Period of Real Cosine Function

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Theorem

The period of the real cosine function is $2 \pi$.


That is, $2 \pi$ is the smallest value $L \in \R_{>0}$ such that:

$\forall x \in \R: \cos x = \map \cos {x + L}$


Proof

From Sine and Cosine are Periodic on Reals, we have that $\cos$ is a periodic real function.

Let $L$ be that period.

From Cosine of Angle plus Full Angle:

$\map \cos {x + 2 \pi} = \cos x$

So $L = 2 \pi$ satisfies:

$\forall x \in \R: \cos x = \map \cos {x + L}$


It remains to be shown that $2 \pi$ is the smallest such $L \in \R_{>0}$ with this property.

We have that:

\(\ds \cos \dfrac \pi 2\) \(=\) \(\ds 0\) Cosine of Right Angle
\(\ds \map \cos {\dfrac \pi 2 + \pi}\) \(=\) \(\ds 0\) Cosine of Three Right Angles
\(\ds \map \cos {\dfrac \pi 2 + 2 \pi}\) \(=\) \(\ds 0\) Cosine of Angle plus Full Angle

and for no other $x \in \closedint 0 {2 \pi}$ is $\cos x = 0$.

Hence if there is another smaller $L \in \R_{>0}$ with the property that classifies it as the period of the real cosine function, it can only be $\pi$.


But then we note:

\(\ds \cos 0\) \(=\) \(\ds 1\) Cosine of Zero is One
\(\ds \map \cos {0 + \pi}\) \(=\) \(\ds -1\) Cosine of Straight Angle
\(\ds \) \(\ne\) \(\ds \cos 0\)

Hence $\pi$ is not the period of the real cosine function.


Thus $2 \pi$ is the smallest value $L \in \R_{>0}$ such that:

$\forall x \in \R: \cos x = \map \cos {x + L}$

and the result follows.

$\blacksquare$