# Math Help - i identity

1. ## i identity

Demonstrate that $\arccos^i(\cosh1) = e^{-\frac{\pi}{2}}$

I am not a mathematician so sorry if this is not very challenging. Basically I just found this identity to be interesting and wonder how obvious/obscure it is to real mathematicians.

2. Note that $cosh(1)=cos(i)$

$cos^{-1}(cos(i))=i$

Therefore, we have $i^{i}$

Now, show this equals $e^{\frac{-\pi}{2}}$

3. Originally Posted by galactus
Note that $cosh(1)=cos(i)$

$cos^{-1}(cos(i))=i$

Therefore, we have $i^{i}$

Now, show this equals $e^{\frac{-\pi}{2}}$

Yep.

Just to tidy up a bit:

$\cos^{i}(\cosh1)^{-1}=i^i=e^{\frac{-\pi}{2}}$