# §22.6 Elementary Identities

## §22.6(i) Sums of Squares

 22.6.1 ${\mathop{\mathrm{sn}\/}\nolimits^{2}}\left(z,k\right)+{\mathop{\mathrm{cn}\/}% \nolimits^{2}}\left(z,k\right)=k^{2}{\mathop{\mathrm{sn}\/}\nolimits^{2}}\left% (z,k\right)+{\mathop{\mathrm{dn}\/}\nolimits^{2}}\left(z,k\right)=1,$
 22.6.2 $1+{\mathop{\mathrm{cs}\/}\nolimits^{2}}\left(z,k\right)=k^{2}+{\mathop{\mathrm% {ds}\/}\nolimits^{2}}\left(z,k\right)={\mathop{\mathrm{ns}\/}\nolimits^{2}}% \left(z,k\right),$
 22.6.3 ${k^{\prime}}^{2}{\mathop{\mathrm{sc}\/}\nolimits^{2}}\left(z,k\right)+1={% \mathop{\mathrm{dc}\/}\nolimits^{2}}\left(z,k\right)={k^{\prime}}^{2}{\mathop{% \mathrm{nc}\/}\nolimits^{2}}\left(z,k\right)+k^{2},$
 22.6.4 $-k^{2}{k^{\prime}}^{2}{\mathop{\mathrm{sd}\/}\nolimits^{2}}\left(z,k\right)=k^% {2}({\mathop{\mathrm{cd}\/}\nolimits^{2}}\left(z,k\right)-1)={k^{\prime}}^{2}(% 1-{\mathop{\mathrm{nd}\/}\nolimits^{2}}\left(z,k\right)).$

