# §22.6 Elementary Identities

## §22.6(i) Sums of Squares

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

## §22.6(ii) Double Argument

 22.6.5 $\operatorname{sn}\left(2z,k\right)=\frac{2\operatorname{sn}\left(z,k\right)% \operatorname{cn}\left(z,k\right)\operatorname{dn}\left(z,k\right)}{1-k^{2}{% \operatorname{sn}}^{4}\left(z,k\right)},$
 22.6.6 $\operatorname{cn}\left(2z,k\right)=\frac{{\operatorname{cn}}^{2}\left(z,k% \right)-{\operatorname{sn}}^{2}\left(z,k\right){\operatorname{dn}}^{2}\left(z,% k\right)}{1-k^{2}{\operatorname{sn}}^{4}\left(z,k\right)}=\frac{{\operatorname% {cn}}^{4}\left(z,k\right)-{k^{\prime}}^{2}{\operatorname{sn}}^{4}\left(z,k% \right)}{1-k^{2}{\operatorname{sn}}^{4}\left(z,k\right)},$
 22.6.7 $\operatorname{dn}\left(2z,k\right)=\frac{{\operatorname{dn}}^{2}\left(z,k% \right)-k^{2}{\operatorname{sn}}^{2}\left(z,k\right){\operatorname{cn}}^{2}% \left(z,k\right)}{1-k^{2}{\operatorname{sn}}^{4}\left(z,k\right)}=\frac{{% \operatorname{dn}}^{4}\left(z,k\right)+k^{2}{k^{\prime}}^{2}{\operatorname{sn}% }^{4}\left(z,k\right)}{1-k^{2}{\operatorname{sn}}^{4}\left(z,k\right)}.$ ⓘ Symbols: $\operatorname{cn}\left(\NVar{z},\NVar{k}\right)$: Jacobian elliptic function, $\operatorname{dn}\left(\NVar{z},\NVar{k}\right)$: Jacobian elliptic function, $\operatorname{sn}\left(\NVar{z},\NVar{k}\right)$: Jacobian elliptic function, $z$: complex, $k$: modulus and $k^{\prime}$: complementary modulus A&S Ref: 16.18.3 Referenced by: §22.6(ii), Erratum (V1.0.7) for 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 $k^{2}{\operatorname{sn}}^{2}\left(z,k\right){\operatorname{cn}}^{2}\left(z,k\right)$ was given incorrectly as $k^{2}{\operatorname{sn}}^{2}\left(z,k\right){\operatorname{dn}}^{2}\left(z,k\right)$. Reported 2014-02-28 by Svante Janson See also: Annotations for §22.6(ii), §22.6 and Ch.22
 22.6.8 $\displaystyle\operatorname{cd}\left(2z,k\right)$ $\displaystyle=\frac{{\operatorname{cd}}^{2}\left(z,k\right)-{k^{\prime}}^{2}{% \operatorname{sd}}^{2}\left(z,k\right){\operatorname{nd}}^{2}\left(z,k\right)}% {1+k^{2}{k^{\prime}}^{2}{\operatorname{sd}}^{4}\left(z,k\right)},$ 22.6.9 $\displaystyle\operatorname{sd}\left(2z,k\right)$ $\displaystyle=\frac{2\operatorname{sd}\left(z,k\right)\operatorname{cd}\left(z% ,k\right)\operatorname{nd}\left(z,k\right)}{1+k^{2}{k^{\prime}}^{2}{% \operatorname{sd}}^{4}\left(z,k\right)},$ 22.6.10 $\displaystyle\operatorname{nd}\left(2z,k\right)$ $\displaystyle=\frac{{\operatorname{nd}}^{2}\left(z,k\right)+k^{2}{% \operatorname{sd}}^{2}\left(z,k\right){\operatorname{cd}}^{2}\left(z,k\right)}% {1+k^{2}{k^{\prime}}^{2}{\operatorname{sd}}^{4}\left(z,k\right)},$ 22.6.11 $\displaystyle\operatorname{dc}\left(2z,k\right)$ $\displaystyle=\frac{{\operatorname{dc}}^{2}\left(z,k\right)+{k^{\prime}}^{2}{% \operatorname{sc}}^{2}\left(z,k\right){\operatorname{nc}}^{2}\left(z,k\right)}% {1-{k^{\prime}}^{2}{\operatorname{sc}}^{4}\left(z,k\right)},$
 22.6.12 $\displaystyle\operatorname{nc}\left(2z,k\right)$ $\displaystyle=\frac{{\operatorname{nc}}^{2}\left(z,k\right)+{\operatorname{sc}% }^{2}\left(z,k\right){\operatorname{dc}}^{2}\left(z,k\right)}{1-{k^{\prime}}^{% 2}{\operatorname{sc}}^{4}\left(z,k\right)},$ 22.6.13 $\displaystyle\operatorname{sc}\left(2z,k\right)$ $\displaystyle=\frac{2\operatorname{sc}\left(z,k\right)\operatorname{dc}\left(z% ,k\right)\operatorname{nc}\left(z,k\right)}{1-{k^{\prime}}^{2}{\operatorname{% sc}}^{4}\left(z,k\right)},$ 22.6.14 $\displaystyle\operatorname{ns}\left(2z,k\right)$ $\displaystyle=\frac{{\operatorname{ns}}^{4}\left(z,k\right)-k^{2}}{2% \operatorname{cs}\left(z,k\right)\operatorname{ds}\left(z,k\right)% \operatorname{ns}\left(z,k\right)},$ 22.6.15 $\displaystyle\operatorname{ds}\left(2z,k\right)$ $\displaystyle=\frac{k^{2}{k^{\prime}}^{2}+{\operatorname{ds}}^{4}\left(z,k% \right)}{2\operatorname{cs}\left(z,k\right)\operatorname{ds}\left(z,k\right)% \operatorname{ns}\left(z,k\right)},$ 22.6.16 $\displaystyle\operatorname{cs}\left(2z,k\right)$ $\displaystyle=\frac{{\operatorname{cs}}^{4}\left(z,k\right)-{k^{\prime}}^{2}}{% 2\operatorname{cs}\left(z,k\right)\operatorname{ds}\left(z,k\right)% \operatorname{ns}\left(z,k\right)}.$

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

## §22.6(iii) Half Argument

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

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

 22.6.22 ${\operatorname{pq}}^{2}\left(\tfrac{1}{2}z,k\right)=\frac{\operatorname{ps}% \left(z,k\right)+\operatorname{rs}\left(z,k\right)}{\operatorname{qs}\left(z,k% \right)+\operatorname{rs}\left(z,k\right)}=\frac{\operatorname{pq}\left(z,k% \right)+\operatorname{rq}\left(z,k\right)}{1+\operatorname{rq}\left(z,k\right)% }=\frac{\operatorname{pr}\left(z,k\right)+1}{\operatorname{qr}\left(z,k\right)% +1}.$ ⓘ Symbols: $\operatorname{pq}\left(\NVar{z},\NVar{k}\right)$: 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), §22.6 and Ch.22

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

See §22.17.