The equation $x=A\sin\left(\dfrac{(ω_1-ω_2)t}{2}\right)\times \cos\left(\dfrac{(ω_1+ω_2)t}{2}\right)$ can be thought of as consisting of two parts which are multiplied together.

The $\cos\left(\dfrac{(ω_1+ω_2)t}{2}\right)$ term which is an oscillation at a frequency which is the average of $\omega_1$ and $\omega_2$.  

The $A\sin\left(\dfrac{(ω_1-ω_2)t}{2}\right)$ term which oscillates at a lower frequency, $\dfrac{(ω_1-ω_2)}{2}$, than the other term and has an amplitude $A$.  

Adding some numbers, let $A=1,\,\omega_1 = 2\,\pi\,100.5$ and $\omega_2 = 2\,\pi\,99.5$, which is a difference of $1\,\rm Hz$.

Thus $\dfrac{(ω_1+ω_2)}{2} = 2\,\pi\, 100$ and $\dfrac{(ω_1-ω_2)}{2} = 2\,\pi\, \frac 12$ ie oscillations at $100\,\rm Hz$ are changing in amplitude at a frequency of $\frac 12\,\rm Hz$ as shown in the graph below.  

[![enter image description here][1]][1] 

The variation in **amplitude** of the wave of frequency $100\,\rm Hz$ is shown in the right hand diagram.  
The $100\,\rm Hz$ wave cannot be properly drawn because between $t=0\,\rm s$ and $t=1\,\rm s$ one hundred oscillations would have to be displayed.  

What you will note is that there is zero amplitude of the $100\,\rm Hz$ wave at $t=0\,\rm s$, maximum amplitude at $t=\frac 12\,\rm s$, zero amplitude at at $t=1\,\rm s$, maximum amplitude at $t=\frac 32\,\rm s$, zero amplitude at at $t=2\,\rm s$.  
Thus in $2$ seconds (the period of the $\sin \left(2\,\pi\,\frac12\right)$ term, the **amplitude** of the $100\rm \, Hz$ goes through two oscillations ie the beat frequency is $\dfrac{(ω_1-ω_2)}{2} \times 2 = ω_1-ω_2$ which in the case is $100.5-99.5=1\,\rm Hz$.  

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Perhaps what you actually hear will become clearer after you use this rather nice simulation, [Interference in Time and Beats Simulation](https://www.compadre.org/osp/EJSS/4370/203.htm)?  

When first started the simulation sets $f_{\rm A}=300\,\rm Hz$  and $f_{\rm B}=305\,\rm Hz$.  

On running the simulation you will hear a sound wave of frequency $302.5\rm\,Hz$ $pulsating$ at a frequency of $5\rm\,Hz$ which is the beat frequency.  

Keeping frequency $f_{\rm A}$ the same change $f_{\rm B}$ to:  

$\bf 1$ $301\,\rm Hz$ to hear a $1\rm\,Hz$ beat,  

$\bf 2$ $300.1\,\rm Hz$ to hear a $0.1\rm\,Hz$, period $10\,\rm s$, beat, 
 
$\bf 3$ $300.05\,\rm Hz$ to hear a $0.05\rm\,Hz$, period $20\,\rm s$, beat. [Do not worry that the hundredth of a hertz are not displayed as the simulation will still work.]

$\bf 4$ Have fun noting that the lowest available frequency is $40\,\rm Hz$ and the highest $1000\,\rm Hz$ so it is not possible with this simulation to show two ultrasonic frequencies producing an audible beat frequency eg $40,000\,\rm Hz$ and $40,100\,\rm Hz$ producing an audible $100\,\rm Hz$ tone


  [1]: https://i.sstatic.net/CDDRM.jpg