I've this function to transform not using the general formula, but just substituting the known transform (i.e. $\text{rect}(t)\rightarrow \text{sinc}(f)$): $\frac{\sin(6\pi t)}{t}$ I know the transform of the function $\frac{\sin(\pi t)}{\pi t}\rightarrow \text{sinc}(f)$, and i taught to apply the rescaling property, but i get a different result that my professor. He gets $\pi \text{rect}(\frac{f}{6})$, i get the same result multiplied for $\frac{1}{6}$.
Can you explain me the correct way to proceed? Thank you!
Using your definitions of the $\text{sinc}$-function ($\text{sinc}(t)=\sin(\pi t)/(\pi t)$) and of the $\text{rect}$-function (a rectangle of height $1$ in the range $-\frac12<t<\frac12$, and zero otherwise), you have the Fourier correspondence
$$T\;\text{sinc}(Tt)\longleftrightarrow\text{rect}\left(\frac{t}{T}\right)\tag{1}$$
The given function can be written as follows
$$f(t)=\frac{\sin(6\pi t)}{t}=6\pi\;\text{sinc}(6t)\tag{2}$$
Transforming $(2)$ according to $(1)$ (with $T=6$) gives
$$6\pi\;\text{sinc}(6t)\longleftrightarrow\pi\;\text{rect}\left(\frac{t}{6}\right)\tag{3}$$
as suggested by your professor.