$$\text{f(t)}=-3\sin(50\pi t)\cdot 6e^{\frac{-|t|}{2}}$$
First I wrote the sinusoidal in terms of complex exponentials then after a minor cancellation and using the distributive property I got:
$$9ie^{i50\pi t}\cdot e^{\frac{-|t|}{2}} - 9ie^{-i50\pi t}\cdot e^{\frac{-|t|}{2}}$$
How can I proceed from here? I do not know if I can write it in Fourier Transform form somehow or simplify it further by writing it to the power of just one $e$ by using $e^x\cdot e^y=e^{x+y}$ since I have $|t|$ on one of them...
Any suggestions or help would be much appreciated!
Let $f(t)=\sin(a t)$, $g(t)=e^{-b |t|}$, and $h(t)=Af(t)g(t)$.
$$\begin{align} \mathscr{F}\{h\}(\omega)&=\frac1{2\pi}\left(\mathscr{F}\{f\}*\mathscr{F}\{g\}\right)(\omega)\tag1 \end{align}$$
The Fourier transform of $f$ is
$$\mathscr{F}\{f\}(\omega)=i\pi (\delta(\omega-a)-\delta(\omega+a))\tag2$$
and the Fourier transform of $g$ is
$$\mathscr{F}\{g\}(\omega)=\frac{2b}{\omega^2+b^2}\tag3$$
Putting $(1)$-$(3)$ together we find
$$\mathscr{F}\{h\}(\omega)=\frac{iA}{2}\left(\frac{2b}{(\omega-a)^2+b^2}-\frac{2b}{(\omega+a)^2+b^2}\right)$$
Now, set $A=-18$, $a=50\pi$, and $b=1/2$.