Cosine Function is Even

From ProofWiki
Jump to navigation Jump to search

Theorem

For all $z \in \C$:

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

That is, the cosine function is even.


Proof 1

Recall the definition of the cosine function:

\(\ds \cos z\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!}\)
\(\ds \) \(=\) \(\ds 1 - \frac {z^2} {2!} + \frac {z^4} {4!} - \cdots\)


From Even Power is Non-Negative:

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

The result follows.

$\blacksquare$


Proof 2

\(\ds \map \cos {-z}\) \(=\) \(\ds \frac {e^{i \paren {-z} } + e^{-i \paren {-z} } } 2\) Cosine Exponential Formulation
\(\ds \) \(=\) \(\ds \frac {e^{i z} + e^{-i z} } 2\) simplifying
\(\ds \) \(=\) \(\ds \cos z\) Cosine Exponential Formulation

$\blacksquare$


Also see


Sources