Hello Zahara,
The Nyquist theorem gives you the bare minimum number of points to describe the oscillation. More points are better, as you are finding out. Taking your case of 5 Hz, Nyquist frequency 10 Hz, then
t = 0:.1:1;
cos(2*pi*5*t)
ans = 1 -1 1 -1 1 -1 1 -1 1 -1
showing that there is a steady oscillation, but at just two points per oscillation, you can't see the shape of the cosine waveform at all
And
sin(2*pi*5*t)
1.0e-14 * [0 0.0122 -0.0245 -0.1409 -0.0490 0.0612 0.2818 -0.2695 -0.0980 0.1102]
which is essentially 0 since the time points match up with the zeros of the sine function. In practical terms, the Nyquist criterion is no help at all here. This may or may not be why you are obtaining the flat line.
