三角函数精确值 是利用三角函数的公式 将特定的三角函数 值加以化简,并以数学根式 或分数 表示
用根式 或分数 表达的精确三角函数 有时很有用,主要用于简化的解决某些方程式 能进一步化简。
根据尼云定理 ,有理数度数的角的正弦值,其中的有理数仅有0,
±
1
2
{\displaystyle \pm\frac{1}{2}}
,±1。
相同角度的转换表
角度单位
值
转
0
{\displaystyle 0}
1
12
{\displaystyle \frac{1}{12}}
1
8
{\displaystyle \frac{1}{8}}
1
6
{\displaystyle \frac{1}{6}}
1
4
{\displaystyle \frac{1}{4}}
1
2
{\displaystyle \frac{1}{2}}
3
4
{\displaystyle \frac{3}{4}}
1
{\displaystyle 1}
角度
0
∘
{\displaystyle 0^{\circ}}
30
∘
{\displaystyle 30^{\circ}}
45
∘
{\displaystyle 45^{\circ}}
60
∘
{\displaystyle 60^{\circ}}
90
∘
{\displaystyle 90^{\circ}}
180
∘
{\displaystyle 180^{\circ}}
270
∘
{\displaystyle 270^{\circ}}
360
∘
{\displaystyle 360^{\circ}}
弧度
0
{\displaystyle 0}
π
6
{\displaystyle \frac{\pi}{6}}
π
4
{\displaystyle \frac{\pi}{4}}
π
3
{\displaystyle \frac{\pi}{3}}
π
2
{\displaystyle \frac{\pi}{2}}
π
{\displaystyle \pi}
3
π
2
{\displaystyle \frac{3\pi}{2}}
2
π
{\displaystyle 2\pi}
梯度
0
g
{\displaystyle 0^g}
33
1
3
g
{\displaystyle 33\frac{1}{3}^g}
50
g
{\displaystyle 50^g}
66
2
3
g
{\displaystyle 66\frac{2}{3}^g}
100
g
{\displaystyle 100^g}
200
g
{\displaystyle 200^g}
300
g
{\displaystyle 300^g}
400
g
{\displaystyle 400^g}
计算方式
基于常识
例如:0°、30°、45°
单位圆
经由半角公式的计算
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 \frac13\,}
角
例如:10°、20°、7°......等等,非三的倍数的角的精确值。
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 \,}
把它改为
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 \,}
当成未知数,
cos
θ
{\displaystyle \cos \theta \,}
当成常数项
解一元三次方程式 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}}}}
同样地,若角度代未知数,则会得到三分之一角公式 。
经由欧拉公式的计算
cos
θ
n
=
ℜ
(
cos
θ
+
i
sin
θ
n
)
=
1
2
(
cos
θ
+
i
sin
θ
n
+
cos
θ
−
i
sin
θ
n
)
{\displaystyle \cos\frac{\theta}{n} = \Re\left(\sqrt[n]{\cos\theta+i\sin\theta}\right) = \frac{1}{2}\left(\sqrt[n]{\cos\theta+i\sin\theta}+\sqrt[n]{\cos\theta-i\sin\theta}\right)}
sin
θ
n
=
ℑ
(
cos
θ
+
i
sin
θ
n
)
=
1
2
i
(
cos
θ
+
i
sin
θ
n
−
cos
θ
−
i
sin
θ
n
)
{\displaystyle \sin\frac{\theta}{n} = \Im\left(\sqrt[n]{\cos\theta+i\sin\theta}\right) = \frac{1}{2i}\left(\sqrt[n]{\cos\theta+i\sin\theta}-\sqrt[n]{\cos\theta-i\sin\theta}\right)}
例如:
sin
1
∘
=
1
2
i
(
cos
3
∘
+
i
sin
3
∘
3
−
cos
3
∘
−
i
sin
3
∘
3
)
{\displaystyle \sin{1^\circ} = \frac{1}{2i}\left(\sqrt[3]{\cos{3^\circ}+i\sin{3^\circ}}-\sqrt[3]{\cos{3^\circ}-i\sin{3^\circ}}\right)}
=
1
4
2
3
i
{
[
2
(
1
+
3
)
5
+
5
+
2
(
5
−
1
)
(
3
−
1
)
]
+
i
[
2
(
1
−
3
)
5
+
5
+
2
(
5
−
1
)
(
3
+
1
)
]
3
{\displaystyle = \frac{1}{4\sqrt[3]{2}i}\Bigg\{\sqrt[3]{\left[2(1+\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3-1)\right]+i\left[2(1-\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3+1)\right]}}
−
[
2
(
1
+
3
)
5
+
5
+
2
(
5
−
1
)
(
3
−
1
)
]
−
i
[
2
(
1
−
3
)
5
+
5
+
2
(
5
−
1
)
(
3
+
1
)
]
3
}
{\displaystyle -\sqrt[3]{\left[2(1+\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3-1)\right]-i\left[2(1-\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3+1)\right]}\Bigg\}}
[1]
经由和角公式的计算
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)\,}
经由托勒密定理的计算
Chord(36°) = a/b = 1/φ, 根据托勒密定理
例如:18°
根据托勒密定理,在圆内接四边形 ABCD中,
a
2
+
a
b
=
b
2
{\displaystyle a^2+ab=b^2}
(
a
b
)
2
+
a
b
=
1
{\displaystyle \left(\frac{a}{b}\right)^2+\frac{a}{b}=1}
c
r
d
36
∘
=
c
r
d
(
∠
A
D
B
)
=
a
b
=
5
−
1
2
{\displaystyle \mathrm{crd}\ {36^\circ}=\mathrm{crd}\left(\angle\mathrm{ADB}\right)=\frac{a}{b}=\frac{\sqrt{5}-1}{2}}
c
r
d
θ
=
2
sin
θ
2
{\displaystyle \mathrm{crd}\ {\theta}=2\sin{\frac{\theta}{2}}\,}
sin
18
∘
=
5
−
1
4
{\displaystyle \sin{18^\circ}=\frac{\sqrt5-1}{4}}
三角函数精确值列表
由于三角函数的特性,大于45°角度的三角函数值,可以经由自0°~45°的角度的三角函数值的相关的计算取得。
0°:根本
sin
0
=
0
{\displaystyle \sin 0=0\,}
cos
0
=
1
{\displaystyle \cos 0=1\,}
tan
0
=
0
{\displaystyle \tan 0=0\,}
1°:2°的一半
sin
1
∘
=
1
+
3
i
16
4
30
−
8
15
+
3
5
+
8
5
+
5
+
4
10
−
4
6
−
4
2
+
(
4
30
+
8
15
+
3
5
+
8
5
+
5
−
4
10
−
4
6
+
4
2
)
i
3
+
{\displaystyle \sin{1^\circ} = \frac{1+\sqrt{3}i}{16}\sqrt[3]{4\sqrt{30}-8\sqrt{15+3\sqrt{5}}+8\sqrt{5+\sqrt{5}}+4\sqrt{10}-4\sqrt{6}-4\sqrt{2}+\left(4\sqrt{30}+8\sqrt{15+3\sqrt{5}}+8\sqrt{5+\sqrt{5}}-4\sqrt{10}-4\sqrt{6}+4\sqrt{2}\right)i}+}
1
−
3
i
16
4
30
−
8
15
+
3
5
+
8
5
+
5
+
4
10
−
4
6
−
4
2
−
(
4
30
+
8
15
+
3
5
+
8
5
+
5
−
4
10
−
4
6
+
4
2
)
i
3
{\displaystyle \frac{1-\sqrt{3}i}{16}\sqrt[3]{4\sqrt{30}-8\sqrt{15+3\sqrt{5}}+8\sqrt{5+\sqrt{5}}+4\sqrt{10}-4\sqrt{6}-4\sqrt{2}-\left(4\sqrt{30}+8\sqrt{15+3\sqrt{5}}+8\sqrt{5+\sqrt{5}}-4\sqrt{10}-4\sqrt{6}+4\sqrt{2}\right)i}}
[2]
1.5°:正一百二十边形
sin
(
π
120
)
=
sin
(
1.5
∘
)
=
(
2
+
2
)
(
15
+
3
−
10
−
2
5
)
−
(
2
−
2
)
(
30
−
6
5
+
5
+
1
)
16
{\displaystyle \sin\left(\frac{\pi}{120}\right) = \sin\left(1.5^\circ\right) = \frac{\left(\sqrt{2+\sqrt2}\right)\left(\sqrt{15}+\sqrt3-\sqrt{10-2\sqrt5}\right) - \left(\sqrt{2-\sqrt2}\right)\left(\sqrt{30-6\sqrt5}+\sqrt5+1\right)}{16}}
cos
(
π
120
)
=
cos
(
1.5
∘
)
=
(
2
+
2
)
(
30
−
6
5
+
5
+
1
)
+
(
2
−
2
)
(
15
+
3
−
10
−
2
5
)
16
{\displaystyle \cos\left(\frac{\pi}{120}\right) = \cos\left(1.5^\circ\right) = \frac{\left(\sqrt{2+\sqrt2}\right)\left(\sqrt{30-6\sqrt5}+\sqrt5+1\right) + \left(\sqrt{2-\sqrt2}\right)\left(\sqrt{15}+\sqrt3-\sqrt{10-2\sqrt5}\right)}{16}}
1.875°:正九十六边形
sin
(
π
96
)
=
sin
(
1.875
∘
)
=
1
2
2
−
2
+
2
+
2
+
3
{\displaystyle \sin\left(\frac{\pi}{96}\right) = \sin\left(1.875^\circ\right) = \frac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}}
cos
(
π
96
)
=
cos
(
1.875
∘
)
=
1
2
2
+
2
+
2
+
2
+
3
{\displaystyle \cos\left(\frac{\pi}{96}\right) = \cos\left(1.875^\circ\right) = \frac12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}}
2°:6°的三分之一
sin
2
∘
=
1
2
i
(
cos
6
∘
+
i
sin
6
∘
3
−
cos
6
∘
−
i
sin
6
∘
3
)
{\displaystyle \sin{2^\circ} = \frac{1}{2i}\left(\sqrt[3]{\cos{6^\circ}+i\sin{6^\circ}}-\sqrt[3]{\cos{6^\circ}-i\sin{6^\circ}}\right)}
=
1
4
i
{
[
2
(
5
−
5
)
+
3
(
5
+
1
)
]
+
i
[
6
(
5
−
5
)
−
5
−
1
]
3
{\displaystyle = \frac{1}{4i}\Bigg\{\sqrt[3]{\left[\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)\right]+i\left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]}}
−
[
2
(
5
−
5
)
+
3
(
5
+
1
)
]
−
i
[
6
(
5
−
5
)
−
5
−
1
]
3
}
{\displaystyle -\sqrt[3]{\left[\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)\right]-i\left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]}\Bigg\}}
cos
2
∘
=
1
2
(
cos
6
∘
+
i
sin
6
∘
3
+
cos
6
∘
−
i
sin
6
∘
3
)
{\displaystyle \cos{2^\circ} = \frac{1}{2}\left(\sqrt[3]{\cos{6^\circ}+i\sin{6^\circ}}+\sqrt[3]{\cos{6^\circ}-i\sin{6^\circ}}\right)}
=
1
4
{
[
2
(
5
−
5
)
+
3
(
5
+
1
)
]
+
i
[
6
(
5
−
5
)
−
5
−
1
]
3
{\displaystyle = \frac{1}{4}\Bigg\{\sqrt[3]{\left[\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)\right]+i\left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]}}
+
[
2
(
5
−
5
)
+
3
(
5
+
1
)
]
−
i
[
6
(
5
−
5
)
−
5
−
1
]
3
}
{\displaystyle +\sqrt[3]{\left[\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)\right]-i\left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]}\Bigg\}}
2.25°:正八十边形
sin
(
π
80
)
=
sin
(
2.25
∘
)
=
1
2
2
−
2
+
2
+
5
+
5
2
{\displaystyle \sin\left(\frac{\pi}{80}\right) = \sin\left(2.25^\circ\right) =\frac{1}{2} \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}}}}
cos
(
π
80
)
=
cos
(
2.25
∘
)
=
1
2
2
+
2
+
2
+
5
+
5
2
{\displaystyle \cos\left(\frac{\pi}{80}\right) = \cos\left(2.25^\circ\right) =\frac{1}{2} \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}}}}
2.8125°:正六十四边形
sin
(
π
64
)
=
sin
(
2.8125
∘
)
=
1
2
2
−
2
+
2
+
2
+
2
{\displaystyle \sin\left(\frac{\pi}{64}\right) = \sin\left(2.8125^\circ\right) = \frac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}}
cos
(
π
64
)
=
cos
(
2.8125
∘
)
=
1
2
2
+
2
+
2
+
2
+
2
{\displaystyle \cos\left(\frac{\pi}{64}\right) = \cos\left(2.8125^\circ\right) = \frac12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}}
3°:正六十边形
sin
π
60
=
sin
3
∘
=
1
4
8
−
3
−
15
−
10
−
2
5
{\displaystyle \sin\frac{\pi}{60}=\sin 3^\circ=\tfrac{1}{4} \sqrt{8-\sqrt3-\sqrt{15}-\sqrt{10-2\sqrt5}}\,}
cos
π
60
=
cos
3
∘
=
1
4
8
+
3
+
15
+
10
−
2
5
{\displaystyle \cos\frac{\pi}{60}=\cos 3^\circ=\tfrac{1}{4} \sqrt{8+\sqrt3+\sqrt{15}+\sqrt{10-2\sqrt5}}\,}
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-\sqrt3)(3+\sqrt5)-2\right]\left[2-\sqrt{2(5-\sqrt5)}\right]\,}
3.75°:正四十八边形
sin
(
π
48
)
=
sin
(
3.75
∘
)
=
1
2
2
−
2
+
2
+
3
{\displaystyle \sin\left(\frac{\pi}{48}\right) = \sin\left(3.75^\circ\right) = \frac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}}
cos
(
π
48
)
=
cos
(
3.75
∘
)
=
1
2
2
+
2
+
2
+
3
{\displaystyle \cos\left(\frac{\pi}{48}\right) = \cos\left(3.75^\circ\right) = \frac12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}
4°:12°的三分之一
sin
4
∘
=
1
2
i
(
cos
12
∘
+
i
sin
12
∘
3
−
cos
12
∘
−
i
sin
12
∘
3
)
{\displaystyle \sin{4^\circ} = \frac{1}{2i}\left(\sqrt[3]{\cos{12^\circ}+i\sin{12^\circ}}-\sqrt[3]{\cos{12^\circ}-i\sin{12^\circ}}\right)}
=
1
4
i
{
[
6
(
5
+
5
)
+
5
−
1
]
+
i
[
2
(
5
+
5
)
−
3
(
5
−
1
)
]
3
{\displaystyle = \frac{1}{4i}\Bigg\{\sqrt[3]{\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]+i\left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]}}
−
[
6
(
5
+
5
)
+
5
−
1
]
−
i
[
2
(
5
+
5
)
−
3
(
5
−
1
)
]
3
}
{\displaystyle -\sqrt[3]{\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]-i\left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]}\Bigg\}}
cos
4
∘
=
1
2
(
cos
12
∘
+
i
sin
12
∘
3
+
cos
12
∘
−
i
sin
12
∘
3
)
{\displaystyle \cos{4^\circ} = \frac{1}{2}\left(\sqrt[3]{\cos{12^\circ}+i\sin{12^\circ}}+\sqrt[3]{\cos{12^\circ}-i\sin{12^\circ}}\right)}
=
1
4
{
[
6
(
5
+
5
)
+
5
−
1
]
+
i
[
2
(
5
+
5
)
−
3
(
5
−
1
)
]
3
{\displaystyle = \frac{1}{4}\Bigg\{\sqrt[3]{\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]+i\left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]}}
+
[
6
(
5
+
5
)
+
5
−
1
]
−
i
[
2
(
5
+
5
)
−
3
(
5
−
1
)
]
3
}
{\displaystyle +\sqrt[3]{\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]-i\left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]}\Bigg\}}
4.5°:正四十边形
sin
(
π
40
)
=
sin
(
4.5
∘
)
=
1
2
2
−
2
+
5
+
5
2
{\displaystyle \sin\left(\frac{\pi}{40}\right) = \sin\left(4.5^\circ\right) =\frac{1}{2} \sqrt{2-\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}}}
cos
(
π
40
)
=
cos
(
4.5
∘
)
=
1
2
2
+
2
+
5
+
5
2
{\displaystyle \cos\left(\frac{\pi}{40}\right) = \cos\left(4.5^\circ\right) =\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}}}
5°:15°的三分之一、正三十六边形
sin
π
36
=
sin
5
∘
=
2
−
2
3
i
2
2
(
2
−
6
)
3
−
2
−
3
−
(
1
+
3
i
)
2
(
2
−
6
)
3
−
2
−
3
8
{\displaystyle \sin\frac{\pi}{36}=\sin 5^\circ = \frac{2 - 2\sqrt{3}\mathrm{i}}{2 \sqrt[3]{2(\sqrt{2} - \sqrt{6})} - 2-\sqrt{3}} - \frac{(1 + \sqrt{3}\mathrm{i}) \sqrt[3]{2(\sqrt{2} - \sqrt{6})} -2-\sqrt{3}}{8}\,}
5.625°:正三十二边形
sin
(
π
32
)
=
sin
(
5.625
∘
)
=
1
2
2
−
2
+
2
+
2
{\displaystyle \sin\left(\frac{\pi}{32}\right) = \sin\left(5.625^\circ\right) = \frac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}}
cos
(
π
32
)
=
cos
(
5.625
∘
)
=
1
2
2
+
2
+
2
+
2
{\displaystyle \cos\left(\frac{\pi}{32}\right) = \cos\left(5.625^\circ\right) = \frac12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}
6°:正三十边形
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-\sqrt5)}-\sqrt5-1\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-\sqrt5)}+\sqrt3(\sqrt5+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-\sqrt5)}-\sqrt3(\sqrt5-1)\right]\,}
cot
π
30
=
cot
6
∘
=
1
2
(
50
+
22
5
+
3
3
+
15
)
{\displaystyle \cot\frac{\pi}{30}=\cot 6^\circ=\tfrac{1}{2} \left(\sqrt{50+22\sqrt5}+3\sqrt3+\sqrt{15}\right)\,}
sec
π
30
=
sec
6
∘
=
3
−
5
−
2
5
{\displaystyle \sec\frac{\pi}{30}=\sec 6^\circ=\sqrt3-\sqrt{5-2\sqrt5}\,}
csc
π
30
=
csc
6
∘
=
2
+
5
+
15
+
6
5
{\displaystyle \csc\frac{\pi}{30}=\csc 6^\circ=2+\sqrt5+\sqrt{15+6\sqrt5}\,}
7.5°:正二十四边形
sin
π
24
=
sin
7.5
∘
=
1
4
8
−
2
6
−
2
2
{\displaystyle \sin\frac{\pi}{24}=\sin 7.5^\circ=\tfrac{1}{4} \sqrt{8-2\sqrt6-2\sqrt2}\,}
cos
π
24
=
cos
7.5
∘
=
1
4
8
+
2
6
+
2
2
{\displaystyle \cos\frac{\pi}{24}=\cos 7.5^\circ=\tfrac{1}{4} \sqrt{8+2\sqrt6+2\sqrt2}\,}
tan
π
24
=
tan
7.5
∘
=
6
+
2
−
2
−
3
{\displaystyle \tan\frac{\pi}{24}=\tan 7.5^\circ=\sqrt6+\sqrt2-2-\sqrt3\,}
cot
π
24
=
cot
7.5
∘
=
6
+
2
+
2
+
3
{\displaystyle \cot\frac{\pi}{24}=\cot 7.5^\circ=\sqrt6+\sqrt2+2+\sqrt3\,}
sec
π
24
=
sec
7.5
∘
=
16
−
6
6
−
10
2
+
8
3
{\displaystyle \sec\frac{\pi}{24}=\sec 7.5^\circ=\sqrt{16-6\sqrt6-10\sqrt2+8\sqrt3}\,}
csc
π
24
=
csc
7.5
∘
=
16
+
6
6
+
10
2
+
8
3
{\displaystyle \csc\frac{\pi}{24}=\csc 7.5^\circ=\sqrt{16+6\sqrt6+10\sqrt2+8\sqrt3}\,}
9°:正二十边形
sin
π
20
=
sin
9
∘
=
1
4
8
−
2
10
+
2
5
{\displaystyle \sin\frac{\pi}{20}=\sin 9^\circ=\tfrac{1}{4} \sqrt{8-2\sqrt{10+2\sqrt5}}\,}
cos
π
20
=
cos
9
∘
=
1
4
8
+
2
10
+
2
5
{\displaystyle \cos\frac{\pi}{20}=\cos 9^\circ=\tfrac{1}{4} \sqrt{8+2\sqrt{10+2\sqrt5}}\,}
tan
π
20
=
tan
9
∘
=
5
+
1
−
5
+
2
5
{\displaystyle \tan\frac{\pi}{20}=\tan 9^\circ=\sqrt5+1-\sqrt{5+2\sqrt5}\,}
cot
π
20
=
cot
9
∘
=
5
+
1
+
5
+
2
5
{\displaystyle \cot\frac{\pi}{20}=\cot9^\circ=\sqrt5+1+\sqrt{5+2\sqrt5}\,}
10°:正十八边形
tan
10
∘
=
−
−
1
−
3
i
6
−
12
3
+
36
i
3
−
−
1
+
3
i
6
−
12
3
−
36
i
3
+
1
3
{\displaystyle {\tan10^\circ=-\frac{-1-\sqrt{3}{\rm{i}}}{6}\sqrt[3]{-12\sqrt3 + 36{\rm{i}}}-\frac{-1+\sqrt{3}{\rm{i}}}{6}\sqrt[3]{-12\sqrt3 - 36{\rm{i}}} + \frac{1}{\sqrt3}}\,}
11.25°:正十六边形
sin
π
16
=
sin
11.25
∘
=
1
2
2
−
2
+
2
{\displaystyle \sin\frac{\pi}{16}=\sin 11.25^\circ=\frac{1}{2}\sqrt{2-\sqrt{2+\sqrt{2}}}}
cos
π
16
=
cos
11.25
∘
=
1
2
2
+
2
+
2
