Trigonométrie ★★★
Exercices niveau moyen - Tronc commun Sciences BIOF
📌 Exercice 1 : Angles associés ★★★
Simplifier les expressions suivantes :
\(A = \cos\left(\frac{\pi}{2} + x\right) + \sin(\pi - x)\)
\(B = \sin\left(\frac{\pi}{2} - x\right) + \cos(\pi + x)\)
\(C = \tan(\pi - x) + \tan\left(\frac{\pi}{2} + x\right)\)
💡 Indice : Utiliser les formules des angles associés (\(\pi \pm x\), \(\frac{\pi}{2} \pm x\)).
✅ Corrigé
A : \(\cos\left(\frac{\pi}{2} + x\right) = -\sin(x)\) et \(\sin(\pi - x) = \sin(x)\)
Donc \(A = -\sin(x) + \sin(x) = 0\)
Donc \(A = -\sin(x) + \sin(x) = 0\)
B : \(\sin\left(\frac{\pi}{2} - x\right) = \cos(x)\) et \(\cos(\pi + x) = -\cos(x)\)
Donc \(B = \cos(x) - \cos(x) = 0\)
Donc \(B = \cos(x) - \cos(x) = 0\)
C : \(\tan(\pi - x) = -\tan(x)\) et \(\tan\left(\frac{\pi}{2} + x\right) = -\cot(x) = -\frac{1}{\tan(x)}\)
Donc \(C = -\tan(x) - \frac{1}{\tan(x)} = -\left(\tan x + \frac{1}{\tan x}\right)\)
Donc \(C = -\tan(x) - \frac{1}{\tan(x)} = -\left(\tan x + \frac{1}{\tan x}\right)\)
📌 Exercice 2 : Calcul de cosinus et sinus ★★★
On donne \(\sin x = \frac{3}{5}\) avec \(x \in \left[\frac{\pi}{2}; \pi\right]\). Calculer \(\cos x\) et \(\tan x\).
On donne \(\cos x = -\frac{2}{3}\) avec \(x \in \left[\pi; \frac{3\pi}{2}\right]\). Calculer \(\sin x\) et \(\tan x\).
💡 Indice : Utiliser \(\cos^2 x + \sin^2 x = 1\) et déterminer le signe selon le quadrant.
✅ Corrigé
1er cas : \(\sin x = \frac{3}{5}\), \(x \in [\frac{\pi}{2}; \pi]\)
\(\cos^2 x = 1 - \sin^2 x = 1 - \frac{9}{25} = \frac{16}{25}\) → \(\cos x = \pm \frac{4}{5}\)
Sur \([\frac{\pi}{2}; \pi]\), \(\cos x \leq 0\) donc \(\cos x = -\frac{4}{5}\)
\(\tan x = \frac{\sin x}{\cos x} = \frac{3/5}{-4/5} = -\frac{3}{4}\)
\(\cos^2 x = 1 - \sin^2 x = 1 - \frac{9}{25} = \frac{16}{25}\) → \(\cos x = \pm \frac{4}{5}\)
Sur \([\frac{\pi}{2}; \pi]\), \(\cos x \leq 0\) donc \(\cos x = -\frac{4}{5}\)
\(\tan x = \frac{\sin x}{\cos x} = \frac{3/5}{-4/5} = -\frac{3}{4}\)
2ème cas : \(\cos x = -\frac{2}{3}\), \(x \in [\pi; \frac{3\pi}{2}]\)
\(\sin^2 x = 1 - \cos^2 x = 1 - \frac{4}{9} = \frac{5}{9}\) → \(\sin x = \pm \frac{\sqrt{5}}{3}\)
Sur \([\pi; \frac{3\pi}{2}]\), \(\sin x \leq 0\) donc \(\sin x = -\frac{\sqrt{5}}{3}\)
\(\tan x = \frac{\sin x}{\cos x} = \frac{-\sqrt{5}/3}{-2/3} = \frac{\sqrt{5}}{2}\)
\(\sin^2 x = 1 - \cos^2 x = 1 - \frac{4}{9} = \frac{5}{9}\) → \(\sin x = \pm \frac{\sqrt{5}}{3}\)
Sur \([\pi; \frac{3\pi}{2}]\), \(\sin x \leq 0\) donc \(\sin x = -\frac{\sqrt{5}}{3}\)
\(\tan x = \frac{\sin x}{\cos x} = \frac{-\sqrt{5}/3}{-2/3} = \frac{\sqrt{5}}{2}\)
📌 Exercice 3 : Équations trigonométriques simples ★★★
Résoudre dans \(\mathbb{R}\) les équations suivantes :
a) \(\cos x = \frac{\sqrt{3}}{2}\)
b) \(\sin x = -\frac{1}{2}\)
c) \(\tan x = \sqrt{3}\)
💡 Indice : Utiliser les valeurs des angles remarquables (\(\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}\)).
