Compare this transform to the box transform that was worked in class. Explain the difference in their frequency scaling. Plot the two transforms to compare them, although the feature that we're looking for can be observed without plotting.
I'm trying to compare the frequency scalling of the following two Fourier transforms \begin{align} F_{1}(\omega) &= \frac{\cos(\omega a)}{\left(\frac{\pi}{2a} - \omega\right)} - \frac{\cos(\omega a)}{\left(-\frac{\pi}{2a}- \omega\right)} \tag{1}\\ F_{2}(\omega) &= 2\frac{\sin(\omega a)}{\omega} \tag{2}, \end{align} where $F_{1}(\omega)$ is the Fourier transform that I computed of \begin{equation} V(t) = \begin{cases} &|t| \le a, \quad V_{0}\cos \left( \frac{\pi t}{2a}\right)\\ &|t| > a,\qquad \ \ \ 0 \end{cases} \end{equation} and $F_2(\omega)$ is the Fourier transform of a box-function (i.e., non-periodic square-wave function).
This is what the functions $(1)$ [red] and $(2)$[green] look like when plotted.
I can't seem to understand how to compare their frequency scaling or what even that means for that matter. Any hint or input would be appreciated.
Perhaps you are asking for too much subtlety here. You are asked to compare the FT of the semicircle box, $F_1(\omega)$, red, to the FT of the square box of the same width, 2a, the sinc function, $$ F_2(\omega)/2a= \operatorname{sinc} (\omega a) , $$ with the peak at the origin of $F_2(0)/2a= 1$. It might pay to scale the vertical coordinate of the red curve so that its value at the frequency origin is also unity. You then compare the effective widths of the two curves.
First note that if you scale the frequencies up by a factor of s, as shown in WP for s=π, the $F_2$ distribution narrows down, and, e.g. the zeros move toward the origin proportionately to the scaling. Now, consider scaling down, with s=2/π instead. In that case, $$ F_3=F_2(2\omega/\pi)/2a= \operatorname{sinc} (2\omega a /\pi) , $$ and the zeroes spread out, keeping unit value at the origin, for comparison.
Compare this with your $F_1$, now also normalized to unit value at the origin, so you compare apples with apples, $$ F_1(\omega)(\pi/4a)= \frac{ \cos(\omega a) } {1-(2\omega a/\pi)^2} ~ . $$
It is still smoother than $F_3$, but closer to it than to $F_2$.