{\displaystyle \cos\frac{\pi}{16}=\cos 11.25^\circ=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2}}}}
tan
π
16
=
tan
11.25
∘
=
4
+
2
2
−
2
−
1
{\displaystyle \tan\frac{\pi}{16}=\tan 11.25^\circ=\sqrt{4+2\sqrt{2}}-\sqrt{2}-1}
cot
π
16
=
cot
11.25
∘
=
4
+
2
2
+
2
+
1
{\displaystyle \cot\frac{\pi}{16}=\cot 11.25^\circ=\sqrt{4+2\sqrt{2}}+\sqrt{2}+1}
12°:正十五边形
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+\sqrt5)}-\sqrt3(\sqrt5-1)\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+\sqrt5)}+\sqrt5-1\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[\sqrt3(3-\sqrt5)-\sqrt{2(25-11\sqrt5)}\right]\,}
15°:正十二边形
sin
π
12
=
sin
15
∘
=
1
4
2
(
3
−
1
)
{\displaystyle \sin\frac{\pi}{12}=\sin 15^\circ=\tfrac{1}{4}\sqrt2\left(\sqrt3-1\right)\,}
cos
π
12
=
cos
15
∘
=
1
4
2
(
3
+
1
)
{\displaystyle \cos\frac{\pi}{12}=\cos 15^\circ=\tfrac{1}{4}\sqrt2\left(\sqrt3+1\right)\,}
tan
π
12
=
tan
15
∘
=
2
−
3
{\displaystyle \tan\frac{\pi}{12}=\tan 15^\circ=2-\sqrt3\,}
18°:正十边形
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(\sqrt5-1\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\left(5+\sqrt5\right)}\,}
tan
π
10
=
tan
18
∘
=
1
5
5
(
5
−
2
5
)
{\displaystyle \tan\frac{\pi}{10}=\tan 18^\circ=\tfrac{1}{5}\sqrt{5\left(5-2\sqrt5\right)}\,}
20°:正九边形、60°的三分之一
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}}\left(\sqrt[3]{i-\sqrt{3}}-\sqrt[3]{i+\sqrt{3}}\right)}
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}}\left(\sqrt[3]{1+i\sqrt{3}}+\sqrt[3]{1-i\sqrt{3}}\right)}
21°:9°与12°的和
sin
7
π
60
=
sin
21
∘
=
1
4
8
+
3
−
15
−
10
+
2
5
{\displaystyle \sin\frac{7\pi}{60}=\sin 21^\circ=\tfrac{1}{4}\sqrt{8+\sqrt3-\sqrt{15}-\sqrt{10+2\sqrt5}}\,}
cos
7
π
60
=
cos
21
∘
=
1
4
8
−
3
+
15
+
10
+
2
5
{\displaystyle \cos\frac{7\pi}{60}=\cos 21^\circ=\tfrac{1}{4}\sqrt{8-\sqrt3+\sqrt{15}+\sqrt{10+2\sqrt5}}\,}
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-\left(2+\sqrt3\right)\left(3-\sqrt5\right)\right]\left[2-\sqrt{2\left(5+\sqrt5\right)}\right]\,}
360/17°,
(
21
3
17
)
∘
{\displaystyle \mathbf{\left(21\frac{3}{17}\right)^{\circ}}}
,
(
360
17
)
∘
{\displaystyle \mathbf{\left(\frac{360}{17}\right)^{\circ}}}
:正十七边形
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\over17}=\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}}
22.5°:正八边形
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°:12°的二倍
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[\sqrt3(\sqrt5+1)-\sqrt2\sqrt{5-\sqrt5}\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(\sqrt6\sqrt{5-\sqrt5}+\sqrt5+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\sqrt5)}-\sqrt3(3+\sqrt5)\right]\,}
180/7°,
(
25
5
7
)
∘
{\displaystyle \mathbf{\left(25\frac{5}{7}\right)^{\circ}}}
,
(
180
7
)
∘
{\displaystyle \mathbf{\left(\frac{180}{7}\right)^{\circ}}}
:正七边形
cos
π
7
=
cos
180
7
∘
=
cos
25
5
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{5}{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°与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+\sqrt5}-\sqrt2\;(\sqrt5-1)\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+\sqrt5}+\sqrt2\;\left(\sqrt5-1\right)\right]\,}
tan
3
π
20
=
tan
27
∘
=
5
−
1
−
5
−
2
5
{\displaystyle \tan\frac{3\pi}{20}=\tan 27^\circ=\sqrt5-1-\sqrt{5-2\sqrt5}\,}
30°:正六边形
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}\sqrt3\,}
tan
π
6
=
tan
30
∘
=
1
3
3
{\displaystyle \tan\frac{\pi}{6}=\tan 30^\circ=\tfrac{1}{3}\sqrt3\,}
