三角函數精確值 是利用三角函數的公式 將特定的三角函數 值加以化簡，並以數學根式 或分數 表示
用根式 或分數 表達的精確三角函數 有時很有用，主要用於簡化的解決某些方程式 能進一步化簡。
注意 ：以下為相同角度的轉換表：
相同角度的轉換表
角度單位
值
轉
0
{\displaystyle \mathbf {0} }
1
12
{\displaystyle {\tfrac {1}{12}}}
1
8
{\displaystyle {\tfrac {1}{8}}}
1
6
{\displaystyle {\tfrac {1}{6}}}
1
4
{\displaystyle {\tfrac {1}{4}}}
1
2
{\displaystyle {\tfrac {1}{2}}}
3
4
{\displaystyle {\tfrac {3}{4}}}
1
{\displaystyle \mathbf {1} }
角度
0°
30°
45°
60°
90°
180°
270°
360°
弧度
0
{\displaystyle \mathbf {0} }
π
6
{\displaystyle {\tfrac {\pi }{6}}}
π
4
{\displaystyle {\tfrac {\pi }{4}}}
π
3
{\displaystyle {\tfrac {\pi }{3}}}
π
2
{\displaystyle {\tfrac {\pi }{2}}}
π
{\displaystyle \pi }
3
π
2
{\displaystyle {\tfrac {3\pi }{2}}}
2
π
{\displaystyle 2\pi }
梯度
0
g
{\displaystyle 0^{g}}
33
1
3
g
{\displaystyle 33{\tfrac {1}{3}}^{g}}
50
g
{\displaystyle 50^{g}}
66
2
3
g
{\displaystyle 66{\tfrac {2}{3}}^{g}}
100
g
{\displaystyle 100^{g}}
200
g
{\displaystyle 200^{g}}
300
g
{\displaystyle 300^{g}}
400
g
{\displaystyle 400^{g}}
計算方式 基於常識 例如：0°、30°、45°
單位圓 例如：15°、22.5°
sin
(
x
2
)
=
±
1
2
(
1
−
cos
x
)
{\displaystyle \sin \left({\frac {x}{2}}\right)=\pm \,{\sqrt {{\tfrac {1}{2}}(1-\cos x)}}}
cos
(
x
2
)
=
±
1
2
(
1
+
cos
x
)
{\displaystyle \cos \left({\frac {x}{2}}\right)=\pm \,{\sqrt {{\tfrac {1}{2}}(1+\cos x)}}}
利用三倍角公式 求
1
3
{\displaystyle {\frac {1}{3}}\,}
例如：10°、20°、7°......等，非三的倍數的角的精確值。
sin
3
θ
=
3
sin
θ
−
4
sin
3
θ
{\displaystyle \sin 3\theta =3\sin \theta -4\sin ^{3}\theta \,}
cos
3
θ
=
4
cos
3
θ
−
3
cos
θ
{\displaystyle \cos 3\theta =4\cos ^{3}\theta -3\cos \theta \,}
把它改為
sin
θ
=
3
sin
1
3
θ
−
4
sin
3
1
3
θ
{\displaystyle \sin \theta =3\sin {\frac {1}{3}}\theta -4\sin ^{3}{\frac {1}{3}}\theta \,}
cos
θ
=
4
cos
3
1
3
θ
−
3
cos
1
3
θ
{\displaystyle \cos \theta =4\cos ^{3}{\frac {1}{3}}\theta -3\cos {\frac {1}{3}}\theta \,}
把
cos
1
3
θ
{\displaystyle \cos {\frac {1}{3}}\theta \,}
cos
θ
{\displaystyle \cos \theta \,}
一元三次方程式 即可求出
例如：
sin
π
9
=
sin
20
∘
=
−
3
16
+
−
1
256
3
+
−
3
16
−
−
1
256
3
{\displaystyle \sin {\frac {\pi }{9}}=\sin 20^{\circ }={\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}+{\sqrt {-{\frac {1}{256}}}}}}+{\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}-{\sqrt {-{\frac {1}{256}}}}}}}
經由合角公式的計算 例如：21° = 9° + 12°
sin
(
x
±
y
)
=
sin
(
x
)
cos
(
y
)
±
cos
(
x
)
sin
(
y
)
{\displaystyle \sin(x\pm y)=\sin(x)\cos(y)\pm \cos(x)\sin(y)\,}
