Euler's Sine Identity/Proof 3

Theorem

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

Proof

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

\(\ds e^{i z}\)

\(=\)

\(\ds \cos z + i \sin z\)

Euler's Formula

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

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

\(=\)

\(\ds \cos z - i \sin z\)

Euler's Formula: Corollary

\(\ds \leadsto \ \ \)

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

\(=\)

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

$(1) - (2)$

\(\ds \)

\(=\)

\(\ds 2 i \sin z\)

simplifying

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds \sin z\)

$\blacksquare$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$: $\text{(ii)}$: $(4.17)$