Period of Real Cosine Function

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:

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:

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.