| 三角学 |
| --- |
|  |
| * 历史（英语：）
* 三角函数 (反三角函数)
* 广义三角函数（英语：）
 |
| 
参考 |
| * 恒等式
* 精确值
* 三角表（英语：）
* 单位圆
 |
| 
定理 |
| * 正弦
* 餘弦
* 正切
* 餘切
* 勾股定理
 |
| 微积分 |
| * 三角换元法
* 积分 (反三角函数)
* 微分
 |
| * 
* 论
* 编
 |

**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。

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，{\displaystyle \pm {\frac {1}{2}}}，±1。

相同角度的轉換表| 角度單位 | 值 |
| --- | --- |
| 轉 | {\displaystyle 0} | {\displaystyle {\frac {1}{12}}} | {\frac {1}{8}} | {\displaystyle {\frac {1}{6}}} | \frac{1}{4} | {\frac {1}{2}} | \frac{3}{4} | 1 |
| 角度 | {\displaystyle 0^{\circ }} | {\displaystyle 30^{\circ }} | {\displaystyle 45^{\circ }} | {\displaystyle 60^{\circ }} | 90^{\circ} | {\displaystyle 180^{\circ }} | {\displaystyle 270^{\circ }} | {\displaystyle 360^{\circ }} |
| 弧度 | {\displaystyle 0} | \frac{\pi}{6} | {\frac {\pi }{4}} | \frac{\pi}{3} | \frac{\pi}{2} | \pi  | {\frac {3\pi }{2}} | 2\pi  |
| 梯度 | 0^g | {\displaystyle 33{\frac {1}{3}}^{g}} | 50^g | {\displaystyle 66{\frac {2}{3}}^{g}} | 100^g | 200^g | 300^g | 400^g |

計算方式
----

### 基於常識

例如：0°、30°、45°

單位圓單位圓
### 經由半角公式的計算

參見：半角公式
例如：15°、22.5°

\sin \left({\frac {x}{2}}\right)=\pm \,{\sqrt {{\tfrac {1}{2}}(1-\cos x)}}
\cos \left({\frac {x}{2}}\right)=\pm \,{\sqrt {{\tfrac {1}{2}}(1+\cos x)}}
### 利用三倍角公式求{\displaystyle {\frac {1}{3}}\,}角

參見：三倍角公式
例如：10°、20°、7°......等等，非三的倍數的角的精確值。

* \sin 3\theta = 3 \sin \theta- 4 \sin^3\theta \,

* \cos 3\theta = 4 \cos^3\theta - 3 \cos \theta \,

把它改為

* \sin \theta =3\sin {\frac {1}{3}}\theta -4\sin ^{3}{\frac {1}{3}}\theta \,
* \cos \theta =4\cos ^{3}{\frac {1}{3}}\theta -3\cos {\frac {1}{3}}\theta \,

把\cos {\frac {1}{3}}\theta \,當成未知數，\cos \theta \,當成常數項
解一元三次方程式即可求出

例如：\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}}}}}}

同樣地，若角度代未知數，則會得到三分之一角公式。

### 经由欧拉公式的计算

更多：歐拉公式
* \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 {\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^{\circ }}={\frac {1}{2i}}\left({\sqrt[ {3}]{\cos {3^{\circ }}+i\sin {3^{\circ }}}}-{\sqrt[ {3}]{\cos {3^{\circ }}-i\sin {3^{\circ }}}}\right)
{\displaystyle ={\frac {1}{4{\sqrt[{3}]{2}}i}}{\Bigg \{}{\sqrt[{3}]{\left[2(1+{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}-1)\right]+i\left[2(1-{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}+1)\right]}}}
{\displaystyle -{\sqrt[{3}]{\left[2(1+{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}-1)\right]-i\left[2(1-{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}+1)\right]}}{\Bigg \}}}
### 經由和角公式的計算

例如：21° = 9° + 12°

\sin(x\pm y)=\sin(x)\cos(y)\pm \cos(x)\sin(y)\,
\cos(x\pm y)=\cos(x)\cos(y)\mp \sin(x)\sin(y)\,
### 經由托勒密定理的計算

更多：托勒密定理和弦 (幾何)
Chord(36°) = a/b = 1/φ, 根据托勒密定理
例如：18°

根據托勒密定理，在圓內接四邊形ABCD中，

{\displaystyle a^{2}+ab=b^{2}}
{\displaystyle \left({\frac {a}{b}}\right)^{2}+{\frac {a}{b}}=1}
{\displaystyle \mathrm {crd} \ {36^{\circ }}=\mathrm {crd} \left(\angle \mathrm {ADB} \right)={\frac {a}{b}}={\frac {{\sqrt {5}}-1}{2}}}
{\mathrm {crd}}\ {\theta }=2\sin {{\frac {\theta }{2}}}\,
{\displaystyle \sin {18^{\circ }}={\frac {{\sqrt {5}}-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\,
\cos 0=1\,
\tan 0=0\,
### 1°：2°的一半

\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}}+
{\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}}
### 1.5°：正一百二十边形

{\displaystyle \sin \left({\frac {\pi }{120}}\right)=\sin \left(1.5^{\circ }\right)={\frac {\left({\sqrt {2+{\sqrt {2}}}}\right)\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)-\left({\sqrt {2-{\sqrt {2}}}}\right)\left({\sqrt {30-6{\sqrt {5}}}}+{\sqrt {5}}+1\right)}{16}}}
{\displaystyle \cos \left({\frac {\pi }{120}}\right)=\cos \left(1.5^{\circ }\right)={\frac {\left({\sqrt {2+{\sqrt {2}}}}\right)\left({\sqrt {30-6{\sqrt {5}}}}+{\sqrt {5}}+1\right)+\left({\sqrt {2-{\sqrt {2}}}}\right)\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)}{16}}}
### 1.875°：正九十六边形

{\displaystyle \sin \left({\frac {\pi }{96}}\right)=\sin \left(1.875^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}}}
{\displaystyle \cos \left({\frac {\pi }{96}}\right)=\cos \left(1.875^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}}}
{\displaystyle \tan \left({\frac {\pi }{96}}\right)=\tan \left(1.875^{\circ }\right)={\frac {\sqrt {2-{\sqrt {{\sqrt {{\sqrt {{\sqrt {3}}+2}}+2}}+2}}}}{\sqrt {{\sqrt {{\sqrt {{\sqrt {{\sqrt {3}}+2}}+2}}+2}}+2}}}}
### 2°：6°的三分之一

\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)
{\displaystyle ={\frac {1}{4i}}{\Bigg \{}{\sqrt[{3}]{\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]+i\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]}}}
{\displaystyle -{\sqrt[{3}]{\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]-i\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]}}{\Bigg \}}}
\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)
{\displaystyle ={\frac {1}{4}}{\Bigg \{}{\sqrt[{3}]{\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]+i\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]}}}
{\displaystyle +{\sqrt[{3}]{\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]-i\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]}}{\Bigg \}}}
### 2.25°：正八十边形

{\displaystyle \sin \left({\frac {\pi }{80}}\right)=\sin \left(2.25^{\circ }\right)={\frac {\sqrt {-2{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+8}}{4}}}
{\displaystyle \cos \left({\frac {\pi }{80}}\right)=\cos \left(2.25^{\circ }\right)={\frac {\sqrt {2{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+8}}{4}}}
{\displaystyle \tan \left({\frac {\pi }{80}}\right)=\tan \left(2.25^{\circ }\right)={\frac {\sqrt {-{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}}}
{\displaystyle \cot \left({\frac {\pi }{80}}\right)=\cot \left(2.25^{\circ }\right)={\frac {\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}{\sqrt {-{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}}}
{\displaystyle \sec \left({\frac {\pi }{80}}\right)=\sec \left(2.25^{\circ }\right)={\frac {2{\sqrt {2}}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}}}
{\displaystyle \csc \left({\frac {\pi }{80}}\right)=\csc \left(2.25^{\circ }\right)={\frac {2{\sqrt {2}}}{\sqrt {-{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}}}
### 2.8125°：正六十四边形

{\displaystyle \sin \left({\frac {\pi }{64}}\right)=\sin \left(2.8125^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}}}}
{\displaystyle \cos \left({\frac {\pi }{64}}\right)=\cos \left(2.8125^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}}}}
### 3°：正六十边形

\sin {\frac {\pi }{60}}=\sin 3^{\circ }={\tfrac {1}{4}}{\sqrt {8-{\sqrt 3}-{\sqrt {15}}-{\sqrt {10-2{\sqrt 5}}}}}\,
\cos {\frac {\pi }{60}}=\cos 3^{\circ }={\tfrac {1}{4}}{\sqrt {8+{\sqrt 3}+{\sqrt {15}}+{\sqrt {10-2{\sqrt 5}}}}}\,
\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]\,
### 3.75°：正四十八边形

