**Proof** Using [[Compound Angle Identity]], $\sin(a+b) = \sin(a)\cos(b)+\sin(b)\cos(a)$. Hence $\sin(n+1) + \sin(n-1) = 2 \cos(1)\sin(n)$. BWOC suppose $\sin(n) \to l$, using [[Shift Rule for Limits]], $2l = (2\cos 1)l$ which yields a contradiction since $\cos 1 \neq 1$.