This is just an example of aliasing. Yes, looking only at the samples n it is impossible to distinguish between 2a\cos(\pi n) = ae^{i\pi n} + ae^{-i\pi n} and 2a\cos(\pi n) + 2b\sin(\pi n) = (a-ib)e^{i\pi n} + (a+ib)e^{-i\pi n} for any b, because the sine term vanishes on all the samples (equivalently, e^{\pm i \pi n} are aliased). You must assume something about what happens in between the samples to determine b. For example, if you want the interpolant that minimizes the mean-square slope you get b=0, since the mean-square slope is:
which is obviously minimized for b=0, whereas a is fixed by the sample observations.
Interpolating inherently involves choices, because there are infinitely many functions that interpolate between a discrete set of samples. Trigonometric interpolation is no different. Fortunately, there are reasonable criteria (like bandlimited interpolants with minimal mean-square slope) that lead to a simple choice of interpolant.
In a sense, yes — if you sample a DFT exactly at the Nyquist limit then the bandlimited assumption is not sufficient to determine the interpolant. You need an additional criterion like the minimal-slope condition above. But I don’t see this as a big problem; after all, the band-limited assumption itself is usually just an approximation.
(For the classic form of the Nyquist–Shannon sampling theorem, you have infinitely many samples — a DTFT, not a DFT — giving a continuous set of frequencies in [-\pi,\pi] and corresponding amplitudes. In this case, the Nyquist frequency \pm \pi is a set of measure zero and doesn’t contribute anything to the interpolated signal as long as the Fourier transform is regular (no delta functions). Hence the bandlimited assumption is sufficient by itself to determine a unique interpolant/reconstruction.)