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。

根據尼雲定理，有理數度數的角的正弦值，其中的有理數僅有 0， $\pm {\frac {1}{2}}$ ，±1。



計算方式

基於常識

例如：0°、30°、45°

經由半角公式的計算

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 ${\frac {1}{3}}\,$ 角

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。

經由歐拉公式的計算

 
* $\cos {\frac {\theta }{n}}=\Re \left({\sqrt[{n}]{\cos \theta +i\sin \theta }}\right)={\frac {1}{2}}\left({\sqrt[{n}]{\cos \theta +i\sin \theta }}+{\sqrt[{n}]{\cos \theta -i\sin \theta }}\right)$

* $\sin {\frac {\theta }{n}}=\Im \left({\sqrt[{n}]{\cos \theta +i\sin \theta }}\right)={\frac {1}{2i}}\left({\sqrt[{n}]{\cos \theta +i\sin \theta }}-{\sqrt[{n}]{\cos \theta -i\sin \theta }}\right)$

例如：

　
: $\sin {1^{\circ }}={\frac {1}{2i}}\left({\sqrt[{3}]{\cos {3^{\circ }}+i\sin {3^{\circ }}}}-{\sqrt[{3}]{\cos {3^{\circ }}-i\sin {3^{\circ }}}}\right)$

　
: $={\frac {1}{4{\sqrt[{3}]{2}}i}}{\Bigg \{}{\sqrt[{3}]{\left[2(1+{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}-1)\right]+i\left[2(1-{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}+1)\right]}}$

　
: $-{\sqrt[{3}]{\left[2(1+{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}-1)\right]-i\left[2(1-{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}+1)\right]}}{\Bigg \}}$

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

例如：18°

根據托勒密定理，在圓內接四邊形 ABCD 中，

　
: $a^{2}+ab=b^{2}$

　
: $\left({\frac {a}{b}}\right)^{2}+{\frac {a}{b}}=1$

　
: $\mathrm {crd} \ {36^{\circ }}=\mathrm {crd} \left(\angle \mathrm {ADB} \right)={\frac {a}{b}}={\frac {{\sqrt {5}}-1}{2}}$

　
: $\mathrm {crd} \ {\theta }=2\sin {\frac {\theta }{2}}\,$

　
: $\sin {18^{\circ }}={\frac {{\sqrt {5}}-1}{4}}$



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

　
: $\sin 0=0\,$

　
: $\cos 0=1\,$

　
: $\tan 0=0\,$

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°：正一百二十邊形

　
: $\sin \left({\frac {\pi }{120}}\right)=\sin \left(1.5^{\circ }\right)={\frac {\left({\sqrt {2+{\sqrt {2}}}}\right)\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)-\left({\sqrt {2-{\sqrt {2}}}}\right)\left({\sqrt {30-6{\sqrt {5}}}}+{\sqrt {5}}+1\right)}{16}}$

　
: $\cos \left({\frac {\pi }{120}}\right)=\cos \left(1.5^{\circ }\right)={\frac {\left({\sqrt {2+{\sqrt {2}}}}\right)\left({\sqrt {30-6{\sqrt {5}}}}+{\sqrt {5}}+1\right)+\left({\sqrt {2-{\sqrt {2}}}}\right)\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)}{16}}$

1.875°：正九十六邊形

　
: $\sin \left({\frac {\pi }{96}}\right)=\sin \left(1.875^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}}$

　
: $\cos \left({\frac {\pi }{96}}\right)=\cos \left(1.875^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}}$

　
: $\tan \left({\frac {\pi }{96}}\right)=\tan \left(1.875^{\circ }\right)={\frac {\sqrt {2-{\sqrt {{\sqrt {{\sqrt {{\sqrt {3}}+2}}+2}}+2}}}}{\sqrt {{\sqrt {{\sqrt {{\sqrt {{\sqrt {3}}+2}}+2}}+2}}+2}}}$

2°：6° 的三分之一

　
: $\sin {2^{\circ }}={\frac {1}{2i}}\left({\sqrt[{3}]{\cos {6^{\circ }}+i\sin {6^{\circ }}}}-{\sqrt[{3}]{\cos {6^{\circ }}-i\sin {6^{\circ }}}}\right)$

　
: $={\frac {1}{4i}}{\Bigg \{}{\sqrt[{3}]{\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]+i\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]}}$

　
: $-{\sqrt[{3}]{\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]-i\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]}}{\Bigg \}}$

　
: $\cos {2^{\circ }}={\frac {1}{2}}\left({\sqrt[{3}]{\cos {6^{\circ }}+i\sin {6^{\circ }}}}+{\sqrt[{3}]{\cos {6^{\circ }}-i\sin {6^{\circ }}}}\right)$

　
: $={\frac {1}{4}}{\Bigg \{}{\sqrt[{3}]{\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]+i\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]}}$

　
: $+{\sqrt[{3}]{\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]-i\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]}}{\Bigg \}}$

2.25°：正八十邊形

　
: $\sin \left({\frac {\pi }{80}}\right)=\sin \left(2.25^{\circ }\right)={\frac {\sqrt {-2{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+8}}{4}}$

　
: $\cos \left({\frac {\pi }{80}}\right)=\cos \left(2.25^{\circ }\right)={\frac {\sqrt {2{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+8}}{4}}$

　
: $\tan \left({\frac {\pi }{80}}\right)=\tan \left(2.25^{\circ }\right)={\frac {\sqrt {-{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}}$

　
: $\cot \left({\frac {\pi }{80}}\right)=\cot \left(2.25^{\circ }\right)={\frac {\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}{\sqrt {-{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}}$

　
: $\sec \left({\frac {\pi }{80}}\right)=\sec \left(2.25^{\circ }\right)={\frac {2{\sqrt {2}}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}}$

　
: $\csc \left({\frac {\pi }{80}}\right)=\csc \left(2.25^{\circ }\right)={\frac {2{\sqrt {2}}}{\sqrt {-{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}}$

2.8125°：正六十四邊形

　
: $\sin \left({\frac {\pi }{64}}\right)=\sin \left(2.8125^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}}}$

　
: $\cos \left({\frac {\pi }{64}}\right)=\cos \left(2.8125^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}}}$

3°：正六十邊形

　
: $\sin {\frac {\pi }{60}}=\sin 3^{\circ }={\tfrac {1}{4}}{\sqrt {8-{\sqrt {3}}-{\sqrt {15}}-{\sqrt {10-2{\sqrt {5}}}}}}\,$

　
: $\cos {\frac {\pi }{60}}=\cos 3^{\circ }={\tfrac {1}{4}}{\sqrt {8+{\sqrt {3}}+{\sqrt {15}}+{\sqrt {10-2{\sqrt {5}}}}}}\,$

　
: $\tan {\frac {\pi }{60}}=\tan 3^{\circ }={\tfrac {1}{4}}\left[(2-{\sqrt {3}})(3+{\sqrt {5}})-2\right]\left[2-{\sqrt {2(5-{\sqrt {5}})}}\right]\,$

3.75°：正四十八邊形

　
: $\sin \left({\frac {\pi }{48}}\right)=\sin \left(3.75^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}$

　
: $\cos \left({\frac {\pi }{48}}\right)=\cos \left(3.75^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}$

4°：12° 的三分之一

　
: $\sin {4^{\circ }}={\frac {1}{2i}}\left({\sqrt[{3}]{\cos {12^{\circ }}+i\sin {12^{\circ }}}}-{\sqrt[{3}]{\cos {12^{\circ }}-i\sin {12^{\circ }}}}\right)$

　
: $={\frac {1}{4i}}{\Bigg \{}{\sqrt[{3}]{\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]+i\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]}}$

　
: $-{\sqrt[{3}]{\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]-i\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]}}{\Bigg \}}$

　
: $\cos {4^{\circ }}={\frac {1}{2}}\left({\sqrt[{3}]{\cos {12^{\circ }}+i\sin {12^{\circ }}}}+{\sqrt[{3}]{\cos {12^{\circ }}-i\sin {12^{\circ }}}}\right)$

　
: $={\frac {1}{4}}{\Bigg \{}{\sqrt[{3}]{\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]+i\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]}}$

　
: $+{\sqrt[{3}]{\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]-i\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]}}{\Bigg \}}$

4.5°：正四十邊形

　
: $\sin \left({\frac {\pi }{40}}\right)=\sin \left(4.5^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {\frac {5+{\sqrt {5}}}{2}}}}}}}$