33°:15°与18°的和
sin
11
π
60
=
sin
33
∘
=
1
4
8
−
3
−
15
+
10
−
2
5
{\displaystyle \sin\frac{11\pi}{60}=\sin 33^\circ=\tfrac{1}{4}\sqrt{8-\sqrt3-\sqrt{15}+\sqrt{10-2\sqrt5}}\,}
cos
11
π
60
=
cos
33
∘
=
1
4
8
+
3
+
15
−
10
−
2
5
{\displaystyle \cos\frac{11\pi}{60}=\cos 33^\circ=\tfrac{1}{4}\sqrt{8+\sqrt3+\sqrt{15}-\sqrt{10-2\sqrt5}}\,}
tan
11
π
60
=
tan
33
∘
=
1
4
(
2
3
−
5
−
1
)
(
2
5
+
2
5
+
3
+
5
)
{\displaystyle \tan\frac{11\pi}{60}=\tan 33^\circ=\tfrac{1}{4}\left(2\sqrt3-\sqrt5-1\right)\left(2\sqrt{5+2\sqrt5}+3+\sqrt5\right)\,}
cot
11
π
60
=
cot
33
∘
=
1
4
(
2
3
+
5
+
1
)
(
2
5
+
2
5
−
3
−
5
)
{\displaystyle \cot\frac{11\pi}{60}=\cot33^\circ=\tfrac{1}{4}\left(2\sqrt3+\sqrt5+1\right)\left(2\sqrt{5+2\sqrt5}-3-\sqrt5\right)\,}
36°:正五边形
sin
π
5
=
sin
36
∘
=
1
4
[
2
(
5
−
5
)
]
{\displaystyle \sin\frac{\pi}{5}=\sin 36^\circ=\tfrac14\left[\sqrt{2\left(5-\sqrt5\right)}\right]\,}
cos
π
5
=
cos
36
∘
=
1
+
5
4
=
1
2
φ
{\displaystyle \cos\frac{\pi}{5}=\cos 36^\circ=\frac{1+\sqrt5}{4}=\tfrac{1}{2}\varphi\,}
tan
π
5
=
tan
36
∘
=
5
−
2
5
{\displaystyle \tan\frac{\pi}{5}=\tan 36^\circ=\sqrt{5-2\sqrt5}\,}
39°:18°与21°的和
sin
13
π
60
=
sin
39
∘
=
1
4
8
−
3
+
15
+
10
+
2
5
{\displaystyle \sin\frac{13\pi}{60}=\sin 39^\circ=\tfrac1{4}\sqrt{8-\sqrt3+\sqrt{15}+\sqrt{10+2\sqrt5}}\,}
cos
13
π
60
=
cos
39
∘
=
1
4
8
+
3
−
15
+
10
+
2
5
{\displaystyle \cos\frac{13\pi}{60}=\cos 39^\circ=\tfrac1{4}\sqrt{8+\sqrt3-\sqrt{15}+\sqrt{10+2\sqrt5}}\,}
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=\tfrac14\left[\left(2-\sqrt3\right)\left(3-\sqrt5\right)-2\right]\left[2-\sqrt{2\left(5+\sqrt5\right)}\right]\,}
42°:21°的2倍
sin
7
π
30
=
sin
42
∘
=
6
5
+
5
−
5
+
1
8
{\displaystyle \sin\frac{7\pi}{30}=\sin 42^\circ=\frac{\sqrt6\sqrt{5+\sqrt5}-\sqrt5+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{\sqrt2\sqrt{5+\sqrt5}+\sqrt3\left(\sqrt5-1\right)}{8}\,}
tan
7
π
30
=
tan
42
∘
=
1
2
(
3
+
15
−
10
+
2
5
)
{\displaystyle \tan\frac{7\pi}{30}=\tan 42^\circ=\frac1{2}\left(\sqrt3+\sqrt{15}-\sqrt{10+2\sqrt5}\right)\,}
cot
7
π
30
=
cot
42
∘
=
1
2
(
3
3
−
15
+
50
−
22
5
)
{\displaystyle \cot\frac{7\pi}{30}=\cot 42^\circ=\frac1{2}\left(3\sqrt3-\sqrt{15}+\sqrt{50-22\sqrt5}\right)\,}
sec
7
π
30
=
sec
42
∘
=
5
+
2
5
−
3
{\displaystyle \sec\frac{7\pi}{30}=\sec 42^\circ=\sqrt{5+2\sqrt5}-\sqrt3\,}
sec
7
π
30
=
sec
42
∘
=
15
−
6
5
+
5
−
2
{\displaystyle \sec\frac{7\pi}{30}=\sec 42^\circ=\sqrt{15-6\sqrt5}+\sqrt5-2\,}
45°:正方形
sin
π
4
=
sin
45
∘
=
2
2
=
1
2
{\displaystyle \sin\frac{\pi}{4}=\sin 45^\circ=\frac{\sqrt2}{2}=\frac{1}{\sqrt2}\,}
cos
π
4
=
cos
45
∘
=
2
2
=
1
2
{\displaystyle \cos\frac{\pi}{4}=\cos 45^\circ=\frac{\sqrt2}{2}=\frac{1}{\sqrt2}\,}
tan
π
4
=
tan
45
∘
=
1
{\displaystyle \tan\frac{\pi}{4}=\tan 45^\circ=1}
54°:27°与27°的和
sin
3
π
10
=
sin
54
∘
=
5
+
1
4
{\displaystyle \sin\frac{3\pi}{10}=\sin 54^\circ=\frac{\sqrt5+1}{4}\,\!}
cos
3
π
10
=
cos
54
∘
=
10
−
2
5
4
{\displaystyle \cos\frac{3\pi}{10}=\cos 54^\circ=\frac{\sqrt{10-2\sqrt{5}}}{4}}
tan
3
π
10
=
tan
54
∘
=
25
+
10
5
5
{\displaystyle \tan\frac{3\pi}{10}=\tan 54^\circ=\frac{\sqrt{25+10\sqrt{5}}}{5}\,}
cot
3
π
10
=
cot
54
∘
=
5
−
2
5
{\displaystyle \cot\frac{3\pi}{10}=\cot 54^\circ=\sqrt{5-2\sqrt{5}}\,}
60°:等边三角形
sin
π
3
=
sin
60
∘
=
3
2
{\displaystyle \sin\frac{\pi}{3}=\sin 60^\circ=\frac{\sqrt3}{2}\,}
cos
π
3
=
cos
60
∘
=
1
