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
$$\begin{align}
\mathscr{F}\left(A\cos (\omega_0t+\theta)\right)&=\frac A2e^{-i\theta \omega/\omega_0}\left(\delta(\omega+\omega_0)+ \delta(\omega-\omega_0)\right)\\\\
&=\frac A2\left(e^{i\theta}\delta(\omega+\omega_0)+ e^{-i\theta}\delta(\omega-\omega_0)\right)\\\\
\end{align}$$
$$\bbox[5px,border:2px solid #C0A000]{\mathscr{F}\left(A\cos (\omega_0t+\theta)\right)=\frac A2\left(e^{i\theta}\delta(\omega+\omega_0)+ e^{-i\theta}\delta(\omega-\omega_0)\right)}$$
The answer by Hans is good and explains under what conditions this makes sense. The OP asked me to post my comment as an answer. To justify this, I'll try to expand it with a bit more detail.
There are various conventions for defining the Fourier transform. I will assume we are working with the following:
$$\hat{f}(\omega) = \int_{-\infty}^{\infty}f(t) e^{-i \omega t} dt$$
Then, applying the definition to the function $g(t) = e^{i \omega_0 t}f(t)$, we can manipulate formally as follows:
$$\begin{aligned}
\hat{g}(\omega) &= \int_{-\infty}^{\infty}f(t)e^{i\omega_0 t}e^{-i \omega t} dt \\
& = \int_{-\infty}^{\infty}f(t) e^{-i(\omega - \omega_0)t} dt \\
&= \hat{f}(\omega - \omega_0)
\end{aligned}$$
Now, under what conditions does that above formal manipulation make sense?
Note that the integral defining the Fourier transform, namely
$$\hat{f}(\omega) = \int_{-\infty}^{\infty}f(t)e^{-i \omega t} dt$$
converges whenever $\int_{-\infty}^{\infty}|f(t)| dt$ conveges, in other words, whenever $f$ is absolutely integrable.
Now, if $\omega_0$ is real, then $|e^{i\omega_0 t}| = 1$, so
$$\int_{-\infty}^{\infty}f(t)e^{i \omega_0 t}e^{-i \omega t} dt$$
also converges whenever $f$ is absolutely integrable.
However, if $\omega_0$ is not real, then we can write $\omega_0 = a_0 + i b_0$, where $a_0$ and $b_0$ are both real, and $a_0$ is nonzero. Then $e^{i\omega_0 t} = e^{(a_0 + ib_0)t} = e^{a_0 t} e^{ib_0 t}$, so
$$|e^{i\omega_0 t}| = |e^{a_0 t}||e^{i b_0 t}| = |e^{a_0 t}| = e^{a_0 t}$$
Therefore, convergence of the integral
$$\int_{-\infty}^{\infty}f(t)e^{i \omega_0 t}e^{-i \omega t} dt$$
requires convergence of
$$\int_{-\infty}^{\infty}|f(t)|e^{a_0 t} dt$$
Now, $e^{a_0 t}$ grows exponentially in either the positive $t$ or negative $t$ direction, depending on the sign of $a_0$. Consequently, if the integral is to converge, we require $f$ to decay exponentially in that direction in order to compensate.
The conclusion is that when $\omega_0$ is real, the integral defining $\mathcal F(f(t) e^{i\omega t})$ always makes sense for any absolutely integrable function $f$. But when $\omega_0$ is not real, this is no longer the case. So one has to be careful about what conditions on $f$ and/or the real part of $\omega_0$ are required so that the integral will converge. The answer by Hans summarizes these conditions nicely.
Best Answer
In general, the fourier transform of a continuous time signal $x(t)$ is given by:
$$\begin{align*} X(\omega) &= \int_{-\infty}^{\infty} x(t) e^{-i\omega t} dt \end{align*}$$
But, please note that the signal $x(t)$ must be absolutely integrable over all time i.e.,
$$\begin{align*} \int_{-\infty}^{\infty} |x(t)| dt \space < \space \infty \end{align*}$$
The function $e^{i\omega_0 t}$ however, is not absolutely integrable and the fourier transform does not converge: $$\begin{align*} \int_{-\infty}^{\infty} |e^{i\omega_0 t}| dt \space = \space \int_{-\infty}^{\infty} 1. dt \space= \space \infty \end{align*}$$
However, to find the Fourier Transform (F.T) of the above function, we can use the duality property of F.T i.e., $$\begin{align*} if \space \space &F\{x(t)\} \space \space = \space \space X(\omega), \space then\\ & F\{X(t)\} \space = \space 2\pi x(-\omega) \end{align*}$$
Now, consider the Dirac delta function (it's not a function, really), $\delta(t)$. $$\begin{align*} &F\{\delta(t)\} \space \space = \space \space 1, \space then\\ & F\{1\} \space = \space 2\pi \delta(-\omega) \space = 2\pi\delta(\omega) \space \space \space (1) \end{align*}$$
Ok, so far so good. Duality alone will not fetch the desired result. We now use the modulation property of F.T. $$\begin{align*} if \space \space &F\{x(t)\} \space \space = \space \space X(\omega), \space then\\ & F\{x(t).e^{i\omega_0t}\} \space = \space X(\omega-\omega_0) \end{align*}$$
Therefore, from eqn (1), $$\begin{align*} & F\{1.e^{i\omega_0t}\} \space = \space F\{e^{i\omega_0t}\} \space = \space 2\pi\delta(\omega-\omega_0) \end{align*}$$