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Exercice 1

📌 Consigne : Compléter le tableau ci-dessous en déterminant le cosinus, le sinus et la tangente de chaque angle.

✅ Corrigé

a) Cosinus :
\(\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}\)
\(\cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}\)
\(\cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}\)
\(\cos\left(\frac{55\pi}{3}\right) = \cos\left(\frac{55\pi}{3} - 18\pi\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\)

b) Sinus :
\(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\)
\(\sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}\)
\(\sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}\)
\(\sin\left(\frac{55\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\)

c) Tangente :
\(\tan\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{3}\)
\(\tan\left(-\frac{\pi}{3}\right) = -\sqrt{3}\)
\(\tan\left(\frac{4\pi}{3}\right) = \sqrt{3}\)
\(\tan\left(\frac{55\pi}{3}\right) = \sqrt{3}\)

📌 Exercice 2

Simplifier les expressions suivantes :

\[ A = \cos^2\left(\frac{\pi}{11}\right) + \cos^2\left(\frac{3\pi}{11}\right) + \cos^2\left(\frac{5\pi}{22}\right) + \cos^2\left(\frac{9\pi}{22}\right) \]

\[ B = \sin\left(\frac{\pi}{13}\right) + \sin\left(\frac{3\pi}{13}\right) + \sin\left(\frac{14\pi}{13}\right) + \sin\left(\frac{16\pi}{13}\right) \]

\[ C = \tan\left(\frac{\pi}{7}\right) + \tan\left(\frac{3\pi}{7}\right) + \tan\left(\frac{4\pi}{7}\right) + \tan\left(\frac{6\pi}{7}\right) \]

✅ Corrigé

A :
\(\cos^2\left(\frac{5\pi}{22}\right) = \cos^2\left(\frac{\pi}{2} - \frac{3\pi}{11}\right) = \sin^2\left(\frac{3\pi}{11}\right)\)
\(\cos^2\left(\frac{9\pi}{22}\right) = \cos^2\left(\frac{\pi}{2} - \frac{\pi}{11}\right) = \sin^2\left(\frac{\pi}{11}\right)\)

Donc \(A = \cos^2\left(\frac{\pi}{11}\right) + \sin^2\left(\frac{\pi}{11}\right) + \cos^2\left(\frac{3\pi}{11}\right) + \sin^2\left(\frac{3\pi}{11}\right) = 1 + 1 = 2\)

B :
\(\sin\left(\frac{14\pi}{13}\right) = \sin\left(\pi + \frac{\pi}{13}\right) = -\sin\left(\frac{\pi}{13}\right)\)
\(\sin\left(\frac{16\pi}{13}\right) = \sin\left(\pi + \frac{3\pi}{13}\right) = -\sin\left(\frac{3\pi}{13}\right)\)
Donc \(B = \sin\left(\frac{\pi}{13}\right) + \sin\left(\frac{3\pi}{13}\right) - \sin\left(\frac{\pi}{13}\right) - \sin\left(\frac{3\pi}{13}\right) = 0\)

C :
\(\tan\left(\frac{6\pi}{7}\right) = \tan\left(\pi - \frac{\pi}{7}\right) = -\tan\left(\frac{\pi}{7}\right)\)
\(\tan\left(\frac{4\pi}{7}\right) = \tan\left(\pi - \frac{3\pi}{7}\right) = -\tan\left(\frac{3\pi}{7}\right)\)
Donc \(C = \tan\left(\frac{\pi}{7}\right) + \tan\left(\frac{3\pi}{7}\right) - \tan\left(\frac{3\pi}{7}\right) - \tan\left(\frac{\pi}{7}\right) = 0\)

📌 Exercice 3

Simplifier les expressions suivantes :

\[ A = \sin(x + 5\pi) + \sin\left(\frac{9\pi}{2} - x\right) + \sin(13\pi - x) + \sin\left(\frac{17\pi}{2} + x\right) \]

\[ B = \cos(x + 9\pi) + \cos(13\pi - x) + \cos(x + 28\pi) + \cos x \]

\[ C = \tan(x + 11\pi) - \tan\left(\frac{15\pi}{2} - x\right) - \frac{1}{\cos x \cdot \sin x} \]

✅ Corrigé

A :
\(\sin(x + 5\pi) = \sin(x + \pi) = -\sin x\)
\(\sin\left(\frac{9\pi}{2} - x\right) = \sin\left(\frac{\pi}{2} - x\right) = \cos x\)
\(\sin(13\pi - x) = \sin(\pi - x) = \sin x\)
\(\sin\left(\frac{17\pi}{2} + x\right) = \sin\left(\frac{\pi}{2} + x\right) = \cos x\)
Donc \(A = -\sin x + \cos x + \sin x + \cos x = 2\cos x\)

B :
\(\cos(x + 9\pi) = \cos(x + \pi) = -\cos x\)
\(\cos(13\pi - x) = \cos(\pi - x) = -\cos x\)
\(\cos(x + 28\pi) = \cos x\)
Donc \(B = -\cos x - \cos x + \cos x + \cos x = 0\)

C :
\(\tan(x + 11\pi) = \tan(x + \pi) = \tan x\)
\(\tan\left(\frac{15\pi}{2} - x\right) = \tan\left(\frac{3\pi}{2} - x\right) = \frac{1}{\tan x}\)
Donc \(C = \tan x - \frac{1}{\tan x} - \frac{1}{\cos x \sin x}\)
Or \(\frac{1}{\cos x \sin x} = \frac{2}{\sin 2x}\) et \(\tan x - \frac{1}{\tan x} = \frac{\sin^2 x - \cos^2 x}{\sin x \cos x} = -\frac{2\cos 2x}{\sin 2x}\)
Donc \(C = 0\)

📌 Exercice 4

1) On suppose que \(x \in \left[\frac{3\pi}{2}, 2\pi\right]\) tel que \(\cos x = \frac{2}{5}\). Calculer \(\sin x\) et \(\tan x\).

2) On suppose que \(x \in \left[\pi, \frac{3\pi}{2}\right]\) tel que \(\tan x = \frac{1}{4}\). Calculer \(\cos x\) et \(\sin x\).

✅ Corrigé

1) \(\cos x = \frac{2}{5}\), \(x \in \left[\frac{3\pi}{2}, 2\pi\right]\)
\(\sin^2 x = 1 - \cos^2 x = 1 - \frac{4}{25} = \frac{21}{25}\)
\(\sin x = \pm \frac{\sqrt{21}}{5}\)
Sur \(\left[\frac{3\pi}{2}, 2\pi\right]\), \(\sin x \leq 0\) donc \(\sin x = -\frac{\sqrt{21}}{5}\)
\(\tan x = \frac{\sin x}{\cos x} = -\frac{\sqrt{21}}{2}\)

2) \(\tan x = \frac{1}{4}\), \(x \in \left[\pi, \frac{3\pi}{2}\right]\)
On a \(\cos^2 x = \frac{1}{1 + \tan^2 x} = \frac{1}{1 + \frac{1}{16}} = \frac{16}{17}\)
\(\cos x = \pm \frac{4}{\sqrt{17}}\)
Sur \(\left[\pi, \frac{3\pi}{2}\right]\), \(\cos x \leq 0\) donc \(\cos x = -\frac{4}{\sqrt{17}}\)
\(\sin x = \tan x \cdot \cos x = \frac{1}{4} \times \left(-\frac{4}{\sqrt{17}}\right) = -\frac{1}{\sqrt{17}}\)

النشاط السابق Calcul trigonométrique

النشاط التالي TRIGONOMÉTRIE : FORMULAIRE