I en regulær n -kant er a = π/n den halve sentralvinkel og b = π(1/2 - 1/n) den halve, indre vinkel . Eksakte trigonometriske konstanter er eksakte verdier som brukes for å uttrykke vinkler nøyaktig. Alle konstantene er utledet fra forholdet mellom to sider i en trekant.
Alle eksakte verdier av sinus, cosinus og tangens til vinkler med 3-graders inkrementer er det mulig å utlede ved å bruke identitetene for halve vinkler, dobbelte vinkler og sum/differanse med verdiene for 0°, 30°, 36°, og 45°. Det tilsvarer at de er konstruerbare tall og basert på konstruksjon av regulære mangekanter . Disse spesielle vinklene som er listet, er de halve sentralvinklene i de tilsvarende mangekantene. Det er kun mulig å finne eksakte verdier for vinkler på formen m π {\displaystyle \pi } / n heltall slik at det går an å konstruere et polygoner med n eller m sider.
Konstantene oppgis på eksakt form, dvs. ved hjelp av røtter og brøker , uten avrunding til desimaltall , som kan lede til unøyaktigheter dersom man bruker de i videre beregninger. Mange av verdiene er irrasjonelle . Dersom man evaulerer funksjonene sin x {\displaystyle \sin x} cos x {\displaystyle \cos x} 1 / 2
Velkjente konstanter Eksakte verdier på formen ( cos θ , sin θ ) {\displaystyle (\cos \theta ,\sin \theta )} enhetssirkelen ; alle disse er et multiplum av 30° og 45° ( π {\displaystyle \pi } / 6 π {\displaystyle \pi } / 4 Følgende konstanter kan utledes for verdier ut fra en sekstendeling av enhetssirkelen ; disse gjelder for verdiene man får av å dele en sirkel i åtte eller tolv like deler. Én hel omdreining er gitt ved 360° eller 2 π {\displaystyle 2\pi }
Dreining Grader Radianer Sinus Cosinus Tangens 0 0° 0 0 1 0 1 / 12 30° π {\displaystyle \pi } / 6 1 / 2 √3 / 2 √3 / 3 1 / 8 45° π {\displaystyle \pi } / 4 √2 / 2 √2 / 2 1 1 / 6 60° π {\displaystyle \pi } / 3 √3 / 2 1 / 2 √3 1 / 4 90° π {\displaystyle \pi } / 2 1 0 1 / 3 120° 2 π {\displaystyle \pi } / 3 √3 / 2 −1 / 2 −√3 3 / 8 135° 3 π {\displaystyle \pi } / 4 √2 / 2 −√2 / 2 −1 5 / 12 150° 5 π {\displaystyle \pi } / 6 1 / 2 −√3 / 2 −√3 / 3 1 / 2 180° π {\displaystyle \pi } 0 −1 0 7 / 12 210° 7 π {\displaystyle \pi } / 6 −1 / 2 −√3 / 2 √3 / 3 5 / 8 225° 5 π {\displaystyle \pi } / 4 −√2 / 2 −√2 / 2 1 2 / 3 240° 4 π {\displaystyle \pi } / 3 −√3 / 2 −1 / 2 √3 3 / 4 270° 3 π {\displaystyle \pi } / 2 −1 0 5 / 6 300° 5 π {\displaystyle \pi } / 3 −√3 / 2 1 / 2 −√3 7 / 8 315° 7 π {\displaystyle \pi } / 4 −√2 / 2 √2 / 2 −1 11 / 12 330° 11 π {\displaystyle \pi } / 6 −1 / 2 √3 / 2 −√3 / 3 1 360° 2 π {\displaystyle \pi } 0 1 0
Andre verdier Verdier for vinkler utenfor området [0°, 45°] kan utledes fra disse verdiene ved bruk av formlene for symmetri i trigonometriske identiteter . Merk at 1° = π/180 radianer .
