Writing $z=e^{j\omega}$ and using partial fraction expansion, you can rewrite $X(z)$ as
$$X(z)=\frac{a}{z-a}+\frac{1}{1-az}\tag{1}$$
The two terms in $(1)$ are DTFTs (or $\mathcal{Z}$-transforms) of basic sequences:
$$\frac{a}{z-a}\Longleftrightarrow a^nu[n-1]\\
\frac{1}{1-az}\Longleftrightarrow a^{-n}u[-n]\tag{2}$$
where $u[n]$ is the unit step, and where $|a|<1$ has been taken into account. Combining the expressions in $(2)$ results in
$$x[n]=a^nu[n-1]+a^{-n}u[-n]=a^{|n|}$$
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
$$\begin{align}
\mathscr{F}\left(A\cos (\omega_0t+\theta)\right)&=\frac A2e^{-i\theta \omega/\omega_0}\left(\delta(\omega+\omega_0)+ \delta(\omega-\omega_0)\right)\\\\
&=\frac A2\left(e^{i\theta}\delta(\omega+\omega_0)+ e^{-i\theta}\delta(\omega-\omega_0)\right)\\\\
\end{align}$$
$$\bbox[5px,border:2px solid #C0A000]{\mathscr{F}\left(A\cos (\omega_0t+\theta)\right)=\frac A2\left(e^{i\theta}\delta(\omega+\omega_0)+ e^{-i\theta}\delta(\omega-\omega_0)\right)}$$
Best Answer
First, an example with simple exponential
We start off by a simplified example with the determining the Discrete-Time Fourier Transform (DTFT) of the exponential function given by $$x_1[n]=e^{i\Omega_0n} \tag{1}.$$
Using the Fourier Transform $$\mathcal F\{x_1[n] \} \tag{2}=X_1(\Omega)=\sum_{k=-\infty}^{\infty}x_1[k]e^{-i\Omega k}$$
we can calculate $\mathcal F \{ x_1[n]\}$ by substituting $x_1[n]$ and simplify the expression as follows:
$$\sum_{k=-\infty}^{\infty}e^{i\Omega_0 k}e^{-i\Omega k}=\sum_{k=-\infty}^{\infty}e^{i(\Omega_0-\Omega)k} \tag{3}$$
Then, by defining $\Omega'=\Omega - \Omega_0$ (observe the changed order) so that
$$\sum_{k=-\infty}^{\infty}1\cdot e^{-i\Omega'k} \tag{4}$$
we cleary see that it's now just the DTFT of a constant $1$, i.e. $\mathcal F \{1 \}$, which is given by $$\mathcal F \{1 \}=2\pi \delta(\Omega') \tag{5}$$ and $2\pi \delta(\Omega-\Omega_0)$ in our case. Now, why is this? Where does the $2\pi$ come from?
So we have assumed that $\mathcal F \{1 \} = 2\pi\delta(\Omega)$, which defines the constant $1$ as a function $x_2[n]=1$. If we use the Inverse Discrete-Time Fourier Transform (IDTFT)
$$\mathcal F^{-1} \{ X_2(\Omega) \} = \frac{1}{2\pi} \int_{-\pi}^{\pi} X_2(\Omega)e^{i\Omega n} d\Omega \tag{6}$$
and substitute $X_2(\Omega)$ for our assumed $2\pi\delta(\Omega)$ we get
$$\frac{1}{2\pi} \int_{-\pi}^{\pi} 2\pi\delta(\Omega)e^{i\Omega n} d\Omega \tag{7}$$
which, since $\delta(\Omega)$ is "on" or $1$ only for $\Omega=0$, evaulates to
$$\frac{2\pi}{2\pi} e^{i\Omega 0} = 1. \tag{8}$$
We have thereby shown that the DTFT of a constant $1$ is equal to $2\pi\delta(\Omega)$.
If we now continue from equation $(4)$, we have that the resulting DTFT of an exponential $x_1[n]=e^{i\Omega_0n}$ becomes
$$\mathcal F\{x_1[n] \} = \sum_{k=-\infty}^{\infty}1\cdot e^{-i\Omega'k} = 2\pi\delta(\Omega') = 2\pi\delta(\Omega - \Omega_0) \tag{9}.$$
Secondly, show Discrete-Time Fourier Transform of a sine
We have the discrete-time function $$y[n]=\sin(\Omega_0n+\phi) \tag{10}.$$
Observe that the $\phi$ causes a time-shift and can be handled separately. Therefore consider, for now, only the function
$$x[n]=\sin(\Omega_0n) \tag{11}.$$
Using Euler's formula we can write the sine as exponentials in the form
$$x[n]=\frac{1}{2i} \left ( e^{i\Omega_0n} - e^{-i\Omega_0n} \right ) \tag{12}$$
Using the DTFT of $x[n]$ we get
$$\mathcal F\{ x[n] \} = \sum_{k=-\infty}^{\infty}\frac{1}{2i} \left ( e^{i\Omega_0k} - e^{-i\Omega_0k} \right ) e^{-i\Omega k} \tag{13}$$
which can be rewritten as
$$\frac{1}{2i} \left [ \sum_{k=-\infty}^{\infty}e^{-i(\Omega-\Omega_0)k} - \sum_{k=-\infty}^{\infty}e^{-i(\Omega+\Omega_0)k} \right] \tag{14}.$$
We can now see that each term in equation $(14)$ is on the form in equation $(9)$ and therefore we get
$$\frac{1}{2i} \left [ 2\pi\delta(\Omega-\Omega_0) - 2\pi\delta(\Omega+\Omega_0 \right] \tag{15}$$
and
$$i\pi \left [ \delta(\Omega+\Omega_0) - \delta(\Omega-\Omega_0) \right] \tag{16}$$
if we shift the order of the terms to compensate the negative sign when $i$ jumps up from the denominator.
To consider the time-shift $\phi$ we mutltiply our results with $e^{i\Omega\phi}$ and finally get
$$e^{i\Omega\phi} \cdot i\pi \left [ \delta(\Omega+\Omega_0) - \delta(\Omega-\Omega_0) \right] \tag{17}$$
I'm not a hundred percent certain of my answer and I don't fully understand how to get the general form in the questions asked. If someone has edits to suggest, please do so.
The solutions are inspired and based on the following YouTube videos: