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The Jacobi sine function can be defined using two definitions. The first is $operatorname{sn}(u,m)=sin(phi)$ where $u=int_0^{phi}frac{dtheta}{sqrt{1-msin^2theta}}$.

Alternativey it may be defined as the function satisfying the following equation $$(operatorname{sn}')^2+ (operatorname{sn}^2−1)(1−k^2operatorname{sn}^2) = 0.$$

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.

We let $m=k^2$, $k'=sqrt{1-k^2}$ and introduce $$K=int_0^1frac{mathrm dt}{sqrt{(1-t^2)(1-k^2t^2)}}=int_0^{frac {pi}2}frac{mathrm dvarphi}{sqrt{1-k^2sin^2varphi}}$$
and
$$K'=int_0^1frac{mathrm dt}{sqrt{(1-t^2)(1-k'^2t^2)}}.$$

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

Proof: By changing the subject $k^2t^2=1-k'^2t^2$,
begin{align*}
int_0^frac 1kfrac{mathrm dt}{sqrt{(1-t^2)(1-k^2t^2)}}&=underbrace{int_0^1frac{mathrm dt}{sqrt{(1-t^2)(1-k^2t^2)}}}_{=K}+underbrace{iint_1^frac 1kfrac{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,
$$operatorname{sn}(u+2K)=-operatorname{sn}u,quadoperatorname{cn}(u+2K)=-operatorname{cn}u,quad operatorname{dn}(u+2K)=operatorname{dn}u.tag{3}$$
Add $2K$ again,
$$operatorname{sn}(u+4K)=operatorname{sn}u,quadoperatorname{cn}(u+4K)=operatorname{cn}u.tag{4}$$
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)=-ioperatorname{cs}u.tag{8}
end{align*}
Add $K'i$ again,
begin{align*}
operatorname{sn}(u+2K'i)=operatorname{sn}u,quadoperatorname{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.