Calculer un cosinus et un sinus à l'aide des formules d'addition Exercice

\cos\left(\dfrac{7\pi}{12}\right) = \dfrac{\sqrt{2}-\sqrt{6}}{4}

\cos\left(\dfrac{7\pi}{12}\right) = \dfrac{\sqrt{6}-\sqrt{2}}{4}

\cos\left(\dfrac{7\pi}{12}\right) = \dfrac{\sqrt{2}+\sqrt{6}}{4}

\cos\left(\dfrac{7\pi}{12}\right) = \dfrac{\sqrt{6}+\sqrt{2}}{4}

\cos\left(\dfrac{\pi}{12}\right) = \dfrac{\sqrt{2}+\sqrt{6}}{4}

\cos\left(\dfrac{\pi}{12}\right) = \dfrac{\sqrt{2}-\sqrt{6}}{4}

\cos\left(\dfrac{\pi}{12}\right) = \dfrac{\sqrt{2}-\sqrt{6}}{4}

\cos\left(\dfrac{\pi}{12}\right) = \dfrac{\sqrt{6}-\sqrt{2}}{4}

\cos\left(\dfrac{13\pi}{12}\right) = -\dfrac{\sqrt{6}+\sqrt{2}}{4}

\cos\left(\dfrac{13\pi}{12}\right) = \dfrac{\sqrt{6}+\sqrt{2}}{4}

\cos\left(\dfrac{13\pi}{12}\right) =\dfrac{\sqrt{6}-\sqrt{2}}{4}

\cos\left(\dfrac{13\pi}{12}\right) = \dfrac{\sqrt{2}-\sqrt{6}}{4}

\sin\left(\dfrac{7\pi}{12}\right) = \dfrac{\sqrt{6}+\sqrt{2}}{4}

\sin\left(\dfrac{7\pi}{12}\right) = \dfrac{\sqrt{6}-\sqrt{2}}{4}

\sin\left(\dfrac{7\pi}{12}\right) = \dfrac{\sqrt{2}-\sqrt{6}}{4}

\sin\left(\dfrac{7\pi}{12}\right) = -\dfrac{\sqrt{6}+\sqrt{2}}{4}

\sin\left(\dfrac{11\pi}{12}\right) = \dfrac{\sqrt{6}-\sqrt{2}}{4}

\sin\left(\dfrac{11\pi}{12}\right) = \dfrac{\sqrt{6}+\sqrt{2}}{4}

\sin\left(\dfrac{11\pi}{12}\right) = \dfrac{\sqrt{2}-\sqrt{6}}{4}

\sin\left(\dfrac{11\pi}{12}\right) = -\dfrac{\sqrt{6}+\sqrt{2}}{4}

\sin\left(-\dfrac{\pi}{12}\right) = \dfrac{\sqrt{2}-\sqrt{6}}{4}

\sin\left(-\dfrac{\pi}{12}\right) = \dfrac{\sqrt{2}+\sqrt{6}}{4}

\sin\left(-\dfrac{\pi}{12}\right) = \dfrac{\sqrt{6}-\sqrt{2}}{4}

\sin\left(-\dfrac{\pi}{12}\right) =- \dfrac{\sqrt{2}+\sqrt{6}}{4}

\sin\left(\dfrac{5\pi}{12}\right) = \dfrac{\sqrt{6} + \sqrt{2}}{4}

\sin\left(\dfrac{5\pi}{12}\right) = \dfrac{\sqrt{6} - \sqrt{2}}{4}

\sin\left(\dfrac{5\pi}{12}\right) = -\dfrac{\sqrt{6} + \sqrt{2}}{4}

\sin\left(\dfrac{5\pi}{12}\right) = \dfrac{\sqrt{2} - \sqrt{6}}{4}