{\displaystyle \sin \left({\frac {\pi }{48}}\right)=\sin \left(3.75^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}
{\displaystyle \cos \left({\frac {\pi }{48}}\right)=\cos \left(3.75^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}
### 4°：12°的三分之一

\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)
{\displaystyle ={\frac {1}{4i}}{\Bigg \{}{\sqrt[{3}]{\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]+i\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]}}}
{\displaystyle -{\sqrt[{3}]{\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]-i\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]}}{\Bigg \}}}
\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)
{\displaystyle ={\frac {1}{4}}{\Bigg \{}{\sqrt[{3}]{\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]+i\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]}}}
{\displaystyle +{\sqrt[{3}]{\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]-i\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]}}{\Bigg \}}}
### 4.5°：正四十边形

{\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}}}}}}}}
{\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 {\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

{\displaystyle \sin \left({\frac {\pi }{32}}\right)=\sin \left(5.625^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}}
{\displaystyle \cos \left({\frac {\pi }{32}}\right)=\cos \left(5.625^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}}
### 6°：正三十边形

\sin {\frac {\pi }{30}}=\sin 6^{\circ }={\tfrac {1}{8}}\left[{\sqrt {6(5-{\sqrt 5})}}-{\sqrt 5}-1\right]\,
\cos {\frac {\pi }{30}}=\cos 6^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2(5-{\sqrt 5})}}+{\sqrt 3}({\sqrt 5}+1)\right]\,
\tan {\frac {\pi }{30}}=\tan 6^{\circ }={\tfrac {1}{2}}\left[{\sqrt {2(5-{\sqrt 5})}}-{\sqrt 3}({\sqrt 5}-1)\right]\,
\cot {\frac {\pi }{30}}=\cot 6^{\circ }={\tfrac {1}{2}}\left({\sqrt {50+22{\sqrt 5}}}+3{\sqrt 3}+{\sqrt {15}}\right)\,
\sec {\frac {\pi }{30}}=\sec 6^{\circ }={\sqrt 3}-{\sqrt {5-2{\sqrt 5}}}\,
\csc {\frac {\pi }{30}}=\csc 6^{\circ }=2+{\sqrt 5}+{\sqrt {15+6{\sqrt 5}}}\,
### 7.5°：正二十四边形

\sin {\frac {\pi }{24}}=\sin 7.5^{\circ }={\tfrac {1}{4}}{\sqrt {8-2{\sqrt 6}-2{\sqrt 2}}}\,
\cos {\frac {\pi }{24}}=\cos 7.5^{\circ }={\tfrac {1}{4}}{\sqrt {8+2{\sqrt 6}+2{\sqrt 2}}}\,
\tan {\frac {\pi }{24}}=\tan 7.5^{\circ }={\sqrt 6}+{\sqrt 2}-2-{\sqrt 3}\,
\cot {\frac {\pi }{24}}=\cot 7.5^{\circ }={\sqrt 6}+{\sqrt 2}+2+{\sqrt 3}\,
\sec {\frac {\pi }{24}}=\sec 7.5^{\circ }={\sqrt {16-6{\sqrt 6}-10{\sqrt 2}+8{\sqrt 3}}}\,
\csc {\frac {\pi }{24}}=\csc 7.5^{\circ }={\sqrt {16+6{\sqrt 6}+10{\sqrt 2}+8{\sqrt 3}}}\,
### 9°：正二十边形

更多：二十边形
\sin {\frac {\pi }{20}}=\sin 9^{\circ }={\tfrac {1}{4}}{\sqrt {8-2{\sqrt {10+2{\sqrt 5}}}}}\,
\cos {\frac {\pi }{20}}=\cos 9^{\circ }={\tfrac {1}{4}}{\sqrt {8+2{\sqrt {10+2{\sqrt 5}}}}}\,
\tan {\frac {\pi }{20}}=\tan 9^{\circ }={\sqrt 5}+1-{\sqrt {5+2{\sqrt 5}}}\,
\cot {\frac {\pi }{20}}=\cot 9^{\circ }={\sqrt 5}+1+{\sqrt {5+2{\sqrt 5}}}\,
### 10°：正十八边形

