Sine and Cosine are Periodic on Reals/Cosine

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Theorem

The cosine function is periodic on the set of real numbers $\R$:

$\exists L \in \R_{\neq 0}: \forall x \in \R: \cos x = \map \cos {x + L}$


SineCos.png

Proof

From Real Cosine Function has Zeroes, the cosine function has at least one positive zero.

Therefore there exists a Greatest Lower Bound $\eta \in \R_{>0}$ to the set of positive zeroes.

Since Cosine Function is Continuous, $\eta$ is a zero.

Because Cosine Function is Even:

$\cos \eta = \map \cos {-\eta} = 0$

By definition of greatest lower bound:

$\cos x \neq 0$ for $-\eta < x < \eta$

Because Cosine of Zero is One, it follows from the Intermediate Value Theorem that:

$\cos x > 0$ for $-\eta < x < \eta$


From Sum of Squares of Sine and Cosine:

$\cos^2 x + \sin^2 x = 1$

Hence as $\cos \eta = 0$ it follows that:

$\sin^2 \eta = 1$

So either $\sin \eta = 1$ or $\sin \eta = -1$.

But from Derivative of Sine Function:

$\map {\dfrac \d {\d x} } {\sin x} = \cos x$

On the interval $\openint {-\eta} \eta$, it has been shown that $\cos x > 0$.

Thus by Derivative of Monotone Function, $\sin x$ is increasing on $\closedint {-\eta} \eta$.

Since Sine of Zero is Zero it follows that:

$\sin \eta > 0$

So it must be the case that:

$\sin \eta = 1$


Now we apply Sine of Sum and Cosine of Sum:

\(\ds \map \sin {x + \eta}\) \(=\) \(\ds \sin x \cos \eta + \cos x \sin \eta\) Sine of Sum
\(\ds \) \(=\) \(\ds \cos x\)
\(\ds \map \cos {x + \eta}\) \(=\) \(\ds \cos x \cos \eta - \sin x \sin \eta\)
\(\ds \) \(=\) \(\ds -\sin x\) Cosine of Sum

Hence it follows that:

\(\ds \map \cos {x + 4 \eta}\) \(=\) \(\ds -\map \sin {x + 3 \eta}\)
\(\ds \) \(=\) \(\ds -\map \cos {x + 2 \eta}\)
\(\ds \) \(=\) \(\ds \map \sin {x + \eta}\)
\(\ds \) \(=\) \(\ds \cos x\)

Thus $\cos$ is periodic on $\R$ with a period $L \leq 4 \eta$.

$\blacksquare$


Sources

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