# Sine and Cosine are Periodic on Reals

## Theorem

The sine and cosine functions are periodic on the set of real numbers $\R$:

- $(1): \quad \map \cos {x + 2 \pi} = \cos x$
- $(2): \quad \map \sin {x + 2 \pi} = \sin x$

### Corollary

- $\map \cos {x + \pi} = -\cos x$
- $\map \sin {x + \pi} = -\sin x$

- $\cos x$ is strictly positive on the interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$ and strictly negative on the interval $\openint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$

- $\sin x$ is strictly positive on the interval $\openint 0 \pi$ and strictly negative on the interval $\openint \pi {2 \pi}$

### Zeroes of Sine and Cosine

- $(1): \quad \forall n \in \Z: x = \paren {n + \dfrac 1 2} \pi \implies \cos x = 0$
- $(2): \quad \forall n \in \Z: x = n \pi \implies \sin x = 0$

## 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:

- $\exists \xi > 0: \forall x \in \openint {-\xi} \xi: \cos x > 0$

Aiming for a contradiction, suppose $\cos x$ 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 Constantâ€Ž, $\cos x$ 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:

- $\map \sin {x + \eta} = \sin x \cos \eta + \cos x \sin \eta = \cos x$
- $\map \cos {x + \eta} = \cos x \cos \eta - \sin x \sin \eta = -\sin x$

Hence it follows, after some algebra, that:

- $\map \sin {x + 4 \eta} = \sin x$
- $\map \cos {x + 4 \eta} = \cos x$

Thus $\sin$ and $\cos$ are periodic on $\R$ with period $4 \eta$.

$\blacksquare$

## Pi

Given that the period of $\sin$ and $\cos$ is $4 \eta$, we define the real number $\pi$ (called **pi**, pronounced **pie**) as:

- $\pi := 2 \eta$

See Pi.

## Note

Given that we have *defined* sine and cosine in terms of a power series, it is a plausible proposition to define $\pi$ using the same language.

$\pi$ is, of course, the famous irrational constant $3.14159 \ldots$.

## Sources

- 1960: Walter Ledermann:
*Complex Numbers*... (previous) ... (next): $\S 2$. Geometrical Representations - 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 16.4$ - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Entry:**period, periodic**