Euler's Cosine Identity/Real Domain/Proof 2

Theorem

$\cos x = \dfrac {e^{i x} + e^{-i x} } 2$

$e^{i x} = \cos x + i \sin x$

Then, starting from the right hand side:

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

$=$

$\ds \frac {\cos x + i \sin x + \map \cos {-x} + i \map \sin {-x} } 2$

$\ds $

$=$

$\ds \frac {\cos x + \map \cos {-x} } 2$

$\ds $

$=$

$\ds \frac {2 \cos x} 2$

$\ds $

$=$

$\ds \cos x$

$\blacksquare$