　
: $\cos \left({\frac {\pi }{40}}\right)=\cos \left(4.5^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {\frac {5+{\sqrt {5}}}{2}}}}}}}$

5°：15° 的三分之一、正三十六邊形

　
: $\sin {\frac {\pi }{36}}=\sin 5^{\circ }={\frac {2-2{\sqrt {3}}\mathrm {i} }{2{\sqrt[{3}]{2({\sqrt {2}}-{\sqrt {6}})}}-2-{\sqrt {3}}}}-{\frac {(1+{\sqrt {3}}\mathrm {i} ){\sqrt[{3}]{2({\sqrt {2}}-{\sqrt {6}})}}-2-{\sqrt {3}}}{8}}\,$

5.625°：正三十二邊形

　
: $\sin \left({\frac {\pi }{32}}\right)=\sin \left(5.625^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}$

　
: $\cos \left({\frac {\pi }{32}}\right)=\cos \left(5.625^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}$

6°：正三十邊形

　
: $\sin {\frac {\pi }{30}}=\sin 6^{\circ }={\tfrac {1}{8}}\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]\,$

　
: $\cos {\frac {\pi }{30}}=\cos 6^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]\,$

　
: $\tan {\frac {\pi }{30}}=\tan 6^{\circ }={\tfrac {1}{2}}\left[{\sqrt {2(5-{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]\,$

　
: $\cot {\frac {\pi }{30}}=\cot 6^{\circ }={\tfrac {1}{2}}\left({\sqrt {50+22{\sqrt {5}}}}+3{\sqrt {3}}+{\sqrt {15}}\right)\,$

　
: $\sec {\frac {\pi }{30}}=\sec 6^{\circ }={\sqrt {3}}-{\sqrt {5-2{\sqrt {5}}}}\,$

　
: $\csc {\frac {\pi }{30}}=\csc 6^{\circ }=2+{\sqrt {5}}+{\sqrt {15+6{\sqrt {5}}}}\,$

7.5°：正二十四邊形

　
: $\sin {\frac {\pi }{24}}=\sin 7.5^{\circ }={\tfrac {1}{4}}{\sqrt {8-2{\sqrt {6}}-2{\sqrt {2}}}}\,$

　
: $\cos {\frac {\pi }{24}}=\cos 7.5^{\circ }={\tfrac {1}{4}}{\sqrt {8+2{\sqrt {6}}+2{\sqrt {2}}}}\,$

　
: $\tan {\frac {\pi }{24}}=\tan 7.5^{\circ }={\sqrt {6}}+{\sqrt {2}}-2-{\sqrt {3}}\,$

　
: $\cot {\frac {\pi }{24}}=\cot 7.5^{\circ }={\sqrt {6}}+{\sqrt {2}}+2+{\sqrt {3}}\,$

　
: $\sec {\frac {\pi }{24}}=\sec 7.5^{\circ }={\sqrt {16-6{\sqrt {6}}-10{\sqrt {2}}+8{\sqrt {3}}}}\,$

　
: $\csc {\frac {\pi }{24}}=\csc 7.5^{\circ }={\sqrt {16+6{\sqrt {6}}+10{\sqrt {2}}+8{\sqrt {3}}}}\,$

9°：正二十邊形

　
: $\sin {\frac {\pi }{20}}=\sin 9^{\circ }={\tfrac {1}{4}}{\sqrt {8-2{\sqrt {10+2{\sqrt {5}}}}}}\,$

　
: $\cos {\frac {\pi }{20}}=\cos 9^{\circ }={\tfrac {1}{4}}{\sqrt {8+2{\sqrt {10+2{\sqrt {5}}}}}}\,$

　
: $\tan {\frac {\pi }{20}}=\tan 9^{\circ }={\sqrt {5}}+1-{\sqrt {5+2{\sqrt {5}}}}\,$

　
: $\cot {\frac {\pi }{20}}=\cot 9^{\circ }={\sqrt {5}}+1+{\sqrt {5+2{\sqrt {5}}}}\,$

10°：正十八邊形

　
: ${\tan 10^{\circ }=-{\frac {-1-{\sqrt {3}}{\rm {i}}}{6}}{\sqrt[{3}]{-12{\sqrt {3}}+36{\rm {i}}}}-{\frac {-1+{\sqrt {3}}{\rm {i}}}{6}}{\sqrt[{3}]{-12{\sqrt {3}}-36{\rm {i}}}}+{\frac {1}{\sqrt {3}}}}\,$

