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 .
I en regulær n -kant er a = π/n den halve sentralvinkel og b = π(1/2 - 1/n) den halve, indre vinkel . 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 radianer ), der m og n er 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}
argumenter , er de eneste mulige rasjonale løsningene 0, ±1 og ±1 / 2
Velkjente konstanter
rediger
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
rediger
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
rediger
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
rediger
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
rediger
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
rediger
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
rediger
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}}\,}
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°
rediger
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
rediger
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°
rediger
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°
rediger
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}}}}\,}
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°
rediger
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]\,}
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°
rediger
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°
rediger
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}}\,}
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\,}
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 .
(en) Eric W. Weisstein , Constructible polygon MathWorld .(en) Eric W. Weisstein , Trigonometry angles MathWorld .(en) Eric W. Weisstein , Niven's Theorem 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 . Conway, John H.; Radin, Charles; Radun, Lorenzo (1998). «On angles whose squared trigonometric functions are rational» . . [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 . MR 1706614 Girstmair, Kurt (1997). «Some linear relations between values of trigonometric functions at k*pi/n». Acta Arithmetica . 81: 387–398. MR 1472818 Gurak, 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 2240647 Servi, L. D. (2003). «Nested square roots of 2». Am. Math. Monthly (4 utg.). 110: 326–330. doi :10.2307/3647881 . MR 1984573 JSTOR 3647881 Eksterne lenker
rediger
![{\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)
