Solving the functional equations $f(g(x)) = \sin x$ and $g(f(x)) = \cos x$

Question

Solving the functional equations $f(g(x)) = \sin x$ and $g(f(x)) = \cos x$ .

I've managed to reduce an expression for $f(x)$ as:

$f(\cos x) = \sin(f(x))$

But I have no idea how to proceed. I would appreciate some help !

Answer 1

(Too long for a comment.)

If we assume that $g(x)$ is nice enough to have a Taylor series, then there is no solution.

Going in the other direction (i.e. replacing $x$ with $g(x)$ in the cosine equation and reduce using the sine equation), we get

$$ g(\sin x) = \cos g(x). $$

Putting $g(x) = a_0 + a_1 x + a_2 x^2 + \text{etc.}$, we get

\begin{align} a_0 &= \cos a_0 &&(1) \\ a_1 &= -a_1 S &&(x) \\ a_2 &= -\frac{{a_1}^2 C}{2} - a_2 S &&(x^2) \\ -\frac{a_1}{6} + a_3 &= \frac{{a_1}^3 S}{6} - a_1 a_2 C - a_3 S &&(x^3) \\ -\frac{a_2}{3} + a_4 &= \text{(terms with $a_1$, $a_2$, $a_3$)} - a_4 S &&(x^4) \\ \frac{a_1}{120} - \frac{a_3}{2} + a_5 &= \text{(terms with $a_1$, $a_2$, $a_3$, $a_4$)} - a_5 S &&(x^5) \\ \frac{2 a_2}{45} - \frac{2 a_4}{3} + a_6 &= \text{(terms with $a_1$, $a_2$, $a_3$, $a_4$, $a_5$)} - a_6 S &&(x^6) \\ &\text{etc.} \end{align}

where $C = \cos a_0$ and $S = \sin a_0$.

We see that $a_0$ is the Dottie number, the unique real fixed point of the cosine function (so that $C = 0.739...$ and $S = 0.673...$). All subsequent coefficients vanish, as the equation for the coefficient of $x^n$ (for $n \ge 1$) reduces to

$$a_n = -a_n S.$$

Thus $g(x) = a_0$, and the sine equation becomes $f(a_0) = \sin x$, a contradiction.

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This of course doesn't rule out the possibility of solutions where $g(x)$ doesn't have a Taylor series.