(3 \, t) + 3 \, \cos(t) \right )\" ou \"\cos^4 t = \displaystyle \frac 1 8 \left ( \cos(4 \, t) + 4\, \cos(2\, t) + 3 \right )\".<\/span><\/p>\n<\/div>\n

V\u00e9rifiez que vous savez le faire !<\/p>\n

Question 2<\/b>
\nSimplifier le r\u00e9sultat pr\u00e9c\u00e9dent pour \"\cos^{2 n} (t)\"<\/div>\n
\n

\"\cos^{2n} t = \displaystyle \frac 1 {2 ^{2n} } \sum _{k = 0} ^{2n} \binom {2 n} k \cos \left ( {(2 k - 2n) \, t} \right ) \"
\nOn transforme
\n\"A_n = \displaystyle \sum _{k = n + 1 } ^{2n} \binom {2 n} k \cos\left ( {(2 k - 2n) \, t} \right )\"
\nen posant \"j = 2 n - k\"
\n\"A_n = \displaystyle \sum _{j = 0} ^{n - 1} \binom {2 n} {2 n - j} \cos\left ( {(4 n - 2 j - 2n) \, t} \right )\"
\npar propri\u00e9t\u00e9 du coefficient du bin\u00f4me
\n\"A_n = \displaystyle \sum _{j = 0} ^{n - 1} \binom {2 n} {j} \cos \left ( {(2 n - 2 j) \, t} \right )\"
\net par parit\u00e9 de la fonction \"\cos\"
\n\"A_n = \displaystyle \sum _{j = 0} ^{n - 1} \binom {2 n} {j} \cos\left ( {(2 j - 2 n) \, t}\right )\".
\nPuis en consid\u00e9rant le terme pour \"k = n\" de la somme :
\n\"\displaystyle \binom {2 n} {n} \cos \left ( {(2 n - 2 n) \, t} \right ) = \binom {2 n} {n}\"
\n\"\cos^{2n} t = \displaystyle \frac 1 {2 ^{2n} } \binom {2 n} {n}\" \" \quad \quad \displaystyle + \, \frac 1 {2 ^{2n - 1} } \; \sum _{k = 0} ^{n - 1} \binom {2 n} k \cos \left ( {(2 k - 2n) \, t} \right )\"<\/p>\n

Question 3<\/b>
\nSimplifier l’\u00e9criture de \"\cos^{2 n + 1} (t)\".<\/p>\n

Correction<\/strong> : Valeur de \"\cos^{2n +1}(t)\"
\n\"\cos^{2n +1 } t =\" \" \displaystyle \frac 1 {2 ^{2n +1 } } \sum _{k = 0} ^{2n+1 } \binom {2 n +1 } k \cos\left ( {(2 k - 2n - 1) \, t}\right ) \"
\nOn transforme
\n\"B_n = \displaystyle \sum _{k = n + 1 } ^{2n +1} \binom {2 n +1 } k \cos \left ( {(2 k - 2n - 1 ) \, t}\right )\"
\nen posant \"j = 2 n +1 - k\"
\navec le m\u00eame type de calculs que dans le cas pair :
\n\"B_n = \displaystyle \sum _{j = 0 } ^{n } \binom {2 n +1 } {j} \cos \left ( {(2 j - 2n - 1 ) \, t}\right )\"
\ndonc
\n\"\cos^{2n+1} t =\" \"\quad \displaystyle \frac 1 {2 ^{2n} } \sum _{k = 0} ^{n} \binom {2 n +1 } k \cos \left ( {(2 k - 2n - 1 ) \, t}\right ) \"<\/p>\n

Ne prenez pas de retard et d\u00e9couvrez d\u00e8s maintenant les chapitres au programme de Maths \u00e0 venir en MPSI, PTSI et PCSI :<\/p>\n