I made a proof using isometry, but I'm not sure. Here is the proof:
Let A be a 2×2 matrix whose first row is equal to (cos(1), -sin(1)) and second row is equal to (sin(1), cos(1)). And define the sequence {X(n)} by
X(1)=(cos(1), sin(1)), X(n+1)=A^nX(1).
Since A is an isometry, norm(X(n+1)-X(n))=norm(X(2)-X(1)) =/0 for all natural numbers n. Hence {X(n)} is not a Cauchy sequnce.
One of {cos(n)}, {sin(n)} diverges because {x(n)} converges iff {cos(n)} and {sin(n)} both converges. Suppose f(n) be the one that diverges. Then sqrt[1-(f(n)^2] also diverges. This sequnce is equal to the other one. Hence, {cos(n)}, {sin(n)} both diverge.
Let's prove $\{ \sin(n) \}$ diverges by way of contradiction.
Suppose that $\lim{\sin n}=a$. We know the equality: $$ \sin(n+1) - \sin (n-1) = 2 \cos(n) \sin (1) $$ taking limit $n \to \infty$ from both sides we obtain: $$ a-a=0=2 \sin(1) \lim_{n\to \infty}{\cos(n)} $$ thus $\lim_{n\to\infty}{\cos(n)}=0$. Again, we have: $$ \sin(2n)=2 \sin(n) \cos(n) $$ taking limit: $$ a =2 a \cdot 0 = 0 $$
But $\sin^2n+\cos^2n=1$, taking limit: $$ 0+0=1 $$ which is a contradiction.