## §22.6(ii) Double Argument

 22.6.5 $\mathop{\mathrm{sn}\/}\nolimits\left(2z,k\right)=\frac{2\mathop{\mathrm{sn}\/}% \nolimits\left(z,k\right)\mathop{\mathrm{cn}\/}\nolimits\left(z,k\right)% \mathop{\mathrm{dn}\/}\nolimits\left(z,k\right)}{1-k^{2}{\mathop{\mathrm{sn}\/% }\nolimits^{4}}\left(z,k\right)},$
 22.6.6 $\mathop{\mathrm{cn}\/}\nolimits\left(2z,k\right)=\frac{{\mathop{\mathrm{cn}\/}% \nolimits^{2}}\left(z,k\right)-{\mathop{\mathrm{sn}\/}\nolimits^{2}}\left(z,k% \right){\mathop{\mathrm{dn}\/}\nolimits^{2}}\left(z,k\right)}{1-k^{2}{\mathop{% \mathrm{sn}\/}\nolimits^{4}}\left(z,k\right)}=\frac{{\mathop{\mathrm{cn}\/}% \nolimits^{4}}\left(z,k\right)-{k^{\prime}}^{2}{\mathop{\mathrm{sn}\/}% \nolimits^{4}}\left(z,k\right)}{1-k^{2}{\mathop{\mathrm{sn}\/}\nolimits^{4}}% \left(z,k\right)},$
 22.6.7 $\mathop{\mathrm{dn}\/}\nolimits\left(2z,k\right)=\frac{{\mathop{\mathrm{dn}\/}% \nolimits^{2}}\left(z,k\right)-k^{2}{\mathop{\mathrm{sn}\/}\nolimits^{2}}\left% (z,k\right){\mathop{\mathrm{cn}\/}\nolimits^{2}}\left(z,k\right)}{1-k^{2}{% \mathop{\mathrm{sn}\/}\nolimits^{4}}\left(z,k\right)}=\frac{{\mathop{\mathrm{% dn}\/}\nolimits^{4}}\left(z,k\right)+k^{2}{k^{\prime}}^{2}{\mathop{\mathrm{sn}% \/}\nolimits^{4}}\left(z,k\right)}{1-k^{2}{\mathop{\mathrm{sn}\/}\nolimits^{4}% }\left(z,k\right)}.$ Symbols: $\mathop{\mathrm{cn}\/}\nolimits\left(\NVar{z},\NVar{k}\right)$: Jacobian elliptic function, $\mathop{\mathrm{dn}\/}\nolimits\left(\NVar{z},\NVar{k}\right)$: Jacobian elliptic function, $\mathop{\mathrm{sn}\/}\nolimits\left(\NVar{z},\NVar{k}\right)$: Jacobian elliptic function, $z$: complex, $k$: modulus and $k^{\prime}$: complementary modulus A&S Ref: 16.18.3 Referenced by: §22.6(ii), Equation (22.6.7) Permalink: http://dlmf.nist.gov/22.6.E7 Encodings: TeX, pMML, png Errata (effective with 1.0.7): Originally the term $k^{2}{\mathop{\mathrm{sn}\/}\nolimits^{2}}\left(z,k\right){\mathop{\mathrm{cn}% \/}\nolimits^{2}}\left(z,k\right)$ was given incorrectly as $k^{2}{\mathop{\mathrm{sn}\/}\nolimits^{2}}\left(z,k\right){\mathop{\mathrm{dn}% \/}\nolimits^{2}}\left(z,k\right)$. Reported 2014-02-28 by Svante Janson See also: Annotations for 22.6(ii)
 22.6.8 $\displaystyle\mathop{\mathrm{cd}\/}\nolimits\left(2z,k\right)$ $\displaystyle=\frac{{\mathop{\mathrm{cd}\/}\nolimits^{2}}\left(z,k\right)-{k^{% \prime}}^{2}{\mathop{\mathrm{sd}\/}\nolimits^{2}}\left(z,k\right){\mathop{% \mathrm{nd}\/}\nolimits^{2}}\left(z,k\right)}{1+k^{2}{k^{\prime}}^{2}{\mathop{% \mathrm{sd}\/}\nolimits^{4}}\left(z,k\right)},$ 22.6.9 $\displaystyle\mathop{\mathrm{sd}\/}\nolimits\left(2z,k\right)$ $\displaystyle=\frac{2\mathop{\mathrm{sd}\/}\nolimits\left(z,k\right)\mathop{% \mathrm{cd}\/}\nolimits\left(z,k\right)\mathop{\mathrm{nd}\/}\nolimits\left(z,% k\right)}{1+k^{2}{k^{\prime}}^{2}{\mathop{\mathrm{sd}\/}\nolimits^{4}}\left(z,% k\right)},$ 22.6.10 $\displaystyle\mathop{\mathrm{nd}\/}\nolimits\left(2z,k\right)$ $\displaystyle=\frac{{\mathop{\mathrm{nd}\/}\nolimits^{2}}\left(z,k\right)+k^{2% }{\mathop{\mathrm{sd}\/}\nolimits^{2}}\left(z,k\right){\mathop{\mathrm{cd}\/}% \nolimits^{2}}\left(z,k\right)}{1+k^{2}{k^{\prime}}^{2}{\mathop{\mathrm{sd}\/}% \nolimits^{4}}\left(z,k\right)},$ 22.6.11 $\displaystyle\mathop{\mathrm{dc}\/}\nolimits\left(2z,k\right)$ $\displaystyle=\frac{{\mathop{\mathrm{dc}\/}\nolimits^{2}}\left(z,k\right)+{k^{% \prime}}^{2}{\mathop{\mathrm{sc}\/}\nolimits^{2}}\left(z,k\right){\mathop{% \mathrm{nc}\/}\nolimits^{2}}\left(z,k\right)}{1-{k^{\prime}}^{2}{\mathop{% \mathrm{sc}\/}\nolimits^{4}}\left(z,k\right)},$
 22.6.12 $\displaystyle\mathop{\mathrm{nc}\/}\nolimits\left(2z,k\right)$ $\displaystyle=\frac{{\mathop{\mathrm{nc}\/}\nolimits^{2}}\left(z,k\right)+{% \mathop{\mathrm{sc}\/}\nolimits^{2}}\left(z,k\right){\mathop{\mathrm{dc}\/}% \nolimits^{2}}\left(z,k\right)}{1-{k^{\prime}}^{2}{\mathop{\mathrm{sc}\/}% \nolimits^{4}}\left(z,k\right)},$ 22.6.13 $\displaystyle\mathop{\mathrm{sc}\/}\nolimits\left(2z,k\right)$ $\displaystyle=\frac{2\mathop{\mathrm{sc}\/}\nolimits\left(z,k\right)\mathop{% \mathrm{dc}\/}\nolimits\left(z,k\right)\mathop{\mathrm{nc}\/}\nolimits\left(z,% k\right)}{1-{k^{\prime}}^{2}{\mathop{\mathrm{sc}\/}\nolimits^{4}}\left(z,k% \right)},$ 22.6.14 $\displaystyle\mathop{\mathrm{ns}\/}\nolimits\left(2z,k\right)$ $\displaystyle=\frac{{\mathop{\mathrm{ns}\/}\nolimits^{4}}\left(z,k\right)-k^{2% }}{2\mathop{\mathrm{cs}\/}\nolimits\left(z,k\right)\mathop{\mathrm{ds}\/}% \nolimits\left(z,k\right)\mathop{\mathrm{ns}\/}\nolimits\left(z,k\right)},$ 22.6.15 $\displaystyle\mathop{\mathrm{ds}\/}\nolimits\left(2z,k\right)$ $\displaystyle=\frac{k^{2}{k^{\prime}}^{2}+{\mathop{\mathrm{ds}\/}\nolimits^{4}% }\left(z,k\right)}{2\mathop{\mathrm{cs}\/}\nolimits\left(z,k\right)\mathop{% \mathrm{ds}\/}\nolimits\left(z,k\right)\mathop{\mathrm{ns}\/}\nolimits\left(z,% k\right)},$ 22.6.16 $\displaystyle\mathop{\mathrm{cs}\/}\nolimits\left(2z,k\right)$ $\displaystyle=\frac{{\mathop{\mathrm{cs}\/}\nolimits^{4}}\left(z,k\right)-{k^{% \prime}}^{2}}{2\mathop{\mathrm{cs}\/}\nolimits\left(z,k\right)\mathop{\mathrm{% ds}\/}\nolimits\left(z,k\right)\mathop{\mathrm{ns}\/}\nolimits\left(z,k\right)}.$

