Interference beats for a more general trigonometric sum Suppose I have three frequencies $\alpha,\beta,\gamma$ that are all close in value, and I consider the sum $\sin(\alpha x) +\sin(\beta x) +\sin(\gamma x)$ If there were only summands I could find a million explanations online that shows me how to rewrite this as a sinusoid amplitude-modulated by another sinusoid. But what about the case of three summands? 
Can we write this as $A(x)\sin(\frac{\alpha+\beta+\gamma}{3}x)$ where $A(x)$ is some reasonably simple function? 
 A: Let us expand
$$\sin\left(\dfrac{\alpha x}{3}+\dfrac{(\beta x+\gamma x)}{3}\right)$$
We get
$$\sin \dfrac{\alpha x}{3} \cos \dfrac{(\beta x+\gamma x)}{3} + \cos \dfrac{\alpha x}{3} \sin \dfrac{(\beta x+\gamma x)}{3}$$
Let us now expand
$$\cos \dfrac{(\beta x+\gamma x)}{3}= \cos \dfrac{\beta x}{3} \cos \dfrac{\gamma x}{3} - \sin \dfrac{\beta x}{3} \sin \dfrac{\gamma x}{3}$$
And
$$\sin \dfrac{(\beta x+\gamma x)}{3}= \sin \dfrac{\beta x}{3} \cos \dfrac{\gamma x}{3} + \cos \dfrac{\beta x}{3} \sin \dfrac{\gamma x}{3}$$
Substitution simply gives
$$\sin\left(\dfrac{\alpha x}{3}+\dfrac{(\beta x+\gamma x)}{3}\right)=\\ \sin \dfrac{\alpha x}{3} \cos \dfrac{\beta x}{3} \cos \dfrac{\gamma x}{3} - \sin \dfrac{\alpha x}{3}\sin \dfrac{\beta x}{3} \sin \dfrac{\gamma x}{3} \\
+ \cos \dfrac{\alpha x}{3} \sin \dfrac{\beta x}{3} \cos \dfrac{\gamma x}{3} + \cos \dfrac{\alpha x}{3}\cos \dfrac{\beta x}{3} \sin \dfrac{\gamma x}{3}$$
I do not know what you mean by "reasonably simple", however you can expand as follows:
$$\sin(\dfrac{x}{3})=\sum^\infty_{k=0}\dfrac{(-1)^k\left(\dfrac{x}{3}-\dfrac{\pi}{2}\right)^{2k}}{(2k)!}$$
Maybe this will give you idea on the function $A(x)$, on which I am willing to bet is not "simple", satisfying
$$\sin(\alpha x)+\sin(\beta x)+\sin(\gamma x)=A(x)\sin\left(\dfrac{\alpha+\beta+\gamma}{3}x\right)$$
