Find the limit (or prove it doesn't exist) for $(s_n)$ where $s_n = \cos(n)$. Let $n$ be in radians. (Hint: look at $\sin(n + 2) - \sin(n)$ and $\cos(n + 2) - \cos(n)$.)
From Calculus, I know that this limit doesn't exist. I'm having an issue figuring out how to give a proof using the hint we were provided.
Let us suppose $s_n\to a$. Hence $s_{n+2}-s_n=\cos(n+2)-\cos(n)\to 0$. We know that $$\cos(n+2)-\cos(n)=-2\sin\left(\dfrac{2n+2}{2}\right)\sin(1)=-2\sin(n+1)\sin(1),$$ so $t_n=\sin(n)$ approaches $0$ as $n\to\infty$.
Now, since $$\sin(n+2)-\sin(n)=2\cos\left(\dfrac{2n+2}{2}\right)\sin(1)=2\cos(n+1)\sin(1),$$ we must have $s_n=\cos(n)\to 0$ as well.
Since $\sin^2(n)+\cos^2(n)=1$ for all $n$, we get a contradiction.
