Sound beats mathematical expression I am reading about sound beats. According to the superposition principle, if at a certain position the two waves are given by, let say, 
$$\ y_a(t)=Asin(2\pi f_at)$$
$$\ y_b(t)=-Asin(2\pi f_bt)$$
then the total wave will be
$$\ y_a(t)+y_b(t)=[2Asin(\frac{1}{2}(2\pi)( f_a-f_b)t]cos\frac{1}{2}(2\pi)( f_a+f_b)t$$
Now, my question is, when grouping the terms in this expression, why do you say that the $sin(\frac{1}{2}(2\pi)( f_a-f_b)t$ is the amplitude factor and not the $cos\frac{1}{2}(2\pi)( f_a+f_b)t$. How can you know that it's the sine not the cosine that is varying the amplitude.
Edit: my real question is that why is it the $f_a-f_b$ not the $f_a+f_b$ term that is modulating the amplitude.
Image source: university physics by young and freedman.

 A: It's not particularly relevant that it is the $\sin$ term rather than the $\cos$ term, except in so far as that's how the expansion of the functions work.
Given we start with the expression
$$\begin{align}
y(t) &= y_a(t) + y_b(t)\\
     &= 2A\sin\left(2\pi \frac{f_a - f_b}{2}t\right)
          \cos\left(2\pi \frac{f_a + f_b}{2}t\right)\tag{1}
\end{align}
$$
The thing to look at is what this looks like when $f_a$ is very close to $f_b$, and in particular when $\left|f_a - f_b)\right| \ll (f_a + f_b)$.
In this case it's useful to define
$$
\begin{align}
f &= \frac{f_a + f_b}{2}\\
\Delta f &= \frac{f_a - f_b}{2}
\end{align}
$$
And $\left|\Delta f\right| \ll f$
So our expression now looks like
$$y(t) = 2A\sin(2\pi\Delta f t)\cos(2\pi ft)$$
Depending on how close in frequency $f_a$ & $f_b$ are, $\Delta f$ can be very small indeed: it can be zero in fact.  But $f$ is not going to be zero unless both $f_a$ & $f_b$ are zero (let's quietly assume that $f_a \ge 0$ & $f_b \ge 0$, which we can safely do).
So, if these are audio frequencies, what this is going to sound like is a tone of frequency $f\approx f_a \approx f_b$, changing slowly in volume at a frequency of $\Delta f$ which may be very small.
That's why it's that term which is thought of as modulating the amplitude.
Here are two pictures which, I hope, make it clear why it is the difference term which should be thought of as doing the modulating.  First of all here is a second of $\sin (200 \pi t)$: a $100\,\mathrm{Hz}$ sine wave:

Now here is a second of $\sin(200\pi t) + \sin(204\pi t)$: a waveform made up of $100\,\mathrm{Hz}$ and $102\,\mathrm{Hz}$ sine waves:

I think it is extremely obvious that this is a high-frequency wave being modulated by a low-frequency wave.
(Note that both of these pictures have some artifacts in due to my inadequate plotting skills.)

In the original version of this answer I thought the question was why the $\sin$ term rather than a $\cos$ term is doing the modulating, and I had a rather grotty paragraph showing that you could make it be either.  Here is a cleaned-up version of that, for posterity.
First of all, define $\omega_a = 2\pi f_a$ & $\omega_b = 2\pi f_b$: this is just to get rid of a bunch of annoying factors of $2\pi$: $\omega$ is the angular frequency.  To make things simpler I'll also make $A=1$: again this is just to make this term go away so I have less to write.
Now the original expression looks like
$$\sin(\omega_a t) - \sin(\omega_b t)$$
And if we assume $\omega_a \approx \omega_b$ we can write $\omega_a = \omega +\Delta\omega$ & $\omega_b = \omega - \Delta\omega$.  Then the above becomes
$$\sin(\omega t + \Delta\omega t) - \sin(\omega t - \Delta\omega t)
= 2\sin(\Delta\omega t)\cos(\omega t)$$
by the double-angle formula for $\sin$.
But what if, instead, we consider
$$
\begin{align}
\sin(\omega t + \Delta\omega t) - \sin(\omega t - \Delta\omega t - \pi)
&= \sin(\omega t + \Delta\omega t) + \sin(\omega t - \Delta\omega t)\\
&= 2\cos(\Delta\omega t)\sin(\omega t)
\end{align}
$$
This is just offsetting one of the terms by a constant of $\pi$: acoustically this is the same thing.  But now using the same double-angle formula we find the term doing the modulating is $\cos$ not $\sin$: it doesn't make any difference, in fact.
A: It's convention.  It's how we typically define modulation.
You are absolutely correct that mathematically, it doesn't matter if you are modulating around $\frac{f_a-f_b}{2}$ or $\frac{f_a+f_b}{2}$.  However, when we talk about modulation, we will always refer to the slower frequency as the one that is modulating the higher frequency.
In this particular example, the difference is arbitrary. In other cases, it will not be.  When we look at how we use modulation in practical applications, its virtually always using a low frequency signal to modulate a high frequency carrier (with a fixed amplitude).  Doing so provides lots of convenient mathematical properties which we leverage.  Doing it the other way (modulating a low frequency carrier with a high frequency signal) just doesn't provide the behaviors we want.  It provides other behaviors.
For example, one well known use of modulation is to modulate the audio-frequency signals from a radio station into a frequency band which can be easily transmitted through airspace.  In such cases, the signal is in the low kHz, and the carrier is in high kHz to MHz.
So when your wave question showing beats can be written in a form which looks like modulation, we'll always declare the low frequency side to be modulating the amplitude.
