Optics - Three polarizers with 1 moving

1. The problem statement, all variables and given/known data

Thee linear polarizers are in sequence. Let first and last polarizer be crossed (perpendicular to each-other) and the middle polarizer rotate with angular frequency ##\omega##. Show that under such circumstances, ##I## is given by: $$I = I_0 \frac{1}{16}(1-\cos (4\omega t))$$.



The light that strikes the first polarizer is unpolarized.



Please, see the following link (yes, my paint skills are horrible! I tried to make it in inkspace, but I don't know how to work with that program yet!): http://ift.tt/1ooRRRx



2. Relevant equations



Malus Law, which is given by ##I(\theta) = I_0 \cos^2 (\theta)##



3. The attempt at a solution



If you saw the following link, you noticed two angles, ##\theta## and ##\alpha##.



I think ##\theta=\omega t ## and ##\alpha = 90 - \omega t##



So, after the first polarizer:



$$I_1= \frac{I_0}{2}$$



After the second polarizer:



$$I_2= I_1 \cos^2 (\omega t)$$



After the third polarizer:



$$I_3 = I_2 \cos^2 (90-\omega t) \Leftrightarrow I_3 = I_1 \cos^2 (\omega t) \cos^2 (90-\omega t) $$



I can play around with that but I never arrive at the ##I = I_0 \frac{1}{16}(1-\cos (4\omega t))##.



Any ideas?