Euler's Cosine Identity/Real Domain/Proof 3

Theorem

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

$\text {(1)}: \quad$

$\ds e^{i x}$

$=$

$\ds \cos x + i \sin x$

$\text {(2)}: \quad$

$\ds e^{-i x}$

$=$

$\ds \cos x - i \sin x$

$\ds \leadsto \ \ $

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

$=$

$\ds \paren {\cos x + i \sin x} + \paren {\cos x - i \sin x}$

$(1) + (2)$

$\ds $

$=$

$\ds 2 \cos x$

simplifying

$\ds \leadsto \ \ $

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

$=$

$\ds \cos x$

$\blacksquare$

1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 2$. Geometrical Representations: $(2.22)$
1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $22 \ \text{(a)}$