When would the time shift properties of Dirac Delta function become invalid?

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The time shift properties of Fourier transform is as below:-

$\mathscr{F}[x(t-t_0)]= e^{-2 i \pi f t_0} \mathscr{F}[x(t)]$

However, I tried to apply the property to a constant function $x(t)=1$, and I got this:-

$\mathscr{F}[1]= e^{-2 i \pi f t_0} \mathscr{F}[1]$

$\delta (f)= e^{-2 i \pi f t_0} \delta (f)$

Why is that? I think $e^{-2 i \pi f t_0} \delta (f)$ is not equal to $\delta (f)$?

If I take $x(t)=cos(2 \pi f_c t + \phi _c)$, I would also get an answer that is not same as the expected Fourier transformed result.

Why is that? Many thanks!


I added the part of $x(t)=cos(2 \pi f_c t + \phi _c)$ here.

Now we have (by using time shift properties):

$\mathscr{F}[cos(2 \pi f_c t + \phi _c)]= \mathscr{F}[cos(2 \pi f_c (t - \frac{-\phi _c}{2 \pi f_c}))]= e^{-2 i \pi f \frac{-\phi _c}{2 \pi f_c}} \mathscr{F}[cos(2 \pi f_c t)]= e^{\frac{i f \phi _c}{f_c}} \mathscr{F}[cos(2 \pi f_c t)]= e^{\frac{i f \phi _c}{f_c}} (\frac{1}{2} \delta(f-f_c) + \frac{1}{2} \delta(f+f_c))=\frac{1}{2} e^{\frac{i f \phi _c}{f_c}} (\delta(f-f_c) + \delta(f+f_c))$

But from some reference book, I saw $\mathscr{F}[cos(2 \pi f_c t + \phi _c)]=\frac{1}{2}(e^{i \phi _c} \delta(f-f_c) + e^{-i \phi _c} \delta(f+f_c))$

Are they in fact same? Or I shouldn't apply the time shift properties here?

Many thanks!