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:
| \(\ds \cos \dfrac \pi 2\) | \(=\) | \(\ds 0\) | 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.
$\blacksquare$