Is $(^{1/2}) = \Omega(^{\sin(n)})$?
First, let's unpack this. We say that $O(\sqrt{n}) = \Omega(n^{\sin n})$ if any function $f(n)$ which satisfies $f(n) = O(\sqrt{n})$ also satisfies $f(n) = \Omega(n^{\sin n})$.
In particular, if $O(\sqrt{n}) = \Omega(n^{\sin n})$ then $\sqrt{n} = \Omega(n^{\sin n})$. If $\sqrt{n} = \Omega(n^{\sin n})$ then according to the definition, there exist $C,N&gt;0$ such that $\sqrt{n} \geq Cn^{\sin n}$ for all $n \geq N$. Let us assume that this is the case.
In what follows, I assume that the argument in $\sin n$ is measured in radians.
Since $2\pi$ is irrational, the sequence $n \bmod 2\pi$ is equidistributed in $[0,2\pi)$. In particular, there are infinitely many $n$ such that $n \bmod 2\pi \in (0.4\pi,0.6\pi)$, and so $\sin n \geq 0.95$. For each such $n \geq N$, we have
$$ \sqrt{n} \geq Cn^{\sin n} \geq Cn^{0.95} \Longrightarrow n \leq C^{1/0.45}.$$
Since there are infinitely many $n$ such that $\sin n \geq 0.95$, we can find one such $n$ which exceeds both $N$ and $C^{1/0.45}$, and so reach a contradiction. This shows that $\sqrt{n}$ is not $\Omega(n^{\sin n})$.