I have calculated the following Fourier transform:
$F(\omega)=\frac{2}{\omega^2+1}$
for $f(t)= e^{-|t|}$
Now I need to find the Fourier transform for
$f_3(t)=cos(2t)e^{-{|t|}}$
Only using the properties of Fourier transform.
I am struggling to see which property is required here. Initially I tried the linearity property but I think this is wrong as $cos(2t)$ is not a constant. I tried to find the answer on maple and found it to be: $\frac{2\omega^2+10}{\omega^4-6\omega^2+25}$
I can't see any property that would give me this.. I thought perhaps since they are two functions of $t$ I should use the Convolution theorem, but that would require me to know the Fourier transform of $cos(2t)$ which is undefined, from what Maple tells me.
I'm out of ideas, I feel as though I am missing something. Any help would be appreciated.
Let us observe that $$ \cos 2t= \frac 1 2 \left( e^{2it} + e^{-2it} \right) = \int_{-\infty}^{\infty}\frac{d\omega}{2\pi}e^{i\omega t}\left( \pi \delta(\omega - 2)+\pi\delta(\omega+ 2)\right),$$ where $\delta(\omega)$ denotes the Dirac delta function.
From this, it is clear that the Fourier transform of $\cos 2t$ is $\pi \delta(\omega - 2)+\pi\delta(\omega+ 2)$.
Now you can apply the convolution theorem.
Alternative approach, without using convolutions or the delta function:
The Fourier transform of $\cos 2t f(t)$ is \begin{align*} \int_{-\infty}^\infty dt e^{-i\omega t}\cos 2tf(t) & = \int_{-\infty}^\infty dt \ e^{-i\omega t}\times \tfrac 1 2 \left( e^{2it} + e^{-2it}\right) f(t) \\ & = \tfrac 1 2 \int_{-\infty}^\infty dt e^{-i(\omega - 2) t}f(t) + \tfrac 1 2 \int_{-\infty}^\infty dt e^{-i(\omega + 2) t}f(t) \\ & = \tfrac 1 2 \widetilde f(\omega - 2) + \tfrac 1 2 \widetilde f(\omega + 2). \end{align*}