更多：十八边形
{\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 {1}{{\sqrt 3}}}}\,
### 11.25°：正十六边形

{\displaystyle \sin {\frac {\pi }{16}}=\sin 11.25^{\circ }={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2}}}}}}}
{\displaystyle \cos {\frac {\pi }{16}}=\cos 11.25^{\circ }={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}
{\displaystyle \tan {\frac {\pi }{16}}=\tan 11.25^{\circ }={\sqrt {4+2{\sqrt {2}}}}-{\sqrt {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

更多：十五边形
\sin {\frac {\pi }{15}}=\sin 12^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2(5+{\sqrt 5})}}-{\sqrt 3}({\sqrt 5}-1)\right]\,
\cos {\frac {\pi }{15}}=\cos 12^{\circ }={\tfrac {1}{8}}\left[{\sqrt {6(5+{\sqrt 5})}}+{\sqrt 5}-1\right]\,
\tan {\frac {\pi }{15}}=\tan 12^{\circ }={\tfrac {1}{2}}\left[{\sqrt 3}(3-{\sqrt 5})-{\sqrt {2(25-11{\sqrt 5})}}\right]\,
### 15°：正十二边形

更多：十二边形
{\displaystyle \sin {\frac {\pi }{12}}=\sin 15^{\circ }={\frac {1}{4}}{\sqrt {2}}\left({\sqrt {3}}-1\right)\,}
{\displaystyle \cos {\frac {\pi }{12}}=\cos 15^{\circ }={\frac {1}{4}}{\sqrt {2}}\left({\sqrt {3}}+1\right)\,}
\tan {\frac {\pi }{12}}=\tan 15^{\circ }=2-{\sqrt 3}\,
{\displaystyle \cot {\frac {\pi }{12}}=\cot 15^{\circ }=2+{\sqrt {3}}\,}
### 18°：正十边形

更多：十边形
\sin {\frac {\pi }{10}}=\sin 18^{\circ }={\tfrac {1}{4}}\left({\sqrt 5}-1\right)={\tfrac {1}{2}}\varphi ^{{-1}}\,
{\displaystyle \cos {\frac {\pi }{10}}=\cos 18^{\circ }={\tfrac {1}{4}}{\sqrt {2\left(5+{\sqrt {5}}\right)}}\,}
{\displaystyle \tan {\frac {\pi }{10}}=\tan 18^{\circ }={\tfrac {1}{5}}{\sqrt {5\left(5-2{\sqrt {5}}\right)}}\,}
### 20°：正九边形、60°的三分之一

更多：九边形
\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}}}}}}=
{\displaystyle 2^{-{\frac {4}{3}}}\left({\sqrt[{3}]{i-{\sqrt {3}}}}-{\sqrt[{3}]{i+{\sqrt {3}}}}\right)}
\cos {\frac {\pi }{9}}=\cos 20^{\circ }=
{\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 {\frac {7\pi }{60}}=\sin 21^{\circ }={\tfrac {1}{4}}{\sqrt {8+{\sqrt 3}-{\sqrt {15}}-{\sqrt {10+2{\sqrt 5}}}}}\,
\cos {\frac {7\pi }{60}}=\cos 21^{\circ }={\tfrac {1}{4}}{\sqrt {8-{\sqrt 3}+{\sqrt {15}}+{\sqrt {10+2{\sqrt 5}}}}}\,
{\displaystyle \tan {\frac {7\pi }{60}}=\tan 21^{\circ }={\tfrac {1}{4}}\left[2-\left(2+{\sqrt {3}}\right)\left(3-{\sqrt {5}}\right)\right]\left[2-{\sqrt {2\left(5+{\sqrt {5}}\right)}}\right]\,}
### 360/17°，{\mathbf {\left(21{\frac {3}{17}}\right)^{{\circ }}}}，{\mathbf {\left({\frac {360}{17}}\right)^{{\circ }}}}：正十七边形

更多：十七边形
\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}}
### 22.5°：正八边形

更多：八边形
{\displaystyle \sin {\frac {\pi }{8}}=\sin 22.5^{\circ }={\tfrac {1}{2}}\left({\sqrt {2-{\sqrt {2}}}}\right)}
{\displaystyle \cos {\frac {\pi }{8}}=\cos 22.5^{\circ }={\tfrac {1}{2}}\left({\sqrt {2+{\sqrt {2}}}}\right)\,}
\tan {\frac {\pi }{8}}=\tan 22.5^{\circ }={\sqrt {2}}-1\,
### 24°：12°的二倍

