Euler's Sine Identity

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Theorem

For any complex number $z$:

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

where:

$\exp z$ denotes the exponential function
$\sin z$ denotes the complex sine function
$i$ denotes the inaginary unit.


Real Domain

This result is often presented and proved separately for arguments in the real domain:


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


Proof 1

Recall the definition of the sine function:

\(\ds \sin z\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!}\)
\(\ds \) \(=\) \(\ds z - \frac {z^3} {3!} + \frac {z^5} {5!} - \frac {z^7} {7!} + \cdots + \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!} + \cdots\)


Recall the definition of the exponential as a power series:

\(\ds e^z\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}\)
\(\ds \) \(=\) \(\ds 1 + \frac z {1!} + \frac {z^2} {2!} + \frac {z^3} {3!} + \cdots + \frac {z^n} {n!} + \cdots\)


Then, starting from the right hand side:

\(\ds \frac {e^{i z} - e^{-i x} } {2 i}\) \(=\) \(\ds \frac 1 {2 i} \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {i z}^n} {n!} - \sum_{n \mathop = 0}^\infty \frac {\paren {-i z}^n} {n!} }\)
\(\ds \) \(=\) \(\ds \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^n - \paren {-i z}^n} {n!} }\)
\(\ds \) \(=\) \(\ds \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^{2 n} - \paren {-i z}^{2 n} } {\paren {2 n}!} + \frac {\paren {i z}^{2 n + 1} - \paren {-i z}^{2 n + 1} } {\paren {2 n + 1}!} }\) split into even and odd $n$
\(\ds \) \(=\) \(\ds \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n + 1} - \paren {-i z}^{2 n + 1} } {\paren {2 n + 1}!}\) as $\paren {-i z}^{2 n} = \paren {i z}^{2 n}$
\(\ds \) \(=\) \(\ds \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \frac {2 \paren {i z}^{2 n + 1} } {\paren {2 n + 1}!}\) as $\paren {-1}^{2 n + 1} = -1$
\(\ds \) \(=\) \(\ds \frac 1 i \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n + 1} } {\paren {2 n + 1}!}\) cancel $2$
\(\ds \) \(=\) \(\ds \frac 1 i \sum_{n \mathop = 0}^\infty \frac {i \paren {-1}^n z^{2 n + 1} } {\paren {2 n + 1}!}\) as $i^{2 n + 1} = i \paren {-1}^n $
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1!} }\) cancel $i$
\(\ds \) \(=\) \(\ds \sin z\)

$\blacksquare$


Proof 2

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

$\blacksquare$


Proof 3

\(\text {(1)}: \quad\) \(\ds e^{i z}\) \(=\) \(\ds \cos z + i \sin z\) 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
\(\ds \leadsto \ \ \) \(\ds \frac {e^{i z} - e^{-i z} } {2 i}\) \(=\) \(\ds \sin z\)

$\blacksquare$


Also presented as

Euler's Sine Identity can also be presented as:

$\sin z = \dfrac 1 2 i \paren {e^{-i z} - e^{i z} }$


Also see


Source of Name

This entry was named for Leonhard Paul Euler.


Sources

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