I'm pretty sure that we can't do any better with trig identities; it looks like any such expression would have a term vaguely like $\sqrt{1 + r \cos \theta}$, and then the square root ruins the nice intuition.
In lieu of that I'll offer a derivation of the group velocity that uses no fancy math at all! Hopefully that's what you wanted, even if you didn't directly ask it.
First off, let's figure out what group velocity is. Apparently, group velocity is the velocity of the envelope of a wave. But what is the envelope of a wave? I can create a wave with any initial position and velocity, so it can be an arbitrarily weird shape. There might be no discernible envelope at all.
So let's back up and find some examples. When we talk about the envelope of a wave, we mean some curve you draw around an oscillation, like this.
In order to do this, there must be a well defined oscillation to draw the envelope around. That means that our wave must be made up of individual frequencies that are close to one central frequency. To make things convenient, let's write that schematically as
$$\text{wave} = \sum_{k'} \sin(k'x - \omega(k') t) \text{ for a bunch of } k' \approx k$$
However, if we just have one frequency, the wave is just an infinite sinusoid $\sin(kx)$. This doesn't have an envelope, strictly speaking, because it just goes on forever at the same amplitude. We must have waves of other frequencies, which will constructively and destructively interfere with each other, to actually get an envelope.
So we've concluded that the envelope is defined by where a bunch of sinusoids making up our wave constructively or destructively interfere. Their phases are, as a function of space and time,
$$\phi(k') = k'x - \omega(k') t$$
Now let's assume for simplicity that the top of the envelope is at $x = 0$ at time $t = 0$. That means that the waves must constructively interfere there, so all the $\phi(k')$ are about the same.
As time goes on, the envelope will move, but the peak will still be where the phases of the component waves are the same. That means
$$\text{peak of envelope satisfies } \frac{d\phi(k')}{dk'} = 0$$
Performing the differentiation, we have
$$x - \frac{d\omega(k')}{dk'} t = 0$$
Since we said $k' \approx k$, let's drop the primes and rearrange for
$$\frac{x}{t} = \frac{d \omega(k)}{dk}$$
But $x/t$ is exactly the speed of the peak of the envelope, so this is the group velocity.
It doesn't really play a role (in a way), or at least not as far as physical results go. Whenever someone says
we consider a plane wave of the form $f(x) = Ae^{i(kx-\omega t)}$,
what they are really saying is something like
we consider an oscillatory function of the form $f_\mathrm{re}(x) = |A|\cos(kx-\omega t +\varphi)$, but:
- we can represent that in the form $f_\mathrm{re}(x) = \mathrm{Re}(A e^{i(kx-\omega t)})=\frac12(A e^{i(kx-\omega t)}+A^* e^{-i(kx-\omega t)})$, because of Euler's formula;
- everything that follows in our analysis works equally well for the two components $A e^{i(kx-\omega t)}$ and $A^* e^{-i(kx-\omega t)}$;
- everything in our analysis is linear, so it will automatically work for sums like the sum of $A e^{i(kx-\omega t)}$ and its conjugate in $f_\mathrm{re}(x)$;
- plus, everything is just really, really damn convenient if we use complex exponentials, compared to the trigonometric hoop-jumping we'd need to do if we kept the explicit cosines;
- so, in fact, we're just going to pretend that the real quantity of interest is $f(x) = Ae^{i(kx-\omega t)}$, in the understanding that you obtain the physical results by taking the real part (i.e. adding the conjugate and dividing by two) once everything is done;
- and, actually, we might even forget to take the real part at the end, because it's boring, but we'll trust you to keep it in the back of your mind that it's only the real part that physically matters.
This looks a bit like the authors are trying to cheat you, or at least like they are abusing the notation, but in practice it works really well, and using exponentials really does save you a lot of pain.
That said, if you are careful with your writing it's plenty possible to avoid implying that $f(x) = Ae^{i(kx-\omega t)}$ is a physical quantity, but many authors are pretty lazy and they are not as careful with those distinctions as they might.
(As an important caveat, though: this answer applies to quantities which must be real to make physical sense. It does not apply to quantum-mechanical wavefunctions, which must be complex-valued, and where saying $\Psi(x,t) = e^{i(kx-\omega t)}$ really does specify a complex-valued wavefuntion.)
Best Answer
The electric field is actually a real quantity. The complex notation is just a mathematical trick, we use to simplify the calculations. This trick is fine as long as we are dealing with linear systems, where the fields are multiplied by scalar numbers or added to scalar numbers. Once we leave this regime and calculate e.g. intensities $I\propto |E|^2$, we should convert the electric field to a real number before doing the calculation. E.g. the plane wave $$ E = E_0 e^{i(kx - wt)} $$ would yields $|E|^2 = |E_0|^2$ irrespectively of position and time, while $$ |\Re\{E\}|^2 = |E_0|^2 \cos^2(kx - wt) $$ accounts for the oscillations.