# Sine Exponential Formulation/Real Domain

## Theorem

For any real number $x \in \R$:

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

where:

$e^{i x}$ denotes the exponential function
$\sin x$ denotes the real sine function
$i$ denotes the inaginary unit.

## Proof 1

Recall the definition of the sine function:

 $\displaystyle \sin x$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$ $\displaystyle$ $=$ $\displaystyle x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7} {7!} + \cdots + \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} + \cdots$

Recall the definition of the exponential as a power series:

 $\displaystyle \exp x$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ $\displaystyle$ $=$ $\displaystyle 1 + \frac x {1!} + \frac {x^2} {2!} + \frac {x^3} {3!} + \cdots + \frac {x^n} {n!} + \cdots$

Then, starting from the right hand side:

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

$\blacksquare$

## Proof 2

Recall Euler's Formula:

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

Then, starting from the right hand side:

 $\displaystyle \frac {e^{i x} - e^{-i x} }{2 i}$ $=$ $\displaystyle \frac {\paren {\cos x + i \sin x} - \paren {\cos \paren {-x} + i \sin \paren {-x} } } {2 i}$ $\displaystyle$ $=$ $\displaystyle \frac {\paren {\cos x + i \sin x - \cos x - i \sin \paren {-x} } } {2 i}$ Cosine Function is Even $\displaystyle$ $=$ $\displaystyle \frac {i \sin x - i \sin \paren {-x} } {2 i}$ $\displaystyle$ $=$ $\displaystyle \frac {i \sin x - i \paren {-\sin \paren {-x} } } {2 i}$ Sine Function is Odd $\displaystyle$ $=$ $\displaystyle \frac {2 i \sin x} {2 i}$ $\displaystyle$ $=$ $\displaystyle \sin x$

$\blacksquare$

## Proof 3

 $\text {(1)}: \quad$ $\displaystyle e^{i x}$ $=$ $\displaystyle \cos x + i \sin x$ Euler's Formula $\text {(2)}: \quad$ $\displaystyle e^{-i x}$ $=$ $\displaystyle \cos x - i \sin x$ Euler's Formula: Corollary $\displaystyle \leadsto \ \$ $\displaystyle e^{i x} - e^{-i x}$ $=$ $\displaystyle \paren {\cos x + i \sin x} - \paren {\cos x - i \sin x}$ $(1) - (2)$ $\displaystyle$ $=$ $\displaystyle 2 i \sin x$ simplifying $\displaystyle \leadsto \ \$ $\displaystyle \frac {e^{i x} - e^{-i x} } {2 i}$ $=$ $\displaystyle \sin x$

$\blacksquare$

## Also presented as

This result can also be presented as:

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