Is your text Hecht and Zajac? If so, and if the newest edition is similar to my old edition, the paragraph previous to the one that conatins your expression explains that the expression results from the superposition of 2 linearly polarized light waves, i.e., both
E_{0x} cos \left (kz - wt \right) \hat{i}
and
E_{0y}cos \left( kz - wt + \epsilon \right) \hat{j}
are the electric fields of linearly polarized (along different axes) light waves.
Depending on the values of E_{0x}, E_{0y}, and \epsilon, the superposition can result in: linear polarization; elliptical polarization; circular polarization.
For example, E_{0x} = E_{0y} = E_{0} and \epsilon = - \pi/2 give circular polarization. This means that circularly polarized light results from the sum 2 linearly polarized waves, and this gives
<br />
\begin{equation*}<br />
\begin{split}<br />
E &= E_{0} \left[ cos \left(kz - wt \right) \hat{i} + cos \left( kz - wt - \pi/2 \right) \hat{j} \right]\\<br />
&= E_{0} \left[cos \left(kz - wt \right) \hat{i} + sin(kz - wt)\hat{j} \right]<br />
\end{split}<br />
\end{equation*}<br />As you say, after going through a polarizer with transmission axis given by the unit vector \hat{n},
<br />
\begin{equation*}<br />
\begin{split}<br />
E' &= E \cdot \hat{n} \hat{n}\\<br />
&= E_{0} \left[ cos \left(kz - wt \right) cos \theta + sin \left(kz - wt \right) sin \theta \right] \hat{n}\\<br />
&= E_{0} cos \left(kz - wt - \theta \right) \hat{n}<br />
\end{split}<br />
\end{equation*}<br />
Again, as you say this gives an average intensity independent of theta.
Interesting.
Before doing the calculation, I didn't know the answer.
Regards,
George
