Série d'exercices corrigés - Tronc commun Sciences BIOF
📌 Exercice 1 ★★★
Soit \(ABC\) un triangle tel que \(\widehat{BAC} = 60^\circ\) et \(\widehat{ABC} = 40\) grad.
1) Calculer en radian la mesure de l'angle \(\widehat{ACB}\).
2) Dans le plan orienté, on considère \(MNP\) un triangle isocèle en \(M\) tel que \((\widehat{NP},\widehat{PM}) = \frac{\pi}{7} + 2k\pi\). Calculer les mesures des angles orientés \((\widehat{PN},\widehat{PM})\) et \((\widehat{MN},\widehat{MP})\).
✅ Corrigé détaillé
1) Conversion : \(40\) grad = \(40 \times \frac{\pi}{200} = \frac{\pi}{5}\) rad.
\(60^\circ = \frac{\pi}{3}\) rad.
Somme des angles d'un triangle = \(\pi\) rad.
Donc \(\widehat{ACB} = \pi - \frac{\pi}{3} - \frac{\pi}{5} = \pi - \frac{5\pi+3\pi}{15} = \pi - \frac{8\pi}{15} = \frac{7\pi}{15}\) rad.
\(60^\circ = \frac{\pi}{3}\) rad.
Somme des angles d'un triangle = \(\pi\) rad.
Donc \(\widehat{ACB} = \pi - \frac{\pi}{3} - \frac{\pi}{5} = \pi - \frac{5\pi+3\pi}{15} = \pi - \frac{8\pi}{15} = \frac{7\pi}{15}\) rad.
2 ) Triangle isocèle en \(M\) → \(MN = MP\).
L'angle \((\widehat{NP},\widehat{PM})\) est l'angle extérieur. Dans un triangle isocèle, les angles à la base sont égaux.
\( \widehat{NMP} = \frac{\pi}{7} \). Les angles à la base valent \(\frac{\pi - \frac{\pi}{7}}{2} = \frac{6\pi/7}{2} = \frac{3\pi}{7}\).
\((\widehat{PN},\widehat{PM}) = \pi - \frac{3\pi}{7} = \frac{4\pi}{7}\).
\((\widehat{MN},\widehat{MP}) = \frac{3\pi}{7}\).
L'angle \((\widehat{NP},\widehat{PM})\) est l'angle extérieur. Dans un triangle isocèle, les angles à la base sont égaux.
\( \widehat{NMP} = \frac{\pi}{7} \). Les angles à la base valent \(\frac{\pi - \frac{\pi}{7}}{2} = \frac{6\pi/7}{2} = \frac{3\pi}{7}\).
\((\widehat{PN},\widehat{PM}) = \pi - \frac{3\pi}{7} = \frac{4\pi}{7}\).
\((\widehat{MN},\widehat{MP}) = \frac{3\pi}{7}\).
📌 Exercice 2 ★★★
On considère sur le cercle trigonométrique les points \(A\left(\frac{17\pi}{4}\right)\), \(B\left(\frac{45\pi}{4}\right)\), \(C\left(-\frac{13\pi}{3}\right)\), \(D\left(-\frac{31\pi}{6}\right)\).
1) Déterminer l'abscisse curviligne principale de chacun des points.
2) Représenter les points sur le cercle trigonométrique.
✅ Corrigé détaillé
A) \(\frac{17\pi}{4} = \frac{16\pi}{4} + \frac{\pi}{4} = 4\pi + \frac{\pi}{4}\) → mesure principale \(\frac{\pi}{4}\).
B) \(\frac{45\pi}{4} = \frac{44\pi}{4} + \frac{\pi}{4} = 11\pi + \frac{\pi}{4}\) → \(11\pi = 10\pi + \pi\) → mesure principale \(\pi + \frac{\pi}{4} = \frac{5\pi}{4}\).
C) \(-\frac{13\pi}{3} = -\frac{12\pi}{3} - \frac{\pi}{3} = -4\pi - \frac{\pi}{3}\) → mesure principale \(-\frac{\pi}{3}\).
D) \(-\frac{31\pi}{6} = -\frac{30\pi}{6} - \frac{\pi}{6} = -5\pi - \frac{\pi}{6}\) → mesure principale \(-\frac{\pi}{6}\).
