Durée : 1 heure. Barème sur 20.
Exercice 1 (5 pts)
- Rappeler les formules de \(\cos(a+b)\) et \(\sin(a+b)\). (2 pts)
- En déduire la valeur exacte de \(\cos\dfrac{\pi}{12}\). (3 pts)
Correction
- \(\cos(a+b) = \cos a\cos b - \sin a\sin b\) et \(\sin(a+b) = \sin a\cos b + \cos a\sin b\).
- \(\dfrac{\pi}{12} = \dfrac{\pi}{3} - \dfrac{\pi}{4}\), donc \(\cos\dfrac{\pi}{12} = \dfrac12\times\dfrac{\sqrt2}{2} + \dfrac{\sqrt3}{2}\times\dfrac{\sqrt2}{2} = \dfrac{\sqrt2 + \sqrt6}{4}\).
Exercice 2 (5 pts)
Sachant que \(\sin x = \dfrac{12}{13}\) avec \(x\in\left]0\,;\dfrac{\pi}{2}\right[\) :
- Calculer \(\cos x\). (2 pts)
- Calculer \(\sin 2x\) et \(\cos 2x\). (3 pts)
Correction
- \(\cos x = \dfrac{5}{13}\).
- \(\sin 2x = 2\times\dfrac{12}{13}\times\dfrac{5}{13} = \dfrac{120}{169}\) ; \(\cos 2x = 1 - 2\sin^2 x = 1 - \dfrac{288}{169} = -\dfrac{119}{169}\).
Exercice 3 (5 pts)
Résoudre dans \([0\,;2\pi[\) : \(2\sin^2 x + \sin x - 1 = 0\).
Correction
\(X = \sin x\) : \(2X^2 + X - 1 = 0\) donne \(X = \dfrac12\) ou \(X = -1\). \(\sin x = \dfrac12\) : \(x = \dfrac{\pi}{6}\) ou \(\dfrac{5\pi}{6}\). \(\sin x = -1\) : \(x = \dfrac{3\pi}{2}\).
Exercice 4 (5 pts)
Montrer que, pour tout réel \(x\) tel que \(\sin x \ne 0\) :
\[\frac{1 - \cos 2x}{\sin 2x} = \tan x\]
Correction
\(1 - \cos 2x = 2\sin^2 x\) et \(\sin 2x = 2\sin x\cos x\) ; le quotient vaut \(\dfrac{2\sin^2 x}{2\sin x\cos x} = \tan x\) (pour \(\cos x \ne 0\)).