How can we interpret the negative side of a double-sided frequency spectrum
It is useful to think of observable (real-valued) waves as the sum of two complex waves:

it doesn't require a nonlinear operation to drop the imaginary part, like \$\cos x=\Re(e^{ix})\$ does.
it doesn't require an (arbitrary) definition whether an observed sine wave has increasing or decreasing phase (because \$\sin x=-\sin -x\$).
it explains mixing behaviour

The latter is the most important point for practical purposes: if you mix two sine waves by multiplying them, you end up with two sines, one with the sum of the frequencies, one with the difference, and a phase shift:
$$
(\sin x)(\sin y)=\frac{\cos(x-y)-\cos(x+y)}{2}
$$
This makes a lot more sense if you substitute
$$
\begin{eqnarray}
\sin u&amp;=&amp;\frac{e^{-iu}-e^{iu}}{2}i\\
\cos u&amp;=&amp;\frac{e^{-iu}+e^{iu}}{2}
\end{eqnarray}
$$
to get
$$
\begin{eqnarray}
(\frac{e^{ix}-e^{-ix}}{2}i)(\frac{e^{iy}-e^{-iy}}{2}i)&amp;=&amp;-\frac{(e^{ix}-e^{-ix})(e^{iy}-e^{-iy})}{4}\\
&amp;=&amp;-\frac{e^{ix}e^{iy}-e^{-ix}e^{iy}-e^{ix}e^{-iy}+e^{-ix}e^{-iy}}{4}\\
&amp;=&amp;-\frac{e^{i(x+y)}-e^{i(-x+y)}-e^{i(x-y)}+e^{i(-x-y)}}{4}\\
&amp;=&amp;-\frac{e^{i(x+y)}-e^{-i(x-y)}-e^{i(x-y)}+e^{-i(x+y)}}{4}\\
&amp;=&amp;-\frac{e^{i(x+y)}+e^{-i(x+y)}-e^{-i(x-y)}-e^{i(x-y)}}{4}\\
&amp;=&amp;-\frac{(e^{i(x+y)}+e^{-i(x+y)})-(e^{-i(x-y)}+e^{i(x-y)})}{4}\\
&amp;=&amp;-\frac{\cos(x+y)-\cos(x-y)}{2}\\
&amp;=&amp;\frac{\cos(x-y)-\cos(x+y)}{2}
\end{eqnarray}
$$
That way it is also way more intuitive how the spectrum of the multiplication of two signals is the convolution of the signal's spectra: the input signals have two peaks each, and the output signal has four.
This system also remains consistent when you operate on complex-valued signals, e.g. digital baseband, where it is possible to mix with a complex-valued sine and get a frequency shift without moving half the signal's energy to a mirror image.
The mixing behaviour is observable in the real world, so a model that adequately explains it is useful.