ចំនួនថេរត្រីកោណមាត្រពិត

${\displaystyle \sin 0=0\,}$

${\displaystyle \cos 0=1\,}$

${\displaystyle \tan 0=0\,}$

${\displaystyle \cot 0=\infty \,}$​មិនកំនត់

${\displaystyle \sin {\frac {\pi }{60}}=\sin 3^{\circ }={\frac {2(1-{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}+1)}{16}}\,}$

${\displaystyle \cos {\frac {\pi }{60}}=\cos 3^{\circ }={\frac {2(1+{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}-1)}{16}}\,}$

${\displaystyle \tan {\frac {\pi }{60}}=\tan 3^{\circ }={\frac {\left((2-{\sqrt {3}})(3+{\sqrt {5}})-2\right)\left(2-{\sqrt {2(5-{\sqrt {5}})}}\right)}{4}}\,}$

${\displaystyle \cot {\frac {\pi }{60}}=\cot 3^{\circ }={\frac {\left((2+{\sqrt {3}})(3+{\sqrt {5}})-2\right)\left(2+{\sqrt {2(5-{\sqrt {5}})}}\right)}{4}}\,}$

${\displaystyle \sin {\frac {\pi }{30}}=\sin 6^{\circ }={\frac {{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1}{8}}\,}$

${\displaystyle \cos {\frac {\pi }{30}}=\cos 6^{\circ }={\frac {{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)}{8}}\,}$

${\displaystyle \tan {\frac {\pi }{30}}=\tan 6^{\circ }={\frac {{\sqrt {2(5-{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)}{2}}\,}$

${\displaystyle \cot {\frac {\pi }{30}}=\cot 6^{\circ }={\frac {{\sqrt {3}}(3+{\sqrt {5}})+{\sqrt {2(25+11{\sqrt {5}})}}}{2}}\,}$

${\displaystyle \sin {\frac {\pi }{20}}=\sin 9^{\circ }={\frac {{\sqrt {2}}({\sqrt {5}}+1)-2{\sqrt {5-{\sqrt {5}}}}}{8}}\,}$

${\displaystyle \cos {\frac {\pi }{20}}=\cos 9^{\circ }={\frac {{\sqrt {2}}({\sqrt {5}}+1)+2{\sqrt {5-{\sqrt {5}}}}}{8}}\,}$

${\displaystyle \tan {\frac {\pi }{20}}=\tan 9^{\circ }={\sqrt {5}}+1-{\sqrt {5+2{\sqrt {5}}}}\,}$

${\displaystyle \cot {\frac {\pi }{20}}=\cot 9^{\circ }={\sqrt {5}}+1+{\sqrt {5+2{\sqrt {5}}}}\,}$

${\displaystyle \sin {\frac {\pi }{15}}=\sin 12^{\circ }={\frac {{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)}{8}}\,}$

${\displaystyle \cos {\frac {\pi }{15}}=\cos 12^{\circ }={\frac {{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1}{8}}\,}$

${\displaystyle \tan {\frac {\pi }{15}}=\tan 12^{\circ }={\frac {{\sqrt {3}}(3-{\sqrt {5}})-{\sqrt {2(25-11{\sqrt {5}})}}}{2}}\,}$

${\displaystyle \cot {\frac {\pi }{15}}=\cot 12^{\circ }={\frac {{\sqrt {3}}({\sqrt {5}}+1)+{\sqrt {2(5+{\sqrt {5}})}}}{2}}\,}$

${\displaystyle \sin {\frac {\pi }{12}}=\sin 15^{\circ }={\frac {{\sqrt {2}}({\sqrt {3}}-1)}{4}}\,}$

${\displaystyle \cos {\frac {\pi }{12}}=\cos 15^{\circ }={\frac {{\sqrt {2}}({\sqrt {3}}+1)}{4}}\,}$

${\displaystyle \tan {\frac {\pi }{12}}=\tan 15^{\circ }=2-{\sqrt {3}}\,}$

${\displaystyle \cot {\frac {\pi }{12}}=\cot 15^{\circ }=2+{\sqrt {3}}\,}$

${\displaystyle \sin {\frac {\pi }{10}}=\sin 18^{\circ }={\frac {{\sqrt {5}}-1}{4}}={\frac {\varphi -1}{2}}={\frac {1}{2\varphi }}\,}$

${\displaystyle \cos {\frac {\pi }{10}}=\cos 18^{\circ }={\frac {\sqrt {2(5+{\sqrt {5}})}}{4}}\,}$

${\displaystyle \tan {\frac {\pi }{10}}=\tan 18^{\circ }={\frac {\sqrt {5(5-2{\sqrt {5}})}}{5}}\,}$

${\displaystyle \cot {\frac {\pi }{10}}=\cot 18^{\circ }={\sqrt {5+2{\sqrt {5}}}}\,}$

