Fourier transform of ${\Pi}_{a}(t)\cos(2{\pi}f_{0}t)$ I want to find the following: $\mathfrak{F}[{\Pi}_{a}(t)\cos(2{\pi}f_{0}t)]$ 

I first did: $\mathfrak{F}[{\Pi}_{a}(t)] = 2a\text{ sinc}(2af)$ 
then: $\mathfrak{F}[\cos(2{\pi}f_{0}t)] = \frac{1}{2}[\delta(f-f_{0})+\delta(f+f_{0})]$ 

So: $$\begin{align}\mathfrak{F}[{\Pi}_{a}(t)\cos(2{\pi}f_{0}t)] &amp;= \mathfrak{F}[{\Pi}_{a}(t)]\ast \mathfrak{F}[\cos(2{\pi}f_{0}t)]\\&amp;=2a\text{ sinc}(2af)\ast\frac{1}{2}[\delta(f-f_{0})+\delta(f+f_{0})]\\&amp;=a\text{ sinc}(2af)\ast\delta(f-f_{0})+a\text{ sinc}(2af)\ast\delta(f+f_{0})\\&amp;=a\text{ sinc}(2af)+a\text{ sinc}(2af)=2a\text{ sinc}(2af)\end{align}$$

I would like to know if it is correct and if there is an alternative way to find the result easier.
Your last line is wrong. Note that you have

$$H(f)\star\delta(f-f_0)=H(f-f_0)\tag{1}$$

for any $H(f)$.

So your last line should be

$$\begin{align*}
&amp;a\,\text{sinc}(2af)\star\delta(f-f_{0})+a\text{ sinc}(2af)\star\delta(f+f_{0})=\\&amp;a\,\text{sinc}(2a(f-f_0))+a\,\text{sinc}(2a(f+f_0))\tag{2}
\end{align*}$$

You basically get copies of your original spectrum centered at $f_0$ and $-f_0$.