\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 {\frac {2\pi }{15}}=\cos 24^{\circ }={\tfrac {1}{8}}\left({\sqrt 6}{\sqrt {5-{\sqrt 5}}}+{\sqrt 5}+1\right)\,
\tan {\frac {2\pi }{15}}=\tan 24^{\circ }={\tfrac {1}{2}}\left[{\sqrt {2(25+11{\sqrt 5})}}-{\sqrt 3}(3+{\sqrt 5})\right]\,
### 180/7°，{\mathbf {\left(25{\frac {5}{7}}\right)^{{\circ }}}}，{\mathbf {\left({\frac {180}{7}}\right)^{{\circ }}}}：正七边形

更多：七边形
\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 {\frac {3\pi }{20}}=\sin 27^{\circ }={\tfrac {1}{8}}\left[2{\sqrt {5+{\sqrt 5}}}-{\sqrt 2}\;({\sqrt 5}-1)\right]\,
{\displaystyle \cos {\frac {3\pi }{20}}=\cos 27^{\circ }={\tfrac {1}{8}}\left[2{\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}\;\left({\sqrt {5}}-1\right)\right]\,}
\tan {\frac {3\pi }{20}}=\tan 27^{\circ }={\sqrt 5}-1-{\sqrt {5-2{\sqrt 5}}}\,
### 30°：正六边形

更多：六边形
\sin {\frac {\pi }{6}}=\sin 30^{\circ }={\tfrac {1}{2}}\,
\cos {\frac {\pi }{6}}=\cos 30^{\circ }={\tfrac {1}{2}}{\sqrt 3}\,
\tan {\frac {\pi }{6}}=\tan 30^{\circ }={\tfrac {1}{3}}{\sqrt 3}\,
### 33°：15°与18°的和

\sin {\frac {11\pi }{60}}=\sin 33^{\circ }={\tfrac {1}{4}}{\sqrt {8-{\sqrt 3}-{\sqrt {15}}+{\sqrt {10-2{\sqrt 5}}}}}\,
\cos {\frac {11\pi }{60}}=\cos 33^{\circ }={\tfrac {1}{4}}{\sqrt {8+{\sqrt 3}+{\sqrt {15}}-{\sqrt {10-2{\sqrt 5}}}}}\,
{\displaystyle \tan {\frac {11\pi }{60}}=\tan 33^{\circ }={\tfrac {1}{4}}\left(2{\sqrt {3}}-{\sqrt {5}}-1\right)\left(2{\sqrt {5+2{\sqrt {5}}}}+3+{\sqrt {5}}\right)\,}
{\displaystyle \cot {\frac {11\pi }{60}}=\cot 33^{\circ }={\tfrac {1}{4}}\left(2{\sqrt {3}}+{\sqrt {5}}+1\right)\left(2{\sqrt {5+2{\sqrt {5}}}}-3-{\sqrt {5}}\right)\,}
### 36°：正五边形

更多：五边形
{\displaystyle \sin {\frac {\pi }{5}}=\sin 36^{\circ }={\tfrac {1}{4}}\left[{\sqrt {2\left(5-{\sqrt {5}}\right)}}\right]\,}
\cos {\frac {\pi }{5}}=\cos 36^{\circ }={\frac {1+{\sqrt 5}}{4}}={\tfrac {1}{2}}\varphi \,
\tan {\frac {\pi }{5}}=\tan 36^{\circ }={\sqrt {5-2{\sqrt 5}}}\,
### 39°：18°与21°的和

\sin {\frac {13\pi }{60}}=\sin 39^{\circ }={\tfrac 1{4}}{\sqrt {8-{\sqrt 3}+{\sqrt {15}}+{\sqrt {10+2{\sqrt 5}}}}}\,
\cos {\frac {13\pi }{60}}=\cos 39^{\circ }={\tfrac 1{4}}{\sqrt {8+{\sqrt 3}-{\sqrt {15}}+{\sqrt {10+2{\sqrt 5}}}}}\,
{\displaystyle \tan {\frac {13\pi }{60}}=\tan 39^{\circ }={\tfrac {1}{4}}\left[\left(2-{\sqrt {3}}\right)\left(3-{\sqrt {5}}\right)-2\right]\left[2-{\sqrt {2\left(5+{\sqrt {5}}\right)}}\right]\,}
### 42°：21°的2倍