${\displaystyle \sin {\frac {7\pi }{60}}=\sin 21^{\circ }={\frac {2({\sqrt {3}}+1){\sqrt {5-{\sqrt {5}}}}-{\sqrt {2}}({\sqrt {3}}-1)(1+{\sqrt {5}})}{16}}\,}$

${\displaystyle \cos {\frac {7\pi }{60}}=\cos 21^{\circ }={\frac {2({\sqrt {3}}-1){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {3}}+1)(1+{\sqrt {5}})}{16}}\,}$

${\displaystyle \tan {\frac {7\pi }{60}}=\tan 21^{\circ }={\frac {\left(2-(2+{\sqrt {3}})(3-{\sqrt {5}})\right)\left(2-{\sqrt {2(5+{\sqrt {5}})}}\right)}{4}}\,}$

${\displaystyle \cot {\frac {7\pi }{60}}=\cot 21^{\circ }={\frac {\left(2-(2-{\sqrt {3}})(3-{\sqrt {5}})\right)\left(2+{\sqrt {10{\sqrt {5}}}}\right)}{4}}\,}$

${\displaystyle \sin {\frac {\pi }{8}}=\sin 22.5^{\circ }={\frac {\sqrt {2-{\sqrt {2}}}}{2}}\,}$

${\displaystyle \cos {\frac {\pi }{8}}=\cos 22.5^{\circ }={\frac {\sqrt {2+{\sqrt {2}}}}{2}}\,}$

${\displaystyle \tan {\frac {\pi }{8}}=\tan 22.5^{\circ }={\sqrt {2}}-1\,}$

${\displaystyle \cot {\frac {\pi }{8}}=\cot 22.5^{\circ }={\sqrt {2}}+1\,}$

${\displaystyle \sin {\frac {2\pi }{15}}=\sin 24^{\circ }={\frac {{\sqrt {3}}({\sqrt {5}}+1)-{\sqrt {2}}{\sqrt {5-{\sqrt {5}}}}}{8}}\,}$

${\displaystyle \cos {\frac {2\pi }{15}}=\cos 24^{\circ }={\frac {{\sqrt {6}}{\sqrt {5-{\sqrt {5}}}}+{\sqrt {5}}+1}{8}}\,}$

${\displaystyle \tan {\frac {2\pi }{15}}=\tan 24^{\circ }={\frac {{\sqrt {50+22{\sqrt {5}}}}-{\sqrt {3}}(3+{\sqrt {5}})}{2}}\,}$

${\displaystyle \cot {\frac {2\pi }{15}}=\cot 24^{\circ }={\frac {{\sqrt {2}}{\sqrt {5-{\sqrt {5}}}}+{\sqrt {3}}({\sqrt {5}}-1)}{2}}\,}$

${\displaystyle \sin {\frac {3\pi }{20}}=\sin 27^{\circ }={\frac {({\sqrt {5}}+1){\sqrt {5+{\sqrt {5}}}}-{\sqrt {2}}({\sqrt {5}}-1)}{8}}\,}$

${\displaystyle \cos {\frac {3\pi }{20}}=\cos 27^{\circ }={\frac {({\sqrt {5}}+1){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)}{8}}\,}$

${\displaystyle \tan {\frac {3\pi }{20}}=\tan 27^{\circ }={\sqrt {5}}-1-{\sqrt {5-2{\sqrt {5}}}}\,}$

${\displaystyle \cot {\frac {3\pi }{20}}=\cot 27^{\circ }={\sqrt {5}}-1+{\sqrt {5-2{\sqrt {5}}}}\,}$

${\displaystyle \sin {\frac {\pi }{6}}=\sin 30^{\circ }={\frac {1}{2}}\,}$

${\displaystyle \cos {\frac {\pi }{6}}=\cos 30^{\circ }={\frac {\sqrt {3}}{2}}\,}$

${\displaystyle \tan {\frac {\pi }{6}}=\tan 30^{\circ }={\frac {\sqrt {3}}{3}}\,}$

${\displaystyle \cot {\frac {\pi }{6}}=\cot 30^{\circ }={\sqrt {3}}\,}$

${\displaystyle \sin {\frac {11\pi }{60}}=\sin 33^{\circ }={\frac {2({\sqrt {3}}-1){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}(1+{\sqrt {3}})({\sqrt {5}}-1)}{16}}\,}$

${\displaystyle \cos {\frac {11\pi }{60}}=\cos 33^{\circ }={\frac {2({\sqrt {3}}+1){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}(1-{\sqrt {3}})({\sqrt {5}}-1)}{16}}\,}$

${\displaystyle \tan {\frac {11\pi }{60}}=\tan 33^{\circ }={\frac {\left(2-(2-{\sqrt {3}})(3+{\sqrt {5}})\right)\left(2+{\sqrt {2(5-{\sqrt {5}})}}\right)}{4}}\,}$

