I'm having trouble on figuring out how find the Fourier Transform of the following function, and I'm not allowed to use the straight up definition of the Fourier Transform but rather use it's properties and Fourier Transform table.
$$f(t) = 5 \frac{\sin(3\pi t)\sin(5\pi t)}{t^2}$$
Here is my attempt at the solution,
First I see that it may help to use the Product to Sum Trig Identity defined as:
$$ \sin(u)\sin(v) = \frac{1}{2}[\cos(u-v)-\cos(u+v)] $$
So applying that identity to f(t) we can rewrite it as:
$$ \frac{5}{2t^2}[\cos(-2\pi t)-\cos(8\pi t)] $$
That's it I'm stuck I was thinking of rewriting $t^2$ to $t^2/t^4$ and possibly use the duality property or as I previously learned from someone on my other Fourier Transform question to use the Frequency Differentiation property. Those cos() functions I don't know what to do with them. What I plan to do is take the Fourier Transform of each individual one because of the Linearity property.
Write $$\frac{\sin{(3\pi t)}\sin{(5\pi t})}{t^2}=\frac{\sin{(3\pi t)}}{t}\frac{\sin(5\pi t)}{t}.$$
Now $\mathcal{F}(fg)=(2\pi)^{-1}(\mathcal{F}(f)\ast\mathcal{F}(g))$, up to constants, where $\ast$ denotes the convolution operator.
Lastly $\mathcal{F}\left(\frac{\sin{at}}{t}\right)=\pi \chi_a(\omega)=\begin{cases} \pi & (|\omega|\lt a) \\ 0 & (|\omega|\gt a). \end{cases}$ The convolution between the $\chi$'s shouldn't be too hard to calculate.