cos
(
x
±
y
)
=
cos
(
x
)
cos
(
y
)
∓
sin
(
x
)
sin
(
y
)
{\displaystyle \cos(x\pm y)=\cos(x)\cos(y)\mp \sin(x)\sin(y)\,}
經由托勒密定理的計算 Chord(36°) = a/b = 1/f, from 托勒密定理 例如：18°
c
r
d
36
∘
=
c
r
d
(
∠
A
D
B
)
=
a
b
=
2
1
+
5
{\displaystyle \mathrm {crd} \ {36^{\circ }}=\mathrm {crd} \left(\angle \mathrm {ADB} \right)={\frac {a}{b}}={\frac {2}{1+{\sqrt {5}}}}}
c
r
d
θ
=
2
sin
θ
2
{\displaystyle \mathrm {crd} \ {\theta }=2\sin {\frac {\theta }{2}}\,}
sin
18
∘
=
1
1
+
5
=
1
4
(
5
−
1
)
{\displaystyle \sin {18^{\circ }}={\frac {1}{1+{\sqrt {5}}}}={\tfrac {1}{4}}\left({\sqrt {5}}-1\right)}
三角函數精確值列表 由於三角函數的特性，大於45°角度的三角函數值，可以經由自0°～ 45°的角度的三角函數值的相關的計算取得。
0°: 根本
sin
0
=
0
{\displaystyle \sin 0=0\,}
cos
0
=
1
{\displaystyle \cos 0=1\,}
tan
0
=
0
{\displaystyle \tan 0=0\,}
3°: 正60邊形
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]\,}
6°: 正30邊形
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]\,}
9°: 正20邊形
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}}}}\,}
10°
tan
10
∘
=
−
−
1
−
3
i
6
−
12
3
+
36
i
3
−
−
1
+
3
i
6
−
12
3
−
36
i
3
+
3
3
{\displaystyle {}_{\tan 10^{\circ }=-{\frac {-1-{\sqrt {3}}{\rm {i}}}{6}}{\sqrt[{3}]{-12{\sqrt {3}}+36{\rm {i}}}}-{\frac {-1+{\sqrt {3}}{\rm {i}}}{6}}{\sqrt[{3}]{-12{\sqrt {3}}-36{\rm {i}}}}+{\frac {\sqrt {3}}{3}}}\,}
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]\,}
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}}\,}
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}})}}\,}
20°: 正九邊形 和 60°的三分之一(
1
3
{\displaystyle {\frac {1}{3}}\,}
sin
π
9
=
sin
20
∘
=
−
3
16
+
−
1
256
3
+
−
3
16
−
−
1
256
3
=
{\displaystyle \sin {\frac {\pi }{9}}=\sin 20^{\circ }={\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}+{\sqrt {-{\frac {1}{256}}}}}}+{\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}-{\sqrt {-{\frac {1}{256}}}}}}=}
2
−
4
3
(
i
−
3
3
−
i
+
3
3
)
{\displaystyle 2^{-{\frac {4}{3}}}({\sqrt[{3}]{i-{\sqrt {3}}}}-{\sqrt[{3}]{i+{\sqrt {3}}}})}
cos
π
9
=
cos
20
∘
=
{\displaystyle \cos {\frac {\pi }{9}}=\cos 20^{\circ }=}
2
−
4
3
(
1
+
i
3
3
+
1
−
i
3
3
)
{\displaystyle 2^{-{\frac {4}{3}}}({\sqrt[{3}]{1+i{\sqrt {3}}}}+{\sqrt[{3}]{1-i{\sqrt {3}}}})}
21°: 9° 與 12°的和
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]\,}
(
21
3
17
∘
)
,
{\displaystyle (21{\frac {3}{17}}^{\circ }),\,}
正17邊形
cos
2
π
17
=
−
1
+
17
+
34
−
2
17
+
2
17
+
3
17
−
34
−
2
17
−
2
34
+
2
17
16
.