0°: fundamental sin 0 = 0 {\displaystyle \sin 0=0\,} cos 0 = 1 {\displaystyle \cos 0=1\,} tan 0 = 0 {\displaystyle \tan 0=0\,} cot 0 er undefinert {\displaystyle \cot 0{\mbox{ er undefinert}}\,} 3°: 60-sidet polygon sin π 60 = sin 3 ∘ = 1 16 [ 2 ( 1 − 3 ) 5 + 5 + 2 ( 5 − 1 ) ( 3 + 1 ) ] {\displaystyle \sin {\frac {\pi }{60}}=\sin 3^{\circ }={\tfrac {1}{16}}\left[2(1-{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}+1)\right]\,} cos π 60 = cos 3 ∘ = 1 16 [ 2 ( 1 + 3 ) 5 + 5 + 2 ( 5 − 1 ) ( 3 − 1 ) ] {\displaystyle \cos {\frac {\pi }{60}}=\cos 3^{\circ }={\tfrac {1}{16}}\left[2(1+{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}-1)\right]\,} tan π 60 = tan 3 ∘ = 1 4 [ ( 2 − 3 ) ( 3 + 5 ) − 2 ] [ 2 − 2 ( 5 − 5 ) ] {\displaystyle \tan {\frac {\pi }{60}}=\tan 3^{\circ }={\tfrac {1}{4}}\left[(2-{\sqrt {3}})(3+{\sqrt {5}})-2\right]\left[2-{\sqrt {2(5-{\sqrt {5}})}}\right]\,} cot π 60 = cot 3 ∘ = 1 4 [ ( 2 + 3 ) ( 3 + 5 ) − 2 ] [ 2 + 2 ( 5 − 5 ) ] {\displaystyle \cot {\frac {\pi }{60}}=\cot 3^{\circ }={\tfrac {1}{4}}\left[(2+{\sqrt {3}})(3+{\sqrt {5}})-2\right]\left[2+{\sqrt {2(5-{\sqrt {5}})}}\right]\,} 6°: 30-sidet polygon sin π 30 = sin 6 ∘ = 1 8 [ 6 ( 5 − 5 ) − 5 − 1 ] {\displaystyle \sin {\frac {\pi }{30}}=\sin 6^{\circ }={\tfrac {1}{8}}\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]\,} cos π 30 = cos 6 ∘ = 1 8 [ 2 ( 5 − 5 ) + 3 ( 5 + 1 ) ] {\displaystyle \cos {\frac {\pi }{30}}=\cos 6^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]\,} tan π 30 = tan 6 ∘ = 1 2 [ 2 ( 5 − 5 ) − 3 ( 5 − 1 ) ] {\displaystyle \tan {\frac {\pi }{30}}=\tan 6^{\circ }={\tfrac {1}{2}}\left[{\sqrt {2(5-{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]\,} cot π 30 = cot 6 ∘ = 1 2 [ 3 ( 3 + 5 ) + 2 ( 25 + 11 5 ) ] {\displaystyle \cot {\frac {\pi }{30}}=\cot 6^{\circ }={\tfrac {1}{2}}\left[{\sqrt {3}}(3+{\sqrt {5}})+{\sqrt {2(25+11{\sqrt {5}})}}\right]\,} 9°: 20-sidet polygon sin π 20 = sin 9 ∘ = 1 8 [ 2 ( 5 + 1 ) − 2 5 − 5 ] {\displaystyle \sin {\frac {\pi }{20}}=\sin 9^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2}}({\sqrt {5}}+1)-2{\sqrt {5-{\sqrt {5}}}}\right]\,} cos π 20 = cos 9 ∘ = 1 8 [ 2 ( 5 + 1 ) + 2 5 − 5 ] {\displaystyle \cos {\frac {\pi }{20}}=\cos 9^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2}}({\sqrt {5}}+1)+2{\sqrt {5-{\sqrt {5}}}}\right]\,} tan π 20 = tan 9 ∘ = 5 + 1 − 5 + 2 5 {\displaystyle \tan {\frac {\pi }{20}}=\tan 9^{\circ }={\sqrt {5}}+1-{\sqrt {5+2{\sqrt {5}}}}\,} cot π 20 = cot 9 ∘ = 5 + 1 + 5 + 2 5 {\displaystyle \cot {\frac {\pi }{20}}=\cot 9^{\circ }={\sqrt {5}}+1+{\sqrt {5+2{\sqrt {5}}}}\,} 12°: 15-sidet polygon sin π 15 = sin 12 ∘ = 1 8 [ 2 ( 5 + 5 ) − 3 ( 5 − 1 ) ] {\displaystyle \sin {\frac {\pi }{15}}=\sin 12^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]\,} cos π 15 = cos 12 ∘ = 1 8 [ 6 ( 5 + 5 ) + 5 − 1 ] {\displaystyle \cos {\frac {\pi }{15}}=\cos 12^{\circ }={\tfrac {1}{8}}\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]\,} tan π 15 = tan 12 ∘ = 1 2 [ 3 ( 3 − 5 ) − 2 ( 25 − 11 5 ) ] {\displaystyle \tan {\frac {\pi }{15}}=\tan 12^{\circ }={\tfrac {1}{2}}\left[{\sqrt {3}}(3-{\sqrt {5}})-{\sqrt {2(25-11{\sqrt {5}})}}\right]\,} cot π 15 = cot 12 ∘ = 1 2 [ 3 ( 5 + 1 ) + 2 ( 5 + 5 ) ] {\displaystyle \cot {\frac {\pi }{15}}=\cot 12^{\circ }={\tfrac {1}{2}}\left[{\sqrt {3}}({\sqrt {5}}+1)+{\sqrt {2(5+{\sqrt {5}})}}\right]\,} 15°: dodekagon sin π 12 = sin 15 ∘ = 1 4 2 ( 3 − 1 ) {\displaystyle \sin {\frac {\pi }{12}}=\sin 15^{\circ }={\tfrac {1}{4}}{\sqrt {2}}({\sqrt {3}}-1)\,} cos π 12 = cos 15 ∘ = 1 4 2 ( 3 + 1 ) {\displaystyle \cos {\frac {\pi }{12}}=\cos 15^{\circ }={\tfrac {1}{4}}{\sqrt {2}}({\sqrt {3}}+1)\,} tan π 12 = tan 15 ∘ = 2 − 3 {\displaystyle \tan {\frac {\pi }{12}}=\tan 15^{\circ }=2-{\sqrt {3}}\,} cot π 12 = cot 15 ∘ = 2 + 3 {\displaystyle \cot {\frac {\pi }{12}}=\cot 15^{\circ }=2+{\sqrt {3}}\,} 18°: dekagon sin π 10 = sin 18 ∘ = 1 4 ( 5 − 1 ) = 1 2 φ − 1 {\displaystyle \sin {\frac {\pi }{10}}=\sin 18^{\circ }={\tfrac {1}{4}}\left({\sqrt {5}}-1\right)={\tfrac {1}{2}}\varphi ^{-1}\,} cos π 10 = cos 18 ∘ = 1 4 2 ( 5 + 5 ) {\displaystyle \cos {\frac {\pi }{10}}=\cos 18^{\circ }={\tfrac {1}{4}}{\sqrt {2(5+{\sqrt {5}})}}\,} tan π 10 = tan 18 ∘ = 1 5 5 ( 5 − 2 5 ) {\displaystyle \tan {\frac {\pi }{10}}=\tan 18^{\circ }={\tfrac {1}{5}}{\sqrt {5(5-2{\sqrt {5}})}}\,} cot π 10 = cot 18 ∘ = 5 + 2 5 {\displaystyle \cot {\frac {\pi }{10}}=\cot 18^{\circ }={\sqrt {5+2{\sqrt {5}}}}\,} 21°: summen 9° + 12° sin 7 π 60 = sin 21 ∘ = 1 16 [ 2 ( 3 + 1 ) 5 − 5 − 2 ( 3 − 1 ) ( 1 + 5 ) ] {\displaystyle \sin {\frac {7\pi }{60}}=\sin 21^{\circ }={\tfrac {1}{16}}\left[2({\sqrt {3}}+1){\sqrt {5-{\sqrt {5}}}}-{\sqrt {2}}({\sqrt {3}}-1)(1+{\sqrt {5}})\right]\,} cos 7 π 60 = cos 21 ∘ = 1 16 [ 2 ( 3 − 1 ) 5 − 5 + 2 ( 3 + 1 ) ( 1 + 5 ) ] {\displaystyle \cos {\frac {7\pi }{60}}=\cos 21^{\circ }={\tfrac {1}{16}}\left[2({\sqrt {3}}-1){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {3}}+1)(1+{\sqrt {5}})\right]\,} tan 7 π 60 = tan 21 ∘ = 1 4 [ 2 − ( 2 + 3 ) ( 3 − 5 ) ] [ 2 − 2 ( 5 + 5 ) ] {\displaystyle \tan {\frac {7\pi }{60}}=\tan 21^{\circ }={\tfrac {1}{4}}\left[2-(2+{\sqrt {3}})(3-{\sqrt {5}})\right]\left[2-{\sqrt {2(5+{\sqrt {5}})}}\right]\,} cot 7 π 60 = cot 21 ∘ = 1 4 [ 2 − ( 2 − 3 ) ( 3 − 5 ) ] [ 2 + 2 ( 5 + 5 ) ] {\displaystyle \cot {\frac {7\pi }{60}}=\cot 21^{\circ }={\tfrac {1}{4}}\left[2-(2-{\sqrt {3}})(3-{\sqrt {5}})\right]\left[2+{\sqrt {2(5+{\sqrt {5}})}}\right]\,} 22.5°: oktogon sin π 8 = sin 22.5 ∘ = 1 2 ( 2 − 2 ) , {\displaystyle \sin {\frac {\pi }{8}}=\sin 22.5^{\circ }={\tfrac {1}{2}}({\sqrt {2-{\sqrt {2}}}}),} cos π 8 = cos 22.5 ∘ = 1 2 ( 2 + 2 ) {\displaystyle \cos {\frac {\pi }{8}}=\cos 22.5^{\circ }={\tfrac {1}{2}}({\sqrt {2+{\sqrt {2}}}})\,} tan π 8 = tan 22.5 ∘ = 2 − 1 {\displaystyle \tan {\frac {\pi }{8}}=\tan 22.5^{\circ }={\sqrt {2}}-1\,} cot π 8 = cot 22.5 ∘ = 2 + 1 {\displaystyle \cot {\frac {\pi }{8}}=\cot 22.5^{\circ }={\sqrt {2}}+1\,} 24°: summen 12° + 12° sin 2 π 15 = sin 24 ∘ = 1 8 [ 3 ( 5 + 1 ) − 2 5 − 5 ] {\displaystyle \sin {\frac {2\pi }{15}}=\sin 24^{\circ }={\tfrac {1}{8}}\left[{\sqrt {3}}({\sqrt {5}}+1)-{\sqrt {2}}{\sqrt {5-{\sqrt {5}}}}\right]\,} cos 2 π 15 = cos 24 ∘ = 1 8 ( 6 5 − 5 + 5 + 1 ) {\displaystyle \cos {\frac {2\pi }{15}}=\cos 24^{\circ }={\tfrac {1}{8}}\left({\sqrt {6}}{\sqrt {5-{\sqrt {5}}}}+{\sqrt {5}}+1\right)\,} tan 2 π 15 = tan 24 ∘ = 1 2 [ 2 ( 25 + 11 5 ) − 3 ( 3 + 5 ) ] {\displaystyle \tan {\frac {2\pi }{15}}=\tan 24^{\circ }={\tfrac {1}{2}}\left[{\sqrt {2(25+11{\sqrt {5}})}}-{\sqrt {3}}(3+{\sqrt {5}})\right]\,} cot 2 π 15 = cot 24 ∘ = 1 2 [ 2 5 − 5 + 3 ( 5 − 1 ) ] {\displaystyle \cot {\frac {2\pi }{15}}=\cot 24^{\circ }={\tfrac {1}{2}}\left[{\sqrt {2}}{\sqrt {5-{\sqrt {5}}}}+{\sqrt {3}}({\sqrt {5}}-1)\right]\,} 27°: summen 12° + 15° sin 3 π 20 = sin 27 ∘ = 1 8 [ 2 5 + 5 − 2 ( 5 − 1 ) ] {\displaystyle \sin {\frac {3\pi }{20}}=\sin 27^{\circ }={\tfrac {1}{8}}\left[2{\sqrt {5+{\sqrt {5}}}}-{\sqrt {2}}\;({\sqrt {5}}-1)\right]\,} cos 3 π 20 = cos 27 ∘ = 1 8 [ 2 5 + 5 + 2 ( 5 − 1 ) ] {\displaystyle \cos {\frac {3\pi }{20}}=\cos 27^{\circ }={\tfrac {1}{8}}\left[2{\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}\;({\sqrt {5}}-1)\right]\,} tan 3 π 20 = tan 27 ∘ = 5 − 1 − 5 − 2 5 {\displaystyle \tan {\frac {3\pi }{20}}=\tan 27^{\circ }={\sqrt {5}}-1-{\sqrt {5-2{\sqrt {5}}}}\,} cot 3 π 20 = cot 27 ∘ = 5 − 1 + 5 − 2 5 {\displaystyle \cot {\frac {3\pi }{20}}=\cot 27^{\circ }={\sqrt {5}}-1+{\sqrt {5-2{\sqrt {5}}}}\,} 30°: heksagon sin π 6 = sin 30 ∘ = 1 2 {\displaystyle \sin {\frac {\pi }{6}}=\sin 30^{\circ }={\tfrac {1}{2}}\,} cos π 6 = cos 30 ∘ = 1 2 3 {\displaystyle \cos {\frac {\pi }{6}}=\cos 30^{\circ }={\tfrac {1}{2}}{\sqrt {3}}\,} tan π 6 = tan 30 ∘ = 1 3 3 {\displaystyle \tan {\frac {\pi }{6}}=\tan 30^{\circ }={\tfrac {1}{3}}{\sqrt {3}}\,} cot π 6 = cot 30 ∘ = 3 {\displaystyle \cot {\frac {\pi }{6}}=\cot 30^{\circ }={\sqrt {3}}\,} 33°: summen 15° + 18° sin 11 π 60 = sin 33 ∘ = 1 16 [ 2 ( 3 − 1 ) 5 + 5 + 2 ( 1 + 3 ) ( 5 − 1 ) ] {\displaystyle \sin {\frac {11\pi }{60}}=\sin 33^{\circ }={\tfrac {1}{16}}\left[2({\sqrt {3}}-1){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}(1+{\sqrt {3}})({\sqrt {5}}-1)\right]\,} cos 11 π 60 = cos 33 ∘ = 1 16 [ 2 ( 3 + 1 ) 5 + 5 + 2 ( 1 − 3 ) ( 5 − 1 ) ] {\displaystyle \cos {\frac {11\pi }{60}}=\cos 33^{\circ }={\tfrac {1}{16}}\left[2({\sqrt {3}}+1){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}(1-{\sqrt {3}})({\sqrt {5}}-1)\right]\,} tan 11 π 60 = tan 33 ∘ = 1 4 [ 2 − ( 2 − 3 ) ( 3 + 5 ) ] [ 2 + 2 ( 5 − 5 ) ] {\displaystyle \tan {\frac {11\pi }{60}}=\tan 33^{\circ }={\tfrac {1}{4}}\left[2-(2-{\sqrt {3}})(3+{\sqrt {5}})\right]\left[2+{\sqrt {2(5-{\sqrt {5}})}}\right]\,} cot 11 π 60 = cot 33 ∘ = 1 4 [ 2 − ( 2 + 3 ) ( 3 + 5 ) ] [ 2 − 2 ( 5 − 5 ) ] {\displaystyle \cot {\frac {11\pi }{60}}=\cot 33^{\circ }={\tfrac {1}{4}}\left[2-(2+{\sqrt {3}})(3+{\sqrt {5}})\right]\left[2-{\sqrt {2(5-{\sqrt {5}})}}\right]\,} 36°: pentagon sin π 5 = sin 36 ∘ = 1 4 [ 2 ( 5 − 5 ) ] {\displaystyle \sin {\frac {\pi }{5}}=\sin 36^{\circ }={\tfrac {1}{4}}[{\sqrt {2(5-{\sqrt {5}})}}]\,} cos π 5 = cos 36 ∘ = 1 + 5 4 = 1 2 φ {\displaystyle \cos {\frac {\pi }{5}}=\cos 36^{\circ }={\frac {1+{\sqrt {5}}}{4}}={\tfrac {1}{2}}\varphi \,} tan π 5 = tan 36 ∘ = 5 − 2 5 {\displaystyle \tan {\frac {\pi }{5}}=\tan 36^{\circ }={\sqrt {5-2{\sqrt {5}}}}\,} cot π 5 = cot 36 ∘ = 1 5 [ 5 ( 5 + 2 5 ) ] {\displaystyle \cot {\frac {\pi }{5}}=\cot 36^{\circ }={\tfrac {1}{5}}[{\sqrt {5(5+2{\sqrt {5}})}}]\,} 39°: summen 18° + 21° sin 13 π 60 = sin 39 ∘ = 1 16 [ 2 ( 1 − 3 ) 5 − 5 + 2 ( 3 + 1 ) ( 5 + 1 ) ] {\displaystyle \sin {\frac {13\pi }{60}}=\sin 39^{\circ }={\tfrac {1}{16}}[2(1-{\sqrt {3}}){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {3}}+1)({\sqrt {5}}+1)]\,} cos 13 π 60 = cos 39 ∘ = 1 16 [ 2 ( 1 + 3 ) 5 − 5 + 2 ( 3 − 1 ) ( 5 + 1 ) ] {\displaystyle \cos {\frac {13\pi }{60}}=\cos 39^{\circ }={\tfrac {1}{16}}[2(1+{\sqrt {3}}){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {3}}-1)({\sqrt {5}}+1)]\,} tan 13 π 60 = tan 39 ∘ = 1 4 [ ( 2 − 3 ) ( 3 − 5 ) − 2 ] [ 2 − 2 ( 5 + 5 ) ] {\displaystyle \tan {\frac {13\pi }{60}}=\tan 39^{\circ }={\tfrac {1}{4}}\left[(2-{\sqrt {3}})(3-{\sqrt {5}})-2\right]\left[2-{\sqrt {2(5+{\sqrt {5}})}}\right]\,} cot 13 π 60 = cot 39 ∘ = 1 4 [ ( 2 + 3 ) ( 3 − 5 ) − 2 ] [ 2 + 2 ( 5 + 5 ) ] {\displaystyle \cot {\frac {13\pi }{60}}=\cot 39^{\circ }={\tfrac {1}{4}}\left[(2+{\sqrt {3}})(3-{\sqrt {5}})-2\right]\left[2+{\sqrt {2(5+{\sqrt {5}})}}\right]\,} 42°: summen 21° + 21° sin 7 π 30 = sin 42 ∘ = 6 5 + 5 − 5 + 1 8 {\displaystyle \sin {\frac {7\pi }{30}}=\sin 42^{\circ }={\frac {{\sqrt {6}}{\sqrt {5+{\sqrt {5}}}}-{\sqrt {5}}+1}{8}}\,} cos 7 π 30 = cos 42 ∘ = 2 5 + 5 + 3 ( 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}}\,} tan 7 π 30 = tan 42 ∘ = 3 ( 5 + 1 ) − 2 5 + 5 2 {\displaystyle \tan {\frac {7\pi }{30}}=\tan 42^{\circ }={\frac {{\sqrt {3}}({\sqrt {5}}+1)-{\sqrt {2}}{\sqrt {5+{\sqrt {5}}}}}{2}}\,} cot 7 π 30 = cot 42 ∘ = 2 ( 25 − 11 5 ) + 3 ( 3 − 5 ) 2 {\displaystyle \cot {\frac {7\pi }{30}}=\cot 42^{\circ }={\frac {{\sqrt {2(25-11{\sqrt {5}})}}+{\sqrt {3}}(3-{\sqrt {5}})}{2}}\,} 45°: kvadrat sin π 4 = sin 45 ∘ = 2 2 = 1 2 {\displaystyle \sin {\frac {\pi }{4}}=\sin 45^{\circ }={\frac {\sqrt {2}}{2}}={\frac {1}{\sqrt {2}}}\,} cos π 4 = cos 45 ∘ = 2 2 = 1 2 {\displaystyle \cos {\frac {\pi }{4}}=\cos 45^{\circ }={\frac {\sqrt {2}}{2}}={\frac {1}{\sqrt {2}}}\,} tan π 4 = tan 45 ∘ = 1 {\displaystyle \tan {\frac {\pi }{4}}=\tan 45^{\circ }=1\,} cot π 4 = cot 45 ∘ = 1 {\displaystyle \cot {\frac {\pi }{4}}=\cot 45^{\circ }=1\,} 60°: trekant sin π 3 = sin 60 ∘ = 1 2 3 {\displaystyle \sin {\frac {\pi }{3}}=\sin 60^{\circ }={\tfrac {1}{2}}{\sqrt {3}}\,} cos π 3 = cos 60 ∘ = 1 2 {\displaystyle \cos {\frac {\pi }{3}}=\cos 60^{\circ }={\tfrac {1}{2}}\,} tan π 3 = tan 60 ∘ = 3 {\displaystyle \tan {\frac {\pi }{3}}=\tan 60^{\circ }={\sqrt {3}}\,} cot π 3 = cot 60 ∘ = 1 3 3 {\displaystyle \cot {\frac {\pi }{3}}=\cot 60^{\circ }={\tfrac {1}{3}}{\sqrt {3}}\,} der φ {\displaystyle \varphi } det gylne snitt .
