We do. The LHC accelerates two protons, each with 3.5 TeV of energy, giving a total of 7 TeV in the CoM frame (The energies are from the initial phase of the previous LHC run. Later in the run this was increased to 8 TeV and the combination of the two dataset was what discovered the Higgs boson. The energies are roughly doubling now for Run II, to 13 TeV).
The main reason for this is as you mentioned, the energy involved. In any frame, we have the following invariant quantity, $s = (p_1 + p_2)^2$ which its square root, $\sqrt{s}$, gives the Centre of Mass (CoM) energy for the experiment, and here $p_i$ represents the momentum four-vector for each particle i. In a collision where two particles are moving in opposite directions with equal energy we have the following:
$$ s \equiv (p_1 + p_2) \cdot (p_1 + p_2) = (E + E, \mathbf{p}_1 − \mathbf{p}_2 ) \cdot (E + E, \mathbf{p}_1 − \mathbf{p}_2 ) = (2E , 0 ) \cdot (2E, 0) = 4E^2$$
and now the CoM energy is given by the square root of this quantity, $$\to E_{CoM} = \sqrt{s} = 2E$$
In a an experiment where one of the particles is at rest (has mass $m_t$) and the other is travelling with momentum $\mathbf{p}$ (and has mass $m_b$) we have the following:
$$ s \equiv (p_1 + p_2) \cdot (p_1 + p_2) = (E_b + m_t, \mathbf{p}_b) \cdot (E_b + m_t, \mathbf{p}_b) = E_b^2 + m_t^2 + 2E_b m_t − p_b^2 = m_t^2 + m_b^2 + 2E_b m_t $$
Assuming the masses are negligible, we have the fixed target (FT) CoM energy,
$$\to E^{\text{FT}}_{CoM} = \sqrt{s} = \sqrt{2E_b m_t}$$
Thus we would need much more energy input in a fixed target experiment to achieve the same energies as in the case with two co-moving particles.
EDIT: Regarding a comment below which I think arises from confusion of what the CoM frame is. $\sqrt{s}$ gives the CoM energy in both cases. This is useful because we can now compare between a fixed target experiment and an experiment where both particles are accelerated at the same speed but in different directions.
So, say my collider has the capability to produce a magnetic field which at its maximum, can accelerate a charged particle so as to have energy of 3.5 TeV. Now in the case that we have two particles with the same energy going in opposite directions, we will give a total CoM energy of 7 Tev, following the result above. In the second case though, theres only one accelerating particle, hence $\sqrt{s} = \sqrt{2 \times 3.5 \times m_t}$ and since E $\ll$ m, this is always less than in the first case.
So be careful, because both experiments can be transformed into a CoM frame. In the CoM frame $|\mathbf{p_1}| = -|\mathbf{p_2}|$. Note this is true in both experiments, even in the second case where one of the particles is stationary. Well, the whole point is that we can use the above formulas so we can skip transforming to the CoM frame; we can compute this quantity directly.
I have the same reservations as you about the derivation that you found. I prefer this treatment ...
For a photon
$$E=hf=\frac{hc}\lambda.$$
We can show, using classical electromagnetic theory, that a 'slice' of an electromagnetic wave carrying energy $E$ in its fields also carries momentum $E/c$. Presumably a photon's momentum, $p$, is related to its energy in the same way. [We can reach the same conclusion by taking a photon to be a particle moving at speed $c$, and therefore massless, for which we can put $m=0$ in $E^2-c^2p^2=c^4m^2$.]
So...
$$p=\frac Ec\ \ \ \ \ \ \text {therefore}\ \ \ \ \ \ p=\frac h\lambda.$$
De Broglie famously suggested that the same relationship applies for a particle of matter, which, he supposed, has a wave aspect to it. This was an hypothesis rather than a logical deduction from established Physics, though de Broglie did note fragments of support for it, such as the light it seems to shed on Bohr's condition for a stable electron orbit in an atom: $pr=nh/2\pi$... If an electron wave fits without discontinuities round a circle of radius 𝑟 centred on the nucleus, then $n\lambda=2\pi r$. Substituting $h/p$ for $\lambda$ gives the Bohr condition!