2
{\displaystyle \cos\frac{\pi}{3}=\cos 60^\circ=\frac{1}{2}\,}
tan
π
3
=
tan
60
∘
=
3
{\displaystyle \tan\frac{\pi}{3}=\tan 60^\circ=\sqrt3\,}
cot
π
3
=
cot
60
∘
=
3
3
=
1
3
{\displaystyle \cot\frac{\pi}{3}=\cot 60^\circ=\frac{\sqrt3}{3}=\frac{1}{\sqrt3}\,}
67.5°:7.5°与60°的和
sin
3
π
8
=
sin
67.5
∘
=
1
2
2
+
2
{\displaystyle \sin\frac{3\pi}{8}=\sin 67.5^\circ=\tfrac{1}{2}\sqrt{2+\sqrt{2}}\,}
cos
3
π
8
=
cos
67.5
∘
=
1
2
2
−
2
{\displaystyle \cos\frac{3\pi}{8}=\cos 67.5^\circ=\tfrac{1}{2}\sqrt{2-\sqrt{2}}\,}
tan
3
π
8
=
tan
67.5
∘
=
2
+
1
{\displaystyle \tan\frac{3\pi}{8}=\tan 67.5^\circ=\sqrt{2}+1\,}
cot
3
π
8
=
cot
67.5
∘
=
2
−
1
{\displaystyle \cot\frac{3\pi}{8}=\cot 67.5^\circ=\sqrt{2}-1\,}
72°:36°与36°的和
sin
2
π
5
=
sin
72
∘
=
1
4
2
(
5
+
5
)
{\displaystyle \sin\frac{2\pi}{5}=\sin 72^\circ=\tfrac{1}{4}\sqrt{2\left(5+\sqrt5\right)}\,}
cos
2
π
5
=
cos
72
∘
=
1
4
(
5
−
1
)
{\displaystyle \cos\frac{2\pi}{5}=\cos 72^\circ=\tfrac{1}{4}\left(\sqrt5-1\right)\,}
tan
2
π
5
=
tan
72
∘
=
5
+
2
5
{\displaystyle \tan\frac{2\pi}{5}=\tan 72^\circ=\sqrt{5+2\sqrt 5}\,}
cot
2
π
5
=
cot
72
∘
=
1
5
5
(
5
−
2
5
)
{\displaystyle \cot\frac{2\pi}{5}=\cot 72^\circ=\tfrac{1}{5}\sqrt{5\left(5-2\sqrt5\right)}\,}
75°: 30°与45°的和
sin
5
π
12
=
sin
75
∘
=
1
4
(
6
+
2
)
{\displaystyle \sin\frac{5\pi}{12}=\sin 75^\circ=\tfrac{1}{4}\left(\sqrt6+\sqrt2\right)\,}
cos
5
π
12
=
cos
75
∘
=
1
4
(
6
−
2
)
{\displaystyle \cos\frac{5\pi}{12}=\cos 75^\circ=\tfrac{1}{4}\left(\sqrt6-\sqrt2\right)\,}
tan
5
π
12
=
tan
75
∘
=
2
+
3
{\displaystyle \tan\frac{5\pi}{12}=\tan 75^\circ=2+\sqrt3\,}
cot
5
π
12
=
cot
75
∘
=
2
−
3
{\displaystyle \cot\frac{5\pi}{12}=\cot 75^\circ=2-\sqrt3\,}
90°:根本
sin
π
2
=
sin
90
∘
=
1
{\displaystyle \sin \frac{\pi}{2}=\sin 90^\circ=1\,}
cos
π
2
=
cos
90
∘
=
0
{\displaystyle \cos \frac{\pi}{2}=\cos 90^\circ=0\,}
cot
π
2
=
cot
90
∘
=
0
{\displaystyle \cot \frac{\pi}{2}=\cot 90^\circ=0\,}
列表
在下表中,
i
{\displaystyle i}
为虚数单位 ,
ω
=
exp
(
π
i
3
)
=
−
1
2
+
1
2
i
3
{\displaystyle \omega=\exp(\frac{\pi i}{3})=-\frac{1}{2}+\frac{1}{2}i\sqrt{3}}
。
n
{\displaystyle n}
sin
(
2
π
n
)
{\displaystyle \sin\left(\frac{2\pi}{n}\right)}
cos
(
2
π
n
)
{\displaystyle \cos\left(\frac{2\pi}{n}\right)}
tan
(
2
π
n
)
{\displaystyle \tan\left(\frac{2\pi}{n}\right)}
1
0
{\displaystyle 0}
1
{\displaystyle 1}
0
{\displaystyle 0}
2
0
{\displaystyle 0}
−
1
{\displaystyle -1}
0
{\displaystyle 0}
3
1
2
3
{\displaystyle \frac{1}{2}\sqrt{3}}
−
1
2
{\displaystyle -\frac{1}{2}}
−
3
{\displaystyle -\sqrt{3}}
4
1
{\displaystyle 1}
0
{\displaystyle 0}
±
∞
{\displaystyle \pm\infty}
5
1
4
(
10
+
2
5
)
{\displaystyle \frac{1}{4}\left(\sqrt{10+2\sqrt{5}}\right)}
1
4
(
5
−
1
)
{\displaystyle \frac{1}{4}\left(\sqrt{5}-1\right)}
5
+
2
5
{\displaystyle \sqrt{5+2\sqrt{5}}}
6
1
2
3
{\displaystyle \frac{1}{2}\sqrt{3}}
1
2
{\displaystyle \frac{1}{2}}
3
{\displaystyle \sqrt{3}}
7
1
2
1
3
(
7
−
ω
2
7
+
21
−
3
2
3
−
ω
7
−
21
−
3
2
3
)
{\displaystyle \frac{1}{2}\sqrt{\frac{1}{3}\left(7-\omega^2\sqrt[3]{\frac{7+21\sqrt{-3}}{2}}-\omega\sqrt[3]{\frac{7-21\sqrt{-3}}{2}}\right)}}
1
6
(
−
1
+
7
+
21
−
3
2
3
+
7
−
21
−
3
2
3
)
{\displaystyle \frac{1}{6}\left(-1+\sqrt[3]{\frac{7+21\sqrt{-3}}{2}}+\sqrt[3]{\frac{7-21\sqrt{-3}}{2}}\right)}
8
1
2
2
{\displaystyle \frac{1}{2}\sqrt{2}}
1
2
2
{\displaystyle \frac{1}{2}\sqrt{2}}
1
{\displaystyle 1}
9
1
4
(
4
(
−
1
+
−
