Period of Real Sine Function

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Theorem

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


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

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


Proof

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

Let $L$ be that period.

From Sine of Angle plus Full Angle:

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

So $L = 2 \pi$ satisfies:

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


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

We have that:

\(\ds \sin 0\) \(=\) \(\ds 0\) Sine of Zero is Zero
\(\ds \map \sin {0 + \pi}\) \(=\) \(\ds 0\) Sine of Straight Angle
\(\ds \cos \map \cos {0 + 2 \pi}\) \(=\) \(\ds 0\) Sine of Angle plus Full Angle

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

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


But then we note:

\(\ds \sin \dfrac \pi 2\) \(=\) \(\ds 1\) Sine of Right Angle
\(\ds \map \sin {\dfrac \pi 2 + \pi}\) \(=\) \(\ds -1\) Sine of Three Right Angles
\(\ds \) \(\ne\) \(\ds \sin \dfrac \pi 2\)

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


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

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

and the result follows.

$\blacksquare$


Sources

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