We'll now look in more detail at the de Broglie wave for a particle moving at constant velocity, $\mathbf u$. De Broglie assumed that the Planck relationship between energy and frequency applies to the particle, that is
$$\ \ \ \ \ \ \ \ \ E=hf\ \ \ \ \ \ \ \text{so}\ \ \ \ \ \ \ \gamma m c^2=hf\ \ \ \ \ \ \ \ \ \text{in which}\ \ \ \ \ \ \ \ \ \ \gamma=\frac1{\sqrt{1-u^2/c^2}},$$
as well as
$$p=\frac h \lambda\ \ \ \ \ \ \ \text{that is}\ \ \ \ \ \ \ \gamma mu=\frac h{\lambda}$$
Dividing $\gamma m c^2=hf$ by $\gamma mu=\frac h{\lambda}$,
$$v=\frac {c^2} u$$
in which $v=f \lambda$ and represents then phase speed of the de Broglie wave.
At first sight this is upsetting: not only is $v$ different from $u$, but since $u<c$ we have $v>c$. But now we note that at any time a particle has a more or less well defined position in space, whereas a wave of sharply defined frequency has an infinite extent. But many such waves of not quite the same frequency and wavelength as each other can interfere to form a wave packet of finite extent. The wave energy travels, not at the phase speed $v$ of a constituent wave, but at the speed of the packet, the so-called group speed, $v_{\text{group}}$. It is this group speed that must be less than $c$.
To calculate the group speed neatly, we use wave variables $\omega$ and $k$, defined by
$$\omega= 2\pi f\ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ k=\frac{2\pi}\lambda$$
Thus, phase speed, $v$, can be written as
$$v=f \lambda=\frac{\omega}k$$
We can show that for any wave packet...
$$v_{\text{group}} =\frac{d\omega}{dk}$$
With our new notation, and writing $\hbar=\frac h{2\pi}$, the Planck and de Broglie relationships become
$$E=\hbar \omega\ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ p=\hbar k.$$
Special relativity gives us
$$E^2-c^2 p^2=c^4 m^2\ \ \ \ \ \ \ \text{so}\ \ \ \ \ \ \ \hbar^2 \omega^2-c^2 \hbar^2 k^2=c^4 m^2$$
From which we deduce that
$$\frac{d\omega}{dk}=c^2 \frac k \omega\ \ \ \ \ \ \ \ \ \ \text{that is}\ \ \ \ \ \ \ \ \ \ v_{\text{group}}=\frac{c^2}v=u$$
The group speed is therefore equal to the particle speed, a satisfying result. de Broglie referred to the particle as having a pilot wave. He did not subscribe to a probability interpretation of the wave.
Best Answer
In relativity, the energy $E$ of a particle is related to its mass $m$ and momentum $p$ by \begin{equation} E = \sqrt{m^2c^4 + p^2 c^2 } \end{equation} Now let's think about the non-relativistic limit, $c\rightarrow \infty$. To do this, we will expand the square root
\begin{equation} E = m c^2 \sqrt{1 + \frac{p^2}{m^2 c^2}} = mc^2 + \frac{p^2}{2m} + \cdots \end{equation} where the $\cdots$ refer to terms that vanish in the limit $c\rightarrow\infty$.
The first term is the famous equation $E=mc^2$. In non-relativistic physics, the mass is a constant, so this is just a constant term in the energy we can ignore.
The second term $E=\frac{p^2}{2m}$ is the non-relativistic expression for kinetic energy.
You can see that non-relativistic physics is a good approximation to relativistic physics when we can ignore the higher order terms. Thinking back to how we expanded the square root, this amounts to the condition $p c \ll m c^2$.