\sin {\frac {7\pi }{30}}=\sin 42^{\circ }={\frac {{\sqrt 6}{\sqrt {5+{\sqrt 5}}}-{\sqrt 5}+1}{8}}\,
{\displaystyle \cos {\frac {7\pi }{30}}=\cos 42^{\circ }={\frac {{\sqrt {2}}{\sqrt {5+{\sqrt {5}}}}+{\sqrt {3}}\left({\sqrt {5}}-1\right)}{8}}\,}
{\displaystyle \tan {\frac {7\pi }{30}}=\tan 42^{\circ }={\frac {1}{2}}\left({\sqrt {3}}+{\sqrt {15}}-{\sqrt {10+2{\sqrt {5}}}}\right)\,}
{\displaystyle \cot {\frac {7\pi }{30}}=\cot 42^{\circ }={\frac {1}{2}}\left(3{\sqrt {3}}-{\sqrt {15}}+{\sqrt {50-22{\sqrt {5}}}}\right)\,}
\sec {\frac {7\pi }{30}}=\sec 42^{\circ }={\sqrt {5+2{\sqrt 5}}}-{\sqrt 3}\,
\sec {\frac {7\pi }{30}}=\sec 42^{\circ }={\sqrt {15-6{\sqrt 5}}}+{\sqrt 5}-2\,
### 45°：正方形

更多：正方形
\sin {\frac {\pi }{4}}=\sin 45^{\circ }={\frac {{\sqrt 2}}{2}}={\frac {1}{{\sqrt 2}}}\,
\cos {\frac {\pi }{4}}=\cos 45^{\circ }={\frac {{\sqrt 2}}{2}}={\frac {1}{{\sqrt 2}}}\,
\tan {\frac {\pi }{4}}=\tan 45^{\circ }=1
### 54°：27°与27°的和

{\displaystyle \sin {\frac {3\pi }{10}}=\sin 54^{\circ }={\frac {{\sqrt {5}}+1}{4}}\,\!}
{\displaystyle \cos {\frac {3\pi }{10}}=\cos 54^{\circ }={\frac {\sqrt {10-2{\sqrt {5}}}}{4}}}
{\displaystyle \tan {\frac {3\pi }{10}}=\tan 54^{\circ }={\frac {\sqrt {25+10{\sqrt {5}}}}{5}}\,}
{\displaystyle \cot {\frac {3\pi }{10}}=\cot 54^{\circ }={\sqrt {5-2{\sqrt {5}}}}\,}
### 60°：等边三角形

{\displaystyle \sin {\frac {\pi }{3}}=\sin 60^{\circ }={\frac {\sqrt {3}}{2}}\,}
{\displaystyle \cos {\frac {\pi }{3}}=\cos 60^{\circ }={\frac {1}{2}}\,}
{\displaystyle \tan {\frac {\pi }{3}}=\tan 60^{\circ }={\sqrt {3}}\,}
{\displaystyle \cot {\frac {\pi }{3}}=\cot 60^{\circ }={\frac {\sqrt {3}}{3}}={\frac {1}{\sqrt {3}}}\,}
### 67.5°：7.5°与60°的和

{\displaystyle \sin {\frac {3\pi }{8}}=\sin 67.5^{\circ }={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2}}}}\,}
{\displaystyle \cos {\frac {3\pi }{8}}=\cos 67.5^{\circ }={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2}}}}\,}
{\displaystyle \tan {\frac {3\pi }{8}}=\tan 67.5^{\circ }={\sqrt {2}}+1\,}
{\displaystyle \cot {\frac {3\pi }{8}}=\cot 67.5^{\circ }={\sqrt {2}}-1\,}
### 72°：36°的二倍

{\displaystyle \sin {\frac {2\pi }{5}}=\sin 72^{\circ }={\tfrac {1}{4}}{\sqrt {2\left(5+{\sqrt {5}}\right)}}\,}
{\displaystyle \cos {\frac {2\pi }{5}}=\cos 72^{\circ }={\tfrac {1}{4}}\left({\sqrt {5}}-1\right)\,}
{\displaystyle \tan {\frac {2\pi }{5}}=\tan 72^{\circ }={\sqrt {5+2{\sqrt {5}}}}\,}
{\displaystyle \cot {\frac {2\pi }{5}}=\cot 72^{\circ }={\tfrac {1}{5}}{\sqrt {5\left(5-2{\sqrt {5}}\right)}}\,}
### 75°: 30°与45°的和

