The discrete Fourier transform implicitly regards the input data as periodic. The length of the input data is an integer multiple of the frequencies that the output data correspond to. If you generate the input data from a sinusoidal that has a different period that doesn't evenly divide the length of the data, then you're effectively creating a jump at the point where the first data point follows the last in the periodic continuation. A jump has appreciable Fourier components at all frequencies, and this is superimposed on your signal.
Nevertheless, you do get a signal with a clear maximum around the input frequency. I tried this using this online FFT calculator. In your case, as you calculated, we'd expect the maximum to be around the $21$st and $22$nd output data point (with the first data point corresponding to frequency $0$), and indeed, the first $40$ output values are:
30.322000,0.000000
30.391603,-0.332305
30.619103,-0.635705
30.983132,-0.983397
31.541176,-1.314554
32.258933,-1.685996
33.213385,-2.076468
34.383753,-2.509669
35.850294,-2.989665
37.679693,-3.538320
39.950279,-4.172808
42.781442,-4.897551
46.400390,-5.801462
51.088228,-6.910079
57.343876,-8.338977
66.017176,-10.282208
78.721064,-13.088380
98.974578,-17.487260
136.068697,-25.448952
225.318492,-44.421047
727.024048,-150.774758
-543.414483,118.240910
-191.999756,43.727082
-114.594042,27.275221
-80.661116,20.008091
-61.661208,15.918785
-49.536603,13.289953
-41.141647,11.459233
-35.002598,10.087933
-30.321248,9.041060
-26.630074,8.216618
-23.664950,7.534937
-21.230810,6.967688
-19.199079,6.488785
-17.492157,6.083581
-16.004951,5.723743
-14.733225,5.416510
-13.616822,5.138707
-12.640339,4.905611
-11.768659,4.665358
So it seems you must have been doing something wrong when you got spikes all over the place at roughly the same amplitude.
You can actually calculate the output in closed form in this case by writing the sinusoidals in terms of complex exponentials and summing the resulting partial geometric series.
Edit:
Here's that last suggestion written out.
You have the time-domain signal $\sin2\pi\nu t$ with $\nu=440\text{Hz}$, and you sample it at the frequency $f=44100\text{Hz}$, so you have the samples
$$
s_k=\sin\frac{2\pi\nu k}f=\frac1{2\mathrm i}\left(\exp\left(\frac{2\pi\mathrm i\nu k}f\right)-\exp\left(-\frac{2\pi\mathrm i\nu k}f\right)\right)
$$
for $0\le k\lt n$, with $n=2048$. Performing a discrete Fourier transform on these samples yields
\begin{eqnarray*}
\tilde s_j
&=&
\sum_{k=0}^{n-1}s_k\exp\left(\frac{2\pi\mathrm i jk}n\right)
\\
&=&
\sum_{k=0}^{n-1}\frac1{2\mathrm i}\left(\exp\left(\frac{2\pi\mathrm i\nu k}f\right)-\exp\left(-\frac{2\pi\mathrm i\nu k}f\right)\right)\exp\left(\frac{2\pi\mathrm i jk}n\right)
\\
&=&
\frac1{2\mathrm i}\sum_{k=0}^{n-1}\left(\exp\left(2\pi\mathrm i\left(\frac jn+\frac\nu f\right)k\right)-\exp\left(2\pi\mathrm i\left(\frac jn-\frac\nu f\right)k\right)\right)
\\
&=&
\frac1{2\mathrm i}\sum_{k=0}^{n-1}\left(\omega_{j+}^k-\omega_{j-}^k\right)
\\
&=&
\frac1{2\mathrm i}\left(\frac{1-\omega_{j+}^n}{1-\omega_{j+}}-\frac{1-\omega_{j-}^n}{1-\omega_{j-}}\right)\end{eqnarray*}
with $\omega_{j\pm}=2\pi\mathrm i\left(\frac jn\pm\frac\nu f\right)$. The second term dominates and is maximal at $\frac jn=\frac\nu f$ (which in your case is at the fractional value $j=\frac{n\nu}f$). It's a phase factor times
$$
\frac{\sin n\pi\left(\frac jn-\frac\nu f\right)}{\sin\pi\left(\frac jn-\frac\nu f\right)}\;,
$$
which is a discrete periodic analogue of the $\operatorname{sinc}$ function. Fitting to this functional form might improve your estimate of the frequency from the output data of the Fourier transform.
Specifically for Python, you obtain the frequencies of the DFT by using numpy.fft.fftfreq(n,d) where n is the sample size and d is the sample spacing $\Delta t$, that is, the intervals at which you sample a signal: $t=\{0,\Delta t,2\Delta t,...,(N-1)\Delta t\}$. The frequency spacing will be $1/(N\Delta t)$. Indeed
$$\hat{x}_{\frac{n}{N\Delta t}}=\sum_{k=0}^{N-1}x_{k \Delta t} e^{-2\pi i(k\Delta t)(\frac{n}{N\Delta t })}=\sum_{k=0}^{N-1}x_{k \Delta t} e^{-2\pi i\frac{kn}{N}}$$
which is the DFT of the signal.
Reference here.
Best Answer
The Fourier transform of a signal $v(t)$ is
$$V(f)=\int_{-\infty}^{\infty} v(t)e^{-j2\pi f}\;\mathrm{d}t$$
If you just substitute $f=0$ in the above equation, you obtain
$$V(0)=\int_{-\infty}^{\infty} v(t)\;\mathrm{d}t$$
which is just a scaled version of the mean of the signal. Any nonzero-mean signal will have a zero-frequency component in its Fourier transform.