Calculate aliasing of $x_a(t) = \cos{(2\pi300t)} + \cos(2\pi600t)$ when sampled with $F_s = 1000$ I'm asked to sample the signal
$$x_a(t) = \cos{(2\pi300t)} + \cos(2\pi600t)$$
with sampling frequency $F_s = 1000$ and plot the magnitude spectrum for the resulting sampled signal. 

My thinking is that the frequency of $300$ does not change, so it just results in the normalized frequency $\frac{3}{10} = 0.3$.

Since $600$ is above the nyquist frequency, aliasing arises. So we get the frequencies $\frac{6}{10} = 0.6$ and $\frac{6}{10} - 1 = - 0.4.$

So in total I would plot peaks at the frequencies $\pm0.3, \pm0.4, \pm0.6$, but the answer is supposedly: 

 

Why? Where is my thinking wrong? And what is the method to always get the right aliasing peaks?
Peter said it all in his comment, but I'll try to explain it here in a bit more detail.

Your thinking is correct, but incomplete. Of course, the component at $f=300$ (unit depends on normalization of $t$) will not cause aliasing, so there will be no additional component below Nyquist, but what you forget is that sampling makes the spectrum periodic with period $f_s$ (sampling frequency). So the $f=\pm 300$ component results in components at $nf_s\pm 300$, $n\in\mathbb{Z}$. For $n=1$ you get the components at $f=1000\pm 300$, one of which (the one at $f=700$) is the one you see in the figure at normalized frequency $0.7$.