Sine and Cosine are Periodic on Reals/Pi/Proof 2
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Theorem
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
- $\cos 0 = 1$
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 $\sin$.
$\blacksquare$