What is the angle for perfect light blockage through a polarizer?

[Cade]

Homework Statement

Unpolarized light passes through a polarizer oriented at 33 degrees from the horizontal axis.

1) At what angle (relative to the horizontal axis) would you place a polarizer so that no light passes through it?
2) At what angle (relative to the horizontal axis) would you place a polarizer so that only a third of the light from the

Homework Equations

Malus' law
I/I0 = cos(θ)$^{2}$

The Attempt at a Solution

1) Light into the second polarizer is polarized at an angle of 33 degrees. No light will pass through the second polarizer if the difference in angles is 90°, so the second polarizer should be at an angle of 90° + 33° = 123°.

2)
$\frac{1}{3}= cos(θ)^{2}$ where θ is the difference in angles between the two polarizers.
$θ difference = arccos(\sqrt{\frac{1}{3}})$
$θ_{final} - θ{_initial} = arccos(\sqrt{\frac{1}{3}})$
$θ_{final} - 33° = arccos(\sqrt{\frac{1}{3}})$
θ = 54.7° + 33° = 87.7° from the horizontal axis

But my instructor says that the left-hand side of the equation is $θ_{final} - θ{_initial}$, which results in θ = 54.7° - 33°, so I'm confused now... have I done these two correctly?

 

[HallsofIvy]

This is a physics problem, not math so I am moving it.

 

[Cade]

I'm sorry, I must have been looking at the wrong section when I posted it.