In the original formula we have that $$\omega_1t-k_1x = A$$say, and $$\omega_2t-k_2x = B$$say, by hypothesis for a specific point at time $t$ and position $x$. This is a point of constant phase (for the $A$ wave and $B$ wave respectively.) To determine the velocity of a (sine or cosine) wave from first principles one wants to know the velocity of that point: how far does a point of constant phase move in time t? This gives the answer of the phase velocity
$$
v_p=\frac{\omega}{k}
$$
So the component waves are moving with phase velocities: $v_{p1}$ and $v_{p2}$ respectively. Using the above values for $A$ and $B$ the middle derivation is an application of the trigonometric identity:
$$
sin A + sin B = 2 sin (1/2(A+B)) \cdot cos (1/2(A-B)).
$$
This gives your expression for $s = s(x,t)$. So how can one talk about a point of constant phase here to obtain the group velocity as it is a product of sin and cos?
Well the trick indeed is to recognise the two separate wave components and treat these (for now) as two separate waves and calculate their (phase) velocity - ie the rate of movement of points of constant phase in each "wave".
For the sine wave ie the envelope we would get $\frac{\overline{\omega}}{\overline{k}}$.
Now for the cosine wave. The short answer is that we are looking for its phase velocity also, namely $\frac{\Delta{\omega}}{\Delta k}$.
However what your professor has done here, is to calculate from first principles the velocity of that cosine wave. That is to ask for the definition of a point of constant phase, viz: $\frac{\Delta{\omega}}{2}t - \frac{\Delta k}{2} x = const$ and then to determine the velocity (by differentiation, etc) of this point, again resulting in
$$
v_g=\frac{\Delta\omega}{\Delta k}
$$
It is hard to know what exact the professor has done, but from what I have understood made an effort to help the situation.
Q1. A Gaussian function is chosen to represent a freely expanding wave function for several reasons:
(i) The Gaussian function represents a normal probability distribution function. Since $|\psi(x)|^2$ represents a probability distribution function for a huge class of particles moving in a similar way, and having momentum within a certain range, the large numbers theorem points towards a Gaussian function, to represent the wave function of particles within a region of space determined by the width, $\sigma$, of the Gaussian function. The $\sigma$ also happens to be the standard deviation, i.e. the uncertainty in the position of the particle.
(ii) The expectation value of the position of the particle turns out to be the value of $b$ in your Gaussian wave packet.
(iii) The Gaussian wave packet also contains the wavy bit shown by the phase factor
$e^{ikx}$.
This is what makes Gaussian wave packets an excellent representation of particles, which are known to be located within some region of width $w$, but nevertheless they are all moving as plane waves while the packet travels along in space.
(iv) The wave packet is made of an infinitely large number of momentum values in a momentum width which relates to $\sigma$ by a Fourier transformation.
Q2. So to put some order in all these, let us consider the general Gaussian function
$\psi(x)=\psi_0e^{-A(x-x_0)^2+Bx+C}$
which has a Fourier transformation
$\psi(k)=\psi_0\sqrt{\frac{\pi}{A}}e^{\frac{B^2}{4A}+Bx_0+C}$.
The normalisation constant $\psi_0$ is given by an integration in the range [-$\infty, +\infty$] and has the value $\sqrt{\frac{A}{\pi}}$ but leave it as $\psi_0$.
Q3. To make contact with your professor, try to apply these results using the following:
$A=\frac{1}{2\sigma^2}$
$B=ik$
$x_0=b$
$C=0$
Best Answer
Jerry Schirmer's answer applies in an infinite space. If you put the system in a box there is no problem: the normalized wavefunction is $\mathrm{e}^{ikx}/\sqrt{V}$. This is the usual theoretical device to make everything nice and well behaved. Then we take the limit $V\rightarrow\infty$ at the end of the day and, if we've done our job correctly, all of the $V$ dependence should drop out of the final answer. It's not too unreasonable either, since we don't actually know that the universe is infinite. Infinite for all practical purposes might as well just be a big box.
EDIT in response to comments:
Assume we operate in a box of volume $V$ (in any number of dimensions - interpret as length, area or volume as appropriate). We want to find the normalisation factor $N$ for the wavefunction $\psi = N \mathrm{e}^{i\vec{k}\cdot\vec{x}}$. So we calculate the norm of the wavefunction, which must be equal to one:
$$ \begin{array}{lcl} 1 &=& \int\mathrm{d}x\ \psi^\star \psi \\ &=& \int\mathrm{d}x\ N^\star N \\ &=& \left| N \right|^2 V \end{array}$$
So $N$ has a magnitude of $1/\sqrt{V}$ and an arbitrary phase which can be chosen to make it real and positive.