Hilbert transform of sinusoid -- apparent contradiction
The error lies in the assumption that if $g(t)$ is the Hilbert transform of $f(t)$, then the Hilbert transform of $f(-t)$ must be $g(-t)$. This is not the case.

Let $f^-(t)=f(-t)$. Then we have

$$g(t)=\mathcal{H}\{f\}(t)=\frac{1}{\pi}\text{p.v.}\int_{-\infty}^{\infty}\frac{f(\tau)}{t-\tau}d\tau\tag{1}$$

and

$$\begin{align}\mathcal{H}\{f^-\}(t)&amp;=\frac{1}{\pi}\text{p.v.}\int_{-\infty}^{\infty}\frac{f(-\tau)}{t-\tau}d\tau\\&amp;=\frac{1}{\pi}\text{p.v.}\int_{-\infty}^{\infty}\frac{f(\tau)}{t+\tau}d\tau\\&amp;=-\mathcal{H}\{f\}(-t)\\&amp;=-g(-t)\end{align}\tag{2}$$

So if $f(t)=\cos(\omega t)$ and $g(t)=\mathcal{H}\{f\}(t)=\sin(\omega t)$, then the Hilbert transform of $f^-(t)=\cos(-\omega t)$ equals $-g(-t)=-\sin(-\omega t)=\sin(\omega t)$. Happiness again.

EDIT:

In response to your comment about substituting for $\omega$ instead of reversing time, note that

$$\mathcal{H}\{\cos(\omega_0 t)\}=\sin(\omega_0 t)$$

is only valid for $\omega_0&gt;0$. This can most easily be seen in the frequency domain, where the Hilbert transform corresponds to a multiplication with $-j\;\text{sign}(\omega)$. The Fourier transform of $f(t)=\cos(\omega_0t)$ is given by

$$F(\omega)=\pi\left[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)\right]$$

If $\omega_0&gt;0$, multiplication by $-j\;\text{sign}(\omega)$ gives the Fourier transform of $\sin(\omega_0t)$. However, if $\omega_0&lt;0$, we obtain the Fourier transform of $-\sin(\omega_0t)$, because then $\delta(\omega+\omega_0)$ appears at positive frequencies instead of $\delta(\omega-\omega_0)$.