Euler's Sine Identity/Real Domain/Proof 1

Theorem

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

Proof

Recall the definition of the sine function:

\(\ds \sin x\)

\(=\)

\(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}\)

\(\ds \)

\(=\)

\(\ds x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7} {7!} + \cdots + \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} + \cdots\)

Recall the definition of the exponential as a power series:

\(\ds e^x\)

\(=\)

\(\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}\)

\(\ds \)

\(=\)

\(\ds 1 + \frac x {1!} + \frac {x^2} {2!} + \frac {x^3} {3!} + \cdots + \frac {x^n} {n!} + \cdots\)

Then, starting from the right hand side:

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

\(=\)

\(\ds \frac 1 {2 i} \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {i x}^n} {n!} - \sum_{n \mathop = 0}^\infty \frac {\paren {-i x}^n} {n!} }\)

\(\ds \)

\(=\)

\(\ds \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i x}^n - \paren {-i x}^n} {n!} }\)

\(\ds \)

\(=\)

\(\ds \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i x}^{2 n} - \paren {-i x}^{2 n} } {\paren {2 n}!} + \frac {\paren {i x}^{2 n + 1} - \paren {-i x}^{2 n + 1} } {\paren {2 n + 1}!} }\)

split into even and odd $n$

\(\ds \)

\(=\)

\(\ds \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \frac {\paren {i x}^{2 n + 1} - \paren {-i x}^{2 n + 1} } {\paren {2 n + 1}!}\)

as $\paren {-i x}^{2 n} = \paren {i x}^{2 n}$

\(\ds \)

\(=\)

\(\ds \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \frac {2 \paren {i x}^{2 n + 1} } {\paren {2 n + 1}!}\)

as $\paren {-1}^{2 n + 1} = -1$

\(\ds \)

\(=\)

\(\ds \frac 1 i \sum_{n \mathop = 0}^\infty \frac {\paren {i x}^{2 n + 1} } {\paren {2 n + 1}!}\)

cancel $2$

\(\ds \)

\(=\)

\(\ds \frac 1 i \sum_{n \mathop = 0}^\infty \frac {i \paren {-1}^n x^{2 n + 1} } {\paren {2 n + 1}!}\)

as $i^{2 n + 1} = i \paren {-1})^n $

\(\ds \)

\(=\)

\(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1!} }\)

cancel $i$

\(\ds \)

\(=\)

\(\ds \sin x\)

$\blacksquare$