Time-differentiation property of Fourier transform for $\sin(\omega_0t)$ and $\cos(\omega_0t)$
It does work. Note that the Fourier transforms you stated in your question are not the transforms of $\cos(t)$ and $\sin(t)$, but of $\cos(\omega_0t)$ and $\sin(\omega_0t)$.

Let's Assume we know the transform of $\sin(\omega_0t)$:

$$\mathcal{F}\{\sin(\omega_0t)\}=\frac{\pi}{j}[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)]\tag{1}$$

Now we take the derivative of $\sin(\omega_0t)$:

$$[\sin(\omega_0t)]'=\omega_0\cos(\omega_0t)\tag{2}$$

Using $(1)$ and the differentiation property of the Fourier transform, the Fourier transform of $(2)$ must be

$$\mathcal{F}\{\omega_0\cos(\omega_0t)\}=j\omega\cdot \frac{\pi}{j}[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)]\tag{3}$$

Noting that for any function $f(\omega)$ that is continuous at $\omega=\omega_0$ we have

$$f(\omega)\delta(\omega-\omega_0)=f(\omega_0)\delta(\omega-\omega_0)\tag{4}$$

and, consequently, we can rewrite $(3)$ as

$$\begin{align}\mathcal{F}\{\omega_0\cos(\omega_0t)\}&amp;= \pi[\omega_0\delta(\omega-\omega_0)-(-\omega_0)\delta(\omega+\omega_0)]\\&amp;=\omega_0\pi[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)]\end{align}\tag{5}$$

as expected.