Se også Litteratur (en) Eric W. Weisstein , Constructible polygon i MathWorld .(en) Eric W. Weisstein , Trigonometry angles i MathWorld .π/3 (60°) — π/6 (30°) — π/12 (15°) — π/24 (7.5°) π/4 (45°) — π/8 (22.5°) — π/16 (11.25°) — π/32 (5.625°) π/5 (36°) — π/10 (18°) — π/20 (9°) π/7 — π/14 π/9 (20°) — π/18 (10°) π/11 π/13 π/15 (12°) — π/30 (6°) π/17 π/19 π/23 (en) Eric W. Weisstein , Niven's Theorem i MathWorld .Bracken, Paul; Cizek, Jiri (2002). «Evaluation of quantum mechanical perturbation sums in terms of quadratic surds and their use in approximation of zeta(3)/pi^3». Int. J. Quantum Chemistry (1 utg.). 90: 42–53. doi :10.1002/qua.1803. CS1-vedlikehold: Flere navn: forfatterliste (link) Conway, John H.; Radin, Charles; Radun, Lorenzo (1998). «On angles whose squared trigonometric functions are rational». . CS1-vedlikehold: Flere navn: forfatterliste (link) [arXiv] Conway, John H.; Radin, Charles; Radun Lorenzo (1999). «On angles whose squared trigonometric functions are rational». Disc. Comput. Geom. (3 utg.). 22: 321–332. doi :10.1007/PL00009463. CS1-vedlikehold: Flere navn: forfatterliste (link) MR 1706614Girstmair, Kurt (1997). «Some linear relations between values of trigonometric functions at k*pi/n». Acta Arithmetica . 81: 387–398. MR 1472818Gurak, S. (2006). «On the minimal polynomial of gauss periods for prime powers». Mathematics of Computation (256 utg.). 75: 2021–2035. Bibcode :2006MaCom..75.2021G. doi :10.1090/S0025-5718-06-01885-0. MR 2240647Servi, L. D. (2003). «Nested square roots of 2». Am. Math. Monthly (4 utg.). 110: 326–330. doi :10.2307/3647881. MR 1984573JSTOR 3647881Eksterne lenker Constructible Regular Polygons Naming polygons Decimal expansion of sine of * degrees, OEIS Oppslagsverk/autoritetsdata MathWorld
![{\displaystyle \sin {\frac {\pi }{60}}=\sin 3^{\circ }={\tfrac {1}{16}}\left[2(1-{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}+1)\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c6d0ff8a889e7f8b754988fcbd6ba98380ed570)
![{\displaystyle \cos {\frac {\pi }{60}}=\cos 3^{\circ }={\tfrac {1}{16}}\left[2(1+{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}-1)\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd184460e47162735bde15cf93ad6f515fab79ef)
![{\displaystyle \tan {\frac {\pi }{60}}=\tan 3^{\circ }={\tfrac {1}{4}}\left[(2-{\sqrt {3}})(3+{\sqrt {5}})-2\right]\left[2-{\sqrt {2(5-{\sqrt {5}})}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9afc8cae3b0181ce3ca0fa38cd9737a9096fc5de)
![{\displaystyle \cot {\frac {\pi }{60}}=\cot 3^{\circ }={\tfrac {1}{4}}\left[(2+{\sqrt {3}})(3+{\sqrt {5}})-2\right]\left[2+{\sqrt {2(5-{\sqrt {5}})}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b010ae9dc1e68b6ef1be95008a95a4f19089ad3c)
![{\displaystyle \sin {\frac {\pi }{30}}=\sin 6^{\circ }={\tfrac {1}{8}}\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e32eed64add3521d7d7c9bf695482e8c3add4c36)
![{\displaystyle \cos {\frac {\pi }{30}}=\cos 6^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f90aff5eb92b8553f8dfd11cd76f3f30bd223560)
![{\displaystyle \tan {\frac {\pi }{30}}=\tan 6^{\circ }={\tfrac {1}{2}}\left[{\sqrt {2(5-{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca95e17f7bfec511a0d9029c1aa1a0c8e5cc5630)