✅ Corrigé
a) \(\cos x = \frac{\sqrt{3}}{2} = \cos\left(\frac{\pi}{6}\right)\)
\(x = \frac{\pi}{6} + 2k\pi\) ou \(x = -\frac{\pi}{6} + 2k\pi\), \(k \in \mathbb{Z}\)
\(x = \frac{\pi}{6} + 2k\pi\) ou \(x = -\frac{\pi}{6} + 2k\pi\), \(k \in \mathbb{Z}\)
b) \(\sin x = -\frac{1}{2} = \sin\left(-\frac{\pi}{6}\right)\)
\(x = -\frac{\pi}{6} + 2k\pi\) ou \(x = \pi - \left(-\frac{\pi}{6}\right) + 2k\pi = \frac{7\pi}{6} + 2k\pi\), \(k \in \mathbb{Z}\)
\(x = -\frac{\pi}{6} + 2k\pi\) ou \(x = \pi - \left(-\frac{\pi}{6}\right) + 2k\pi = \frac{7\pi}{6} + 2k\pi\), \(k \in \mathbb{Z}\)
c) \(\tan x = \sqrt{3} = \tan\left(\frac{\pi}{3}\right)\)
\(x = \frac{\pi}{3} + k\pi\), \(k \in \mathbb{Z}\)
\(x = \frac{\pi}{3} + k\pi\), \(k \in \mathbb{Z}\)
📌 Exercice 4 : Équations avec transformations ★★★
Résoudre dans \(\mathbb{R}\) les équations suivantes :
a) \(\cos(2x) = \frac{1}{2}\)
b) \(\sin(3x) = \sin x\)
c) \(\cos(2x) + \cos x = 0\)
💡 Indice : Pour c), utiliser \(\cos(2x) = 2\cos^2 x - 1\).
✅ Corrigé
a) \(\cos(2x) = \frac{1}{2} = \cos\left(\frac{\pi}{3}\right)\)
\(2x = \frac{\pi}{3} + 2k\pi\) ou \(2x = -\frac{\pi}{3} + 2k\pi\)
Donc \(x = \frac{\pi}{6} + k\pi\) ou \(x = -\frac{\pi}{6} + k\pi\), \(k \in \mathbb{Z}\)
\(2x = \frac{\pi}{3} + 2k\pi\) ou \(2x = -\frac{\pi}{3} + 2k\pi\)
Donc \(x = \frac{\pi}{6} + k\pi\) ou \(x = -\frac{\pi}{6} + k\pi\), \(k \in \mathbb{Z}\)
b) \(\sin(3x) = \sin x\)
\(3x = x + 2k\pi\) ou \(3x = \pi - x + 2k\pi\)
Soit \(2x = 2k\pi\) → \(x = k\pi\)
ou \(4x = \pi + 2k\pi\) → \(x = \frac{\pi}{4} + \frac{k\pi}{2}\)
\(3x = x + 2k\pi\) ou \(3x = \pi - x + 2k\pi\)
Soit \(2x = 2k\pi\) → \(x = k\pi\)
ou \(4x = \pi + 2k\pi\) → \(x = \frac{\pi}{4} + \frac{k\pi}{2}\)
c) \(\cos(2x) + \cos x = 0\) → \(2\cos^2 x - 1 + \cos x = 0\) → \(2\cos^2 x + \cos x - 1 = 0\)
On pose \(X = \cos x\) : \(2X^2 + X - 1 = 0\) → \(\Delta = 1 + 8 = 9\)
\(X = \frac{-1 \pm 3}{4}\) → \(X = \frac{1}{2}\) ou \(X = -1\)
\(\cos x = \frac{1}{2}\) → \(x = \pm \frac{\pi}{3} + 2k\pi\)
\(\cos x = -1\) → \(x = \pi + 2k\pi\)
On pose \(X = \cos x\) : \(2X^2 + X - 1 = 0\) → \(\Delta = 1 + 8 = 9\)
\(X = \frac{-1 \pm 3}{4}\) → \(X = \frac{1}{2}\) ou \(X = -1\)
\(\cos x = \frac{1}{2}\) → \(x = \pm \frac{\pi}{3} + 2k\pi\)
\(\cos x = -1\) → \(x = \pi + 2k\pi\)
📌 Exercice 5 : Simplification d'expressions ★★★
Simplifier les expressions suivantes :
\(A = \cos^2 x - \sin^2 x + \cos^2 x + \sin^2 x\)
\(B = \frac{\sin x}{1 + \cos x} + \frac{\sin x}{1 - \cos x}\)
\(C = \frac{1}{\cos x} - \tan x \cdot \sin x\)
💡 Indice : Mettre au même dénominateur pour B et utiliser \(\tan x = \frac{\sin x}{\cos x}\).
✅ Corrigé
A : \(\cos^2 x - \sin^2 x + \cos^2 x + \sin^2 x = \cos^2 x + \cos^2 x = 2\cos^2 x\)
(car \(-\sin^2 x + \sin^2 x = 0\))
(car \(-\sin^2 x + \sin^2 x = 0\))
B : \(\frac{\sin x}{1 + \cos x} + \frac{\sin x}{1 - \cos x} = \frac{\sin x(1 - \cos x) + \sin x(1 + \cos x)}{(1 + \cos x)(1 - \cos x)}\)
\(= \frac{\sin x - \sin x \cos x + \sin x + \sin x \cos x}{1 - \cos^2 x} = \frac{2\sin x}{\sin^2 x} = \frac{2}{\sin x}\)
\(= \frac{\sin x - \sin x \cos x + \sin x + \sin x \cos x}{1 - \cos^2 x} = \frac{2\sin x}{\sin^2 x} = \frac{2}{\sin x}\)
C : \(\frac{1}{\cos x} - \tan x \cdot \sin x = \frac{1}{\cos x} - \frac{\sin x}{\cos x} \cdot \sin x = \frac{1}{\cos x} - \frac{\sin^2 x}{\cos x} = \frac{1 - \sin^2 x}{\cos x} = \frac{\cos^2 x}{\cos x} = \cos x\)
📌 Exercice 6 : Formules d'addition ★★★
Calculer \(\cos 75^\circ\) et \(\sin 75^\circ\) en utilisant les formules d'addition.
Calculer \(\cos 15^\circ\) et \(\sin 15^\circ\) en utilisant les formules d'addition.
💡 Indice : \(75^\circ = 45^\circ + 30^\circ\) et \(15^\circ = 45^\circ - 30^\circ\).
✅ Corrigé
\(\cos 75^\circ\) : \(\cos(45^\circ + 30^\circ) = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ\)
\(= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}\)
\(= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}\)
\(\sin 75^\circ\) : \(\sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ\)
\(= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}\)
\(= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}\)
\(\cos 15^\circ\) : \(\cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ\)
\(= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}\)
\(= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}\)
\(\sin 15^\circ\) : \(\sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ\)
\(= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}\)
\(= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}\)
📌 Exercice 7 : Inéquations trigonométriques ★★★
Résoudre dans \([0; 2\pi]\) les inéquations suivantes :
a) \(\cos x \leq \frac{1}{2}\)
b) \(\sin x > \frac{\sqrt{3}}{2}\)
c) \(\tan x \geq 1\)
💡 Indice : Utiliser le cercle trigonométrique pour déterminer les intervalles.
✅ Corrigé
a) \(\cos x \leq \frac{1}{2}\) → sur \([0; 2\pi]\) : \(x \in [\frac{\pi}{3}; \frac{5\pi}{3}]\)
b) \(\sin x > \frac{\sqrt{3}}{2}\) → sur \([0; 2\pi]\) : \(x \in ]\frac{\pi}{3}; \frac{2\pi}{3}[\)
c) \(\tan x \geq 1\) → sur \([0; 2\pi]\) en excluant \(\frac{\pi}{2}\) et \(\frac{3\pi}{2}\) :
\(x \in [\frac{\pi}{4}; \frac{\pi}{2}[ \cup [\frac{5\pi}{4}; \frac{3\pi}{2}[\)
\(x \in [\frac{\pi}{4}; \frac{\pi}{2}[ \cup [\frac{5\pi}{4}; \frac{3\pi}{2}[\)