# What real function is (e^(ix)-e^(-ix))/(ie^(ix)+ie^(-ix)) equal to?

Oct 24, 2015

$\tan \left(x\right)$

#### Explanation:

${e}^{i x} = \cos \left(x\right) + i \sin \left(x\right)$

$\cos \left(- x\right) = \cos \left(x\right)$

$\sin \left(- x\right) = - \sin \left(x\right)$

So:

${e}^{i x} - {e}^{- i x} = \left(\cos \left(x\right) + i \sin \left(x\right)\right) - \left(\cos \left(- x\right) + i \sin \left(- x\right)\right)$

$= \left(\cos \left(x\right) + i \sin \left(x\right)\right) - \left(\cos \left(x\right) - i \sin \left(x\right)\right) = 2 i \sin \left(x\right)$

And:

${e}^{i x} + {e}^{- i x} = \left(\cos \left(x\right) + i \sin \left(x\right)\right) + \left(\cos \left(- x\right) + i \sin \left(- x\right)\right)$

$= \left(\cos \left(x\right) + i \sin \left(x\right)\right) + \left(\cos \left(x\right) - i \sin \left(x\right)\right) = 2 \cos \left(x\right)$

So:

$\frac{{e}^{i x} - {e}^{- i x}}{i {e}^{i x} + i {e}^{- i x}} = \frac{2 i \sin \left(x\right)}{2 i \cos \left(x\right)} = \sin \frac{x}{\cos} \left(x\right) = \tan \left(x\right)$