Euler's Sine Identity/Proof 2

Theorem

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

Proof

Recall Euler's Formula:

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

Then, starting from the right hand side:

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

\(=\)

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

\(\ds \)

\(=\)

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

Cosine Function is Even

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

Sine Function is Odd

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \sin z\)

$\blacksquare$