Sine and Cosine are Periodic on Reals/Cosine

Theorem
The cosine function is periodic on the set of real numbers $\R$:
 * $\map \cos {x + 2 \pi} = \cos x$


 * SineCos.png

Proof
From Cosine of Zero is One we have that $\cos 0 = 1$.

From Cosine Function is Even we have that $\cos x = \map \cos {-x}$.

As the Cosine Function is Continuous, it follows that:
 * $\exists \xi > 0: \forall x \in \openint {-\xi} \xi: \cos x > 0$

$\cos x$ were positive everywhere on $\R$.

From Derivative of Cosine Function:
 * $\map {D_{xx} } {\cos x} = \map {D_x} {-\sin x} = -\cos x$

Thus $-\cos x$ would always be negative.

Thus from Second Derivative of Concave Real Function is Non-Positive, $\cos x$ would be concave everywhere on $\R$.

But from Real Cosine Function is Bounded, $\cos x$ is bounded on $\R$.

By Differentiable Bounded Concave Real Function is Constant‎, $\cos x$ would then be a constant function.

This contradicts the fact that $\cos x$ is not a constant function.

Thus by Proof by Contradiction $\cos x$ can not be positive everywhere on $\R$.

Therefore, there must exist a smallest positive $\eta \in \R_{>0}$ such that $\cos \eta = 0$.

By definition, $\cos \eta = \map \cos {-\eta} = 0$ and $\cos x > 0$ for $-\eta < x < \eta$.

Now we show that $\sin \eta = 1$.

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 $\map {D_x} {\sin x} = \cos x$.

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

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

Since $\sin 0 = 0$ it follows that $\sin \eta > 0$.

So it must be that $\sin \eta = 1$.

Now we apply Sine of Sum and Cosine of Sum:

Hence it follows that:

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

We then define $\pi \in \R$ as:


 * $\pi := 2 \eta$