📌 Exercice 3 ★★★
1) Calculer les rapports trigonométriques des angles : \(-\frac{\pi}{4}, \frac{5\pi}{6}, \frac{4\pi}{3}, 11\pi, \frac{19\pi}{3}\).
2) Sachant que \(\cos\frac{\pi}{12} = \frac{\sqrt{6}+\sqrt{2}}{4}\), calculer \(\sin\frac{\pi}{12}\), puis les rapports de \(\frac{11\pi}{12}\) et \(\frac{13\pi}{12}\).
3) Soit \(\alpha \in [\frac{\pi}{2}; \pi]\) tel que \(\sin\alpha = \frac{1}{5}\). Calculer \(\cos\alpha\) et \(\tan\alpha\).
4) Soit \(\beta \in [-\frac{\pi}{2}; 0]\) tel que \(\tan\beta = -2\). Calculer \(\cos\beta\) et \(\sin\beta\).
✅ Corrigé détaillé
1) Tableau des valeurs :
\(\sin(-\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}\), \(\cos(-\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\), \(\tan(-\frac{\pi}{4}) = -1\)
\(\sin\frac{5\pi}{6} = \frac{1}{2}\), \(\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}\), \(\tan\frac{5\pi}{6} = -\frac{\sqrt{3}}{3}\)
\(\sin\frac{4\pi}{3} = -\frac{\sqrt{3}}{2}\), \(\cos\frac{4\pi}{3} = -\frac{1}{2}\), \(\tan\frac{4\pi}{3} = \sqrt{3}\)
\(\sin 11\pi = 0\), \(\cos 11\pi = -1\), \(\tan 11\pi = 0\)
\(\sin\frac{19\pi}{3} = \sin(6\pi + \frac{\pi}{3}) = \frac{\sqrt{3}}{2}\), \(\cos\frac{19\pi}{3} = \frac{1}{2}\), \(\tan\frac{19\pi}{3} = \sqrt{3}\)
\(\sin(-\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}\), \(\cos(-\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\), \(\tan(-\frac{\pi}{4}) = -1\)
\(\sin\frac{5\pi}{6} = \frac{1}{2}\), \(\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}\), \(\tan\frac{5\pi}{6} = -\frac{\sqrt{3}}{3}\)
\(\sin\frac{4\pi}{3} = -\frac{\sqrt{3}}{2}\), \(\cos\frac{4\pi}{3} = -\frac{1}{2}\), \(\tan\frac{4\pi}{3} = \sqrt{3}\)
\(\sin 11\pi = 0\), \(\cos 11\pi = -1\), \(\tan 11\pi = 0\)
\(\sin\frac{19\pi}{3} = \sin(6\pi + \frac{\pi}{3}) = \frac{\sqrt{3}}{2}\), \(\cos\frac{19\pi}{3} = \frac{1}{2}\), \(\tan\frac{19\pi}{3} = \sqrt{3}\)
2) \(\sin^2\frac{\pi}{12} = 1 - \cos^2\frac{\pi}{12} = 1 - \frac{(\sqrt{6}+\sqrt{2})^2}{16} = 1 - \frac{6+2+2\sqrt{12}}{16} = 1 - \frac{8+4\sqrt{3}}{16} = \frac{8-4\sqrt{3}}{16} = \frac{2-\sqrt{3}}{4}\)
\(\sin\frac{\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4}\) (car positif)
\(\frac{11\pi}{12} = \pi - \frac{\pi}{12}\) → \(\sin\frac{11\pi}{12} = \sin\frac{\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4}\), \(\cos\frac{11\pi}{12} = -\cos\frac{\pi}{12} = -\frac{\sqrt{6}+\sqrt{2}}{4}\)
\(\frac{13\pi}{12} = \pi + \frac{\pi}{12}\) → \(\sin\frac{13\pi}{12} = -\sin\frac{\pi}{12} = -\frac{\sqrt{6}-\sqrt{2}}{4}\), \(\cos\frac{13\pi}{12} = -\cos\frac{\pi}{12} = -\frac{\sqrt{6}+\sqrt{2}}{4}\)
\(\sin\frac{\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4}\) (car positif)
\(\frac{11\pi}{12} = \pi - \frac{\pi}{12}\) → \(\sin\frac{11\pi}{12} = \sin\frac{\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4}\), \(\cos\frac{11\pi}{12} = -\cos\frac{\pi}{12} = -\frac{\sqrt{6}+\sqrt{2}}{4}\)
\(\frac{13\pi}{12} = \pi + \frac{\pi}{12}\) → \(\sin\frac{13\pi}{12} = -\sin\frac{\pi}{12} = -\frac{\sqrt{6}-\sqrt{2}}{4}\), \(\cos\frac{13\pi}{12} = -\cos\frac{\pi}{12} = -\frac{\sqrt{6}+\sqrt{2}}{4}\)
3 ) \(\cos^2\alpha = 1 - \frac{1}{25} = \frac{24}{25}\) → \(\cos\alpha = -\frac{2\sqrt{6}}{5}\) (car \(\alpha \in [\frac{\pi}{2};\pi]\), cos négatif).
