Assuming we denote Fourier transforms as follows:
$\mathcal{F}[f(t)](\omega)=\tilde{f}(\omega)$, then we have the following identity:
$\mathcal{F}[\frac{df}{dt}](\omega)=i\omega \tilde{f}(\omega)$
In addition, we know that the fourier transforms of sines and cosines are given by
$\mathcal{F}[\sin(\omega_0 t)]=\frac{i}{2}\left(\delta(\omega+\omega_0)-\delta(\omega-\omega_0)\right)$
and
$\mathcal{F}[\cos(\omega_0 t)]=\frac{1}{2}\left(\delta(\omega+\omega_0)+\delta(\omega-\omega_0)\right)$
However, if we apply the first identity:
$\mathcal{F}[\cos(\omega_0 t)]=\mathcal{F}[\frac{1}{\omega_0}\frac{d}{dt}\sin(\omega_0 t)]=\frac{-\omega}{2\omega_0}\left(\delta(\omega+\omega_0)-\delta(\omega-\omega_0)\right)$
which looks nothing like the first formula for the fourier transform of the cosine. I realize it's zero everywhere except at $\pm \omega_0$, so the prefactor isn't an issue, but the signs are wrong.
Where is the error?
EDIT: corrected sign on initial identify as per first comment
$$\mathcal{F}[\frac{df}{dt}](\omega)=i\omega \tilde{f}(\omega) $$
and
$$\mathcal{F}[\cos(\omega_0 t)]=\mathcal{F}[\frac{1}{\omega_0}\frac{d}{dt}\sin(\omega_0 t)]=\frac{1}{\omega_0}(i\omega)\frac{i}{2}\left(\delta(\omega+\omega_0)-\delta(\omega-\omega_0)\right)=\frac{-\omega}{2\omega_0}\left(\delta(\omega+\omega_0)-\delta(\omega-\omega_0)\right)$$
$$\frac{-\omega}{2\omega_0}\left(\delta(\omega+\omega_0)-\delta(\omega-\omega_0)\right)=\left(\frac{-\omega}{2\omega_0}\delta(\omega+\omega_0)-\frac{-\omega}{2\omega_0}\delta(\omega-\omega_0)\right)=\left(\frac{\omega_0}{2\omega_0}\delta(\omega+\omega_0)+\frac{\omega_0}{2\omega_0}\delta(\omega-\omega_0)\right)=\frac{1}{2}\left(\delta(\omega+\omega_0)+\delta(\omega-\omega_0)\right)$$