ランデン変換
ガウス変換は
sin  φ = 2 1 + k sin  θ 1 + 1 − 4 k ( 1 + k ) 2 sin 2  θ {\displaystyle \sin \phi ={\frac {{\frac {2}{1+k}}\sin \theta }{1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}}} cos  φ d φ = 2 1 + k cos  θ 1 + 1 − 4 k ( 1 + k ) 2 sin 2  θ d θ + 2 1 + k ( 4 k ( 1 + k ) 2 sin 2  θ cos  θ ) 1 − 4 k ( 1 + k ) 2 sin 2  θ ( 1 + 1 − 4 k ( 1 + k ) 2 sin 2  θ ) 2 d θ = 2 1 + k cos  θ 1 − 4 k ( 1 + k ) 2 sin 2  θ ( 1 + 1 − 4 k ( 1 + k ) 2 sin 2  θ ) d θ {\displaystyle {\begin{aligned}\cos \phi {d\phi }&={\frac {{\frac {2}{1+k}}\cos \theta }{1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}}{d\theta }+{\frac {{\frac {2}{1+k}}\left({\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta \cos \theta \right)}{{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\left(1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)^{2}}}{d\theta }\\&={\frac {{\frac {2}{1+k}}\cos \theta }{{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\left(1+{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)}}{d\theta }\end{aligned}}}
sin  β {\displaystyle \sin \beta } を陽にすると