Yes, it suffices that $a_n=o(n)$.

Denote $h(n)=|\sum_{j=1}^n \cos a_j|+|\sum_{j=1}^n \sin a_j|$.

For any fixed integer $M>0$ there exists arbitrarily large $n$ for which $a_{n+M}-a_n<1/M^2$. Thus by triangle inequality and 1-Lipschitя property of functions $\cos$ and $\sin$ we have $$h(n+M)+h(n)\geqslant |\cos a_{n+1}+\ldots+\cos a_{n+M}|+|\sin a_{n+1}+\ldots+\sin a_{n+M}|\\\geqslant M|\cos a_{n+1}|-M/M^2+M|\sin a_{n+1}|-M/M^2\geqslant 2M-2$$ that yields that $h(n+M)$ or $h(n)$ is at least $M-1$. Your claim follows immediately.

Yes, it suffices that $a_n=o(n)$.

Denote $h(n)=|\sum_{j=1}^n \cos a_j|+|\sum_{j=1}^n \sin a_j|$.

For any fixed integer $M>0$ there exists arbitrarily large $n$ for which $a_{n+M}-a_n<1/M^2$. Thus by triangle inequality and 1-Lipschitz property of functions $\cos$ and $\sin$ we have $$h(n+M)+h(n)\geqslant |\cos a_{n+1}+\ldots+\cos a_{n+M}|+|\sin a_{n+1}+\ldots+\sin a_{n+M}|\\\geqslant M|\cos a_{n+1}|-M/M^2+M|\sin a_{n+1}|-M/M^2\geqslant 2M-2$$ that yields that $h(n+M)$ or $h(n)$ is at least $M-1$. Your claim follows immediately.