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I know a few properties of $\sin$ and $\cos$, for example:

  • $\sin^2+\cos^2=1$

  • $\sin (a+b) = \sin a\cos b+\cos a\sin b$.

  • $\cos (a+b) = \cos a\cos b-\sin a\sin b$.
  • $\sin (x+\delta) = \sin x$ for some real $\delta$.
  • $\cos (x+\delta) = \cos x$ for the same real $\delta$.

Obviously, the list is incomplete.

My question is: Which set of properties characterize $\sin$ and $\cos$ (i.e, if $s,s',c,c'$ satisfy the properties, then $s=s', c=c'$) ?

I'm looking for a set of such properties that doesn't reference differentiability, continuity, or the number $\pi$.

E: I've given up on the requirement of not assuming continuity (although, if someone does get a proof without assuming it, that person will get a nice bounty $:$-$)$ ), however, I still don't want differentiability or $\pi$ in this characterization.

Just to be clear, I'm looking for a proof of something like

The continuous solutions to $f^2+g^2=1$, $f(a+b)=f(a)g(b)+g(a)f(b)$ and $g(a+b)=g(a)g(b)-f(a)f(b)$ are unique.

YoTengoUnLCD
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  • Is $sin(arccosx)-cos(arcsinx)=0$ acceptable? Otherwise I will take off my comment... – imranfat Apr 27 '16 at 22:36
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    I highly doubt you will be able to find any reasonable characterization that does not include continuity. Note, for instance, that all the conditions you have written down are preserved by precomposing with any group-homomorphism $\mathbb{R}/\delta\mathbb{Z}\to\mathbb{R}/\delta\mathbb{Z}$. – Eric Wofsey Apr 27 '16 at 22:36
  • @imranfat That has a few domain issues, I believe (Remark: using $f(g^{-1})-g(f^{-1})=0$ assumes bijectivity in some domain, which is another property for the list :-P). – YoTengoUnLCD Apr 27 '16 at 22:38
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    In contrast, if you do assume continuity, then your first three conditions already characterize $\sin$ and $\cos$ up to scaling the domain. – Eric Wofsey Apr 27 '16 at 22:39
  • @YoTengoUnLCD. Yes, it is not true for all values of $x$. If you make a graph, it is only true for $x$ values in $[-1,1]$ – imranfat Apr 27 '16 at 22:39
  • @EricWofsey How would you deduce, e.g., differentiability of $\sin$ and $\cos$ from the first three conditions plus continuity? Does this require a geometric argument such as this one, or can it be done rigorously? –  Apr 27 '16 at 23:03
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    @Bungo: The first three conditions say $\cos x+i\sin x$ is a homomorphism $\mathbb{R}\to S^1$, and it is not too hard to show every such continuous homomorphism has the form $e^{iax}$ for some $a\in\mathbb{R}$, essentially because $\mathbb{Q}$ is dense in $\mathbb{R}$. – Eric Wofsey Apr 27 '16 at 23:08
  • I suppose you could probably avoid assuming continuity per se by instead assuming some sort of local monotonicity. – Eric Wofsey Apr 27 '16 at 23:11
  • @EricWofsey Interesting comments, if you check the list I posted, you'll see all the equations relate $\sin, \cos$ in some way, and I wanted this characterization to be like that. I hope there's a route to this. Else, I guess I'll have to go back to continuity. – YoTengoUnLCD Apr 27 '16 at 23:15
  • @EricWofsey Wouldn't it be circular reasoning to deduce properties about $\sin$ and $\cos$ from $e^{iax}$? I guess it depends on how one interprets the OP's question. My interpretation was, assuming we know nothing of $\sin$ or $\cos$ (and therefore nothing of $e^{ix}$), can we unambiguously define $\sin$ and $\cos$ (e.g. by deducing their Taylor series representations) from the given properties? –  Apr 27 '16 at 23:25
  • Well, that's not really important anyway. If we give a list of properties, prove that the functions that satisfy those properties are unique, and that $\cos,\sin$ satisfy them (with whatever definition we choose), we're done. – YoTengoUnLCD Apr 28 '16 at 00:16
  • Why the downvote? – YoTengoUnLCD Apr 30 '16 at 22:55
  • Last requirement in the quote should contain a minus. I was wondering why I could deduce that $g(x)=1$ from the first and last equation. – wythagoras May 02 '16 at 19:56
  • For conciseness, you could pack the relations using the handy $\text{cis}(x)=\cos(x)+i\sin(x)$ notation. (Caution, I am $not$ referring to the imaginary exponential $e^{ix}$.) Then, $|\text{cis}(x)|=1$, $\text{cis}(a+b)=\text{cis}(a)\text{cis}(b)$, $\text{cis}(\delta)=1$. –  May 02 '16 at 20:00
  • @wythagoras Whoops, thanks! – YoTengoUnLCD May 02 '16 at 20:01
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    You'll have a hard time distinguishing ${\cos(x),\sin(x)}$ from ${\cos(cx), \sin(cx)}$ for constants $c$ without $\pi$ or some calculus: for example $\lim_{x \to 0} g(x)/x = 1$ (which is essentially the derivative at one point). – Robert Israel May 02 '16 at 20:06
  • @RobertIsrael Hmm, is there no way to avoid both of those conditions? Eric Wofsey said the three conditions I quoted already characterized $s,c$ (given continuity). My hope was to prove the periodicity of these functions (from this sought for definition) and define $\pi$ as the minimal period of them. – YoTengoUnLCD May 02 '16 at 20:14
  • You misunderstood @EricWofsey. The three conditions $f^2 + g^2 = 1$, $f(a+b) = f(a) g(b) + g(a) f(b)$, $g(a+b) = g(a) g(b) - f(a) f(b)$ are true for $\sin(cx)$ and $\cos(cx)$ as well as for $\sin(x)$ and $\cos(x)$. – Robert Israel May 02 '16 at 22:10
  • @RobertIsrael A quick question I forgot to ask: Why does $\pi$ solve that issue? – YoTengoUnLCD May 03 '16 at 22:03
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    $\sin(\pi) = 0$. – Robert Israel May 03 '16 at 22:03
  • Wouldn't any such proof just be a re-statement of the definition of $\sin x$ and $\cos x$, possibly with a restriction on the domain to avoid multiplicative identity. –  May 07 '16 at 00:19
  • $f \in C^\infty(\mathbb R,\mathbb R) ; \text{and} ;\forall n, \forall x \in \mathbb R, |f^{(n)}(x)| \leq 1 ; \text{and} ; f'(0)=1$ is enough to say $f=\sin$ – Gabriel Romon May 07 '16 at 16:36
  • $f$ analytic, $f''+f=0$, $f(0)=0$, $f'(0)=1$. – Dimitris May 07 '16 at 16:55

