Sine Function is Odd

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Theorem

For all $z \in \C$:

$\map \sin {-z} = -\sin z$

That is, the sine function is odd.


Proof

Recall the definition of the sine function:

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


From Sign of Odd Power, we have that:

$\forall n \in \N: -\paren {z^{2 n + 1} } = \paren {-z}^{2 n + 1}$

The result follows directly.

$\blacksquare$


Also see


Sources

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