Erweitert man den Bruch ${\displaystyle A={\frac {\Gamma \left({\frac {1}{24}}\right)\,\Gamma \left({\frac {11}{24}}\right)}{\Gamma \left({\frac {5}{24}}\right)\,\Gamma \left({\frac {7}{24}}\right)}}}$ mit ${\displaystyle \Gamma \left({\frac {13}{24}}\right)\Gamma \left({\frac {23}{24}}\right)}$, so ist ${\displaystyle A={\frac {\Gamma \left({\frac {1}{24}}\right)\,\Gamma \left({\frac {23}{24}}\right)\;\cdot \;\Gamma \left({\frac {11}{24}}\right)\,\Gamma \left({\frac {13}{24}}\right)}{\Gamma \left({\frac {5}{24}}\right)\,\Gamma \left({\frac {13}{24}}\right)\,\cdot \,\Gamma \left({\frac {7}{24}}\right)\,\Gamma \left({\frac {23}{24}}\right)}}}$.

Nach dem Eulerschen Ergänzungssatz ist ${\displaystyle \Gamma \left({\frac {1}{24}}\right)\Gamma \left({\frac {23}{24}}\right)={\frac {\pi }{\sin \left({\frac {\pi }{24}}\right)}}}$ und ${\displaystyle \Gamma \left({\frac {11}{24}}\right)\Gamma \left({\frac {13}{24}}\right)={\frac {\pi }{\cos \left({\frac {\pi }{24}}\right)}}}$.

Daher ist ${\displaystyle A={\frac {\pi ^{2}}{\sin \left({\frac {\pi }{24}}\right)\cos \left({\frac {\pi }{24}}\right)}}\cdot {\frac {1}{\Gamma \left({\frac {5}{24}}\right)\,\Gamma \left({\frac {13}{24}}\right)\,\cdot \,\Gamma \left({\frac {7}{24}}\right)\,\Gamma \left({\frac {23}{24}}\right)}}}$.

Erweitert man nun mit ${\displaystyle \Gamma \left({\frac {15}{24}}\right)\Gamma \left({\frac {21}{24}}\right)}$, so ist ${\displaystyle A={\frac {2\pi ^{2}}{\sin \left({\frac {\pi }{12}}\right)}}\cdot {\frac {\Gamma \left({\frac {15}{24}}\right)\cdot \Gamma \left({\frac {21}{24}}\right)}{\Gamma \left({\frac {5}{24}}\right)\,\Gamma \left({\frac {13}{24}}\right)\,\Gamma \left({\frac {21}{24}}\right)\,\cdot \,\Gamma \left({\frac {7}{24}}\right)\,\Gamma \left({\frac {15}{24}}\right)\,\Gamma \left({\frac {23}{24}}\right)}}}$.

Nach der Gaußschen Multiplikationsformel ${\displaystyle \Gamma (z)\;\Gamma \left(z+{\frac {1}{3}}\right)\,\Gamma \left(z+{\frac {2}{3}}\right)={\frac {2\pi {\sqrt {3}}}{3^{3z}}}\,\,\Gamma (3z)}$ ist dabei

${\displaystyle \Gamma \left({\frac {5}{24}}\right)\Gamma \left({\frac {13}{24}}\right)\Gamma \left({\frac {21}{24}}\right)=\Gamma \left({\frac {5}{24}}\right)\Gamma \left({\frac {5}{24}}+{\frac {1}{3}}\right)\Gamma \left({\frac {5}{24}}+{\frac {2}{3}}\right)={\frac {2\pi {\sqrt {3}}}{3^{\frac {15}{24}}}}\;\,\Gamma \left({\frac {15}{24}}\right)}$ und

${\displaystyle \Gamma \left({\frac {7}{24}}\right)\Gamma \left({\frac {15}{24}}\right)\Gamma \left({\frac {23}{24}}\right)=\Gamma \left({\frac {7}{24}}\right)\Gamma \left({\frac {7}{24}}+{\frac {1}{3}}\right)\Gamma \left({\frac {7}{24}}+{\frac {2}{3}}\right)={\frac {2\pi {\sqrt {3}}}{3^{\frac {21}{24}}}}\;\,\Gamma \left({\frac {21}{24}}\right)}$.

Also ist ${\displaystyle A={\frac {2\pi ^{2}}{\sin \left({\frac {\pi }{12}}\right)}}\cdot {\frac {3^{\frac {15}{24}}\cdot 3^{\frac {21}{24}}}{2\pi {\sqrt {3}}\cdot 2\pi {\sqrt {3}}}}}$, und das ist ${\displaystyle {\frac {\sqrt {3}}{2\cdot \sin \left({\frac {\pi }{12}}\right)}}}$.

Nachdem ${\displaystyle \sin \left({\frac {\pi }{12}}\right)={\sqrt {\frac {1-\cos \left({\frac {\pi }{6}}\right)}{2}}}={\sqrt {\frac {1-{\frac {\sqrt {3}}{2}}}{2}}}}$ ist,

ist ${\displaystyle 2\cdot \sin \left({\frac {\pi }{12}}\right)={\sqrt {2-{\sqrt {3}}}}}$ und somit ${\displaystyle {\frac {1}{2\cdot \sin \left({\frac {\pi }{12}}\right)}}={\sqrt {2+{\sqrt {3}}}}}$ . Also ist ${\displaystyle A={\sqrt {3}}\cdot {\sqrt {2+{\sqrt {3}}}}}$.