Can this expression be simplified and is the real part of the left hand side equal to minus pi?

Question

By starting with some Dirichlet series similar to logarithms, or somewhat similar to logarithm Dirichlet series, I arrived at this expression:

$$\left(\text{ArcSin}\left[\text{Sqrt}[2]*\text{Cos}[\text{Im}[\text{ZetaZero}[2]]*\text{Log}[2]]\left/\left(\frac{1}{18} \left(\sqrt{3} \pi -\sqrt{6} \pi \text{Cos}[\text{Im}[\text{ZetaZero}[1]] \text{Log}[2]]+9 \text{Log}[3]+9 \sqrt{2} \text{Cos}[\text{Im}[\text{ZetaZero}[1]] \text{Log}[2]] \text{Log}[3]\right)\right)\right.\right]*2\right)= -\pi$$

Can it be simplified and is the real part the left hand side equal to $-\pi$ ?

I apologize for the bad Latex formatting.

As a Mathematica command the the real part of the left hand side is:

N[Re[(ArcSin[
 Sqrt[2]*Cos[
    Im[ZetaZero[2]]*Log[2]]/(1/
     18 (Sqrt[3] \[Pi] - 
       Sqrt[6] \[Pi] Cos[Im[ZetaZero[1]] Log[2]] + 9 Log[3] + 
       9 Sqrt[2] Cos[Im[ZetaZero[1]] Log[2]] Log[3]))]*2)], 90]

= -3.1415926535897932384626433832795028841971693993751058209749445923078\ 1640628620899862803483

Edit 31.8.2012:

This is what the Latex should look like:

= -3.1415926535897932384626433832795028841971693993751058209749445923078\ 1640628620899862803483

Answer 1

Notice that for $x>1$, $z =\arcsin(x)$ is complex. Let $z=u+i v$, then $$ x = \sin(u+i v) = \cosh(v) \sin(u) + i \sinh(v) \cos(u) $$ Since $x$ is real $\cos(u) = 0$, and since $x>1$, $\sin(u) > 0$.

Restricting to the principal branch-cut of $\arcsin$, we thus have: $$ \Re\left( \arcsin(x) \right) = u =\frac{\pi}{2} $$ Similarly for $x<-1$, $\Re\left( \arcsin(x) \right) = -\frac{\pi}{2}$.