Differences in transforming a sinc FIR filter from lowpass to highpass using spectral inversion and spectral reversal
This answer is gonna come in installments.
First consider the ideal $\operatorname{sinc}(\cdot)$ filter, which in the frequency domain is an ideal brickwall.  We'll do this in normalized angular frequency, $\omega$.
Let
$$ H_\mathrm{LPF}(e^{j\omega}) = \begin{cases}
1, \qquad  |\omega| &lt; \omega_0 \\
0, \qquad  \omega_0 &lt;  |\omega| \le \pi \\
\end{cases} $$
and define the frequency response outside of $\pm\pi$ to be periodic:
$$ H_\mathrm{LPF}(e^{j(\omega+2\pi)}) = H_\mathrm{LPF}(e^{j\omega}) \qquad \forall \omega \in \mathbb{R} $$
This will have impulse response of
$$ h_\mathrm{LPF}[n] = \frac{\omega_0}{\pi} \operatorname{sinc}\left( \frac{\omega_0}{\pi} n \right) $$
Now, if you make a high-pass filter by subtracting the low-pass from a wire:
$$ H_\mathrm{HPF}(e^{j\omega}) = 1 - H_\mathrm{LPF}(e^{j\omega}) $$
Then the cutoff frequency remains the same $\omega_0$ and the impulse response is
$$\begin{align}
 h_\mathrm{HPF}[n] &amp;= \delta[n] - h_\mathrm{LPF}[n] \\
 &amp; = \delta[n] - \frac{\omega_0}{2} \operatorname{sinc}\left( \frac{\omega_0}{\pi} n \right) \\
\end{align} $$
Now, suppose you make a high-pass filter by heterodyning the low-pass from DC to Nyquist:
$$ H_\mathrm{HPF}(e^{j\omega}) = H_\mathrm{LPF}(e^{j(\pi+\omega)}) $$
Then the cutoff frequency becomes $\pi-\omega_0$ and the impulse response is
$$\begin{align}
 h_\mathrm{HPF}[n] &amp;= (-1)^n h_\mathrm{LPF}[n] \\
 &amp; = (-1)^n \frac{\omega_0}{\pi} \operatorname{sinc}\left( \frac{\omega_0}{\pi} n \right) \\
\end{align} $$
So, if we harmonize the two definitions of the cutoff frequencies, the two brick-wall high-pass filters will be identical in the frequency domain, so they should in the time domain.  But that would mean that
$$\begin{align}
\delta[n] - \frac{\omega_0}{\pi} \operatorname{sinc}\left( \frac{\omega_0}{\pi} n \right) &amp;\stackrel{?}{=} (-1)^n \frac{\pi - \omega_0}{\pi} \operatorname{sinc}\left( \frac{\pi-\omega_0}{\pi} n \right) \\
\\
&amp;= (-1)^n \frac{\pi - \omega_0}{\pi} \frac{\sin\left( \pi \frac{\pi-\omega_0}{\pi} n \right)}{\pi \frac{\pi-\omega_0}{\pi} n} \qquad &amp; (n \ne 0) \\
\\
&amp;= (-1)^n \ \frac{\sin\big( (\pi - \omega_0) n \big)}{\pi n} \qquad &amp; (n \ne 0) \\
\\
&amp;= (-1)^n \ \frac{\sin(\pi n)\cos(\omega_0 n) - \cos(\pi n)\sin(\omega_0 n)}{\pi n} \qquad &amp; (n \ne 0) \\
\\
&amp;= (-1)^n \ \frac{0\cos(\omega_0 n) - (-1)^n\sin(\omega_0 n)}{\pi n} \qquad &amp; (n \ne 0) \\
\\
&amp;= (-1)^n \ \frac{- (-1)^n \ \sin\left(\pi\frac{\omega_0}{\pi} n\right)}{\pi n} \qquad &amp; (n \ne 0) \\
\\
&amp;= \frac{- \sin\left(\pi\frac{\omega_0}{\pi} n\right)}{\pi n} \qquad &amp; (n \ne 0) \\
\\
&amp;= -\frac{\omega_0}{\pi} \frac{\sin\left(\pi\frac{\omega_0}{\pi} n\right)}{\pi \frac{\omega_0}{\pi} n} \qquad &amp; (n \ne 0) \\
\\
&amp;= -\frac{\omega_0}{\pi} \operatorname{sinc}\left(\frac{\omega_0}{\pi} n \right) \qquad &amp; (n \ne 0) \\
\end{align}$$
So I guess it's true, because you can also show that equally exists for the case of $n=0$.