Fourier transform of phase shifted sinusoid Lets say I have a cosine function
$f(t) = Acos(\omega_0t) $.
I find the Fourier transform $\mathfrak{F}[f(t)] = \frac{A}{2}\left [ \delta (\omega - \omega_0 )  + \delta (\omega + \omega_0 )\right ]$ which is two delta spikes symmetric about zero at $\omega_0$ and $-\omega_0$. I think this is correct.
Now, if I phase shift $f(t) $ to get 
$f_s(t) = Acos(\omega_0t+\theta)$
my intuition tells me that the $\mathfrak{F}[f_s(t)] = \mathfrak{F}[f(t)]$ because the phase shift should not impact the frequency content of the signal.
However, when I do the work, I get the following:
$f_s(t)=\frac{A}{2}\left [ e^{i(\omega_0t+\theta)}+e^{-i(\omega_0t+\theta)}\right ]$  (Euler's formula)
So, 
$\mathfrak{F}[f_s(t)]=\frac{A}{2}\left [ \int_{-\infty}^\infty e^{i(\omega_0t+\theta)}e^{-i\omega t}dt + \int_{-\infty}^\infty e^{-i(\omega_0t+\theta)}e^{-i\omega t}dt\right ]$
$\, \,\,\,\,\,\,\,\,\,\,=\frac{A}{2}e^{i\theta}\int_{-\infty}^\infty e^{i(\omega_0-\omega)t}dt+\frac{A}{2}e^{-i\theta}\int_{-\infty}^\infty e^{-i(\omega_0+\omega)t}dt$
$\, \,\,\,\,\,\,\,\,\,\,=Acos(\theta)\left [\delta(\omega+\omega_0) + \delta(\omega-\omega_0) \right ]$
And thus,
$\mathfrak{F}[f_s(t)] \neq \mathfrak{F}[f(t)] $ which goes against my intuition.
Is the math wrong or is my intuition wrong?
 A: If we define the Foruier Transform of $f$ to be 
$$\mathscr{F}\left(f(t)\right)(\omega)=\frac{1}{2\pi}\int_{-\infty}^\infty f(t)e^{i\omega t}\,dt$$
then the Foruier Transform of $f(t+t_0)$ is 
$$\begin{align}
\mathscr{F}\left(f(t+t_0)\right)(\omega)&=\frac{1}{2\pi}\int_{-\infty}^\infty f(t+t_0)e^{i\omega t}\,dt\\\\
&=e^{-i\omega t_0}\frac{1}{2\pi}\int_{-\infty}^\infty f(t)e^{i\omega t}\,dt\\\\
&=\bbox[5px,border:2px solid #C0A000]{e^{-i\omega t_0}\mathscr{F}\left(f(t)\right)(\omega)} \tag 1
\end{align}$$
This general result shows that a time-shift transforms to a phase multiplication.
Now, if $f(t)=A\cos \omega_0t$, then its Fourier Transform is given by
$$\begin{align}
\mathscr{F}\left(A\cos \omega_0t\right)&=\frac{1}{2\pi}\int_{-\infty}^\infty A\cos (\omega_0t)\,e^{i\omega t}\,dt\\\\
&=\frac{1}{2\pi}\int_{-\infty}^\infty A\left(\frac{e^{i\omega_0t}+e^{-i\omega_0t}}{2}\right)\,e^{i\omega t}\,dt\\\\
&=\frac A2\left(\frac{1}{2\pi}\int_{-\infty}^\infty e^{i(\omega+\omega_0)t}\,dt+\frac{1}{2\pi}\int_{-\infty}^\infty e^{i(\omega-\omega_0)t}\,dt\right)\\\\
&=\frac A2\left(\frac{1}{2\pi}2\pi \delta(\omega+\omega_0)+\frac{1}{2\pi}2\pi \delta(\omega-\omega_0)\right)\\\\
&=\bbox[5px,border:2px solid #C0A000]{\frac A2\left(\delta(\omega+\omega_0)+ \delta(\omega-\omega_0)\right)} \tag 2\\\\
\end{align}$$
Finally, if $f(t)=A\cos (\omega_0 t+\theta)=A\cos (\omega_0 (t+t_0))$, where $t_0=\theta /\omega_0$, then we see that this added phase is tantamount to a time shift.  Using $(1)$ and $(2)$ reveals that 
$$\begin{align}
\mathscr{F}\left(A\cos (\omega_0t+\theta)\right)&=\frac A2e^{-i\theta \omega/\omega_0}\left(\delta(\omega+\omega_0)+ \delta(\omega-\omega_0)\right)\\\\
&=\frac A2\left(e^{i\theta}\delta(\omega+\omega_0)+ e^{-i\theta}\delta(\omega-\omega_0)\right)\\\\
\end{align}$$
$$\bbox[5px,border:2px solid #C0A000]{\mathscr{F}\left(A\cos (\omega_0t+\theta)\right)=\frac A2\left(e^{i\theta}\delta(\omega+\omega_0)+ e^{-i\theta}\delta(\omega-\omega_0)\right)}$$
A: That is correct. If you want another way of looking at it you can use the fact that cos(a+b)= cos(a)cos(b)-sin(a)sin(b).
However, I think a more interesting thing to note is that if you look at the average value of $\mathfrak{F}[f_s]$ (and by average value I mean the root mean square), then you end up with $\mathfrak{F}[f].
