How to calculate the perceived frequency of two sinusoidal waves added together? If you add together two sinusoidal waves of different frequencies, how do you calculate the frequency of the resulting function as perceived by a human?
 A: When two sine wave audio signals are added, and the frequencies are sufficiently different, we hear both frequencies, not some sort of average of the two. 
For example, listen to 250 hz and 600 hz sine wave audio signals added together.
You can hear both signals.  It does not sound like a single sine wave signal, of, say 425 hz.  
If the signals have frequencies close together, then we hear beats caused by the interference of the two signals.
For example, listen to 400 hz and 410 hz sine waves added together.  We hear the 10 hz difference in the the signal.
Adding two sinusoidal audio signals does not result in a signal perceived as a single sinusoidal signal.
A: Identical Amplitudes
When two sinusoidal waves of close frequency are played together, we get
$$
\begin{align}
\sin(\omega_1t)+\sin(\omega_2t)
&=2\sin\left(\frac{\omega_1+\omega_2}{2}t\right)\cos\left(\frac{\omega_1-\omega_2}{2}t\right)\\
&=\pm\sqrt{2+2\cos((\omega_1-\omega_2)t)}\;\sin\left(\frac{\omega_1+\omega_2}{2}t\right)\tag{1}
\end{align}
$$
Unless played together, two tones of equal frequency, but different phase sound just the same, so the "$\pm$" goes undetected (the sign flips only when the amplitude is $0$), and what is heard is the average of the two frequencies with an amplitude modulation which has a frequency equal to the difference of the frequencies.
$\hspace{1.5cm}$
The green curve is the sum of two sinusoids with $\omega_1=21$ and $\omega_2=20$; its frequency is $\omega=20.5$. The red curve is the amplitude as given in $(1)$, which has frequency $\omega=|\omega_1-\omega_2|=1$.
Differing Amplitudes
A similar, but more complex and less pronounced, effect occurs if the amplitudes are not the same; let $\alpha_1< \alpha_2$. To simplify the math, consider the wave as a complex character:
$$
\begin{align}
\alpha_1e^{i\omega_1 t}+\alpha_2e^{i\omega_2 t}
&=e^{i\omega_2t}\left(\alpha_1e^{i(\omega_1-\omega_2)t}+\alpha_2\right)\tag{2}
\end{align}
$$
The average frequency, $\omega_2$, is given by $e^{i\omega_2 t}$ (the frequency of the higher amplitude component), and the amplitude and a phase shift is provided by $\alpha_1e^{i(\omega_1-\omega_2)t}+\alpha_2$:
$\hspace{3.5cm}$
The amplitude (the length of the blue line) is
$$
\left|\alpha_1e^{i(\omega_1-\omega_2)t}+\alpha_2\right|=\sqrt{\alpha_1^2+\alpha_2^2+2\alpha_1\alpha_2\cos((\omega_1-\omega_2)t)}\tag{3}
$$
The phase shift (the angle of the blue line) is
$$
\tan^{-1}\left(\frac{\alpha_1\sin((\omega_1-\omega_2)t)}{\alpha_1\cos((\omega_1-\omega_2)t)+\alpha_2}\right)\tag{4}
$$
The maximum phase shift (the angle of the green lines) to either side is
$$
\sin^{-1}\left(\frac{\alpha_1}{\alpha_2}\right)\tag{5}
$$
This phase modulation has the effect of varying the frequency of the resulting sound from 
$$
\omega_2+\frac{\alpha_1(\omega_1-\omega_2)}{\alpha_2+\alpha_1}
=\frac{\alpha_2\omega_2+\alpha_1\omega_1}{\alpha_2+\alpha_1}\tag{6}
$$
(between $\omega_2$ and $\omega_1$) at peak amplitude to
$$
\omega_2-\frac{\alpha_1(\omega_1-\omega_2)}{\alpha_2-\alpha_1}
=\frac{\alpha_2\omega_2-\alpha_1\omega_1}{\alpha_2-\alpha_1}\tag{7}
$$
(on the other side of $\omega_2$ from $\omega_1$) at minimum amplitude.
Equation $(3)$ says that the amplitude varies between $|\alpha_1+\alpha_2|$ and $|\alpha_1-\alpha_2|$ with frequency $|\omega_1-\omega_2|$.
$\hspace{1.5cm}$
The green curve is the sum of two sinusoids with $\alpha_1=1$, $\omega_1=21$ and $\alpha_2=3$, $\omega_2=20$; its frequency varies between $\omega=20.25$ at peak amplitude to $\omega=19.5$ at minimum amplitude. The red curve is the amplitude as given in $(3)$, which has frequency $\omega=|\omega_1-\omega_2|=1$.
Conclusion
When two sinusoidal waves of close frequency are played together, the resulting sound has an average frequency of the higher amplitude component, but with a modulation of the amplitude and phase (beating) that has the frequency of the difference of the frequencies of the component waves. The amplitude of the beat varies between the sum and the difference of those of the component waves, and the phase modulation causes the frequency of the resulting sound to oscillate around the frequency of the higher amplitude component (between the frequencies of the components at peak amplitude, and outside at minimum amplitude).
If the waves have the same amplitude, the phase modulation has the effect of changing the frequency of the resulting sound to be the average of the component frequencies with an instantaneous phase shift of half a wave when the amplitude is $0$.
A: Since I don't have enough reputaion for commenting (I'm sure I will earn a lot with this post 3:> ) I will post this as an answer.
@robjohn is a moderator so I assume he'll do whatever is right with it.
Ad Differing Amplitudes from the Rob's answer.
It is perfectly correct to be surprised that the frequency varies with the amplitude. Mainly because it doesn't. Well at least not if a modulating wave has only positive values as in the above example. Modulating is multiplying not adding so the x-axis crossing points of the original wave remain unchanged. That is how the AM works. I assume that by a variable frequency you've meant roughly cycles per second not only shape deviations in the middle of a cycle.
Another thing is that you have used phasor aritmetics for time-varying waves which I believe won't add up. In your equation one can arbitrary pick ω1 or ω2 for the resulting wave frequency ("average frequency" as you have called it).
Despite the fact this is not the correct equation...
In your rotating circles diagram you've depicted the modulating wave but your calculations are more like it was the wave being added to the e^iω2t.
The amplitude is not equal to the length of the blue line. On your diagram the amplitude is just the length from the center of the α2 circle to the red dot projected on the x-axis.
Phase shift as the angle of the blue line. Again it is not an addition. It is a multiplication. And actually even if it was an addition you would have to place the little circle (α1) on the e^iω2t circle since this would be the wave that would gain the phase and the amplitude offset not the α2 (as it is a DC component and it's neutral regarding phase shift during waves addition).
Yes, I know that in the case of adding two waves of a different amplitude and frequency the ~frequency of the resulting wave varies with the width of the envelope. I'm just saying that it is not the equation for this.
