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This question is in the spirit of this question I asked earlier.

Let $f, g:\mathbb{R}\to\mathbb{R}$ be differentiable functions such that $f(0)=0$, $g(0)=1$, $f'(x)=g(x)$, and $g'(x)=-f(x)$. Is it possible to show that both $f$ and $g$ are periodic just from these conditions (without using the fact that $f(x)=\sin(x)$ and $g(x)=\cos(x)$ are the unique solutions of this pair of conditions)?

Aadi Rane
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    You can essentially piece together an argument based on what I wrote in this answer (and the link). – peek-a-boo Mar 24 '24 at 05:03
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    You can immediately prove that $f(x)^2+g(x)^2=1$, so you are moving about the unit circle $S^1$. You also know that from the point $(x,y)\in S^1$ onwards you are travelling counterclockwise in the direction $(-y,x)$, at unit speed. Basically reverse-engineering the answers here. Yeah, many ways to see you end up with sine and cosine. – Jyrki Lahtonen Mar 24 '24 at 05:12
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    @peek-a-boo: I wrote my own version many years ago. I wonder how many there are across the site. – Martin Argerami Mar 24 '24 at 05:15
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    @MartinArgerami ooh nice. Yea there are lots of nice posts (and approaches overall) to dealing with trigonometric functions in general. Another super slick approach I like (in hindsight) is by defining arctan first, by treating the arclength of a circle as the primitive/starting point. – peek-a-boo Mar 24 '24 at 05:27

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Yes. You have $0=f'f+g'g=\frac12(f^2+g^2)'$. So latter is constant and you get $f^2+g^2=1$. This also gives you $-1\leq f,g\leq1$.

Next you get the addition formulas. Let $h(x)=f(x+y)$. We have $h''(x)=-h(x)$, $h(0)=f(y)$, $h'(0)=g (y)$. Knowing that this IVP has unique solution, one checks that $f(y)g (x)+g (y)f (x)$ is a solution, hence the solution and so $$ f(x+y)=f (x)g (y)+g(x)f (y). $$ And differentiating we get $$ g (x+y)=g (x)g (y)-f(x)f(y). $$

Note that $g>0$ on some interval around $0$. If $g$ is never $0$, this would imply that $f$ is always increasing and bounded, so $\lim_{x \to\infty}g(x)=0$. From $f(2x)=2f(x)g (x)$ this leads to the contradiction that $f\to0$. It follows that there exists $b>0$ with $g(b)=0$. Using the continuity of $g$, that $g(0)=1$, and taking the infimum of all positive zeroes, we may assume that $b$ is the least positive zero.

We have $f(b)^2=1$. As $f(0)=0$ and $f$ is increasing up to $b$, $f(b)=1$. It follows that $$ f(2b)=2f(b)g (b)=0,\qquad g (2b)=-f (b)^2=-1. $$ And $$ f(4b)=2f(2b)g (2b)=0,\qquad g(4b)=g(2b)^2-f(2b)^2=1-0=1. $$ Finally, $$ f(x+4b)=f(x)g(4b)+g(x)f(4b)=f(x)-0=f(x). $$ Differentiating, $$ g (x+4b)=g (x). $$

Martin Argerami
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