Is there a way to simplify this equation Is there a way to simplify this equation?
$$
CE = 2 FD \sin \left( \arctan \left( \frac{AF}{FD} \right) - \arccos\left( \frac{AB}{\sqrt{AF^2+FD^2}}\right) \right) + \sqrt{AF^2+FD^2-AB^2}
$$ 

Edit:
So here's a bit more information regarding the problem. I want to know, beforehand, if I can transport safely some IKEA packages in my car (or in any of the family cars) before going to the store and actually trying it.

Plane Z is the "maximum safe" plane to place the package (further up and I might injure my neck). Some of the packages are longer than the length of Plane Z but if I tilt the package sideways (as shown in the below image) I might be able to fit it.

I have $\overline{AF}$ (Package Length), $\overline{FD}$ (Package Width) and $\overline{AB}$ (with of the car trunk) 
What I want to know is the length of $\overline{CE}$

So here's the math I came with (my math is only high school level, so please bear with me). 
$$AD^2 = AB^2 + BD^2 <=> AD^2 = AF^2 + FD^2$$
$$BD = \sqrt{AF^2 + FD^2 - AB^2}$$
$$ θ = γ - α $$
$$\tan(γ) = \frac{AF}{FD} <=> \cos(α) = \frac{AB}{\sqrt{AF^2+FD^2}}$$
$$γ = \arctan \left( \frac{AF}{FD} \right) <=> α = \arccos \left( \frac{AB}{\sqrt{AF^2+FD^2}}\right)$$
$$ θ = \arctan \left( \frac{AF}{FD} \right) - \arccos\left( \frac{AB}{\sqrt{AF^2+FD^2}}\right) $$

$$CE = CB + BD + DE = 2CB + BD$$
$$CE = 2CB + \sqrt{AF^2 + FD^2 - AB^2}$$
$$CE = 2 FD \sin(θ) + \sqrt{AF^2 + FD^2 - AB^2}$$
$$
CE = 2 FD \sin \left( \arctan \left( \frac{AF}{FD} \right) - \arccos\left( \frac{AB}{\sqrt{AF^2+FD^2}}\right) \right) + \sqrt{AF^2+FD^2-AB^2}
$$ 
 A: Your equation does not make sense in terms of units.  If $a$, $b$, $c$ are lengths (let's say in $m$), then $c/(a^2 + b^2)$ is in $m^{-1}$.  But a function such as arccos must operate on pure numbers, not quantities that have units attached.
Also, although it's technically OK because they're dimensionless, it's quite rare to see two angles (which is what the results of an arctan and an arccos would be) being multiplied to make a new angle (which you then take the sine of).  What could be the geometric interpretation of this?
EDIT: After the edits, it does make sense.  Expanding out the sin, with Maple's help I get
$$ CE = \dfrac{(AF^2 - FD^2) \sqrt{AF^2+FD^2-AB^2} + 2 AB\; AF\; FD}{AF^2 + FD^2}$$
A: In advance I apologize for my folly.
Wolfram Alpha is puzzled :)
You can see results here
But i think, we can try more simplify, if it is possible, i hope we will succeed.
(sorry for my bad English, i'm Russian student).
We can make the change of variable:
To expand arctan, we need to get a tan of something, but we cant(i hope clearly - why).
I tried to convert the expression according transformation rules of trigonometric functions & more else ways for simplify
& don't get result :(
Sorry, i did all I could...
But, we can do following:
1) z²=a²+b²
X=b sin(arccos(1/√(1-(a/b)²))-2arcsin(√((1-(c/√(a²+b²)))/2)))-√(z²-c²)
X=b sin(arccos(1/√(1-(a/b)²))-2arcsin(√(((|z|-c)/|2z|))))-√(z²-c²)
I repeat again, I did all I could... I'm silly :(
2) Try to prove that this expression can not be simplified further
