[tex]Functia~sinus~are~perioada~2k \pi,~unde~ k \in \mathbb{Z}. \\ \\ Adica~\sin (x+2k \pi)=\sin x~\forall~k \in \mathbb{Z}. \\ \\ Avem~deci:~\sin \frac{25 \pi}{4}= \sin \left( \frac{ \pi}{4}+6 \pi \right)= \sin \left( \frac{ \pi}{4}+2 \cdot 3 \pi \right)= \sin \frac{\pi}{4}= \frac{\sqrt{2}}{2}[/tex]
