What does (e^(ix)+e^(ix))/(2i) equal?

Oct 25, 2015

$\sin \left(x\right) - i \cos \left(x\right)$

but I think you meant to ask...

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 with the question as asked:

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

I think you may have been wanting one of the following results:

$\frac{{e}^{i x} + {e}^{- i x}}{2}$

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

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

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

$\textcolor{w h i t e}{}$

$\frac{{e}^{i x} - {e}^{- i x}}{2 i}$

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

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

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