Are both halves of a symmetrical FFT result required to generate the original waveform?
You can perfectly use, $X_h[k]$, one half of the conjugate symmetric DFT $X_f[k]$ which belongs to a real signal $x[n]$ of length $N$.

Of course in doing so, you must symmetrically pad the half length DFT $X_h[k]$ to the full length DFT $X_f[k]$, which is obviously necessary for an unaliased reconstruction of $x[n]$ from its proper length DFT $X_f[k]$ of length N.

Note that during the creation of the mirrored padded full DFT $X_f[k]$ from the symetric half $X_h[k]$, the first sample $X_h[0]$ is not repeated and the remainig samples $X_h[1]$ to $X_h[M-1]$ are mirrored with their phase being negated (i.e. they are conjugated), where $M$ is the length of the half length DFT $X_h[k]$.

The length $M$ of the half sized DFT $X_h[k]$ is given by the following: If the signal's (or its natural DFT $X_f[k]$'s) length $N$ is even then $M= 1 + \frac N 2 = \frac{N+2}{2}$, otherwise, if $N$ is odd $M = 1 + \frac{N-1}{2} = \frac{N+1}{2} $.

Also when $N$ is even, the last sample of the half sized DFT, $X_h[M]$ is not repeated, and when $N$ is odd then the last sample $X_h[M]$ is conjugate mirrored in formation of the full length DFT $X_f[k]$, as examplified below Matlab/Octave code:

x = [1:8];      % N = 8 , a signal of length N = 8 (even)
X = fft(x,8);   % X[k] is the natural DFT of x[n]
Xh = X(1:5);    % M = (8+2)/2 = 5, also matlab indexing starts from k=1
Xf = [ Xh conj(Xh(end-1:-1:2))];    %N:even =&gt; Xh[0] and Xh[5] not repeated
xr = ifft(Xf,8); % reconstruct the original signal from Xf[k]
xr = 1     2     3     4     5     6     7     8


as expected.