Differentiation of sine in Fourier domain
I assume you mean the derivative with respect to $t$. In that case, the derivative of $\sin(\omega_0t)$ is not $\cos(\omega_0t)$ but $\omega_0\cos(\omega_0t)$. And luckily, this is also obtained via the Fourier transform relation you mentioned in your question:

$$\begin{align}\mathcal{F}\left\{\frac{d}{dt}\sin(\omega_0t)\right\}&amp;=j\omega\cdot \mathcal{F}\left\{\sin(\omega_0t)\right\}\\&amp;=\pi\omega[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)]\\&amp;=\pi[\omega_0\delta(\omega-\omega_0)-(-\omega_0)\delta(\omega+\omega_0)]\\&amp;=\pi\omega_0[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)]\\&amp;=\omega_0\mathcal{F}\{\cos(\omega_0t)\}\end{align}$$

where I've used the fact that $f(\omega)\delta(\omega-\omega_0)=f(\omega_0)\delta(\omega-\omega_0)$ for any function $f(\omega)$ that is continuous at $\omega=\omega_0$. Consequently you have $\omega\delta(\omega+\omega_0)=-\omega_0\delta(\omega+\omega_0)$.