Cosine Function is Even

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Theorem

For all $z \in \C$:

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

That is, the cosine function is even.


Proof 1

Recall the definition of the cosine function:

$\displaystyle \cos z = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {z^{2 n} } {\left({2 n}\right)!} = 1 - \frac {z^2} {2!} + \frac {z^4} {4!} - \cdots$


From Even Power is Non-Negative:

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

The result follows.

$\blacksquare$


Proof 2

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

$\blacksquare$


Also see


Sources