Respuesta :

dalendrk
[tex]n=0\to\sin(2\cdot0\pi)=\sin0=0\\\\n=1\to\sin(2\cdot1\pi)=\sin2\pi=0\\\\n=2\to\sin(2\cdot2\pi)=\sin4\pi=0\\\vdots\\\sin2n\pi=0\ for\ n\in\mathbb{Z}[/tex]