Linearity property of the Fourier Transform

If $F_{1}(\omega)$ and $F_{2}(\omega)$ are the Fourier transforms of $f_{1}(t)$ and $f_{2}(t)$, respectively, then
$$\mathcal{F}\left[a_{1} f_{1}(t)+a_{2} f_{2}(t)\right]=a_{1} F_{1}(\omega)+a_{2} F_{2}(\omega) \tag{1}$$
where $a_{1}$ and $a_{2}$ are constants. This property simply states that the Fourier transform of a linear combination of functions is the same as the linear combination of the transforms of the individual functions. The proof of the linearity property in Eq. (1) is straightforward. By definition,
\begin{aligned}\mathcal{F}\left[a_{1} f_{1}(t)+a_{2} f_{2}(t)\right] &=\int_{-\infty}^{\infty}\left[a_{1} f_{1}(t)+a_{2} f_{2}(t)\right] e^{-j \omega t} d t \\&=\int_{-\infty}^{\infty} a_{1} f_{1}(t) e^{-j \omega t} d t+\int_{-\infty}^{\infty} a_{2} f_{2}(t) e^{-j \omega t} d t \\&=a_{1} F_{1}(\omega)+a_{2} F_{2}(\omega)\end{aligned} \tag{2}
For example, $\sin \omega_{0} t=\frac{1}{2 j}\left(e^{j \omega_{0} t}-e^{-j \omega_{0} t}\right)$. Using the linearity property,
\begin{aligned}F\left[\sin \omega_{0} t\right] &=\frac{1}{2 j}\left[\mathcal{F}\left(e^{j \omega_{0} t}\right)-\mathcal{F}\left(e^{-j \omega_{0} t}\right)\right] \\&=\frac{\pi}{j}\left[\delta\left(\omega-\omega_{0}\right)-\delta\left(\omega+\omega_{0}\right)\right]\end{aligned}