Euler's Sine Identity/Real Domain/Proof 2

Theorem

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

Proof

Recall Euler's Formula:

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

Then, starting from the right hand side:

\(\ds \frac {e^{i x} - e^{-i x} }{2 i}\)

\(=\)

\(\ds \frac {\paren {\cos x + i \sin x} - \paren {\map \cos {-x} + i \map \sin {-x} } } {2 i}\)

\(\ds \)

\(=\)

\(\ds \frac {\paren {\cos x + i \sin x - \cos x - i \map \sin {-x} } } {2 i}\)

Cosine Function is Even

\(\ds \)

\(=\)

\(\ds \frac {i \sin x - i \map \sin {-x} } {2 i}\)

\(\ds \)

\(=\)

\(\ds \frac {i \sin x - i \paren {-\map \sin {-x} } } {2 i}\)

Sine Function is Odd

\(\ds \)

\(=\)

\(\ds \frac {2 i \sin x} {2 i}\)

\(\ds \)

\(=\)

\(\ds \sin x\)

$\blacksquare$