Function such that $f(x) f(\pi/2 - x) = 1$

Question

I'm looking for functions that are smooth ($C^\infty$) between $0 < x < \pi/2$ that satisfy the equation $$f(x)\, f(\pi/2-x) = 1$$ on the inteverval $0<x<\pi/2$. I know that the constant function $f=1$ satisfies this equation, as well as $f=\tan(x)$. Are these the only solutions?

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It seems that there are infinitely many $C^\infty$ functions that work, so long as the power series at $x=\pi/4$ is consistent with the restrictions coming from taking derivatives of the above expression at $\pi/4$. Each of these power series should correspond to an analytic function that satisfies the above equation in a neighborhood of $x=\pi/4$. So the question seems to be how many analytic solutions are there to the above problem?

Answer 1

Taking the logarithm gives $$ \log f(x)+\log f(\pi/2 - x)=0, $$ or $$ \log f\left(\pi/4 + (x-\pi/4)\right)=-\log f\left(\pi/4 - (x - \pi/4)\right). $$ That is, $\log f$ needs to be odd under reflection across $\pi/4$. So let $g(x)$ be any odd function (defined at least for $|x|<\pi/4$); then $$ f(x)=\exp{g(x - \pi/4)} $$ meets the conditions. If $g(x)$ is chosen to be continuous, smooth, or analytic, then $f(x)$ will be an equally "nice" function. For instance, $f(x)=e^{x-\pi/4}$ is a simple analytic example, or $f(x)=e^{(x-\pi/4)^3}$.

Answer 2

Rescaling the interval to $(0,2),$ we want $f(x)f(2-x)=1$. Assuming this $f$ is smooth, and say $f(1)=1$ [one of the two choices there], let $g(x)$ be its restriction to $[0,1]$, and then on $[1,2]$ we have $f(x)=h(x) \equiv 1/g(2-x).$ The derivative matching is automatic: $$h'(x)=\frac{-1}{g(2-x)^2}g'(2-x)\cdot (-1),$$ which since $g(1)=1$ and $g'(2-1)=g'(1),$ gives a match between the left side derivative of $g$ at $x=1$ and the right side derivative of $h$ at $1.$

So it seems $f$ may be chosen arbitrarily nonzero on $(0,1]$ with $f$ smooth on a on that interval (derivative from the left at $1$ existing), and then extended to all of $(0,2)$ as noted.

This is essentially the same as Steven Stadniki's comment, just extended to show there is really no derivative matching issue. [So I made it community wiki.]