I've read it many places that Amplitude Modulation produces sidebands in the frequency domain. But as best as I can imagine it, modulating the amplitude of a fixed-frequency carrier wave just makes that "louder" or "quieter", not higher-frequency or lower-frequency. That is, I believe I could sketch, on graph paper, a path of a wave function that touches a peak or a trough exactly every 1/f increments, regardless of the "volume". Why do the sidebands appear?
[Physics] Understanding the cause of sidebands in Amplitude Modulation
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In amplitude modulation, the frequency of the carrier wave is constant. The frequency spectrum of an AM signal includes sidebands, but those aren't the carrier wave. In your second figure, the carrier wave is the black line. You'll note that the amplitude changes; it increases and decreases in accordance with the modulation, however the frequency of this wave does not change. That is the essence of amplitude modulation. For a carrier wave of constant frequency, the information is encoded in the amplitude of the signal. The presence of sidebands does not imply the frequency is non-constant, merely that the overall signal is not a single pure frequency. The carrier still remains the same throughout and the frequency of the AM wave is connotatively the same as the frequency of the carrier.
Now you can change the amplitude (AKA the power) of this frequency to encode data, and that's how AM radio works.
Correct. Let's say your carrier signal is $\cos\omega t$, and the message signal you want to encode is $\cos\omega_m t$ (where $\omega_m$ means the frequency of the message signal). Then the formula for the modulated signal (the carrier wave with its amplitude varied) is $$x(t)=\cos\omega t\cos\omega_m t$$
From Euler's identity we know $$\cos\omega t=\frac{e^{j\omega t}+e^{-j\omega t}}{2}$$ so $$x(t)=\cos\omega t \cos\omega_m t=\left(\frac{e^{j\omega t}+e^{-j\omega t}}{2}\right)\left(\frac{e^{j\omega_m t}+e^{-j\omega_m t}}{2}\right)$$
Just doing algebraic manipulations this becomes $$x(t) = \frac{e^{j\omega t}e^{j\omega_m t}+e^{j\omega t}e^{-j\omega_m t} + e^{-j\omega t}e^{j\omega_m t}+e^{-j\omega t}e^{-j\omega_m t}}{4}$$
Then using the rule for multiplying exponentials, $e^a e^b = e^{(a+b)}$ this becomes
$$x(t) = \frac{e^{j\omega t+j\omega_m t}+e^{j\omega t-j\omega_m t} + e^{-j\omega t+j\omega_m t}+e^{-j\omega t-j\omega_m t}}{4}$$
or
$$x(t) = \frac{e^{j(\omega+\omega_m) t}+e^{j(\omega-\omega_m) t} + e^{-j(\omega +\omega_m) t}+e^{-j(\omega-\omega_m) t}}{4}$$
or $$x(t) = \frac{e^{j(\omega+\omega_m) t}+ e^{-j(\omega +\omega_m) t}}{4}+\frac{e^{j(\omega-\omega_m) t} +e^{-j(\omega-\omega_m) t}}{4}$$
Recombining the exponentials back into cosines: $$x(t) = \frac{\cos(\omega+\omega_m) t}{2}+\frac{\cos(\omega-\omega_m) t}{2}$$
So what this shows is that just the act of varying the amplitude of the oscillating wave does inherently change the signal from a single frequency to a pair of side-band frequencies.
If you didn't fully modulate the signal (if the message signal were something like $1+a\cos\omega_m t$ with $a<1$) you would find some of the carrier frequency still in the modulated signal. If you use a message signal that contains multiple frequencies, like a voice signal or a real message, instead of a pure sinusoid, you'll find you get replicas of the spectra of the message signal on either side of the carrier, rather than discrete sideband frequencies.
Best Answer
Let your carrier signal be $A_0 \cdot \cos(\omega_c t)$ with amplitude $A_0$ and carrier frequency $\omega_c$. Let your signal be a simple wave, $\phi(t) = A_s \cdot \cos(\omega_s t)$.
Then the modulated signal becomes $$A_0 A_s \cdot \cos(\omega_c t) \cdot \cos(\omega_s t)$$.
In addition, as pointed out by George in the comments, the carrier also gets transmitted.
Using the trigonometric identity $\cos(u) \cdot \cos(v) = \frac{1}{2} [\cos(u-v) + \cos(u+v)]$, you get the final signal: $$\frac{1}{2} A_0 A_s \cdot ( \cos((\omega_c - \omega_s) t) + \cos((\omega_c + \omega_s) t)) + A_0 \cdot \cos(\omega_c t)$$ Hence, the frequency becomes changed, you get the carrier frequency in the middle (at $\omega$) and two side-bands at $\omega_c \pm \omega_s$.
Now, in reality your signal is not a simple cosine, but you could do a Fourier decomposition of the signal and treat each frequency independently. The two frequencies then get smeared out and you get the two sidebands.