\(\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{1/5}{-2\sqrt{6}/5} = -\frac{1}{2\sqrt{6}} = -\frac{\sqrt{6}}{12}\)
\(\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{1/5}{-2\sqrt{6}/5} = -\frac{1}{2\sqrt{6}} = -\frac{\sqrt{6}}{12}\)
4) \(\tan\beta = -2\), \(\beta \in [-\frac{\pi}{2};0]\) → sin négatif, cos positif.
\(\cos^2\beta = \frac{1}{1+\tan^2\beta} = \frac{1}{1+4} = \frac{1}{5}\) → \(\cos\beta = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}\)
\(\sin\beta = \tan\beta \cdot \cos\beta = -2 \times \frac{\sqrt{5}}{5} = -\frac{2\sqrt{5}}{5}\)
\(\cos^2\beta = \frac{1}{1+\tan^2\beta} = \frac{1}{1+4} = \frac{1}{5}\) → \(\cos\beta = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}\)
\(\sin\beta = \tan\beta \cdot \cos\beta = -2 \times \frac{\sqrt{5}}{5} = -\frac{2\sqrt{5}}{5}\)
📌 Exercice 4 ★★★
Montrer que :
1) \(\sin\frac{7\pi}{12} = \cos\frac{\pi}{12}\)
2) \(\cos\frac{5\pi}{12} = \sin\frac{\pi}{12}\)
3) \(\cos^4 x - \sin^4 x = 1 - 2\sin^2 x\)
4) \(\cos^4 x + \sin^4 x = 1 - 2\sin^2 x \cos^2 x\)
5) \(\tan^2 x - \sin^2 x = \tan^2 x \sin^2 x = \frac{\tan^4 x}{1+\tan^2 x}\)
✅ Corrigé détaillé
1 ) \(\sin\frac{7\pi}{12} = \sin(\frac{\pi}{2} + \frac{\pi}{12}) = \cos\frac{\pi}{12}\)
2 ) \(\cos\frac{5\pi}{12} = \cos(\frac{\pi}{2} - \frac{\pi}{12}) = \sin\frac{\pi}{12}\)
3 ) \(\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) = \cos 2x = 1 - 2\sin^2 x\)
4 ) \(\cos^4 x + \sin^4 x = (\cos^2 x + \sin^2 x)^2 - 2\cos^2 x \sin^2 x = 1 - 2\sin^2 x \cos^2 x\)
5 ) \(\tan^2 x - \sin^2 x = \frac{\sin^2 x}{\cos^2 x} - \sin^2 x = \sin^2 x(\frac{1}{\cos^2 x} - 1) = \sin^2 x \cdot \frac{1-\cos^2 x}{\cos^2 x} = \sin^2 x \cdot \frac{\sin^2 x}{\cos^2 x} = \tan^2 x \sin^2 x\)
\(\tan^2 x \sin^2 x = \tan^2 x \cdot \frac{\tan^2 x}{1+\tan^2 x} = \frac{\tan^4 x}{1+\tan^2 x}\)
\(\tan^2 x \sin^2 x = \tan^2 x \cdot \frac{\tan^2 x}{1+\tan^2 x} = \frac{\tan^4 x}{1+\tan^2 x}\)
📌 Exercice 5 ★★★★
1) Calculer : \(A = \cos\frac{\pi}{5} + \cos\frac{2\pi}{5} + \cos\frac{3\pi}{5} + \cos\frac{4\pi}{5}\) et \(B = \sin^2\frac{\pi}{8} + \sin^2\frac{3\pi}{8} + \sin^2\frac{5\pi}{8} + \sin^2\frac{7\pi}{8}\).