{\displaystyle \sin {\frac {5\pi }{12}}=\sin 75^{\circ }={\tfrac {1}{4}}\left({\sqrt {6}}+{\sqrt {2}}\right)\,}
{\displaystyle \cos {\frac {5\pi }{12}}=\cos 75^{\circ }={\tfrac {1}{4}}\left({\sqrt {6}}-{\sqrt {2}}\right)\,}
{\displaystyle \tan {\frac {5\pi }{12}}=\tan 75^{\circ }=2+{\sqrt {3}}\,}
{\displaystyle \cot {\frac {5\pi }{12}}=\cot 75^{\circ }=2-{\sqrt {3}}\,}
### 90°：根本

{\displaystyle \sin {\frac {\pi }{2}}=\sin 90^{\circ }=1\,}
{\displaystyle \cos {\frac {\pi }{2}}=\cos 90^{\circ }=0\,}
{\displaystyle \cot {\frac {\pi }{2}}=\cot 90^{\circ }=0\,}

列表
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在下表中，i為虛數單位，{\displaystyle \omega =\exp({\frac {\pi i}{3}})=-{\frac {1}{2}}+{\frac {1}{2}}i{\sqrt {3}}}。

|  |  |  |  |
| --- | --- | --- | --- |
| n | {\displaystyle \sin \left({\frac {2\pi }{n}}\right)} | {\displaystyle \cos \left({\frac {2\pi }{n}}\right)} | {\displaystyle \tan \left({\frac {2\pi }{n}}\right)} |
| 1 | {\displaystyle 0} | 1 | {\displaystyle 0} |
| 2 | {\displaystyle 0} | -1 | {\displaystyle 0} |
| 3 | {\displaystyle {\frac {1}{2}}{\sqrt {3}}} | {\displaystyle -{\frac {1}{2}}} | {\displaystyle -{\sqrt {3}}} |
| 4 | 1 | {\displaystyle 0} | \pm\infty |
| 5 | {\displaystyle {\frac {1}{4}}\left({\sqrt {10+2{\sqrt {5}}}}\right)} | {\displaystyle {\frac {1}{4}}\left({\sqrt {5}}-1\right)} | {\displaystyle {\sqrt {5+2{\sqrt {5}}}}} |
| 6 | {\displaystyle {\frac {1}{2}}{\sqrt {3}}} | {\frac {1}{2}} | \sqrt{3} |
| 7 | {\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)}}} | {\displaystyle {\frac {1}{6}}\left(-1+{\sqrt[{3}]{\frac {7+21{\sqrt {-3}}}{2}}}+{\sqrt[{3}]{\frac {7-21{\sqrt {-3}}}{2}}}\right)} |  |
| 8 | {\displaystyle {\frac {1}{2}}{\sqrt {2}}} | {\displaystyle {\frac {1}{2}}{\sqrt {2}}} | 1 |
| 9 | {\displaystyle {\frac {1}{4}}\left({\sqrt[{3}]{4(-1+{\sqrt {-3}})}}+{\sqrt[{3}]{4(-1-{\sqrt {-3}})}}\right)} |  |  |
| 10 | {\displaystyle {\frac {1}{4}}\left({\sqrt {10-2{\sqrt {5}}}}\right)} | {\displaystyle {\frac {1}{4}}\left({\sqrt {5}}+1\right)} | {\displaystyle {\sqrt {5-2{\sqrt {5}}}}} |
| 11 |  |  |  |
| 12 | {\frac {1}{2}} | {\displaystyle {\frac {1}{2}}{\sqrt {3}}} | {\displaystyle {\frac {1}{3}}{\sqrt {3}}} |
| 13 |  |  |  |
| 14 | {\displaystyle {\frac {1}{24}}{\sqrt {3\left(112-{\sqrt[{3}]{14336+{\sqrt {-5549064193}}}}-{\sqrt[{3}]{14336-{\sqrt {-5549064193}}}}\right)}}} | {\displaystyle {\frac {1}{24}}{\sqrt {3\left(80+{\sqrt[{3}]{14336+{\sqrt {-5549064193}}}}+{\sqrt[{3}]{14336-{\sqrt {-5549064193}}}}\right)}}} | {\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 | {\displaystyle {\frac {1}{8}}\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)} | {\displaystyle {\frac {1}{8}}\left(1+{\sqrt {5}}+{\sqrt {30-6{\sqrt {5}}}}\right)} | {\displaystyle {\frac {1}{2}}\left(-3{\sqrt {3}}-{\sqrt {15}}+{\sqrt {50+22{\sqrt {5}}}}\right)} |
| 16 | {\displaystyle {\frac {1}{2}}\left({\sqrt {2-{\sqrt {2}}}}\right)} | {\displaystyle {\frac {1}{2}}\left({\sqrt {2+{\sqrt {2}}}}\right)} | {\displaystyle {\sqrt {2}}-1} |
| 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)}}}}} | {\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 | {\displaystyle {\frac {1}{4}}\left({\sqrt[{3}]{4{\sqrt {-1}}-4{\sqrt {3}}}}-{\sqrt[{3}]{4{\sqrt {-1}}+4{\sqrt {3}}}}\right)} | {\displaystyle {\frac {1}{4}}\left({\sqrt[{3}]{4+4{\sqrt {-3}}}}+{\sqrt[{3}]{4-4{\sqrt {-3}}}}\right)} |  |
| 19 |  |  |  |
| 20 | {\displaystyle {\frac {1}{4}}\left({\sqrt {5}}-1\right)} | {\displaystyle {\frac {1}{4}}\left({\sqrt {10+2{\sqrt {5}}}}\right)} | {\displaystyle {\frac {1}{5}}\left({\sqrt {25-10{\sqrt {5}}}}\right)} |
| 21 |  |  |  |
| 22 |  |  |  |
| 23 |  |  |  |
| 24 | {\displaystyle {\frac {1}{4}}\left({\sqrt {6}}-{\sqrt {2}}\right)} | {\displaystyle {\frac {1}{4}}\left({\sqrt {6}}+{\sqrt {2}}\right)} | 2-\sqrt{3} |

