This is a shameless adaptation from my own set of notes, but it does help (me) tie a few things together mentioned in the relevant PDF linked below the question.
Consider the principal value integral \begin{equation}\label{principalval}I = \text{P} \int_{-\infty}^\infty \frac{f\left(x\right) \: dx}{x-x_0} \:,\end{equation} where by shifting $x \rightarrow z$, we assume that $f\left(z\right)$ is analytic except for a finite number of poles, and that $\left|f\right| \rightarrow 0$ on the upper (or lower) infinite semicircle in the complex plane.
Since the pole $x_0$ lies on the real axis, the integration contour cuts directly through $x_0$. This is handled by regularization of the denominator, which entails introducing a small factor $\delta>0$ as $$I = \text{P} \int_{-\infty}^\infty \frac{f\left(x\right) \: dx}{x-x_0} = \lim_{\delta\rightarrow 0} \int_{-\infty}^\infty \frac{\left(x-x_0\right)f\left(x\right) \: dx}{\left(x-x_0\right)^2 + \delta^2} = \lim_{\delta\rightarrow 0} \oint_C \frac{\left(z-x_0\right)f\left(z\right) \: dz}{\left(z-x_0\right)^2 + \delta^2} \:.$$ After a little complex algebra, find $$I = \lim_{\delta\rightarrow 0} \oint_C \frac{f\left(z\right) \: dz}{z-x_0+i\delta} + \lim_{\delta\rightarrow 0} \oint_C i\delta \frac{f\left(z\right) \: dz}{\left(z-x_0-i\delta\right)\left(z-x_0+i\delta\right)} \:,$$ which indicates one simple pole $z_0 = x_0 + i\delta$ inside the upper-half plane. The first integral in fact excludes the pole, so $x_0$ is skipped in subsequent residue calculations. (Use the $\delta$-term as a reminder to skip $x_0$.) The second integral is solved by standard residue calculus, i.e., let $g\left(z\right)=f\left(z\right)/\left(z-x_0+i\delta\right)$, resulting in $\pi i f\left(x_0\right)$.
Pulling the results together, we write \begin{equation}\label{principalvaltwoterms}I^+ = \pi i f\left(x_0\right) + \lim_{\delta\rightarrow 0} \oint_C \frac{f\left(z\right) \: dz}{z-x_0+i\delta} \:,\end{equation} where if we started with $\delta<0$ instead, the integration contour would flip to the lower-half plane, resulting in $$I^- = -\pi i f\left(x_0\right) + \lim_{\delta\rightarrow 0} \oint_C \frac{f\left(z\right) \: dz}{z-x_0-i\delta} \:.$$ In tighter notation (regardless of path or the sign of $\delta$), one may write \begin{equation}\label{principalvalcomplex}I = \text{P} \int_{-\infty}^\infty \frac{f\left(x\right) \: dx}{x-x_0} = \text{P} \oint_C \frac{f\left(z\right) \: dz}{z-x_0} \:,\end{equation} reminding us to include $x_0$ inside integration contour, but take the residue with a factor of $1/2$.
So for instance, to evaluate $$I = \int_{-\infty}^\infty \frac{\sin x}{x} \: dx \:,$$ the answer is almost too trivial to show off the method, but here it is:
$$I = \text{Im} \left( \pi i \: e^{i \cdot 0} + \lim_{\delta\rightarrow 0} \oint_C \frac{e^{iz} \: dz}{z+i\delta} \right) = \text{Im} \left( \pi i \: e^{i \cdot 0} + 0 \right) = \pi$$
By the same apparatus, we can show the cosine-version to be zero.
In your estimate of $C_R$, you used $|\sin(Re^{i\theta})| \le 1$, which is false in general. The sine goes rapidly to $\infty$ in the vertical direction!
The conventional thing to do is to take something like,
$$
f(z)=\frac{\exp(iz)}{z^2(1+z^2)}
$$
and take the imaginary part in the end.
Best Answer
Since,$$(\forall z\in\mathbb C):\frac{\sin z}z=1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots,$$the function $\frac{\sin z}z$ has a primitive:$$F(z)=z-\frac{z^3}{3\times3!}+\frac{z^5}{5\times5!}-\cdots$$and therefore\begin{align}\lim_{r\to0}\frac1r\int_{C_r}\frac{\sin z}z\,\mathrm dz&=\lim_{r\to0}\frac{F(re^{\pi i})}r-\frac{F(re^{0\times i})}r\\&=\lim_{r\to0}e^{\pi i}\frac{F(re^{\pi i})}{re^{\pi i}}-\frac{F(re^{0\times i})}r\\&=e^{\pi i}-1\text{ (because $F'(0)=1$)}\\&=-2.\end{align}