2) Calculer en fonction de \(\tan x\) : \(C = \sin^2 x\), \(D = \sin x \cos x\), \(E = \cos^4 x - 3\cos x \sin^3 x\), \(F = \frac{\cos^3 x - 2\sin^2 x \cos x}{\cos x + 2\sin x}\).
✅ Corrigé détaillé
A - On utilise \(\cos(\pi - x) = -\cos x\) : \(\cos\frac{3\pi}{5} = -\cos\frac{2\pi}{5}\), \(\cos\frac{4\pi}{5} = -\cos\frac{\pi}{5}\).
Donc \(A = \cos\frac{\pi}{5} + \cos\frac{2\pi}{5} - \cos\frac{2\pi}{5} - \cos\frac{\pi}{5} = 0\).
Donc \(A = \cos\frac{\pi}{5} + \cos\frac{2\pi}{5} - \cos\frac{2\pi}{5} - \cos\frac{\pi}{5} = 0\).
B - \(\sin^2(\frac{\pi}{8}) + \sin^2(\frac{7\pi}{8}) = \sin^2(\frac{\pi}{8}) + \sin^2(\pi - \frac{\pi}{8}) = \sin^2(\frac{\pi}{8}) + \sin^2(\frac{\pi}{8}) = 2\sin^2(\frac{\pi}{8})\)
\(\sin^2(\frac{3\pi}{8}) + \sin^2(\frac{5\pi}{8}) = \sin^2(\frac{3\pi}{8}) + \sin^2(\pi - \frac{3\pi}{8}) = 2\sin^2(\frac{3\pi}{8})\)
Or \(\sin\frac{3\pi}{8} = \cos\frac{\pi}{8}\), donc \(\sin^2\frac{3\pi}{8} = \cos^2\frac{\pi}{8}\).
Donc \(B = 2\sin^2\frac{\pi}{8} + 2\cos^2\frac{\pi}{8} = 2\).
\(\sin^2(\frac{3\pi}{8}) + \sin^2(\frac{5\pi}{8}) = \sin^2(\frac{3\pi}{8}) + \sin^2(\pi - \frac{3\pi}{8}) = 2\sin^2(\frac{3\pi}{8})\)
Or \(\sin\frac{3\pi}{8} = \cos\frac{\pi}{8}\), donc \(\sin^2\frac{3\pi}{8} = \cos^2\frac{\pi}{8}\).
Donc \(B = 2\sin^2\frac{\pi}{8} + 2\cos^2\frac{\pi}{8} = 2\).
2) \(C = \sin^2 x = \frac{\tan^2 x}{1+\tan^2 x}\)
\(D = \sin x \cos x = \frac{\tan x}{1+\tan^2 x}\)
\(E = \cos^4 x - 3\cos x \sin^3 x = \cos^4 x(1 - 3\tan^3 x) = \frac{1 - 3\tan^3 x}{(1+\tan^2 x)^2}\)
\(F = \frac{\cos^3 x - 2\sin^2 x \cos x}{\cos x + 2\sin x} = \frac{\cos^3 x(1 - 2\tan^2 x)}{\cos x(1 + 2\tan x)} = \frac{\cos^2 x(1 - 2\tan^2 x)}{1 + 2\tan x} = \frac{1 - 2\tan^2 x}{(1+\tan^2 x)(1+2\tan x)}\)
\(D = \sin x \cos x = \frac{\tan x}{1+\tan^2 x}\)
\(E = \cos^4 x - 3\cos x \sin^3 x = \cos^4 x(1 - 3\tan^3 x) = \frac{1 - 3\tan^3 x}{(1+\tan^2 x)^2}\)
\(F = \frac{\cos^3 x - 2\sin^2 x \cos x}{\cos x + 2\sin x} = \frac{\cos^3 x(1 - 2\tan^2 x)}{\cos x(1 + 2\tan x)} = \frac{\cos^2 x(1 - 2\tan^2 x)}{1 + 2\tan x} = \frac{1 - 2\tan^2 x}{(1+\tan^2 x)(1+2\tan x)}\)
📌 Exercice 6 ★★★★
Soit \(x \in [0;\pi]\) tel que \(\sin x + \cos x = \frac{7}{5}\).