{\displaystyle \operatorname {cos} {2\pi \over 17}={\frac {-1+{\sqrt {17}}+{\sqrt {34-2{\sqrt {17}}}}+2{\sqrt {17+3{\sqrt {17}}-{\sqrt {34-2{\sqrt {17}}}}-2{\sqrt {34+2{\sqrt {17}}}}}}}{16}}.}
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\,}
24°: 兩倍的 12° 角
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]\,}
25(6/7)°，(180/7)°：正七邊形
cos
π
7
=
cos
180
7
∘
=
cos
25
6
7
∘
=
1
6
+
1
−
3
i
24
28
−
84
3
i
3
+
1
+
3
i
24
28
−
84
3
i
3
{\displaystyle \cos {\frac {\pi }{7}}=\cos {\frac {180}{7}}^{\circ }=\cos 25{\frac {6}{7}}^{\circ }={\frac {1}{6}}+{\frac {1-{\sqrt {3}}i}{24}}{\sqrt[{3}]{28-84{\sqrt {3}}i}}+{\frac {1+{\sqrt {3}}i}{24}}{\sqrt[{3}]{28-84{\sqrt {3}}i}}}
27°: 12° 與 15° 的和
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}}}}\,}
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}}\,}
33°: 15° 與 18° 之和
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]\,}
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}}}}\,}
39°: 18°角加21°角
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]\,}
42°: 21°的兩 倍
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}}\,}
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}
相關 參見 參考文獻 埃里克·韋斯坦因 . Constructible polygon . MathWorld . 埃里克·韋斯坦因 . Trigonometry angles . MathWorld . Bracken, Paul; Cizek, Jiri. 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. 2002, 90 (1): 42–53. doi:10.1002/qua.1803 Conway, John H.; Radin, Charles; Radun, Lorenzo. On angles whose squared trigonometric functions are rational. 1998. arXiv:math-ph/9812019 Conway, John H.; Radin, Charles; Radun, Lorenzo. On angles whose squared trigonometric functions are rational. Disc. Comput. Geom. 1999, 22 (3): 321–332. doi:10.1007/PL00009463 MR 1706614 Girstmair, Kurt. Some linear relations between values of trigonometric functions at k*pi/n. Acta Arithmetica. 1997, 81 : 387–398. MR 1472818 Gurak, S. On the minimal polynomial of gauss periods for prime powers. Mathematics of Computation. 2006, 75 (256): 2021–2035. Bibcode:2006MaCom..75.2021G doi:10.1090/S0025-5718-06-01885-0 MR 2240647 Servi, L. D. Nested square roots of 2. Am. Math. Monthly. 2003, 110 (4): 326–330. doi:10.2307/3647881 MR 1984573 
![{\displaystyle \sin {\frac {\pi }{9}}=\sin 20^{\circ }={\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}+{\sqrt {-{\frac {1}{256}}}}}}+{\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}-{\sqrt {-{\frac {1}{256}}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6019908a9d7a84324bc27625b5e5213830d24d8a)
![{\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 \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 \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 {}_{\tan 10^{\circ }=-{\frac {-1-{\sqrt {3}}{\rm {i}}}{6}}{\sqrt[{3}]{-12{\sqrt {3}}+36{\rm {i}}}}-{\frac {-1+{\sqrt {3}}{\rm {i}}}{6}}{\sqrt[{3}]{-12{\sqrt {3}}-36{\rm {i}}}}+{\frac {\sqrt {3}}{3}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c8918a7d997610fc9cd34748d433e667da40ea6)
![{\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 \sin {\frac {\pi }{9}}=\sin 20^{\circ }={\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}+{\sqrt {-{\frac {1}{256}}}}}}+{\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}-{\sqrt {-{\frac {1}{256}}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e39576a98052e848f2f9291c5d5da2437d6d056)
![{\displaystyle 2^{-{\frac {4}{3}}}({\sqrt[{3}]{i-{\sqrt {3}}}}-{\sqrt[{3}]{i+{\sqrt {3}}}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50a287434c058cc24fc6d48baa56b13bc7a26427)
![{\displaystyle 2^{-{\frac {4}{3}}}({\sqrt[{3}]{1+i{\sqrt {3}}}}+{\sqrt[{3}]{1-i{\sqrt {3}}}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff5fc48cf9638893e79d19d632b6e922b18c3f54)
![{\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 \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 \cos {\frac {\pi }{7}}=\cos {\frac {180}{7}}^{\circ }=\cos 25{\frac {6}{7}}^{\circ }={\frac {1}{6}}+{\frac {1-{\sqrt {3}}i}{24}}{\sqrt[{3}]{28-84{\sqrt {3}}i}}+{\frac {1+{\sqrt {3}}i}{24}}{\sqrt[{3}]{28-84{\sqrt {3}}i}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2efc31802fbf4a8197ef2fb472f3a8d413c88a97)
![{\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 \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 \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)
.