How do we prove that $(\space|\sin n|\space)$ is not convergent ? There is a beautiful proof of the non-convergence of $(\sin n)$ by considering the identities $\sin (n+1)=\cos1\sin n+ \sin 1 \cos n $ ,
$\cos (n+1)=\cos1\cos n- \sin 1 \sin n $ and $ \sin^2 n+\cos^2n=1$ , is there a similar proof for $(\space|\sin n|\space)$ ? (though at least any kind of proof will be helpful)
We can prove that $\left|\sin n\right|$ is not convergent in the same spirit. Assuming that $\left|\sin n\right|\to c\in[0,1]$, then the sequence given by: $$ a_n = \cos(2n) = 1-2\sin^2 n$$ converges to $1-2c^2$. However: $$ a_{n+1}-a_n = -2\sin(1)\sin(2n+1), $$ so assuming that $\{a_n\}_{n\in\mathbb N}$ is a Cauchy sequence we must have $c=0$. However, by assuming that $\sin n$ is close to zero we have that $\sin(n+1)$ cannot be close to zero, since from $\left|\sin n\right|\leq\varepsilon$ it follows that: $$\sin(n+1)=\cos(1)\sin n \pm \sin(1)\sqrt{1-\sin^2 n},$$ $$\left|\sin(n+1)\right| \geq \sin(1)-2\varepsilon>\frac{5}{6}-2\varepsilon.$$
Hint: We know $\sin(x)$ and $|x|$ are continuous functions, and so $|\sin(x)|$ is a continuous function. Can you find this limit $\lim_{x\rightarrow \infty}|\sin(x)|$? It does not exists. Do you want to see the proof? Try to apply Cauchy's condition of infinite limite.
Apply the sequential criterion of continuity and see $|\sin(n)|$ does not exists.