 22.6.17 $\displaystyle\frac{1-\mathop{\mathrm{cn}\/}\nolimits\left(2z,k\right)}{1+% \mathop{\mathrm{cn}\/}\nolimits\left(2z,k\right)}$ $\displaystyle=\frac{{\mathop{\mathrm{sn}\/}\nolimits^{2}}\left(z,k\right){% \mathop{\mathrm{dn}\/}\nolimits^{2}}\left(z,k\right)}{{\mathop{\mathrm{cn}\/}% \nolimits^{2}}\left(z,k\right)},$ 22.6.18 $\displaystyle\frac{1-\mathop{\mathrm{dn}\/}\nolimits\left(2z,k\right)}{1+% \mathop{\mathrm{dn}\/}\nolimits\left(2z,k\right)}$ $\displaystyle=\frac{k^{2}{\mathop{\mathrm{sn}\/}\nolimits^{2}}\left(z,k\right)% {\mathop{\mathrm{cn}\/}\nolimits^{2}}\left(z,k\right)}{{\mathop{\mathrm{dn}\/}% \nolimits^{2}}\left(z,k\right)}.$

## §22.6(iii) Half Argument

 22.6.19 $\displaystyle{\mathop{\mathrm{sn}\/}\nolimits^{2}}\left(\tfrac{1}{2}z,k\right)$ $\displaystyle=\frac{1-\mathop{\mathrm{cn}\/}\nolimits\left(z,k\right)}{1+% \mathop{\mathrm{dn}\/}\nolimits\left(z,k\right)}=\frac{1-\mathop{\mathrm{dn}\/% }\nolimits\left(z,k\right)}{k^{2}(1+\mathop{\mathrm{cn}\/}\nolimits\left(z,k% \right))}=\frac{\mathop{\mathrm{dn}\/}\nolimits\left(z,k\right)-k^{2}\mathop{% \mathrm{cn}\/}\nolimits\left(z,k\right)-{k^{\prime}}^{2}}{k^{2}(\mathop{% \mathrm{dn}\/}\nolimits\left(z,k\right)-\mathop{\mathrm{cn}\/}\nolimits\left(z% ,k\right))},$ 22.6.20 $\displaystyle{\mathop{\mathrm{cn}\/}\nolimits^{2}}\left(\tfrac{1}{2}z,k\right)$ $\displaystyle=\frac{-{k^{\prime}}^{2}+\mathop{\mathrm{dn}\/}\nolimits\left(z,k% \right)+k^{2}\mathop{\mathrm{cn}\/}\nolimits\left(z,k\right)}{k^{2}(1+\mathop{% \mathrm{cn}\/}\nolimits\left(z,k\right))}=\frac{{k^{\prime}}^{2}(1-\mathop{% \mathrm{dn}\/}\nolimits\left(z,k\right))}{k^{2}(\mathop{\mathrm{dn}\/}% \nolimits\left(z,k\right)-\mathop{\mathrm{cn}\/}\nolimits\left(z,k\right))}=% \frac{{k^{\prime}}^{2}(1+\mathop{\mathrm{cn}\/}\nolimits\left(z,k\right))}{{k^% {\prime}}^{2}+\mathop{\mathrm{dn}\/}\nolimits\left(z,k\right)-k^{2}\mathop{% \mathrm{cn}\/}\nolimits\left(z,k\right)},$ 22.6.21 $\displaystyle{\mathop{\mathrm{dn}\/}\nolimits^{2}}\left(\tfrac{1}{2}z,k\right)$ $\displaystyle=\frac{k^{2}\mathop{\mathrm{cn}\/}\nolimits\left(z,k\right)+% \mathop{\mathrm{dn}\/}\nolimits\left(z,k\right)+{k^{\prime}}^{2}}{1+\mathop{% \mathrm{dn}\/}\nolimits\left(z,k\right)}=\frac{{k^{\prime}}^{2}(1-\mathop{% \mathrm{cn}\/}\nolimits\left(z,k\right))}{\mathop{\mathrm{dn}\/}\nolimits\left% (z,k\right)-\mathop{\mathrm{cn}\/}\nolimits\left(z,k\right)}=\frac{{k^{\prime}% }^{2}(1+\mathop{\mathrm{dn}\/}\nolimits\left(z,k\right))}{{k^{\prime}}^{2}+% \mathop{\mathrm{dn}\/}\nolimits\left(z,k\right)-k^{2}\mathop{\mathrm{cn}\/}% \nolimits\left(z,k\right)}.$

If $\{\mbox{p,q,r}\}$ is any permutation of $\{\mbox{c,d,n}\}$, then

 22.6.22 ${\mathop{\mathrm{p\!q}\/}\nolimits^{2}}\left(\tfrac{1}{2}z,k\right)=\frac{% \mathop{\mathrm{p\!s}\/}\nolimits\left(z,k\right)+\mathop{\mathrm{r\!s}\/}% \nolimits\left(z,k\right)}{\mathop{\mathrm{q\!s}\/}\nolimits\left(z,k\right)+% \mathop{\mathrm{r\!s}\/}\nolimits\left(z,k\right)}=\frac{\mathop{\mathrm{p\!q}% \/}\nolimits\left(z,k\right)+\mathop{\mathrm{r\!q}\/}\nolimits\left(z,k\right)% }{1+\mathop{\mathrm{r\!q}\/}\nolimits\left(z,k\right)}=\frac{\mathop{\mathrm{p% \!r}\/}\nolimits\left(z,k\right)+1}{\mathop{\mathrm{q\!r}\/}\nolimits\left(z,k% \right)+1}.$ Symbols: $\mathop{\mathrm{p\!q}\/}\nolimits\left(\NVar{z},\NVar{k}\right)$: generic Jacobian elliptic function, $z$: complex and $k$: modulus Referenced by: §22.6(iii) Permalink: http://dlmf.nist.gov/22.6.E22 Encodings: TeX, pMML, png See also: Annotations for 22.6(iii)

For (22.6.22) and similar results, see Carlson (2004).

See §22.17.