11.25°：正十六邊形

　
: $\sin {\frac {\pi }{16}}=\sin 11.25^{\circ }={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2}}}}}}$

　
: $\cos {\frac {\pi }{16}}=\cos 11.25^{\circ }={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}$

　
: $\tan {\frac {\pi }{16}}=\tan 11.25^{\circ }={\sqrt {4+2{\sqrt {2}}}}-{\sqrt {2}}-1$

　
: $\cot {\frac {\pi }{16}}=\cot 11.25^{\circ }={\sqrt {4+2{\sqrt {2}}}}+{\sqrt {2}}+1$

12°：正十五邊形

　
: $\sin {\frac {\pi }{15}}=\sin 12^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]\,$

　
: $\cos {\frac {\pi }{15}}=\cos 12^{\circ }={\tfrac {1}{8}}\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]\,$

　
: $\tan {\frac {\pi }{15}}=\tan 12^{\circ }={\tfrac {1}{2}}\left[{\sqrt {3}}(3-{\sqrt {5}})-{\sqrt {2(25-11{\sqrt {5}})}}\right]\,$

15°：正十二邊形

　
: $\sin {\frac {\pi }{12}}=\sin 15^{\circ }={\frac {1}{4}}{\sqrt {2}}\left({\sqrt {3}}-1\right)\,$

　
: $\cos {\frac {\pi }{12}}=\cos 15^{\circ }={\frac {1}{4}}{\sqrt {2}}\left({\sqrt {3}}+1\right)\,$

　
: $\tan {\frac {\pi }{12}}=\tan 15^{\circ }=2-{\sqrt {3}}\,$

　
: $\cot {\frac {\pi }{12}}=\cot 15^{\circ }=2+{\sqrt {3}}\,$

18°：正十邊形

　
: $\sin {\frac {\pi }{10}}=\sin 18^{\circ }={\tfrac {1}{4}}\left({\sqrt {5}}-1\right)={\tfrac {1}{2}}\varphi ^{-1}\,$

　
: $\cos {\frac {\pi }{10}}=\cos 18^{\circ }={\tfrac {1}{4}}{\sqrt {2\left(5+{\sqrt {5}}\right)}}\,$

　
: $\tan {\frac {\pi }{10}}=\tan 18^{\circ }={\tfrac {1}{5}}{\sqrt {5\left(5-2{\sqrt {5}}\right)}}\,$

20°：正九邊形、60° 的三分之一

　
: $\sin {\frac {\pi }{9}}=\sin 20^{\circ }={\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}+{\sqrt {-{\frac {1}{256}}}}}}+{\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}-{\sqrt {-{\frac {1}{256}}}}}}=$

　
: $2^{-{\frac {4}{3}}}\left({\sqrt[{3}]{i-{\sqrt {3}}}}-{\sqrt[{3}]{i+{\sqrt {3}}}}\right)$

　
: $\cos {\frac {\pi }{9}}=\cos 20^{\circ }=$

　
: $2^{-{\frac {4}{3}}}\left({\sqrt[{3}]{1+i{\sqrt {3}}}}+{\sqrt[{3}]{1-i{\sqrt {3}}}}\right)$

21°：9° 與 12° 的和

　
: $\sin {\frac {7\pi }{60}}=\sin 21^{\circ }={\tfrac {1}{4}}{\sqrt {8+{\sqrt {3}}-{\sqrt {15}}-{\sqrt {10+2{\sqrt {5}}}}}}\,$

　
: $\cos {\frac {7\pi }{60}}=\cos 21^{\circ }={\tfrac {1}{4}}{\sqrt {8-{\sqrt {3}}+{\sqrt {15}}+{\sqrt {10+2{\sqrt {5}}}}}}\,$

　
: $\tan {\frac {7\pi }{60}}=\tan 21^{\circ }={\tfrac {1}{4}}\left[2-\left(2+{\sqrt {3}}\right)\left(3-{\sqrt {5}}\right)\right]\left[2-{\sqrt {2\left(5+{\sqrt {5}}\right)}}\right]\,$

