ランデン変換
sn  ( u , k ) = 2 1 + 1 − k 2 sn  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) 1 + 1 − 1 − k 2 1 + 1 − k 2 sn 2  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) {\displaystyle \operatorname {sn} \left(u,k\right)={\frac {{\tfrac {2}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} \left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}{1+{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}}} cn  ( u , k ) = 1 − sn 2  ( u , k ) = cn  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) dn  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) 1 + 1 − 1 − k 2 1 + 1 − k 2 sn 2  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) {\displaystyle {\begin{aligned}\operatorname {cn} \left(u,k\right)&={\sqrt {1-\operatorname {sn} ^{2}\left(u,k\right)}}\\&={\frac {\operatorname {cn} \left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)\operatorname {dn} \left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}{1+{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}}\end{aligned}}} dn  ( u , k ) = 1 − k 2 sn 2  ( u , k ) = 1 − 1 − 1 − k 2 1 + 1 − k 2 sn 2  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) 1 + 1 − 1 − k 2 1 + 1 − k 2 sn 2  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) = dn 2  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) − 2 1 − k 2 1 + 1 − k 2 2 1 + 1 − k 2 − dn 2  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) {\displaystyle {\begin{aligned}\operatorname {dn} \left(u,k\right)&={\sqrt {1-k^{2}\operatorname {sn} ^{2}\left(u,k\right)}}\\&={\frac {1-{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}{1+{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}}\\&={\frac {\operatorname {dn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)-{\tfrac {2{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}}{{\tfrac {2}{1+{\sqrt {1-k^{2}}}}}-\operatorname {dn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}}\end{aligned}}}
sin  α = 2 1 + k sin  β cos  β 1 − 4 k ( 1 + k ) 2 sin 2  β {\displaystyle \sin \alpha ={\frac {{\frac {2}{1+k}}\sin \beta \cos \beta }{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\beta }}}} sin 2  α = 4 sin 2  β cos 2  β ( 1 + k ) 2 − 4 k sin 2  β = 1 − cos 2  ( 2 β ) 1 + k 2 + 2 k cos  ( 2 β ) {\displaystyle \sin ^{2}\alpha ={\frac {4\sin ^{2}\beta \cos ^{2}\beta }{(1+k)^{2}-4k\sin ^{2}\beta }}={\frac {1-\cos ^{2}(2\beta )}{1+k^{2}+2k\cos(2\beta )}}} cos 2  ( 2 β ) + 2 k sin 2  α cos  ( 2 β ) + k 2 sin 2  α − 1 + sin 2  α = 0 {\displaystyle \cos ^{2}(2\beta )+2k\sin ^{2}\alpha \cos(2\beta )+k^{2}\sin ^{2}\alpha -1+\sin ^{2}\alpha =0} cos  ( 2 β ) = − k sin 2  α + k 2 sin 4  α − k 2 sin 2  α + 1 − sin 2  α {\displaystyle \cos(2\beta )=-k\sin ^{2}\alpha +{\sqrt {k^{2}\sin ^{4}\alpha -k^{2}\sin ^{2}\alpha +1-\sin ^{2}\alpha }}} sin  β = 1 − cos  ( 2 β ) 2 = 1 2 ( 1 + k ) 2 sin 2  α + ( 1 − k 2 sin 2  φ − 1 − sin 2  φ ) 2 {\displaystyle \sin \beta ={\frac {1-\cos(2\beta )}{2}}={\frac {1}{2}}{\sqrt {\left(1+k\right)^{2}\sin ^{2}\alpha +\left({\sqrt {1-k^{2}\sin ^{2}\phi }}-{\sqrt {1-\sin ^{2}\phi }}\right)^{2}}}}
虚数変換により
sn  ( u , k ) = 2 1 + 1 − k 2 sn  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) 1 + 1 − 1 − k 2 1 + 1 − k 2 sn 2  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) {\displaystyle \operatorname {sn} \left(u,k\right)={\frac {{\tfrac {2}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} \left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}{1+{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}}} cn  ( u , k ) = cn  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) dn  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) 1 + 1 − 1 − k 2 1 + 1 − k 2 sn 2  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) {\displaystyle \operatorname {cn} \left(u,k\right)={\frac {\operatorname {cn} \left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)\operatorname {dn} \left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}{1+{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\operatorname {sn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)}}} dn  ( u , k ) = 1 − 1 − k 2 1 + 1 − k 2 − ( 1 − dn 2  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) ) 1 − 1 − k 2 1 + 1 − k 2 + ( 1 − dn 2  ( 1 + 1 − k 2 2 u , 1 − 1 − k 2 1 + 1 − k 2 ) ) {\displaystyle \operatorname {dn} \left(u,k\right)={\frac {{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}-\left(1-\operatorname {dn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)\right)}{{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}+\left(1-\operatorname {dn} ^{2}\left({\tfrac {1+{\sqrt {1-k^{2}}}}{2}}u,{\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}\right)\right)}}}