Euler's Sine Identity/Proof 2
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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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 i \sin z} {2 i}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sin z\) |
$\blacksquare$