i identity

**rainer** · November 27th 2009, 12:47 PM

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.

**galactus** · November 27th 2009, 01:11 PM

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

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

Therefore, we have $i^{i}$

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

**rainer** · November 28th 2009, 12:28 PM

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:

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

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