1 Answers1

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I am not sure whether this is what you're asking. Although the conditions do not reference differentiability, continuity or the number $\pi$, it is used in the proof.

Theorem: If $f: \mathbb R \rightarrow \mathbb R,g : \mathbb R \rightarrow \mathbb R$ are functions that satisfy the following conditions:

  1. $f,g$ are periodic with some period $4\delta$ with $\delta \neq 0$.
  2. $0 \leq g(x) \leq x$ for all $0 \leq x \leq 2\delta$.
  3. $-x \leq g(x)\leq 0$ for all $2\delta \leq x \leq 4\delta$.
  4. $-x+1 \leq f(x) \leq 1$ for all $0 \leq x \leq 1$.
  5. $x+1 \leq f(x) \leq 1$ for all $x \leq 0$.
  6. $g(x+y)=g(x)f(y)+g(y)f(x)$ for all $x,y \in \mathbb R$.
  7. $f(x+y)+f(x-y)=2f(x)f(y)$ for all $x,y \in \mathbb R$.
  8. $g(x) \geq x \cdot f(x)$ for all $-1 \leq x \leq 1$.
  9. $f(\delta)=0$, $g(\delta)=1$

then $f(x)=\cos(x)$ and $g(x)=\sin(x)$.

Part $(1)$, $(2)$ and $(3)$ imply that $g$ is bounded.

Part $(1)$, $(4)$ and $(5)$ imply that $f$ is bounded.

Part $(2)$ and $(3)$ imply that $\lim_{x \to 0} g(x) = 0$ and that $g(0)=0$.

Part $(4)$ and $(5)$ imply that $\lim_{x \to 0} f(x) = 1$ and that $f(0)=1$.

Part $(6)$ implies that \begin{align*} \lim_{x \to a} g(x) &= \lim_{x \to a} g(x-a)f(a)+g(a)f(x-a) \\ &= \lim_{y \to 0} g(y)f(a)+g(a)f(y) \\ &= f(a) \lim_{y \to 0} g(y)+ g(a) \lim_{y \to 0} f(y) = g(a) \end{align*}

This implies that $g$ is continuous.

Condition $(6)$ and $(9)$ imply that $g(x+\delta)=g(\delta)f(x)=f(x)$, hence $f$ is continuous.

Condition $(7)$ is the d'Alembert functional equation. It is known that the only continuous functions $f$ satisfying it are $f(x)=0, f(x)=1, f(x)=\cos(kx), f(x)=\cosh(kx)$ for some $k \in \mathbb R$. The only one that satisfies condition $(1)$ is $f(x)=\cos(kx)$. This implies that $g(x)=\sin(kx)$.

Condition $(2)$, $(3)$ and $(8)$ give $1 \geq \frac{g(x)}{x} \geq f(x)$ for $-1 \leq x \leq 1$ and this implies $g'(0) = \lim_{x \to 0} \frac{g(x)-g(0)}{x} = \lim_{x \to 0} \frac{g(x)}{x} = 1$ and hence $k=1$, so $f(x)=\cos(x)$ and $g(x)=\sin(x)$.

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wythagoras
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