${\displaystyle \cot {\frac {11\pi }{60}}=\cot 33^{\circ }={\frac {\left(2-(2+{\sqrt {3}})(3+{\sqrt {5}})\right)\left(2-{\sqrt {2(5-{\sqrt {5}})}}\right)}{4}}\,}$

${\displaystyle \sin {\frac {\pi }{5}}=\sin 36^{\circ }={\frac {\sqrt {2(5-{\sqrt {5}})}}{4}}\,}$

${\displaystyle \cos {\frac {\pi }{5}}=\cos 36^{\circ }={\frac {1+{\sqrt {5}}}{4}}={\frac {\varphi }{2}}\,}$

${\displaystyle \tan {\frac {\pi }{5}}=\tan 36^{\circ }={\sqrt {5-2{\sqrt {5}}}}\,}$

${\displaystyle \cot {\frac {\pi }{5}}=\cot 36^{\circ }={\frac {\sqrt {5(5+2{\sqrt {5}})}}{5}}\,}$

${\displaystyle \sin {\frac {13\pi }{60}}=\sin 39^{\circ }={\frac {2(1-{\sqrt {3}}){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {3}}+1)({\sqrt {5}}+1)}{16}}\,}$

${\displaystyle \cos {\frac {13\pi }{60}}=\cos 39^{\circ }={\frac {2(1+{\sqrt {3}}){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {3}}-1)({\sqrt {5}}+1)}{16}}\,}$

${\displaystyle \tan {\frac {13\pi }{60}}=\tan 39^{\circ }={\frac {\left((2-{\sqrt {3}})(3-{\sqrt {5}})-2\right)\left(2-{\sqrt {2(5+{\sqrt {5}})}}\right)}{4}}\,}$

${\displaystyle \cot {\frac {13\pi }{60}}=\cot 39^{\circ }={\frac {\left((2+{\sqrt {3}})(3-{\sqrt {5}})-2\right)\left(2+{\sqrt {2(5+{\sqrt {5}})}}\right)}{4}}\,}$

${\displaystyle \sin {\frac {7\pi }{30}}=\sin 42^{\circ }={\frac {{\sqrt {6}}{\sqrt {5+{\sqrt {5}}}}-{\sqrt {5}}+1}{8}}\,}$

${\displaystyle \cos {\frac {7\pi }{30}}=\cos 42^{\circ }={\frac {{\sqrt {2}}{\sqrt {5+{\sqrt {5}}}}+{\sqrt {3}}({\sqrt {5}}-1)}{8}}\,}$

${\displaystyle \tan {\frac {7\pi }{30}}=\tan 42^{\circ }={\frac {{\sqrt {3}}({\sqrt {5}}+1)-{\sqrt {2}}{\sqrt {5+{\sqrt {5}}}}}{2}}\,}$

${\displaystyle \cot {\frac {7\pi }{30}}=\cot 42^{\circ }={\frac {{\sqrt {2(25-11{\sqrt {5}})}}+{\sqrt {3}}(3-{\sqrt {5}})}{2}}\,}$

${\displaystyle \sin {\frac {\pi }{4}}=\sin 45^{\circ }={\frac {\sqrt {2}}{2}}\,}$

${\displaystyle \cos {\frac {\pi }{4}}=\cos 45^{\circ }={\frac {\sqrt {2}}{2}}\,}$

${\displaystyle \tan {\frac {\pi }{4}}=\tan 45^{\circ }=1\,}$

${\displaystyle \cot {\frac {\pi }{4}}=\cot 45^{\circ }=1\,}$

${\displaystyle \sin {\frac {\pi }{3}}=\sin 60^{\circ }={\frac {\sqrt {3}}{2}}\,}$

${\displaystyle \cos {\frac {\pi }{3}}=\cos 60^{\circ }={\frac {1}{2}}\,}$

${\displaystyle \tan {\frac {\pi }{3}}=\tan 60^{\circ }={\sqrt {3}}\,}$

${\displaystyle \cot {\frac {\pi }{3}}=\cot 60^{\circ }={\frac {\sqrt {3}}{3}}\,}$

មាឌនៃសូលីតដែលបង្កើតដោយពហុកោណមានជ្រុង៥(បញ្ចកោណ) ហើយ ${\displaystyle a\,}$ ជាប្រវែងនៃជ្រុងរបស់បញ្ចកោណ

សំដែងដោយ

${\displaystyle V={\frac {5a^{3}\cos {36^{\circ }}}{\tan ^{2}{36^{\circ }}}}}$

ដោយប្រើ

${\displaystyle \cos 36^{\circ }={\frac {{\sqrt {5}}+1}{4}}\,}$

${\displaystyle \tan 36^{\circ }={\sqrt {5-2{\sqrt {5}}}}\,}$

វាក្លាយទៅជា

${\displaystyle V={\frac {a^{3}(15+7{\sqrt {5}})}{4}}\,}$