Euler's Sine Identity/Proof 1
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Theorem
- $\sin z = \dfrac {e^{i z} - e^{-i z} } {2 i}$
Proof
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 $i$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sin z\) |
$\blacksquare$