Does the elliptic function $\operatorname{cn}\left(\frac{2}{3}K\left(\frac{1}{2}\right)\big|\frac{1}{2}\right)$ have a closed form? Given the complete elliptic integral of the first kind $K(k)$ for the modulus $k$,
can the elliptic function $$\text{cn}\left(\frac{2}{3}K\left(\frac{1}{2}\right)\bigg|\frac{1}{2}\right)$$ be expressed in closed form?

To explain how an algebraic equation for $Z:=\operatorname{cn} \left(\frac{2K}{3}|m\right)$ can be obtained, notice that


*

*The usual parity properties imply that
$$\operatorname{cn} \left(\frac{4K}{3}\biggl|\,m\right)=\operatorname{cn} \left(2K-\frac{2K}{3}\biggl|\,m\right)=-\operatorname{cn} \left(-\frac{2K}{3}\biggl|\,m\right)=-\operatorname{cn} \left(\frac{2K}{3}\biggl|\,m\right)=-Z.$$

*On the other hand the doubling formula implies that
$$\operatorname{cn} \left(\frac{4K}{3}\biggl|\,m\right)=\frac{1-2\operatorname{sn}^2 \left(\frac{2K}{3}|m\right)+m\operatorname{sn}^4 \left(\frac{2K}{3}|m\right)}{1-m\operatorname{sn}^4 \left(\frac{2K}{3}|m\right)}=\frac{2Z^2-1+m\left(1-Z^2\right)^2}{1-m\left(1-Z^2\right)^2}.$$
Thus $Z$ satisfies the equation
$$\frac{2Z^2-1+m\left(1-Z^2\right)^2}{1-m\left(1-Z^2\right)^2}=-Z.$$
Although naively it is of $5$th order, it has an obvious root $Z=-1$, so we can deduce from it a $4$th order equation
$$\boxed{\quad m\left(Z^2-1\right)\left(Z-1\right)^2-2Z+1=0\quad}$$
In particular, the relevant solution for $m=k^2=\frac12$ is given by
$$Z=\frac{1-\sqrt2\cdot\sqrt[4]{3}+\sqrt3}{2}\approx 0.435421.$$
