Série Exercicex \( 1 / \) Montrer que pour tout réel \( x \) on a \( \sqrt{2} \sin \left(x+\frac{3 \pi}{4}\right)=\cos x-\sin x \) \( 2 / \) Soit \( f(x)=\cos 2 x-\sin 2 x+1 \) a/Calculer \( f\left(\frac{\pi}{12}\right) \) b/Vérifier que \( f(x)=2 \sqrt{2} \cos x \sin \left(x+\frac{3 \pi}{4}\right) \) c/Déduire \( \cos \left(\frac{\pi}{12}\right) \) \( 3 / \) a/Montrer que pour tout réel \( x \) de \( ] 0, \frac{\pi}{4}\left[\right. \) on \( a: \quad \frac{2 \cos 2 x}{\cos 2 x-\sin 2 x+1}=1+\tan x \) b/En déduire que tan \( \left(\frac{\pi}{12}\right)=2-\sqrt{3} \)

### Exercice 1 Montrer que pour tout réel \( x \) on a \( \sqrt{2} \sin \left(x+\frac{3 \pi}{4}\right)=\cos x-\sin x \) Pour montrer cette équation, nous allons utiliser les formules trigonométriques de déplacement. La formule de déplacement pour la sinusoïde est : \[ \sin(x + \alpha) = \sin x \cos \alpha + \cos x \sin \alpha \] Donc, pour \( \alpha = \frac{3\pi}{4} \), nous avons : \[ \sin\left(x + \frac{3\pi}{4}\right) = \sin x \cos\left(\frac{3\pi}{4}\right) + \cos x \sin\left(\frac{3\pi}{4}\right) \] Sachant que \( \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} \) et \( \sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} \), nous obtenons : \[ \sin\left(x + \frac{3\pi}{4}\right) = \sin x \left(-\frac{\sqrt{2}}{2}\right) + \cos x \left(\frac{\sqrt{2}}{2}\right) \] \[ \sin\left(x + \frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} \sin x + \frac{\sqrt{2}}{2} \cos x \] \[ \sin\left(x + \frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} (\cos x - \sin x) \] Multiplying both sides by \( \sqrt{2} \), we get: \[ \sqrt{2} \sin\left(x + \frac{3\pi}{4}\right) = \cos x - \sin x \] Thus, the equation is proven. ### Exercice 2 a/ Calculer \( f\left(\frac{\pi}{12}\right) \) Given \( f(x) = \cos 2x - \sin 2x + 1 \), we need to calculate \( f\left(\frac{\pi}{12}\right) \). \[ f\left(\frac{\pi}{12}\right) = \cos\left(2 \cdot \frac{\pi}{12}\right) - \sin\left(2 \cdot \frac{\pi}{12}\right) + 1 \] \[ f\left(\frac{\pi}{12}\right) = \cos\left(\frac{\pi}{6}\right) - \sin\left(\frac{\pi}{6}\right) + 1 \] \[ f\left(\frac{\pi}{12}\right) = \frac{\sqrt{3}}{2} - \frac{1}{2} + 1 \] \[ f\left(\frac{\pi}{12}\right) = \frac{\sqrt{3}}{2} + \frac{1}{2} \] \[ f\left(\frac{\pi}{12}\right) = \frac{\sqrt{3} + 1}{2} \] b/ Vérifier que \( f(x) = 2\sqrt{2} \cos x \sin \left(x + \frac{3\pi}{4}\right) \) We already know that \( \sqrt{2} \sin \left(x + \frac{3\pi}{4}\right) = \cos x - \sin x \). Multiplying both sides by 2, we get: \[ 2\sqrt{2} \sin \left(x + \frac{3\pi}{4}\right) = 2(\cos x - \sin x) \] Now, we need to show that \( \cos 2x - \sin 2x + 1 = 2(\cos x - \sin x) \). Using the double angle formulas: \[ \cos 2x = \cos^2 x - \sin^2 x \] \[ \sin 2x = 2 \sin x \cos x \] Substituting these into the equation: \[ \cos 2x - \sin 2x + 1 = (\cos^2 x - \sin^2 x) - 2 \sin x \cos x + 1 \] \[ = (\cos x + \sin x)(\cos x - \sin x) - 2 \sin x \cos x + 1 \] \[ = \cos x - \sin x -