How to show that sin(n) does not converge using Cauchy criterion

Question

I trying to figure out how to show that:

$a_n = \sin(n)$ (n is Natural number)

Does not converge using Cauchty criterion.

Do you guys have an idea? maybe a Hint?

Thank you.

Answer 1

Edit

It appear to be that I miss the fact that OP asked to use Cauchy criterion, this answer prove it in a different way from what OP asked for, I decide to keep it up because it still may help someone.

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Supposed the negative: $\lim \sin n=k\in[-1,1]$

Now from that we can conclude that $\lim\cos n=\sqrt{1-k^2}$(why?)

Using that fact we get $2k\sqrt{1-k^2}=\lim\sin 2n=\lim\sin n=k\implies k=\{\pm\frac{\sqrt3}2,0\}$

And $\sqrt{1-k^2}=\lim\cos n=\lim\cos^2 n-\sin^2 n=1-2k^2\implies k=0$

Now we only left to calculate $1-2k^2=\lim\cos(n+1):\quad \lim\cos(n+1)=\lim\cos n\cos 1-\sin n\sin 1=(1-2k^2)\cos 1-k\sin 1$

This gives contradiction to $k=0$