One property of Fourier transform is: $\frac{d}{dt}f(t) \xrightarrow{\mathscr{F}} (j\omega) F(\omega)$
We know following transform pairs:
$\sin(\omega_0t + \theta) \xrightarrow{\mathscr{F}} j\pi [e^{-j\theta} \delta(\omega + \omega_0) - e^{j \theta} \delta(\omega - \omega_0)]$
$\cos(\omega_0t + \theta) \xrightarrow{\mathscr{F}} \pi [e^{-j\theta} \delta(\omega + \omega_0) + e^{j \theta} \delta(\omega - \omega_0)]$
Now let's take $f(t) = \sin(\omega_0t + \theta)$.
$\frac{d}{dt}f(t) = \omega_0 \cos(\omega_0t + \theta) = f_2(t)$
Now we take fourier transform of $f_2(t)$ and we get
$f_2(t) \xrightarrow{\mathscr{F}} \omega_0 \pi [e^{-j\theta} \delta(\omega + \omega_0) + e^{j \theta} \delta(\omega - \omega_0)]$
Now calculate fourier transform using property:
$\frac{d}{dt}f(t) \xrightarrow{\mathscr{F}} (j \omega) j\pi [e^{-j\theta} \delta(\omega + \omega_0) - e^{j \theta} \delta(\omega - \omega_0)] = -\omega \pi [e^{-j\theta} \delta(\omega + \omega_0) - e^{j \theta} \delta(\omega - \omega_0)] $
Why transformations are not equal?
And most importantly why do I get $\omega_0$ in one case and $\omega$ in other?
Recall that the distributions $f(x)\delta(x-x')$ and $f(x')\delta (x-x')$ are equivalent. Therefore,
$$f(x)\delta(x-x')=f(x')\delta(x-x') \tag 1$$
Now, let $f(x)=-x$, $x=\omega$ and $x'=-\omega_0$ in $(1)$. Then, we have
$$-\omega \delta(\omega+\omega_0)=\omega_0\delta(\omega+\omega_0) \tag 2$$
And then let $f(x)=x$, $x=\omega$ and $x'=\omega_0$. Then, we have
$$\omega \delta(\omega-\omega_0)=\omega_0\delta(\omega-\omega_0) \tag 3$$
Now, we can easily obtain the questioned equality using $(2)$ and $(3)$.