360/17°， $\mathbf {\left(21{\frac {3}{17}}\right)^{\circ }} $ ， $\mathbf {\left({\frac {360}{17}}\right)^{\circ }} $ ：正十七邊形

　
: $\operatorname {cos} {2\pi  \over 17}={\frac {-1+{\sqrt {17}}+{\sqrt {34-2{\sqrt {17}}}}+2{\sqrt {17+3{\sqrt {17}}-{\sqrt {34-2{\sqrt {17}}}}-2{\sqrt {34+2{\sqrt {17}}}}}}}{16}}$

22.5°：正八邊形

　
: $\sin {\frac {\pi }{8}}=\sin 22.5^{\circ }={\tfrac {1}{2}}\left({\sqrt {2-{\sqrt {2}}}}\right)$

　
: $\cos {\frac {\pi }{8}}=\cos 22.5^{\circ }={\tfrac {1}{2}}\left({\sqrt {2+{\sqrt {2}}}}\right)\,$

　
: $\tan {\frac {\pi }{8}}=\tan 22.5^{\circ }={\sqrt {2}}-1\,$

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 }} $ ：正七邊形

　
: $\cos {\frac {\pi }{7}}=\cos {\frac {180}{7}}^{\circ }=\cos 25{\frac {5}{7}}^{\circ }={\frac {1}{6}}+{\frac {1-{\sqrt {3}}i}{24}}{\sqrt[{3}]{28-84{\sqrt {3}}i}}+{\frac {1+{\sqrt {3}}i}{24}}{\sqrt[{3}]{28-84{\sqrt {3}}i}}$

27°：12° 與 15° 的和

　
: $\sin {\frac {3\pi }{20}}=\sin 27^{\circ }={\tfrac {1}{8}}\left[2{\sqrt {5+{\sqrt {5}}}}-{\sqrt {2}}\;({\sqrt {5}}-1)\right]\,$

　
: $\cos {\frac {3\pi }{20}}=\cos 27^{\circ }={\tfrac {1}{8}}\left[2{\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}\;\left({\sqrt {5}}-1\right)\right]\,$

　
: $\tan {\frac {3\pi }{20}}=\tan 27^{\circ }={\sqrt {5}}-1-{\sqrt {5-2{\sqrt {5}}}}\,$

30°：正六邊形

　
: $\sin {\frac {\pi }{6}}=\sin 30^{\circ }={\tfrac {1}{2}}\,$

　
: $\cos {\frac {\pi }{6}}=\cos 30^{\circ }={\tfrac {1}{2}}{\sqrt {3}}\,$

　
: $\tan {\frac {\pi }{6}}=\tan 30^{\circ }={\tfrac {1}{3}}{\sqrt {3}}\,$

33°：15° 與 18° 的和

　
: $\sin {\frac {11\pi }{60}}=\sin 33^{\circ }={\tfrac {1}{4}}{\sqrt {8-{\sqrt {3}}-{\sqrt {15}}+{\sqrt {10-2{\sqrt {5}}}}}}\,$

　
: $\cos {\frac {11\pi }{60}}=\cos 33^{\circ }={\tfrac {1}{4}}{\sqrt {8+{\sqrt {3}}+{\sqrt {15}}-{\sqrt {10-2{\sqrt {5}}}}}}\,$

　
: $\tan {\frac {11\pi }{60}}=\tan 33^{\circ }={\tfrac {1}{4}}\left(2{\sqrt {3}}-{\sqrt {5}}-1\right)\left(2{\sqrt {5+2{\sqrt {5}}}}+3+{\sqrt {5}}\right)\,$

　
: $\cot {\frac {11\pi }{60}}=\cot 33^{\circ }={\tfrac {1}{4}}\left(2{\sqrt {3}}+{\sqrt {5}}+1\right)\left(2{\sqrt {5+2{\sqrt {5}}}}-3-{\sqrt {5}}\right)\,$

36°：正五邊形

　
: $\sin {\frac {\pi }{5}}=\sin 36^{\circ }={\tfrac {1}{4}}\left[{\sqrt {2\left(5-{\sqrt {5}}\right)}}\right]\,$

　
: $\cos {\frac {\pi }{5}}=\cos 36^{\circ }={\frac {1+{\sqrt {5}}}{4}}={\tfrac {1}{2}}\varphi \,$

