How to show that Jacobi sine function is doubly periodic The Jacobi sine function can be defined using two definitions. The first is $\operatorname{sn}(u,m)=\sin(\phi)$ where $u=\int_0^{\phi}\frac{d\theta}{\sqrt{1-m\sin^2\theta}}$.
Alternativey it may be defined as the function satisfying the following equation $$(\operatorname{sn}')^2+ (\operatorname{sn}^2−1)(1−k^2\operatorname{sn}^2) = 0.$$ 
It is known that this function is supposed to be elliptic thus doubly periodic, I have no idea how this follows from any of the two definitions any help will be much  appreciated.
 A: We let $m=k^2$, $k'=\sqrt{1-k^2}$ and introduce $$K=\int_0^1\frac{\mathrm dt}{\sqrt{(1-t^2)(1-k^2t^2)}}=\int_0^{\frac {\pi}2}\frac{\mathrm d\varphi}{\sqrt{1-k^2\sin^2\varphi}}$$
and
$$K'=\int_0^1\frac{\mathrm dt}{\sqrt{(1-t^2)(1-k'^2t^2)}}.$$

Claim: $$\operatorname{sn}(K+iK')=\dfrac 1k,\quad\operatorname{cn}(K+iK')=-\dfrac {ik'}k,\quad \operatorname{dn}(K+iK')=0.$$

Proof: By changing the subject $k^2t^2=1-k'^2t^2$,
\begin{align*}
\int_0^\frac 1k\frac{\mathrm dt}{\sqrt{(1-t^2)(1-k^2t^2)}}&=\underbrace{\int_0^1\frac{\mathrm dt}{\sqrt{(1-t^2)(1-k^2t^2)}}}_{=K}+\underbrace{i\int_1^\frac 1k\frac{\mathrm dt}{\sqrt{(t^2-1)(1-k^2t^2)}}}_{=iK'}.\tag*{$\square$}\\
\end{align*}
We use addition theorem (which can be derived from the definitions) and the fact
$$\operatorname{sn} K=1,\quad \operatorname{cn} K=0,\quad \operatorname{dn} K= k'$$
to get
$$\operatorname{sn}(u+K)=\frac{\operatorname{cn} u \operatorname{dn} u}{1-k^2 \operatorname{sn}^2u}=\frac{\operatorname{cn} u}{\operatorname{dn}u}=\operatorname{cd}u.\tag{1}$$
Similarly we can get
$$\operatorname{cn}(u+K)=-k'\operatorname{sd}u,\quad \operatorname{dn}(u+K)=k'\operatorname{nd}u.\tag{2}$$
Add $K$ one more time,
$$\operatorname{sn}(u+2K)=-\operatorname{sn}u,\quad\operatorname{cn}(u+2K)=-\operatorname{cn}u,\quad \operatorname{dn}(u+2K)=\operatorname{dn}u.\tag{3}$$
Add $2K$ again,
$$\operatorname{sn}(u+4K)=\operatorname{sn}u,\quad\operatorname{cn}(u+4K)=\operatorname{cn}u.\tag{4}$$
Therefore, $\operatorname{sn} u,\operatorname{cn} u$ are $4K$-periodic while $\operatorname{dn} u$ is $2K$-periodic.

Next, we use addition theorem and the claim to get
\begin{align*}
\operatorname{sn}(u+K+iK')&=k^{-1}\operatorname{dc}u,\\
\operatorname{cn}(u+K+iK')&=-ik'k^{-1}\operatorname{nc}u,\\
\operatorname{dn}(u+K+iK')&=ik'\operatorname{sc}u.\tag{5}
\end{align*}
Add $K+K'i$ once more,
\begin{align*}
\operatorname{sn}(u+2K+2K'i)&=-\operatorname{sn}u,\\
\operatorname{cn}(u+2K+2K'i)&=\operatorname{cn}u,\\
\operatorname{dn}(u+2K+2K'i)&=-\operatorname{dn}u.\tag{6}
\end{align*}
Therefore, $\operatorname{sn} u,\operatorname{dn} u$ are $(4K+4K'i)$-periodic while $\operatorname{cn} u$ is $(2K+2K'i)$-periodic.

It is nasty to compute $+K'i$ directly. Instead, we use
\begin{align*}
\operatorname{sn}(u+K'i)=\operatorname{sn}(u-K+K+K'i)=k^{-1}\operatorname{sn}(u-K)=k^{-1}\operatorname{ns}u.\tag{7}
\end{align*}
We also get
\begin{align*}
\operatorname{cn}(u+K'i)=-ik^{-1}\operatorname{ds}u,\quad \operatorname{dn}(u+K'i)=-i\operatorname{cs}u.\tag{8}
\end{align*}
Add $K'i$ again,
\begin{align*}
\operatorname{sn}(u+2K'i)=\operatorname{sn}u,\quad\operatorname{cn}(u+2K'i)=-\operatorname{cn}u,\quad \operatorname{dn}(u+2K'i)=-\operatorname{dn}u.\tag{9}
\end{align*}
Therefore, $\operatorname{sn} u$ is $2K'i$-periodic while $\operatorname{cn} u,\operatorname{dn} u$ are $4K'i$-periodic.

Combining the above results,


*

*$\operatorname{sn} u$ is $(4K,2K'i)$-periodic;


*$\operatorname{cn} u$ is $(4K,2K+2K'i)$-periodic;


*$\operatorname{dn} u$ is $(2K,4K'i)$-periodic.