1) Calculer \(\sin x \cos x\).
2) Déduire \(\sin^3 x + \cos^3 x\).
3) Déterminer \(\sin x\) et \(\cos x\) puis \(\tan x\).
✅ Corrigé détaillé
1) \((\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + 2\sin x \cos x = \frac{49}{25}\)
Donc \(2\sin x \cos x = \frac{49}{25} - 1 = \frac{24}{25}\) → \(\sin x \cos x = \frac{12}{25}\)
Donc \(2\sin x \cos x = \frac{49}{25} - 1 = \frac{24}{25}\) → \(\sin x \cos x = \frac{12}{25}\)
2) \(\sin^3 x + \cos^3 x = (\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x) = \frac{7}{5}(1 - \frac{12}{25}) = \frac{7}{5} \times \frac{13}{25} = \frac{91}{125}\)
3) \(\sin x\) et \(\cos x\) sont solutions de \(t^2 - \frac{7}{5}t + \frac{12}{25} = 0\)
\(\Delta = \frac{49}{25} - \frac{48}{25} = \frac{1}{25}\) → racines : \(t = \frac{7/5 \pm 1/5}{2}\) → \(t = \frac{4}{5}\) ou \(t = \frac{3}{5}\)
Sur \([0;\pi]\), \(\sin x \geq 0\). \(\sin x = \frac{4}{5}\), \(\cos x = \frac{3}{5}\) ou \(\sin x = \frac{3}{5}\), \(\cos x = \frac{4}{5}\).
\(\tan x = \frac{4}{3}\) ou \(\frac{3}{4}\).
\(\Delta = \frac{49}{25} - \frac{48}{25} = \frac{1}{25}\) → racines : \(t = \frac{7/5 \pm 1/5}{2}\) → \(t = \frac{4}{5}\) ou \(t = \frac{3}{5}\)
Sur \([0;\pi]\), \(\sin x \geq 0\). \(\sin x = \frac{4}{5}\), \(\cos x = \frac{3}{5}\) ou \(\sin x = \frac{3}{5}\), \(\cos x = \frac{4}{5}\).
\(\tan x = \frac{4}{3}\) ou \(\frac{3}{4}\).
📌 Exercice 7 ★★★★
Soit \(x \in [0;\pi]\), \(x \neq \frac{\pi}{2}\). On pose \(A(x) = \frac{1}{\sin^2 x + 2\cos^2 x}\).
1) Calculer \(A(0)\) et \(A(\frac{\pi}{6})\).
2) Vérifier que \(A(\pi - x) = A(x)\). Déduire \(A(\frac{5\pi}{6})\) et \(A(\pi)\).
3) Prouver que \(A(\frac{\pi}{2} - x) = \frac{1}{1 + \tan^2 x}\).
4) Montrer que \(A(x) = \frac{1 + \tan^2 x}{2 + \tan^2 x}\).
5) Déterminer les valeurs de \(x\) telles que \(A(x) = \frac{4}{5}\).
✅ Corrigé détaillé
1) \(A(0) = \frac{1}{0 + 2 \times 1} = \frac{1}{2}\)
\(A(\frac{\pi}{6}) = \frac{1}{\frac{1}{4} + 2 \times \frac{3}{4}} = \frac{1}{\frac{1}{4} + \frac{6}{4}} = \frac{1}{\frac{7}{4}} = \frac{4}{7}\)
\(A(\frac{\pi}{6}) = \frac{1}{\frac{1}{4} + 2 \times \frac{3}{4}} = \frac{1}{\frac{1}{4} + \frac{6}{4}} = \frac{1}{\frac{7}{4}} = \frac{4}{7}\)
2 ) \(\sin(\pi - x) = \sin x\), \(\cos(\pi - x) = -\cos x\) → \(\cos^2(\pi - x) = \cos^2 x\) → \(A(\pi - x) = A(x)\).