![{\displaystyle \cot {\frac {\pi }{30}}=\cot 6^{\circ }={\tfrac {1}{2}}\left[{\sqrt {3}}(3+{\sqrt {5}})+{\sqrt {2(25+11{\sqrt {5}})}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4d4b083760baa9008b86595e91600a410442cd08)
![{\displaystyle \sin {\frac {\pi }{20}}=\sin 9^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2}}({\sqrt {5}}+1)-2{\sqrt {5-{\sqrt {5}}}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06fa0f780bd4389e9346cd20ee41ea5982c3335b)
![{\displaystyle \cos {\frac {\pi }{20}}=\cos 9^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2}}({\sqrt {5}}+1)+2{\sqrt {5-{\sqrt {5}}}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4e1418c61026ab91e3baa4bbf3118d80468dd69)
![{\displaystyle \sin {\frac {\pi }{15}}=\sin 12^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/55fb1a72c84d5dec8bf231d619c534df13c261ef)
![{\displaystyle \cos {\frac {\pi }{15}}=\cos 12^{\circ }={\tfrac {1}{8}}\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4923130317e1553f2cfd21ca42c4e29cd2f1f836)
![{\displaystyle \tan {\frac {\pi }{15}}=\tan 12^{\circ }={\tfrac {1}{2}}\left[{\sqrt {3}}(3-{\sqrt {5}})-{\sqrt {2(25-11{\sqrt {5}})}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a207e48f104ece1821014496f7b32e371956076d)
![{\displaystyle \cot {\frac {\pi }{15}}=\cot 12^{\circ }={\tfrac {1}{2}}\left[{\sqrt {3}}({\sqrt {5}}+1)+{\sqrt {2(5+{\sqrt {5}})}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a365437b27a47971bc1b6ba646a55cc64028d65a)
![{\displaystyle \sin {\frac {7\pi }{60}}=\sin 21^{\circ }={\tfrac {1}{16}}\left[2({\sqrt {3}}+1){\sqrt {5-{\sqrt {5}}}}-{\sqrt {2}}({\sqrt {3}}-1)(1+{\sqrt {5}})\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff732646ec0ca23f9f68a4d1ee73d1795f252c26)
![{\displaystyle \cos {\frac {7\pi }{60}}=\cos 21^{\circ }={\tfrac {1}{16}}\left[2({\sqrt {3}}-1){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {3}}+1)(1+{\sqrt {5}})\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40e7440f71edd44040f9fb2f6ef00da5a9361eeb)
![{\displaystyle \tan {\frac {7\pi }{60}}=\tan 21^{\circ }={\tfrac {1}{4}}\left[2-(2+{\sqrt {3}})(3-{\sqrt {5}})\right]\left[2-{\sqrt {2(5+{\sqrt {5}})}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/966ce4718377b52eba218988bb7863f78a12d062)
![{\displaystyle \cot {\frac {7\pi }{60}}=\cot 21^{\circ }={\tfrac {1}{4}}\left[2-(2-{\sqrt {3}})(3-{\sqrt {5}})\right]\left[2+{\sqrt {2(5+{\sqrt {5}})}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67aff972235eb8fe67a96bce37d05b81afc85442)
![{\displaystyle \sin {\frac {2\pi }{15}}=\sin 24^{\circ }={\tfrac {1}{8}}\left[{\sqrt {3}}({\sqrt {5}}+1)-{\sqrt {2}}{\sqrt {5-{\sqrt {5}}}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/748d70aaab6ed6e9d613d6ef799e612174f96970)
![{\displaystyle \tan {\frac {2\pi }{15}}=\tan 24^{\circ }={\tfrac {1}{2}}\left[{\sqrt {2(25+11{\sqrt {5}})}}-{\sqrt {3}}(3+{\sqrt {5}})\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/751f49b7e0a84e36750d6df47e69f7e5ca2f4ba2)
![{\displaystyle \cot {\frac {2\pi }{15}}=\cot 24^{\circ }={\tfrac {1}{2}}\left[{\sqrt {2}}{\sqrt {5-{\sqrt {5}}}}+{\sqrt {3}}({\sqrt {5}}-1)\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7356b37ae7753d547b8f84a2dae050a2efcdb5f)
