Pengenalan kos(π/4) = sin(π/4) = √2/2, bersama perwakilan hasil darab tak terhingga bagi sin dan kosin membawa kepada hasil darab seperti

1 2 = ∏ k = 0 ∞ ( 1 − 1 ( 4 k + 2 ) 2 ) = ( 1 − 1 4 ) ( 1 − 1 36 ) ( 1 − 1 100 ) ⋯ {\displaystyle {\frac {1}{\sqrt {2}}}=\prod _{k=0}^{\infty }\left(1-{\frac {1}{(4k+2)^{2}}}\right)=\left(1-{\frac {1}{4}}\right)\left(1-{\frac {1}{36}}\right)\left(1-{\frac {1}{100}}\right)\cdots }

dan

2 = ∏ k = 0 ∞ ( 4 k + 2 ) 2 ( 4 k + 1 ) ( 4 k + 3 ) = ( 2 ⋅ 2 1 ⋅ 3 ) ( 6 ⋅ 6 5 ⋅ 7 ) ( 10 ⋅ 10 9 ⋅ 11 ) ( 14 ⋅ 14 13 ⋅ 15 ) ⋯ {\displaystyle {\sqrt {2}}=\prod _{k=0}^{\infty }{\frac {(4k+2)^{2}}{(4k+1)(4k+3)}}=\left({\frac {2\cdot 2}{1\cdot 3}}\right)\left({\frac {6\cdot 6}{5\cdot 7}}\right)\left({\frac {10\cdot 10}{9\cdot 11}}\right)\left({\frac {14\cdot 14}{13\cdot 15}}\right)\cdots }

atau bersamaan dengan,

2 = ∏ k = 0 ∞ ( 1 + 1 4 k + 1 ) ( 1 − 1 4 k + 3 ) = ( 1 + 1 1 ) ( 1 − 1 3 ) ( 1 + 1 5 ) ( 1 − 1 7 ) ⋯ . {\displaystyle {\sqrt {2}}=\prod _{k=0}^{\infty }\left(1+{\frac {1}{4k+1}}\right)\left(1-{\frac {1}{4k+3}}\right)=\left(1+{\frac {1}{1}}\right)\left(1-{\frac {1}{3}}\right)\left(1+{\frac {1}{5}}\right)\left(1-{\frac {1}{7}}\right)\cdots .}

Nombor tersebut boleh dinyatakan dengan mengambil siri Taylor bagi fungsi trigonometri. Contohnya, siri bagi kos(π/4) adalah

1 2 = ∑ k = 0 ∞ ( − 1 ) k ( π 4 ) 2 k ( 2 k ) ! . {\displaystyle {\frac {1}{\sqrt {2}}}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}\left({\frac {\pi }{4}}\right)^{2k}}{(2k)!}}.}

Siri Taylor bagi √(1+x) dengan x = 1 memberikan

2 = ∑ k = 0 ∞ ( − 1 ) k + 1 ( 2 k − 3 ) ! ! ( 2 k ) ! ! = 1 + 1 2 − 1 2 ⋅ 4 + 1 ⋅ 3 2 ⋅ 4 ⋅ 6 − 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ⋅ 8 + ⋯ . {\displaystyle {\sqrt {2}}=\sum _{k=0}^{\infty }(-1)^{k+1}{\frac {(2k-3)!!}{(2k)!!}}=1+{\frac {1}{2}}-{\frac {1}{2\cdot 4}}+{\frac {1\cdot 3}{2\cdot 4\cdot 6}}-{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 8}}+\cdots .}

Penumpuan siri ini boleh dicepatkan dengan penukaran Euler, menghasilkan

2 = ∑ k = 0 ∞ ( 2 k + 1 ) ! ( k ! ) 2 2 3 k + 1 = 1 2 + 3 8 + 15 64 + 35 256 + 315 4096 + 693 16384 + ⋯ . {\displaystyle {\sqrt {2}}=\sum _{k=0}^{\infty }{\frac {(2k+1)!}{(k!)^{2}2^{3k+1}}}={\frac {1}{2}}+{\frac {3}{8}}+{\frac {15}{64}}+{\frac {35}{256}}+{\frac {315}{4096}}+{\frac {693}{16384}}+\cdots .}

Tidak diketahui sama ada √2 boleh diwakilikan dengan rumus BBP-type. Rumus BBP-type digunakan untuk π√2 dan √2 ln(1+√2).