三角函數精確值

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。

相同角度的轉換表

角度單位	值
轉	$0$	${\frac {1}{12}}$	${\frac {1}{8}}$	${\frac {1}{6}}$	${\frac {1}{4}}$	${\frac {1}{2}}$	${\frac {3}{4}}$	$1$
角度	$0^{\circ }$	$30^{\circ }$	$45^{\circ }$	$60^{\circ }$	$90^{\circ }$	$180^{\circ }$	$270^{\circ }$	$360^{\circ }$
弧度	$0$	${\frac {\pi }{6}}$	${\frac {\pi }{4}}$	${\frac {\pi }{3}}$	${\frac {\pi }{2}}$	$\pi$	${\frac {3\pi }{2}}$	$2\pi$
梯度	$0^{g}$	$33{\frac {1}{3}}^{g}$	$50^{g}$	$66{\frac {2}{3}}^{g}$	$100^{g}$	$200^{g}$	$300^{g}$	$400^{g}$

計算方式

基於常識

例如：0°、30°、45°

單位圓

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${\frac {1}{3}}\,$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。

經由歐拉公式的計算

更多資訊：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：

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

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

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

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

更多資訊：托勒密定理和弦函數

Chord(36°) = a/b = 1/φ, 根據托勒密定理

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

更多資訊：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

更多資訊：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

更多資訊：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

更多資訊：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

更多資訊：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

更多資訊：九邊形

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

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

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

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

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

更多資訊：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

更多資訊：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

更多資訊：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

更多資訊：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

更多資訊：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

更多資訊：正方形

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

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

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

48°

$\sin 48^{\circ }={\frac {1}{4}}{\sqrt {7-{\sqrt {5}}+{\sqrt {6(5-{\sqrt {5}})}}}}$

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

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

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

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

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

81°

$\sin 81^{\circ }={\frac {1}{2}}{\sqrt {{\frac {1}{2}}{\Big (}4+{\sqrt {2(5+{\sqrt {5}})}}{\Big )}}}$

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，$\omega =\exp({\frac {\pi i}{3}})=-{\frac {1}{2}}+{\frac {1}{2}}i{\sqrt {3}}$。

$n$	$\sin \left({\frac {2\pi }{n}}\right)$	$\cos \left({\frac {2\pi }{n}}\right)$	$\tan \left({\frac {2\pi }{n}}\right)$
1	$0$	$1$	$0$
2	$0$	$-1$	$0$
3	${\frac {1}{2}}{\sqrt {3}}$	$-{\frac {1}{2}}$	$-{\sqrt {3}}$
4	$1$	$0$	$\pm \infty$
5	${\frac {1}{4}}\left({\sqrt {10+2{\sqrt {5}}}}\right)$	${\frac {1}{4}}\left({\sqrt {5}}-1\right)$	${\sqrt {5+2{\sqrt {5}}}}$
6	${\frac {1}{2}}{\sqrt {3}}$	${\frac {1}{2}}$	${\sqrt {3}}$
7	${\frac {1}{2}}{\sqrt {{\frac {1}{3}}\left(7-\omega ^{2}{\sqrt[{3}]{\frac {7+21{\sqrt {-3}}}{2}}}-\omega {\sqrt[{3}]{\frac {7-21{\sqrt {-3}}}{2}}}\right)}}$	${\frac {1}{6}}\left(-1+{\sqrt[{3}]{\frac {7+21{\sqrt {-3}}}{2}}}+{\sqrt[{3}]{\frac {7-21{\sqrt {-3}}}{2}}}\right)$
8	${\frac {1}{2}}{\sqrt {2}}$	${\frac {1}{2}}{\sqrt {2}}$	$1$
9	${\frac {1}{4}}\left({\sqrt[{3}]{4(-1+{\sqrt {-3}})}}+{\sqrt[{3}]{4(-1-{\sqrt {-3}})}}\right)$
10	${\frac {1}{4}}\left({\sqrt {10-2{\sqrt {5}}}}\right)$	${\frac {1}{4}}\left({\sqrt {5}}+1\right)$	${\sqrt {5-2{\sqrt {5}}}}$
11
12	${\frac {1}{2}}$	${\frac {1}{2}}{\sqrt {3}}$	${\frac {1}{3}}{\sqrt {3}}$
13
14	${\frac {1}{24}}{\sqrt {3\left(112-{\sqrt[{3}]{14336+{\sqrt {-5549064193}}}}-{\sqrt[{3}]{14336-{\sqrt {-5549064193}}}}\right)}}$	${\frac {1}{24}}{\sqrt {3\left(80+{\sqrt[{3}]{14336+{\sqrt {-5549064193}}}}+{\sqrt[{3}]{14336-{\sqrt {-5549064193}}}}\right)}}$	${\sqrt {\frac {112-{\sqrt[{3}]{14336+{\sqrt {-5549064193}}}}-{\sqrt[{3}]{14336-{\sqrt {-5549064193}}}}}{80+{\sqrt[{3}]{14336+{\sqrt {-5549064193}}}}+{\sqrt[{3}]{14336-{\sqrt {-5549064193}}}}}}}$
15	${\frac {1}{8}}\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)$	${\frac {1}{8}}\left(1+{\sqrt {5}}+{\sqrt {30-6{\sqrt {5}}}}\right)$	${\frac {1}{2}}\left(-3{\sqrt {3}}-{\sqrt {15}}+{\sqrt {50+22{\sqrt {5}}}}\right)$
16	${\frac {1}{2}}\left({\sqrt {2-{\sqrt {2}}}}\right)$	${\frac {1}{2}}\left({\sqrt {2+{\sqrt {2}}}}\right)$	${\sqrt {2}}-1$
17	${\frac {1}{4}}{\sqrt {8-{\sqrt {2\left(15+{\sqrt {17}}+{\sqrt {34-2{\sqrt {17}}}}-2{\sqrt {17+3{\sqrt {17}}-{\sqrt {170+38{\sqrt {17}}}}}}\right)}}}}$	${\frac {1}{16}}\left(-1+{\sqrt {17}}+{\sqrt {34-2{\sqrt {17}}}}+2{\sqrt {17+3{\sqrt {17}}-{\sqrt {34-2{\sqrt {17}}}}-2{\sqrt {34+2{\sqrt {17}}}}}}\right)$
18	${\frac {1}{4}}\left({\sqrt[{3}]{4{\sqrt {-1}}-4{\sqrt {3}}}}-{\sqrt[{3}]{4{\sqrt {-1}}+4{\sqrt {3}}}}\right)$	${\frac {1}{4}}\left({\sqrt[{3}]{4+4{\sqrt {-3}}}}+{\sqrt[{3}]{4-4{\sqrt {-3}}}}\right)$
19
20	${\frac {1}{4}}\left({\sqrt {5}}-1\right)$	${\frac {1}{4}}\left({\sqrt {10+2{\sqrt {5}}}}\right)$	${\frac {1}{5}}\left({\sqrt {25-10{\sqrt {5}}}}\right)$
21
22
23
24	${\frac {1}{4}}\left({\sqrt {6}}-{\sqrt {2}}\right)$	${\frac {1}{4}}\left({\sqrt {6}}+{\sqrt {2}}\right)$	$2-{\sqrt {3}}$

相關

更多資訊：三角函數和三角恆等式

更多資訊：en:Generating trigonometric tables

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