Another mathy question from the Signal Processing community... This is the question as posed to the DSP folks. Electrical engineers like to use particular definitions of the Fourier Transform (using Hz, rather than angular frequency) and the sinc function.
Out of consideration to you kind folks, I will try to eliminate a few symbols, like I will, without loss of generality, set the sample rate, $f_\text{s}$, to 1 and I try to use angular frequency.
So let $$ 0 < \omega_0 < \pi $$
and for any real $W$ such that
$$ \omega_0 < W < 2\pi - \omega_0 $$
please prove that
$$  \cos(\omega_0 t + \phi) =  \sum\limits_{n=-\infty}^{\infty}  \cos\left(\omega_0 n  + \phi \right) \, \frac{\sin\big( W(t - n) \big)}{\pi(t - n)} $$
without the use of the Fourier Transform.
I s'pose one of the " Whittaker–Nyquist–Kotelnikov–Shannon" folks did this, but I can't see exactly how this gets extracted out of the Poisson summation formula.
I spent all of my rep on a bounty of a previous question.  Sorry, I don't have much rep to spend here.
well, 21 hours left to get the bounty!!  don't let it go to waste.
UPDATE: the 100 rep bounty has expired.  so i guess it's wasted.
 A: *

*Let 
$$f(x) = \frac{\sin(\pi x)}{\pi x} = \int_{-1/2}^{1/2} e^{2 i\pi \xi x}d\xi, \qquad F(x) = \int_{-\infty}^x f(y)dy, \quad F(-\infty) = 0, \ F(+\infty) = 1$$
If $g$ is bounded and $C^1$ then
$$\lim_{m \to \infty}\int_{k-1/2}^{k+1/2} m f(mx) g(x)dx=\lim_{m \to \infty} F(mx) g(x)|_{k-1/2}^{k+1/2}- \int_{k-1/2}^{k+1/2}  F(mx) g'(x)dx \\ = 1_{k> -1/2} g(k+1/2)-1_{k> 1/2}g(k-1/2)-\int_{k-1/2}^{k+1/2}1_{x > 0} g'(x)dx= g(0) 1_{|k| < 1/2}$$
(ie. $m f(mx) \to \delta(x)$ in the sense of distributions)

*Then see the Dirichlet kernel
$$\sum_{n=-\infty}^\infty e^{2i \pi k n}f(x-n) =\lim_{m \to \infty} \sum_{n=-m}^m e^{2i \pi k n} f(x-n) =\lim_{m \to \infty}\sum_{n=-m}^m \int_{-1/2}^{1/2} e^{2i\pi( \xi x+(k-\xi) n)}d\xi \\= \lim_{m \to \infty} \int_{-1/2}^{1/2} e^{2i\pi \xi x} \frac{\sin(2\pi(m+1/2) (k-\xi))}{\sin(\pi (k-\xi))}d\xi \\=\lim_{m \to \infty} \int_{-k-1/2}^{-k+1/2} e^{2 i\pi (\xi+k) x} \frac{\pi \xi}{\sin(\pi  \xi)}m f(m\xi) d\xi =e^{2i \pi k x} 1_{|k| < 1/2}$$
With a subtelty if $|k| = 1/2$, in that case we need a factor $1/2$.
The whole proves the Fourier series, the Fourier  transform as well as the Shannon sampling theorem !