\(A(\frac{5\pi}{6}) = A(\pi - \frac{\pi}{6}) = A(\frac{\pi}{6}) = \frac{4}{7}\)
\(A(\pi) = A(0) = \frac{1}{2}\)
\(A(\frac{5\pi}{6}) = A(\pi - \frac{\pi}{6}) = A(\frac{\pi}{6}) = \frac{4}{7}\)
\(A(\pi) = A(0) = \frac{1}{2}\)
3) \(A(\frac{\pi}{2} - x) = \frac{1}{\sin^2(\frac{\pi}{2} - x) + 2\cos^2(\frac{\pi}{2} - x)} = \frac{1}{\cos^2 x + 2\sin^2 x} = \frac{1}{\cos^2 x(1 + 2\tan^2 x)} = \frac{1 + \tan^2 x}{1 + 2\tan^2 x}\)
4 ) \(A(x) = \frac{1}{\sin^2 x + 2\cos^2 x} = \frac{1}{\cos^2 x(\tan^2 x + 2)} = \frac{1 + \tan^2 x}{2 + \tan^2 x}\)
5 ) \(A(x) = \frac{4}{5}\) → \(\frac{1 + \tan^2 x}{2 + \tan^2 x} = \frac{4}{5}\) → \(5(1 + \tan^2 x) = 4(2 + \tan^2 x)\) → \(5 + 5\tan^2 x = 8 + 4\tan^2 x\) → \(\tan^2 x = 3\) → \(\tan x = \pm\sqrt{3}\)
Sur \([0;\pi]\), \(x = \frac{\pi}{3}\) ou \(x = \frac{2\pi}{3}\).
Sur \([0;\pi]\), \(x = \frac{\pi}{3}\) ou \(x = \frac{2\pi}{3}\).
📌 Exercice 8 ★★★★
Soit \(x \in [0; \frac{\pi}{2}]\). On pose \(G = \frac{\cos x - \sin x}{\cos x + \sin x}\).
1) Déterminer la valeur de \(G\) pour \(x = 0, \frac{\pi}{4}, \frac{\pi}{3}\).
2) Montrer que \(G = \frac{1 - \tan x}{1 + \tan x}\).
3) Sachant que \(G = \frac{1}{7}\), calculer \(\tan x\), puis \(\cos x\) et \(\sin x\).
✅ Corrigé détaillé
1) \(G(0) = \frac{1 - 0}{1 + 0} = 1\)
\(G(\frac{\pi}{4}) = \frac{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}} = 0\)
\(G(\frac{\pi}{3}) = \frac{\frac{1}{2} - \frac{\sqrt{3}}{2}}{\frac{1}{2} + \frac{\sqrt{3}}{2}} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} = \frac{(1 - \sqrt{3})^2}{1 - 3} = \frac{4 - 2\sqrt{3}}{-2} = \sqrt{3} - 2\)
\(G(\frac{\pi}{4}) = \frac{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}} = 0\)
\(G(\frac{\pi}{3}) = \frac{\frac{1}{2} - \frac{\sqrt{3}}{2}}{\frac{1}{2} + \frac{\sqrt{3}}{2}} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} = \frac{(1 - \sqrt{3})^2}{1 - 3} = \frac{4 - 2\sqrt{3}}{-2} = \sqrt{3} - 2\)
2) \(G = \frac{\cos x - \sin x}{\cos x + \sin x} = \frac{1 - \tan x}{1 + \tan x}\) (en divisant numérateur et dénominateur par \(\cos x\))
3) \(\frac{1 - \tan x}{1 + \tan x} = \frac{1}{7}\) → \(7(1 - \tan x) = 1 + \tan x\) → \(7 - 7\tan x = 1 + \tan x\) → \(6 = 8\tan x\) → \(\tan x = \frac{3}{4}\)
\(\cos x = \frac{1}{\sqrt{1+\tan^2 x}} = \frac{1}{\sqrt{1+\frac{9}{16}}} = \frac{1}{\sqrt{\frac{25}{16}}} = \frac{4}{5}\)
\(\sin x = \tan x \cos x = \frac{3}{4} \times \frac{4}{5} = \frac{3}{5}\)
\(\cos x = \frac{1}{\sqrt{1+\tan^2 x}} = \frac{1}{\sqrt{1+\frac{9}{16}}} = \frac{1}{\sqrt{\frac{25}{16}}} = \frac{4}{5}\)
\(\sin x = \tan x \cos x = \frac{3}{4} \times \frac{4}{5} = \frac{3}{5}\)