Sine and Cosine are Periodic on Reals/Pi

From ProofWiki
Jump to navigation Jump to search


Let $\sin: \R \to \R$ be the real sine function, and let $\cos: \R \to \R$ be the real cosine function.

The real number $\pi$ (called pi, pronounced pie) is uniquely defined as:

$\pi := \dfrac p 2$

where $p \in \R$ is the period of $\sin$ and $\cos$.

Proof 1

From the Real Cosine Function is Periodic and Real Sine Function is Periodic, we have that $\cos x$ and $\sin x$ are periodic on $\R$ with the same period.

If we denote the period of $\cos x$ and $\sin x$ as $p$, it follows that $\pi = \dfrac p 2$ is uniquely defined.


Proof 2

By Cosine of Zero is One:

$\cos 0 = 1$

By Cosine of 2 is Strictly Negative:

$\cos 2 < 0$

Thus by the corollary to the Intermediate Value Theorem there exists an $h \in \openint 0 2$ such that:

$\cos h = 0$

By Sine of Sum for all $x \in \R$:

\(\ds \sin x\) \(=\) \(\ds \map \sin {x - h} \cos h + \map \cos {x -h} \sin h\)
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \map \cos {x -h} \sin h\)

By Cosine of Sum for all $x \in \R$:

\(\ds \cos x\) \(=\) \(\ds \map \cos {x - h} \cos h - \map \sin {x -h} \sin h\)
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds -\map \sin {x - h} \sin h\)

By Sum of Squares of Sine and Cosine:

\(\ds \sin^2 h\) \(=\) \(\ds \cos^2 h + \sin^2 h\)
\(\text {(3)}: \quad\) \(\ds \) \(=\) \(\ds 1\)

Thus for all $x \in \R$:

\(\ds \map \cos {x + 4 h}\) \(=\) \(\ds - \map \sin {x + 3 h} \sin h\) by $(2)$
\(\ds \) \(=\) \(\ds - \map \cos {x + 2 h} \sin^2 h\) by $(1)$
\(\ds \) \(=\) \(\ds \map \cos {x + h} \sin^3 h\) by $(2)$
\(\ds \) \(=\) \(\ds \cos x \sin^4 h\) by $(1)$
\(\ds \) \(=\) \(\ds \cos x\) by $(3)$

In particular, $\cos$ is periodic.

By Nonconstant Periodic Function with no Period is Discontinuous Everywhere, $\cos$ has a period $p \in \R_{>0}$.

In view of $(1)$ and $\sin h \ne 0$, the periodic elements of $\sin$ are exactly those of $\cos$.

Thus $p$ is also the period of $\sin$.



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 $3.14159 \ldots$.