以下恆等式成立：
( 1 + 5 2 ) − π 2 ( 3 + 5 2 ) π − 2 4 = 5 − 1 2 {\displaystyle {{\bigg (}{\frac {1+{\sqrt {5}}}{2}}{\bigg )}}^{\frac {-\pi }{2}}{{\bigg (}{\frac {3+{\sqrt {5}}}{2}}{\bigg )}}^{\frac {\pi -2}{4}}={\frac {{\sqrt {5}}-1}{2}}}
