Calculer un sinus ou un cosinus à l'aide de leurs propriétés de périodicité Exercice

\cos\left(\dfrac{13\pi}{6}\right)=\dfrac{\sqrt{3}}{2} 

\cos\left(\dfrac{13\pi}{6}\right)=-\dfrac{\sqrt{3}}{2} 

\cos\left(\dfrac{13\pi}{6}\right)=\dfrac{1}{2} 

\cos\left(\dfrac{13\pi}{6}\right)=0 

\cos\left(\dfrac{25\pi}{4}\right)=\dfrac{\sqrt{2}}{2} 

\cos\left(\dfrac{25\pi}{4}\right)=-\dfrac{\sqrt{2}}{2} 

\cos\left(\dfrac{25\pi}{4}\right)=\dfrac{1}{2} 

\cos\left(\dfrac{25\pi}{4}\right)=\dfrac{\sqrt{3}}{2} 

\sin\left(\dfrac{-3\pi}{2}\right)= 1

\sin\left(\dfrac{-3\pi}{2}\right)= 0

\sin\left(\dfrac{-3\pi}{2}\right)= \dfrac{\sqrt{2}}{2}

\sin\left(\dfrac{-3\pi}{2}\right)= \dfrac{\sqrt{3}}{2}

\sin\left(\dfrac{37\pi}{6}\right)=\dfrac{1}{2} 

\sin\left(\dfrac{37\pi}{6}\right)=\dfrac{\sqrt{3}}{2} 

\sin\left(\dfrac{37\pi}{6}\right)=-\dfrac{\sqrt{3}}{2} 

\sin\left(\dfrac{37\pi}{6}\right)=\dfrac{\sqrt{2}}{2} 

\sin\left(\dfrac{-29\pi}{3}\right)=\dfrac{\sqrt{3}}{2} 

\sin\left(\dfrac{-29\pi}{3}\right)=\dfrac{\sqrt{2}}{2} 

\sin\left(\dfrac{-29\pi}{3}\right)=\dfrac{1}{2} 

\sin\left(\dfrac{-29\pi}{3}\right)=-\dfrac{\sqrt{3}}{2}