Why can adding more polarization filters increase the amount of light that goes through them? I am having some difficulty accepting the implications of the equation governing the intensity of light passing through polarization filters,
$$ I = I_0 \space\cos^2\theta $$ 
with $\theta$ being the angular difference between the two filters.
Here's the difficulty. If I put two filters at an angle of $\frac{\pi}{2}$, then no light makes it through to the other side. But if I then put another filter in between the original two, then we apply the above equation twice successively, neither time getting a result of zero.
That is, if you have two filters that don't allow any light through, you can force them to allow light through by placing a filter in between. It seems to me that, in general, a filter blocks light, so the result is counterintuitive. 
What happens to the light when the second filter is placed to allow some light to pass through the three-filter system?
 A: Indeed, it can be counterintuitive that adding a polarizing filter can increase the transmitted intensity, when each filter only 'removes' light.
Here's a slightly more intuitive way of thinking about it. A filter doesn't strictly remove light -- what it really does is add a light wave which destructively interferes with part of the incoming light. For example, a horizontal polarizing filter removes the horizontal part by transmitting an additional horizontally polarized wave out of phase by $180^\circ$. (This is true on a microscopic level, too. In the polarizing filter, electrons are driven by the incoming wave and, since they accelerate, they emit radiation of their own.)
Now let's say we add a diagonal polarizing filter in between horizontal and vertical filters. Thinking about the horizontal and vertical filters as "destroying light" in the usual way, you are correct in that none of the original light wave will make it through, regardless of what you put in between. 
But the new diagonal filter adds a new diagonally polarized wave! It's part of this wave that makes it out.
A: This is a known effect. It can be explained classically by the vector decomposition of the transverse electric field vector of the light waves. If you have a vertically polarized wave after the first filter propagating in x-direction with electric field amplitude in z-direction $\vec E_0$and this wave encounters a filter that is oriented at $\theta=\pi/4$ to the vertical, then the electric field amplitude component passing this filter is $E_0 cos(\pi/4)=E_0/\sqrt(2)$. When this wave with electrical field amplitude $E_0/\sqrt(2)$ at an angle of $\pi/4$ encounters the polarizer oriented at $\theta=\pi/2$ it has an electric amplitude component in the direction of this last polarizer $\frac {E_0}{\sqrt(2)\sqrt(2)}=\frac {E_0}{2}$ and its intensity is $I=\frac {I_0}{4}$. Thus, indeed, a substantial light intensity passes through when the middle filter is inserted.
A: "It seems to me that, in general, a filter blocks light, so the result is counterintuitive."
If all a polarization filter does is blocking light then your result would indeed be impossible.  However a polarization filter also forces light to take a particular polarization;  In the instant case the intermediate filter rotates the polarization of the light, with the consequence that it's no longer normal to the last filter.
A: There are two effect on a beam of light from passing through a polarizing filter.


*

*First the intensity is reduced as you write above. (For unpolarized light incident on a polarizer you average over all possible polarization and get $I = I_{up}/2$.)

*Second, the light that passes now has the polarization of the filter it just passed.
The second effect is what makes the trick work.
A: The same experiment can be demonstrated with a swinging rope. Place two slits behind each other (let space between them), the slits both in the same way oriented, take a rope through the slit, fix it at one end and swing the rope, of course in the direction of the slits. Now rotate the second slit. The rope no more will swing behind the second slit. But now if you place a third slit between the other two slits and this under 45°. The rope will swing all other his length and by this the swinging direction will be rotated to 90°.
A: This actually isn't unique to optics. If you take a rope and three solid (e.g. metallic) sheets with slits (one horizontal, one vertical and one diagonal), placed after one another as you place your optical polarizers, then you'll have much the same result.
A wave going through first, let's say horizontal, slit will be polarized horizontally, i.e. the part of the rope on the other side will only move in horizontal direction. Then coming to the diagonal slit, the motion will be forced to be not quite horizontal: the wave will reflect much like if you threw a ball into a tunnel under right angle, like this (the tunnel is analogous to our slit):

As you can understand, average motion of the ball shown above will get both horizontal and vertical velocity components.
Now the third pass through the (vertical) slit is just a repetition of the second one (the diagonal slit pass), but rotated by $45^\circ$.