3
)
3
+
4
(
−
1
−
−
3
)
3
)
{\displaystyle \frac{1}{4}\left(\sqrt[3]{4(-1+\sqrt{-3})}+\sqrt[3]{4(-1-\sqrt{-3})}\right)}
10
1
4
(
10
−
2
5
)
{\displaystyle \frac{1}{4}\left(\sqrt{10-2\sqrt{5}}\right)}
1
4
(
5
+
1
)
{\displaystyle \frac{1}{4}\left(\sqrt{5}+1\right)}
5
−
2
5
{\displaystyle \sqrt{5-2\sqrt{5}}}
11
12
1
2
{\displaystyle \frac{1}{2}}
1
2
3
{\displaystyle \frac{1}{2}\sqrt{3}}
1
3
3
{\displaystyle \frac{1}{3}\sqrt{3}}
13
14
1
24
3
(
112
−
14336
+
−
5549064193
3
−
14336
−
−
5549064193
3
)
{\displaystyle \frac{1}{24}\sqrt{3\left(112-\sqrt[3]{14336+\sqrt{-5549064193}}-\sqrt[3]{14336-\sqrt{-5549064193}}\right)} }
1
24
3
(
80
+
14336
+
−
5549064193
3
+
14336
−
−
5549064193
3
)
{\displaystyle \frac{1}{24}\sqrt{3\left(80+\sqrt[3]{14336+\sqrt{-5549064193}}+\sqrt[3]{14336-\sqrt{-5549064193}}\right)} }
112
−
14336
+
−
5549064193
3
−
14336
−
−
5549064193
3
80
+
14336
+
−
5549064193
3
+
14336
−
−
5549064193
3
{\displaystyle \sqrt{\frac{112-\sqrt[3]{14336+\sqrt{-5549064193}}-\sqrt[3]{14336-\sqrt{-5549064193}}}{80+\sqrt[3]{14336+\sqrt{-5549064193}}+\sqrt[3]{14336-\sqrt{-5549064193}}}}}
15
1
8
(
15
+
3
−
10
−
2
5
)
{\displaystyle \frac{1}{8}\left(\sqrt{15}+\sqrt{3}-\sqrt{10-2\sqrt{5}}\right)}
1
8
(
1
+
5
+
30
−
6
5
)
{\displaystyle \frac{1}{8}\left(1+\sqrt{5}+\sqrt{30-6\sqrt{5}}\right)}
1
2
(
−
3
3
−
15
+
50
+
22
5
)
{\displaystyle \frac{1}{2}\left(-3\sqrt{3}-\sqrt{15}+\sqrt{50+22\sqrt{5}}\right)}
16
1
2
(
2
−
2
)
{\displaystyle \frac{1}{2}\left(\sqrt{2-\sqrt{2}}\right)}
1
2
(
2
+
2
)
{\displaystyle \frac{1}{2}\left(\sqrt{2+\sqrt{2}}\right)}
2
−
1
{\displaystyle \sqrt{2}-1}
17
1
4
8
−
2
(
15
+
17
+
34
−
2
17
−
2
17
+
3
17
−
170
+
38
17
)
{\displaystyle \frac{1}{4}\sqrt{8-\sqrt{2\left(15+\sqrt{17}+\sqrt{34-2\sqrt{17}}-2\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}\right)}}}
1
16
(
−
1
+
17
+
34
−
2
17
+
2
17
+
3
17
−
34
−
2
17
−
2
34
+
2
17
)
{\displaystyle \frac{1}{16}\left(-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}}}\right)}
18
1
4
(
4
−
1
−
4
3
3
−
4
−
1
+
4
3
3
)
{\displaystyle \frac{1}{4}\left(\sqrt[3]{4\sqrt{-1}-4\sqrt{3}}-\sqrt[3]{4\sqrt{-1}+4\sqrt{3}}\right)}
1
4
(
4
+
4
−
3
3
+
4
−
4
−
3
3
)
{\displaystyle \frac{1}{4}\left(\sqrt[3]{4+4\sqrt{-3}}+\sqrt[3]{4-4\sqrt{-3}}\right)}
19
20
1
4
(
5
−
1
)
{\displaystyle \frac{1}{4}\left(\sqrt{5}-1\right)}
1
4
(
10
+
2
5
)
{\displaystyle \frac{1}{4}\left(\sqrt{10+2\sqrt{5}}\right)}
1
5
(
25
−
10
5
)
{\displaystyle \frac{1}{5}\left(\sqrt{25-10\sqrt{5}}\right)}
21
22
23
24
1
4
(
6
−
2
)
{\displaystyle \frac{1}{4}\left(\sqrt{6}-\sqrt{2}\right)}
1
4
(
6
+
2
)
{\displaystyle \frac{1}{4}\left(\sqrt{6}+\sqrt{2}\right)}
2
−
3
{\displaystyle 2-\sqrt{3}}
相关
参见
参考文献
埃里克·韦斯坦因 . Constructible polygon . MathWorld .
埃里克·韦斯坦因 . 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 JSTOR 3647881
注释
↑ 由Wolfram Alpha 验算:[1]
↑ 使用Mathematica验算,代码为N[ArcSin[(1 + Sqrt[3] I)/16 Power[4 Sqrt[30] - 8 Sqrt[15 + 3 Sqrt[5]] + 8 Sqrt[5 + Sqrt[5]] + 4 Sqrt[10] - 4 Sqrt[6] - 4 Sqrt[2] + (4 Sqrt[30] + 8 Sqrt[15 + 3 Sqrt[5]] + 8 Sqrt[5 + Sqrt[5]] - 4 Sqrt[10] - 4 Sqrt[6] + 4 Sqrt[2]) I, (3)^-1] + (1 - Sqrt[3] I)/16 Power[4 Sqrt[30] - 8 Sqrt[15 + 3 Sqrt[5]] + 8 Sqrt[5 + Sqrt[5]] + 4 Sqrt[10] - 4 Sqrt[6] - 4 Sqrt[2] - (4 Sqrt[30] + 8 Sqrt[15 + 3 Sqrt[5]] + 8 Sqrt[5 + Sqrt[5]] - 4 Sqrt[10] - 4 Sqrt[6] + 4 Sqrt[2]) I, (3)^-1]], 100]/Degree结果为1与原角度无误差