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

相同角度的轉換表

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

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°

${\displaystyle \sin \left({\frac {x}{2}}\right)=\pm \,{\sqrt {{\tfrac {1}{2}}(1-\cos x)}}}$

${\displaystyle \cos \left({\frac {x}{2}}\right)=\pm \,{\sqrt {{\tfrac {1}{2}}(1+\cos x)}}}$

利用三倍角公式求 1 3 {\displaystyle {\frac {1}{3}}\,} 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。

• ${\displaystyle \sin 3\theta =3\sin \theta -4\sin ^{3}\theta \,}$

• ${\displaystyle \cos 3\theta =4\cos ^{3}\theta -3\cos \theta \,}$

把它改為

• ${\displaystyle \sin \theta =3\sin {\frac {1}{3}}\theta -4\sin ^{3}{\frac {1}{3}}\theta \,}$
• ${\displaystyle \cos \theta =4\cos ^{3}{\frac {1}{3}}\theta -3\cos {\frac {1}{3}}\theta \,}$

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

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

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

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

例如：

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

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

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

經由和角公式的計算

例如：21° = 9° + 12°

${\displaystyle \sin(x\pm y)=\sin(x)\cos(y)\pm \cos(x)\sin(y)\,}$

${\displaystyle \cos(x\pm y)=\cos(x)\cos(y)\mp \sin(x)\sin(y)\,}$

經由托勒密定理的計算

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，

${\displaystyle a^{2}+ab=b^{2}}$

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

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

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

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

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

${\displaystyle \sin 0=0\,}$

${\displaystyle \cos 0=1\,}$

${\displaystyle \tan 0=0\,}$

1°：2°的一半

${\displaystyle \sin {1^{\circ }}={\frac {1+{\sqrt {3}}i}{16}}{\sqrt[{3}]{4{\sqrt {30}}-8{\sqrt {15+3{\sqrt {5}}}}+8{\sqrt {5+{\sqrt {5}}}}+4{\sqrt {10}}-4{\sqrt {6}}-4{\sqrt {2}}+\left(4{\sqrt {30}}+8{\sqrt {15+3{\sqrt {5}}}}+8{\sqrt {5+{\sqrt {5}}}}-4{\sqrt {10}}-4{\sqrt {6}}+4{\sqrt {2}}\right)i}}+}$

${\displaystyle {\frac {1-{\sqrt {3}}i}{16}}{\sqrt[{3}]{4{\sqrt {30}}-8{\sqrt {15+3{\sqrt {5}}}}+8{\sqrt {5+{\sqrt {5}}}}+4{\sqrt {10}}-4{\sqrt {6}}-4{\sqrt {2}}-\left(4{\sqrt {30}}+8{\sqrt {15+3{\sqrt {5}}}}+8{\sqrt {5+{\sqrt {5}}}}-4{\sqrt {10}}-4{\sqrt {6}}+4{\sqrt {2}}\right)i}}}$^[2]

1.5°：正一百二十邊形

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

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

1.875°：正九十六邊形

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

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

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

2°：6°的三分之一

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

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

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

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

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

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

2.25°：正八十邊形

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

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

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

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

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

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

2.8125°：正六十四邊形

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

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

3°：正六十邊形

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

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

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

3.75°：正四十八邊形

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

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

4°：12°的三分之一

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

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

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

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

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

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

4.5°：正四十邊形

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

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

5°：15°的三分之一、正三十六邊形

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

5.625°：正三十二邊形

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

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

6°：正三十邊形

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

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

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

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

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

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

7.5°：正二十四邊形

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

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

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

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

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

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

9°：正二十邊形

更多資訊：二十邊形

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

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

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

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

10°：正十八邊形

更多資訊：十八邊形

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

11.25°：正十六邊形

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

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

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

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

12°：正十五邊形

更多資訊：十五邊形

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

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

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

15°：正十二邊形

更多資訊：十二邊形

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

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

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

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

18°：正十邊形

更多資訊：十邊形

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

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

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

20°：正九邊形、60°的三分之一

更多資訊：九邊形

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

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

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

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

21°：9°與12°的和

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

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

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

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

更多資訊：十七邊形

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

22.5°：正八邊形

更多資訊：八邊形

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

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

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

24°：12°的二倍

${\displaystyle \sin {\frac {2\pi }{15}}=\sin 24^{\circ }={\tfrac {1}{8}}\left[{\sqrt {3}}({\sqrt {5}}+1)-{\sqrt {2}}{\sqrt {5-{\sqrt {5}}}}\right]\,}$

${\displaystyle \cos {\frac {2\pi }{15}}=\cos 24^{\circ }={\tfrac {1}{8}}\left({\sqrt {6}}{\sqrt {5-{\sqrt {5}}}}+{\sqrt {5}}+1\right)\,}$

${\displaystyle \tan {\frac {2\pi }{15}}=\tan 24^{\circ }={\tfrac {1}{2}}\left[{\sqrt {2(25+11{\sqrt {5}})}}-{\sqrt {3}}(3+{\sqrt {5}})\right]\,}$

180/7°， ( 25 5 7 ) ∘ {\displaystyle \mathbf {\left(25{\frac {5}{7}}\right)^{\circ }} } ， ( 180 7 ) ∘ {\displaystyle \mathbf {\left({\frac {180}{7}}\right)^{\circ }} } ：正七邊形

更多資訊：七邊形

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

27°：12°與15°的和

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

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

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

30°：正六邊形

更多資訊：六邊形

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

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

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

33°：15°與18°的和

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

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

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

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

36°：正五邊形

更多資訊：五邊形

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

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

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

39°：18°與21°的和

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

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

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

42°：21°的2倍

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

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

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

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

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

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

45°：正方形

更多資訊：正方形

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

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

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

48°

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

54°：27°與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°：等邊三角形

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

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

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

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

67.5°：7.5°與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°與45°的和

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

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

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

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

81°

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

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

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

${\displaystyle n}$	${\displaystyle \sin \left({\frac {2\pi }{n}}\right)}$	${\displaystyle \cos \left({\frac {2\pi }{n}}\right)}$	${\displaystyle \tan \left({\frac {2\pi }{n}}\right)}$
1	${\displaystyle 0}$	${\displaystyle 1}$	${\displaystyle 0}$
2	${\displaystyle 0}$	${\displaystyle -1}$	${\displaystyle 0}$
3	${\displaystyle {\frac {1}{2}}{\sqrt {3}}}$	${\displaystyle -{\frac {1}{2}}}$	${\displaystyle -{\sqrt {3}}}$
4	${\displaystyle 1}$	${\displaystyle 0}$	${\displaystyle \pm \infty }$
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	${\displaystyle {\frac {1}{2}}{\sqrt {3}}}$	${\displaystyle {\frac {1}{2}}}$	${\displaystyle {\sqrt {3}}}$
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	${\displaystyle {\frac {1}{2}}{\sqrt {2}}}$	${\displaystyle {\frac {1}{2}}{\sqrt {2}}}$	${\displaystyle 1}$
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	${\displaystyle {\frac {1}{2}}}$	${\displaystyle {\frac {1}{2}}{\sqrt {3}}}$	${\displaystyle {\frac {1}{3}}{\sqrt {3}}}$
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	${\displaystyle {\frac {1}{4}}\left({\sqrt {6}}-{\sqrt {2}}\right)}$	${\displaystyle {\frac {1}{4}}\left({\sqrt {6}}+{\sqrt {2}}\right)}$	${\displaystyle 2-{\sqrt {3}}}$

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
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  JSTOR 3647881