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

Routine arithmetic in $\mathbb{R}$ and $\mathbb{Z}$ is used without comment (Routine Facts, Continued: Typed Arithmetic, Maps and Predicativity). By The Number Pi (i), $\pi=2x_{0}$, where $x_{0}$ is the least positive zero of the cosine of The Least Positive Zero of the Cosine (i); thus $\tfrac{\pi}{2}=x_{0}$, and everything below comes from the behaviour of $\sin$ and $\cos$ at this point, transported along the line by the addition formulas.

Positivity of $\pi$. By The Least Positive Zero of the Cosine (i), $0<x_{0}$, so $\pi=2x_{0}>0$. This is (i).

Values at $\tfrac{\pi}{2}$. Since $\tfrac{\pi}{2}=x_{0}$, we have $\cos\tfrac{\pi}{2}=0$ by The Least Positive Zero of the Cosine (i), and $\sin\tfrac{\pi}{2}=1$ by The Least Positive Zero of the Cosine (ii).

Quarter turn. Let $x\in\mathbb{R}$. The addition formulas Addition Formulas for Sine and Cosine (i) and Addition Formulas for Sine and Cosine (ii), with $a=x$ and $\tfrac{\pi}{2}$ in place of their second argument, combined with the values at $\tfrac{\pi}{2}$ just found, give

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

This is (iii).

Half turn. Writing $x+\pi=\bigl(x+\tfrac{\pi}{2}\bigr)+\tfrac{\pi}{2}$ and applying (iii) first at the real number $x+\tfrac{\pi}{2}$ and then at $x$, we obtain

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

This is (iv).

Values at $\pi$. Taking $x=0$ in (iv) and using $\cos0=1$ and $\sin0=0$ (Values at Zero, Parity and Double-Angle Identities for Sine and Cosine (i)), we get

$$\cos\pi=-\cos0=-1,\qquad \sin\pi=-\sin0=0 .$$

Together with the values at $\tfrac{\pi}{2}$ this proves (ii).

Period. In the same way, writing $x+2\pi=(x+\pi)+\pi$ and applying (iv) first at $x+\pi$ and then at $x$,

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

This is (v).

Integer multiples. Every integer $m$ is a natural number $n$, or $0$, or the negative $-n$ of a natural number $n$, by The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements (vi); we treat the three cases in turn. For $m=n\in\mathbb{N}$ we argue by induction on $n$, proving the two identities for every real $x$ at once. The case $n=1$ is (v). If the identities hold for $n$ and every real $x$, then (v), applied at the real number $x+2\pi n$, gives

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

and likewise for the sine; this is the case $n+1$. For $m=0$ there is nothing to prove. For $m=-n$ with $n\in\mathbb{N}$, the case $n$ just proved, applied at the real number $x-2\pi n$, gives

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

and likewise for the sine. Finally, taking $x=0$ and using Values at Zero, Parity and Double-Angle Identities for Sine and Cosine (i), we get $\cos(2\pi m)=\cos0=1$ and $\sin(2\pi m)=\sin0=0$. This proves (vi).

Positivity on a half period. Let $0<x<\pi$; we distinguish whether $x$ lies before or after $\tfrac{\pi}{2}=x_{0}$. If $x\le x_{0}$, then $\sin x>0$ by The Least Positive Zero of the Cosine (ii). If $\tfrac{\pi}{2}<x<\pi$, then

$$0<x-\tfrac{\pi}{2}<\pi-\tfrac{\pi}{2}=x_{0},$$

so $\cos\bigl(x-\tfrac{\pi}{2}\bigr)>0$ by The Least Positive Zero of the Cosine (i), and (iii), applied at the real number $x-\tfrac{\pi}{2}$, gives

$$\sin x=\sin\bigl(\bigl(x-\tfrac{\pi}{2}\bigr)+\tfrac{\pi}{2}\bigr)=\cos\bigl(x-\tfrac{\pi}{2}\bigr)>0 .$$

This proves (vii).