if we specify the interval of study of our angles as (-180,180] ...
Why would you? Transfer functions are basically complex numbers: \$Z=A+jB=|Z| \angle{\tan(B/A)}\$. Think about the polar representation again.
since for both cases the transfer function will be equivalent to a negative number
No.
- For lower frequencies: \$s=j \omega_1\$ where \$\omega_1 << \omega_n\$ (i.e. way lower) therefore \$\omega_1/\omega_n<<1 \rightarrow\omega_1/\omega_n\approx0\$, so: $$ G(s)=\omega_n^2\Big(\frac{s^2}{\omega_n^2}+2\zeta \frac{s}{\omega_n}+1\Big)\\ \Rightarrow G(j\omega_1)\approx\omega_n^2((j0)^2+2\zeta\ 0+1)=\omega_n^2 $$
So at lower frequencies the transfer function is an always-positive constant:
$$ G\approx\omega_n^2=\omega_n^2+j0=\omega_n^2 \angle{0°} $$
- Likewise, for higher frequencies: \$s=j \omega_2\$ where \$\omega_2 >> \omega_n\$ (i.e. way higher) therefore \$\omega_n/\omega_2<<1 \rightarrow\omega_n/\omega_2\approx0\$, so: $$ G(s)=s^2\Big(1+2\zeta \frac{\omega_n}{s}+\frac{\omega_n^2}{s^2}\Big) \\ \Rightarrow G(j\omega_2)\approx(j\omega_2)^2(1+2\zeta\ 0+ 0)=(j\omega_2)^2=-\omega_2^2 $$
So at higher frequencies the transfer function is an always-negative constant:
$$ G\approx -\omega_n^2=-\omega_n^2+j0=\omega_n^2 \angle{180°} $$
