Let $\operatorname{ns}\left(z,k\right)$ be one of the Jacobi's elliptic functions, and $K(k)$ the complete elliptic integral of the first kind. It's well-known that $\operatorname{ns}(mK(k),k)$ where $m\in\mathbb{Q}/2\mathbb{Z}$ is always an algebraic function with respect to $k$. With its addition formula, $$ \operatorname{ns}\left ( u+v ,k\right )\operatorname{ns} \left ( u-v,k \right ) =\frac{k^2-\operatorname{ns}(u,k)^2\operatorname{ns}(v,k)^2}{ \operatorname{ns}(u,k)^2-\operatorname{ns}(v,k)^2}, $$ place $(u,v)=\left ( \frac{2}{3}K(k), \frac{1}{3}K(k) \right )$ and $(u,v)=\left ( K(k), \frac13K(k) \right )$ respectively, giving \begin{aligned} &\operatorname{ns}\left ( \frac{2}{3}K(k),k \right )^2 =\frac{k^2-\operatorname{ns}\left ( \frac{1}{3}K(k),k \right )^2 }{ 1-\operatorname{ns}\left ( \frac{1}{3}K(k),k \right )^2} \\ &\operatorname{ns}\left ( \frac{1}{3}K(k),k \right ) =\frac{k^2-\operatorname{ns}\left ( \frac{2}{3}K(k),k \right )^2\operatorname{ns}\left ( \frac{1}{3}K(k),k \right )^2}{ \operatorname{ns}\left ( \frac{2}{3}K(k),k \right )^2-\operatorname{ns}\left ( \frac{1}{3}K(k),k \right )^2} \end{aligned} The two equations are linear independent with each other, producing exact expressions of $\operatorname{ns}\left ( \frac{1}{3}K(k),k \right ),\operatorname{ns}\left ( \frac{2}{3}K(k),k \right )$. But the nasty appearances make it hard to do for some calculation. So is there a concise algebraic relation between the two? Thanks for your help.
My ultimate goal is to develop another way to calculate some algebraic number $k_n$ which satisfying $K^\prime(k)/K(k)=\sqrt{n}$. And $n=1,2,3,4,5,6,8,9,10,12,16,18...$ can be evaluated, whereas $k_{7},k_{15}...$ can't because the restriction of Wolfram Cloud(I use it to solve equations). Basic ideas are recorded here. It signifies the need of relations among $\operatorname{ns}\left (mK(k),k \right )$ having the same denominator.
Define the two quantities
$$ a_1 := \text{ns}\left(\frac{K(k)}{3},k\right) \quad\text{ and } \quad a_2 := \text{ns}\left(\frac{2K(k)}{3},k\right). \tag1 $$
The power series for these using the notation $\,k^2 = 16x\,$ is
$$ a_1 = 2 - 6 x - 30 x^2 - 258 x^3 - 2694 x^4 - 31128 x^5 - \dots,\\ a_2 = (2 - 2 x - 10 x^2 - 82 x^3 - 818 x^4 - 9076 x^5 -\dots)/\sqrt{3}. \tag2 $$
The question asked is
The answer is $$ a_2 = \frac{a_1}{\sqrt{2a_1-1}} \quad \text{ and } \quad a_1 = a_2^2 + \sqrt{a_2^4-a_2^2}. \tag3 $$
Other algebraic relations are $$ 0 = a_1^4 - 2a_1^3 + 2k^2a_1 - k^2, \\ 0 = 3a_2^8 - 4(1+k^2)a_2^6 + 6k^2a_2^4 - k^4. \tag4 $$