　
: $\tan {\frac {\pi }{5}}=\tan 36^{\circ }={\sqrt {5-2{\sqrt {5}}}}\,$

39°：18° 與 21° 的和

　
: $\sin {\frac {13\pi }{60}}=\sin 39^{\circ }={\tfrac {1}{4}}{\sqrt {8-{\sqrt {3}}+{\sqrt {15}}+{\sqrt {10+2{\sqrt {5}}}}}}\,$

　
: $\cos {\frac {13\pi }{60}}=\cos 39^{\circ }={\tfrac {1}{4}}{\sqrt {8+{\sqrt {3}}-{\sqrt {15}}+{\sqrt {10+2{\sqrt {5}}}}}}\,$

　
: $\tan {\frac {13\pi }{60}}=\tan 39^{\circ }={\tfrac {1}{4}}\left[\left(2-{\sqrt {3}}\right)\left(3-{\sqrt {5}}\right)-2\right]\left[2-{\sqrt {2\left(5+{\sqrt {5}}\right)}}\right]\,$

42°：21° 的 2 倍

　
: $\sin {\frac {7\pi }{30}}=\sin 42^{\circ }={\frac {{\sqrt {6}}{\sqrt {5+{\sqrt {5}}}}-{\sqrt {5}}+1}{8}}\,$

　
: $\cos {\frac {7\pi }{30}}=\cos 42^{\circ }={\frac {{\sqrt {2}}{\sqrt {5+{\sqrt {5}}}}+{\sqrt {3}}\left({\sqrt {5}}-1\right)}{8}}\,$

　
: $\tan {\frac {7\pi }{30}}=\tan 42^{\circ }={\frac {1}{2}}\left({\sqrt {3}}+{\sqrt {15}}-{\sqrt {10+2{\sqrt {5}}}}\right)\,$

　
: $\cot {\frac {7\pi }{30}}=\cot 42^{\circ }={\frac {1}{2}}\left(3{\sqrt {3}}-{\sqrt {15}}+{\sqrt {50-22{\sqrt {5}}}}\right)\,$

　
: $\sec {\frac {7\pi }{30}}=\sec 42^{\circ }={\sqrt {5+2{\sqrt {5}}}}-{\sqrt {3}}\,$

　
: $\sec {\frac {7\pi }{30}}=\sec 42^{\circ }={\sqrt {15-6{\sqrt {5}}}}+{\sqrt {5}}-2\,$

45°：正方形

　
: $\sin {\frac {\pi }{4}}=\sin 45^{\circ }={\frac {\sqrt {2}}{2}}={\frac {1}{\sqrt {2}}}\,$

　
: $\cos {\frac {\pi }{4}}=\cos 45^{\circ }={\frac {\sqrt {2}}{2}}={\frac {1}{\sqrt {2}}}\,$

　
: $\tan {\frac {\pi }{4}}=\tan 45^{\circ }=1$

54°：27° 與 27° 的和

　
: $\sin {\frac {3\pi }{10}}=\sin 54^{\circ }={\frac {{\sqrt {5}}+1}{4}}\,\!$

　
: $\cos {\frac {3\pi }{10}}=\cos 54^{\circ }={\frac {\sqrt {10-2{\sqrt {5}}}}{4}}$

　
: $\tan {\frac {3\pi }{10}}=\tan 54^{\circ }={\frac {\sqrt {25+10{\sqrt {5}}}}{5}}\,$

　
: $\cot {\frac {3\pi }{10}}=\cot 54^{\circ }={\sqrt {5-2{\sqrt {5}}}}\,$

60°：等邊三角形

　
: $\sin {\frac {\pi }{3}}=\sin 60^{\circ }={\frac {\sqrt {3}}{2}}\,$

　
: $\cos {\frac {\pi }{3}}=\cos 60^{\circ }={\frac {1}{2}}\,$

　
: $\tan {\frac {\pi }{3}}=\tan 60^{\circ }={\sqrt {3}}\,$

　
: $\cot {\frac {\pi }{3}}=\cot 60^{\circ }={\frac {\sqrt {3}}{3}}={\frac {1}{\sqrt {3}}}\,$

67.5°：7.5° 與 60° 的和

　
: $\sin {\frac {3\pi }{8}}=\sin 67.5^{\circ }={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2}}}}\,$