相關
--

更多：三角函數和三角恆等式
更多：en:Generating trigonometric tables

參見
--

* 可作图多边形
* 三角數（英语：）
* 十七邊形

參考文獻
----

* 埃里克·韦斯坦因.. MathWorld.
* 埃里克·韦斯坦因.. MathWorld. 
	+ π/3 (60°)（页面存档备份，存于）—π/6 (30°)（页面存档备份，存于）—π/12 (15°)（页面存档备份，存于）—π/24 (7.5°)（页面存档备份，存于）
	+ π/4 (45°)（页面存档备份，存于）—π/8 (22.5°)（页面存档备份，存于）—π/16 (11.25°)（页面存档备份，存于）—π/32 (5.625°)（页面存档备份，存于）
	+ π/5 (36°)（页面存档备份，存于）—π/10 (18°)（页面存档备份，存于）—π/20 (9°)（页面存档备份，存于）
	+ π/7（页面存档备份，存于）—*π/14*
	+ π/9 (20°)（页面存档备份，存于）—π/18 (10°)（页面存档备份，存于）
	+ π/11（页面存档备份，存于）
	+ π/13（页面存档备份，存于）
	+ π/15 (12°)（页面存档备份，存于）—π/30 (6°)（页面存档备份，存于）
	+ π/17（页面存档备份，存于）
	+ *π/19*
	+ π/23（页面存档备份，存于）
* Bracken, Paul; Cizek, Jiri.. Int. J. Quantum Chemistry. 2002, **90** (1): 42–53. doi:10.1002/qua.1803.
* Conway, John H.; Radin, Charles; Radun, Lorenzo.. 1998. arXiv:math-ph/9812019 可免费查阅.
* Conway, John H.; Radin, Charles; Radun, Lorenzo.. Disc. Comput. Geom. 1999, **22** (3): 321–332. doi:10.1007/PL00009463.  MR1706614
* Girstmair, Kurt.. Acta Arithmetica. 1997, **81**: 387–398.  MR1472818
* Gurak, S.. Mathematics of Computation. 2006, **75** (256): 2021–2035. Bibcode:2006MaCom..75.2021G. doi:10.1090/S0025-5718-06-01885-0.  MR2240647
* Servi, L. D.. Am. Math. Monthly. 2003, **110** (4): 326–330. doi:10.2307/3647881.  MR1984573 JSTOR 3647881