Trigonométrie : corrigé du contrôle de maths 1ère à télécharger en PDF

1 Radians et mesures principales / 3 pts

a) \(150^\circ = \dfrac{150\pi}{180} = \dfrac{5\pi}{6}\) rad. (0,5 pt)

b) \(\dfrac{7 \times 180}{4} = 315^\circ\). (0,5 pt)

c) \(\dfrac{29\pi}{6} - 4\pi = \dfrac{29\pi - 24\pi}{6} = \dfrac{5\pi}{6}\). (1 pt)

d) \(-\dfrac{31\pi}{4} + 8\pi = \dfrac{-31\pi + 32\pi}{4} = \dfrac{\pi}{4}\). (1 pt)

2 Valeurs exactes / 4 pts

a) \(\dfrac{3\pi}{4} = \pi - \dfrac{\pi}{4}\) donc \(\cos\dfrac{3\pi}{4} = -\dfrac{\sqrt{2}}{2}\). (1 pt)

b) \(\dfrac{4\pi}{3} = \pi + \dfrac{\pi}{3}\) donc \(\sin\dfrac{4\pi}{3} = -\sin\dfrac{\pi}{3} = -\dfrac{\sqrt{3}}{2}\). (1 pt)

c) \(\cos\left(-\dfrac{\pi}{6}\right) = \cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}\). (1 pt)

d) \(\sin\dfrac{5\pi}{6} = \sin\left(\pi - \dfrac{\pi}{6}\right) = \sin\dfrac{\pi}{6} = \dfrac{1}{2}\). (1 pt)

3 Cosinus connu, sinus à trouver / 3 pts

a) \(\sin^2 x = 1 - \dfrac{144}{169} = \dfrac{25}{169}\). Sur \(\left[-\dfrac{\pi}{2}\,;\,0\right]\), \(\sin x \leqslant 0\), donc \(\sin x = -\dfrac{5}{13}\). (1,5 pt)

b) \(\cos(\pi - x) = -\cos x = -\dfrac{12}{13}\) ; \(\sin(\pi + x) = -\sin x = \dfrac{5}{13}\) ; \(\sin\left(\dfrac{\pi}{2} - x\right) = \cos x = \dfrac{12}{13}\). (0,5 pt chacun)

4 Résolution d’équations / 4 pts

a) \(\cos\dfrac{3\pi}{4} = -\dfrac{\sqrt{2}}{2}\), donc \(x = \dfrac{3\pi}{4} + 2k\pi\) ou \(x = -\dfrac{3\pi}{4} + 2k\pi\). (1 pt) Dans \(]-\pi\,;\,\pi]\) : \(S = \left\{-\dfrac{3\pi}{4}\,;\,\dfrac{3\pi}{4}\right\}\). (1 pt)

b) \(\sin\dfrac{\pi}{6} = \dfrac{1}{2}\), donc \(x = \dfrac{\pi}{6} + 2k\pi\) ou \(x = \pi - \dfrac{\pi}{6} + 2k\pi = \dfrac{5\pi}{6} + 2k\pi\). (1 pt) Dans \(]-\pi\,;\,\pi]\) : \(S = \left\{\dfrac{\pi}{6}\,;\,\dfrac{5\pi}{6}\right\}\). (1 pt)

5 Parité et périodicité / 3 pts

a) \(f(-x) = \sin(-x)\cos(-x) = (-\sin x)\cos x = -f(x)\). (1,5 pt)

b) \(f(x + \pi) = \sin(x + \pi)\cos(x + \pi) = (-\sin x)(-\cos x) = \sin x\cos x = f(x)\). (1,5 pt)

6 Le mouvement d’une bouée / 3 pts

a) \(h(0) = 0\) ; \(h(2) = 2\sin\dfrac{\pi}{2} = 2\) ; \(h(6) = 2\sin\dfrac{3\pi}{2} = -2\). (1,5 pt)

b) Le sinus est au plus \(1\) : hauteur maximale \(2\) m. La période est \(\dfrac{2\pi}{\pi/4} = 8\) s. (0,5 pt)

c) \(h(t) = 1 \iff \sin\dfrac{\pi t}{4} = \dfrac{1}{2}\), donc \(\dfrac{\pi t}{4} = \dfrac{\pi}{6}\) ou \(\dfrac{5\pi}{6}\) (dans \([0\,;\,\pi]\)), soit \(t = \dfrac{2}{3}\) ou \(t = \dfrac{10}{3}\) secondes. (1 pt)