　
: $\cos {\frac {3\pi }{8}}=\cos 67.5^{\circ }={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2}}}}\,$

　
: $\tan {\frac {3\pi }{8}}=\tan 67.5^{\circ }={\sqrt {2}}+1\,$

　
: $\cot {\frac {3\pi }{8}}=\cot 67.5^{\circ }={\sqrt {2}}-1\,$

72°：36° 的二倍

　
: $\sin {\frac {2\pi }{5}}=\sin 72^{\circ }={\tfrac {1}{4}}{\sqrt {2\left(5+{\sqrt {5}}\right)}}\,$

　
: $\cos {\frac {2\pi }{5}}=\cos 72^{\circ }={\tfrac {1}{4}}\left({\sqrt {5}}-1\right)\,$

　
: $\tan {\frac {2\pi }{5}}=\tan 72^{\circ }={\sqrt {5+2{\sqrt {5}}}}\,$

　
: $\cot {\frac {2\pi }{5}}=\cot 72^{\circ }={\tfrac {1}{5}}{\sqrt {5\left(5-2{\sqrt {5}}\right)}}\,$

75°: 30° 與 45° 的和

　
: $\sin {\frac {5\pi }{12}}=\sin 75^{\circ }={\tfrac {1}{4}}\left({\sqrt {6}}+{\sqrt {2}}\right)\,$

　
: $\cos {\frac {5\pi }{12}}=\cos 75^{\circ }={\tfrac {1}{4}}\left({\sqrt {6}}-{\sqrt {2}}\right)\,$

　
: $\tan {\frac {5\pi }{12}}=\tan 75^{\circ }=2+{\sqrt {3}}\,$

　
: $\cot {\frac {5\pi }{12}}=\cot 75^{\circ }=2-{\sqrt {3}}\,$

90°：根本

　
: $\sin {\frac {\pi }{2}}=\sin 90^{\circ }=1\,$

　
: $\cos {\frac {\pi }{2}}=\cos 90^{\circ }=0\,$

　
: $\cot {\frac {\pi }{2}}=\cot 90^{\circ }=0\,$



列表

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



相關



參見


* 可作圖多邊形

* 三角數

* 十七邊形



參考文獻


* 埃里克・韋斯坦因. Constructible polygon. MathWorld.

* 埃里克・韋斯坦因. Trigonometry angles. MathWorld. 
  
* π/3 (60°)（頁面存檔備份，存於網際網路檔案館）—π/6 (30°)（頁面存檔備份，存於網際網路檔案館）—π/12 (15°)（頁面存檔備份，存於網際網路檔案館）—π/24 (7.5°)（頁面存檔備份，存於網際網路檔案館）
  
* π/4 (45°)（頁面存檔備份，存於網際網路檔案館）—π/8 (22.5°)（頁面存檔備份，存於網際網路檔案館）—π/16 (11.25°)（頁面存檔備份，存於網際網路檔案館）—π/32 (5.625°)（頁面存檔備份，存於網際網路檔案館）
  
* π/5 (36°)（頁面存檔備份，存於網際網路檔案館）—π/10 (18°)（頁面存檔備份，存於網際網路檔案館）—π/20 (9°)（頁面存檔備份，存於網際網路檔案館）
  
* π/7（頁面存檔備份，存於網際網路檔案館）—_π/14_
  
* π/9 (20°)（頁面存檔備份，存於網際網路檔案館）—π/18 (10°)（頁面存檔備份，存於網際網路檔案館）
  
* π/11（頁面存檔備份，存於網際網路檔案館）
  
* π/13（頁面存檔備份，存於網際網路檔案館）
  
* π/15 (12°)（頁面存檔備份，存於網際網路檔案館）—π/30 (6°)（頁面存檔備份，存於網際網路檔案館）
  
* π/17（頁面存檔備份，存於網際網路檔案館）
  
* _π/19_
  
* π/23（頁面存檔備份，存於網際網路檔案館）

* 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.  MR1706614

* Girstmair, Kurt. Some linear relations between values of trigonometric functions at k\*pi/n. Acta Arithmetica. 1997, 81: 387–398.  MR1472818

* 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.  MR2240647

* Servi, L. D. Nested square roots of 2. Am. Math. Monthly. 2003, 110 (4): 326–330. doi:10.2307/3647881.  MR1984573 JSTOR 3647881



注釋