![{\displaystyle \sin {\frac {3\pi }{20}}=\sin 27^{\circ }={\tfrac {1}{8}}\left[2{\sqrt {5+{\sqrt {5}}}}-{\sqrt {2}}\;({\sqrt {5}}-1)\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56f979b3d0d06f3124dd266ee7860dd955e654c1)
![{\displaystyle \cos {\frac {3\pi }{20}}=\cos 27^{\circ }={\tfrac {1}{8}}\left[2{\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}\;({\sqrt {5}}-1)\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3fb2b0c3a9180d29f4f03129dd067a1b992b7332)
![{\displaystyle \sin {\frac {11\pi }{60}}=\sin 33^{\circ }={\tfrac {1}{16}}\left[2({\sqrt {3}}-1){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}(1+{\sqrt {3}})({\sqrt {5}}-1)\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f9448eb3706b6f462f1db32cf35b8fcb21b6b19)
![{\displaystyle \cos {\frac {11\pi }{60}}=\cos 33^{\circ }={\tfrac {1}{16}}\left[2({\sqrt {3}}+1){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}(1-{\sqrt {3}})({\sqrt {5}}-1)\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/556981166d36ff827d365539dc1c3927cd65e376)
![{\displaystyle \tan {\frac {11\pi }{60}}=\tan 33^{\circ }={\tfrac {1}{4}}\left[2-(2-{\sqrt {3}})(3+{\sqrt {5}})\right]\left[2+{\sqrt {2(5-{\sqrt {5}})}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4bddc9e197ce12bebf3121cbd1eed3d3171b4213)
![{\displaystyle \cot {\frac {11\pi }{60}}=\cot 33^{\circ }={\tfrac {1}{4}}\left[2-(2+{\sqrt {3}})(3+{\sqrt {5}})\right]\left[2-{\sqrt {2(5-{\sqrt {5}})}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fbebd82a0521850dbf313d264e60a43a7c8faf85)
![{\displaystyle \sin {\frac {\pi }{5}}=\sin 36^{\circ }={\tfrac {1}{4}}[{\sqrt {2(5-{\sqrt {5}})}}]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8dc80fc11df3f88b3ad049451060b6e4a908ea2a)
![{\displaystyle \cot {\frac {\pi }{5}}=\cot 36^{\circ }={\tfrac {1}{5}}[{\sqrt {5(5+2{\sqrt {5}})}}]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b544dbddd4f75e26906419a0be2b59c1e27c9122)
![{\displaystyle \sin {\frac {13\pi }{60}}=\sin 39^{\circ }={\tfrac {1}{16}}[2(1-{\sqrt {3}}){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {3}}+1)({\sqrt {5}}+1)]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9747c181f024a8e812ac7113ede3a454128a50d8)
![{\displaystyle \cos {\frac {13\pi }{60}}=\cos 39^{\circ }={\tfrac {1}{16}}[2(1+{\sqrt {3}}){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {3}}-1)({\sqrt {5}}+1)]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c2772866ea70f64975dd96e5843ea3dff9392189)
![{\displaystyle \tan {\frac {13\pi }{60}}=\tan 39^{\circ }={\tfrac {1}{4}}\left[(2-{\sqrt {3}})(3-{\sqrt {5}})-2\right]\left[2-{\sqrt {2(5+{\sqrt {5}})}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cca29bb8cfbce994a3b86ad88977c64ebf77aa53)
![{\displaystyle \cot {\frac {13\pi }{60}}=\cot 39^{\circ }={\tfrac {1}{4}}\left[(2+{\sqrt {3}})(3-{\sqrt {5}})-2\right]\left[2+{\sqrt {2(5+{\sqrt {5}})}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d18ae9651f056834e7c47ddb6b860f2fd9b3e54)
