What you suggest is essentially correct. If you have a standing wave (say, a vibrating piano string) vibrating at 440 Hz, you will create the note A (above middle C). You could write the displacement of the string by
$$y(x,t)=y_0 \cos(\omega t-kx)$$
where $x$ is the position of the string, $\omega=2\pi f$ (for frequency $f$) and $k=2\pi /\lambda$ is the wave number. This string will produce a sound equivalent to a single frequency tone $f$=440 Hz A. You also have the relationship $\lambda f=v$ for $v$ the speed of the wave on the string.
If you had a second string, vibrating at whatever the frequency of C is, you would have the same expression for the second string with the correct frequency of C.
The result at your ear (position $\vec{r}$) would be something like a pressure (sound) wave that obeys the principle of superposition,
$$P(\vec{r},t)=P_A \cos(\omega_A t +\vec{k}_A\cdot \vec{r})+P_C \cos(\omega_C t +\vec{k}_C\cdot \vec{r})$$
The slight complication here is the relationship bewteen $\lambda$ and $f$ will no longer be the speed on the string, but rather the speed of sound in air. But it's roughly equivalent - you can add the two waves together.
That's the simple, essentially correct, picture. The complication brought up by CDCM is very real, however - notes from musical instruments are never "pure tones". Even the vibrating strings of the piano produce higher harmonics (at the octave, 5th above that, 4th above that, etc etc etc). So in reality the pressure wave from your "single note A" looks more like
$$P(\vec{r},t)=\sum P_i\cos(\omega_i t-\vec{k}_i\cdot \vec{r})$$
for a set of frequencies $f_i$ that is highly dependent on the instrument used. That set of frequencies is not only why a guitar sounds different from a flute, but why 300 year old Italian string instruments are more valuable than the modern equivalents.
