Proof of Quarter-Turn Identities and Periodicity of Sine and Cosine

Each result cited below is universally quantified over the data in its own statement; it is applied to the data named at the point of use. Throughout, $\mathbb{R}$ is a field, so its addition and multiplication are commutative and associative, multiplication distributes over addition, $s+0=s$ and $s\cdot1=s$, $s+(-s)=0$, and $s\cdot s^{-1}=1$ for $s\ne0$; these are used below in rearranging the sums and products that occur.

Preliminaries. Let $x_{0}$ be the real number of The Least Positive Zero of the Cosine (i), so that

$$\cos x_{0}=0, \qquad \sin x_{0}=1,$$

the first by that clause and the second by The Least Positive Zero of the Cosine (ii). By The Number Pi (i) we have $\pi=2x_{0}$, hence

$$\tfrac{\pi}{2}=\pi\cdot2^{-1}=(2x_{0})\cdot2^{-1}=x_{0}\cdot(2\cdot2^{-1})=x_{0} .$$

Moreover $2=1+1$, so distributivity and the multiplicative identity give

$$\pi=2x_{0}=x_{0}+x_{0}, \qquad 2\pi=(1+1)\pi=\pi+\pi .$$

Claim 1 (clause 1). Let $x\in\mathbb{R}$. Apply Addition Formulas for Sine and Cosine (i) with its first argument taken to be $x$ and its second taken to be $\tfrac{\pi}{2}=x_{0}$:

$$\cos\bigl(x+\tfrac{\pi}{2}\bigr)=\cos x\cos x_{0}-\sin x\sin x_{0}=\cos x\cdot0-\sin x\cdot1=-\sin x,$$

using claim 1 of Zero Products and Elementary Identities in a Field for $\cos x\cdot0=0$, the multiplicative identity for $\sin x\cdot1=\sin x$, and the additive identity. Likewise Addition Formulas for Sine and Cosine (ii) gives

$$\sin\bigl(x+\tfrac{\pi}{2}\bigr)=\sin x\cos x_{0}+\cos x\sin x_{0}=\sin x\cdot0+\cos x\cdot1=\cos x .$$

Claim 2 (clause 2). Let $x\in\mathbb{R}$. Since $\pi=x_{0}+x_{0}=\tfrac{\pi}{2}+\tfrac{\pi}{2}$, associativity of addition gives $x+\pi=\bigl(x+\tfrac{\pi}{2}\bigr)+\tfrac{\pi}{2}$. Applying Claim 1 twice, first at the point $x+\tfrac{\pi}{2}$ and then at $x$,

$$\cos(x+\pi)=-\sin\bigl(x+\tfrac{\pi}{2}\bigr)=-\cos x, \qquad \sin(x+\pi)=\cos\bigl(x+\tfrac{\pi}{2}\bigr)=-\sin x .$$

Claim 3 (clause 3). Let $x\in\mathbb{R}$. Since $2\pi=\pi+\pi$, associativity gives $x+2\pi=(x+\pi)+\pi$, and Claim 2 applied twice gives

$$\cos(x+2\pi)=-\cos(x+\pi)=-(-\cos x)=\cos x, \qquad \sin(x+2\pi)=-\sin(x+\pi)=-(-\sin x)=\sin x,$$

the double negations being the involutivity of the additive inverse in $\mathbb{R}$.

Claim 4 (clause 4). Write $\mathbb{N}$ for the natural numbers and $\iota:\mathbb{N}\to\mathbb{R}$ for the canonical map, so that by The Integers as a Subset of the Real Numbers every $m\in\mathbb{Z}$ is $0$, or $\iota(n)$ for some $n\in\mathbb{N}$, or $-\iota(n)$ for some $n\in\mathbb{N}$. We treat the three cases in turn.

The positive case. Let $A$ be the set of those $n\in\mathbb{N}$ such that

$$\cos\bigl(y+2\pi\iota(n)\bigr)=\cos y \qquad\text{and}\qquad \sin\bigl(y+2\pi\iota(n)\bigr)=\sin y \qquad\text{for every }y\in\mathbb{R}.$$

By claim 1 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field one has $\iota(1)=1$, so $2\pi\iota(1)=2\pi$ and $1\in A$ by Claim 3. Suppose $n\in A$. By that same claim 1, $\iota(n+1)=\iota(n)+1$, so distributivity and the multiplicative identity give

$$2\pi\iota(n+1)=2\pi\bigl(\iota(n)+1\bigr)=2\pi\iota(n)+2\pi,$$

whence, by associativity of addition, $y+2\pi\iota(n+1)=\bigl(y+2\pi\iota(n)\bigr)+2\pi$ for every $y$. Applying Claim 3 at the point $y+2\pi\iota(n)$ and then the hypothesis $n\in A$,

$$\cos\bigl(y+2\pi\iota(n+1)\bigr)=\cos\bigl(y+2\pi\iota(n)\bigr)=\cos y,$$

and likewise for $\sin$; so $n+1\in A$. By the principle of induction, applied to $A$, we have $A=\mathbb{N}$.

The case $m=0$. By claim 1 of Zero Products and Elementary Identities in a Field, $2\pi\cdot0=0$, and $x+0=x$ by the additive identity, so both identities hold trivially.

The negative case. Let $n\in\mathbb{N}$ and $m=-\iota(n)$. By commutativity of multiplication together with claim 2 of Zero Products and Elementary Identities in a Field, $2\pi\bigl(-\iota(n)\bigr)=\bigl(-\iota(n)\bigr)2\pi=-\bigl(\iota(n)\,2\pi\bigr)=-\bigl(2\pi\iota(n)\bigr)$, so $x+2\pi m=x-2\pi\iota(n)$. Put $y=x-2\pi\iota(n)$; then $y+2\pi\iota(n)=x$, and since $n\in A$,

$$\cos x=\cos\bigl(y+2\pi\iota(n)\bigr)=\cos y=\cos(x+2\pi m),$$

and likewise $\sin x=\sin(x+2\pi m)$.

The particular values. Taking $x=0$ and using $0+2\pi m=2\pi m$ together with $\cos0=1$ and $\sin0=0$ from Uniform Convergence, Continuity, Parity and Derivatives of Sine and Cosine (iii) gives $\cos(2\pi m)=1$ and $\sin(2\pi m)=0$ for every $m\in\mathbb{Z}$.

Claim 5 (clause 5). Let $x$ satisfy $0<x<\pi$. The order of $\mathbb{R}$ is a total order, as recorded in the preamble of Elementary Order Arithmetic in an Ordered Field, so either $x\le x_{0}$ or $x_{0}<x$.

If $x\le x_{0}$, then $0<x\le x_{0}$ and The Least Positive Zero of the Cosine (ii) gives $0<\sin x$ directly.

Suppose instead $x_{0}<x$, and put $s=x-x_{0}$. Claim 1 of Elementary Order Arithmetic in an Ordered Field, applied with $c=-x_{0}$, turns $x_{0}<x$ into $0<s$ and turns $x<\pi=x_{0}+x_{0}$ into $s<x_{0}$. Hence $0\le s<x_{0}$, and The Least Positive Zero of the Cosine (i) gives $0<\cos s$. Finally $s+\tfrac{\pi}{2}=(x-x_{0})+x_{0}=x$, so Claim 1, applied at the point $s$, gives

$$\sin x=\sin\bigl(s+\tfrac{\pi}{2}\bigr)=\cos s,$$

and therefore $0<\sin x$. This proves clause 5 and completes the proof.