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