Ah, good question. The radian is actually a "fake unit." What I mean by that is that the radian is defined as the ratio of distance around a circle (arclength) to the radius of a circle - in other words, it's a ratio of one distance to another distance. For an angle of one radian specifically, the arclength $s$ is equal to the radius $r$, so you get
$$1\text{ rad} = \frac{s}{r} = \frac{r}{r} = 1$$
The units of distance (meters or whatever) cancel out, and it turns out that "radian" is just a fancy name for 1!
Incidentally, this also implies that "degree" is just a fancy name for the number $\frac{\pi}{180}$, and "rotation" is just a fancy name for the number $2\pi$.
This actually addresses the edit to your question. Suppose that you had some object oscillating at $\omega = \pi/4\frac{\mathrm{rad}}{\mathrm{s}} = 0.785\frac{\mathrm{rad}}{\mathrm{s}}$, and you wanted to evaluate its position after 10 seconds. To get the cosine term, you would plug the numbers in, getting
$$\cos\bigl(0.785\tfrac{\mathrm{rad}}{\mathrm{s}}\times 10\mathrm{s}\bigr) = \cos(7.85\text{ rad}) = \cos(7.85)$$
and then you would go to a trig table in radians (or your calculator in radian mode) and look up 7.85.
However, suppose that you were measuring $\omega_0$ in degrees per second instead of radians per second. You would instead have
$$\cos(45^\circ/\mathrm{s}\times 10\mathrm{s}) = \cos(450^\circ)$$
If you go look this up in a trig table given in degrees, you will get the same answer as $\cos(7.85)$. Why? Well, remember that the unit "degree" is just code for $\pi/180$, so this is actually equal to
$$\cos\bigl(450\times\tfrac{\pi}{180}\bigr)$$
And $450\times\frac{\pi}{180} = 7.85$, which is just $450^\circ$ converted to radians. So now you have the same value in the cosine, $\cos(7.85)$. Trig tables listed in degrees already have this extra factor of $\frac{\pi}{180}$ built into them as a convenience for you; basically, if you look up any number $\theta$ in a table that uses degrees, what you get is actually the cosine (or sine, or whatever) of $\theta\times\frac{\pi}{180}$.
Best Answer
A Newton is 1 ${\rm kg\cdot \rm m/s^2}$, not 1 ${\rm kg\cdot s^2/m}$. Also, you need to take the square root at the end.