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，\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\frac13\,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。

经由欧拉公式的计算
*\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+\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3-1)\right]+i\left[2(1-\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3+1)\right]}
:::-\sqrt[3]{\left[2(1+\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3-1)\right]-i\left[2(1-\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3+1)\right]}\Bigg\}

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{\sqrt5-1}{4}

三角函数精确值列表
由于三角函数的特性，大于45°角度的三角函数值，可以经由自0°～45°的角度的三角函数值的相关的计算取得。

0°：根本
: \sin 0=0\,
: \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
:\sin\left(\frac{\pi}{120}\right) = \sin\left(1.5^\circ\right) = \frac{\left(\sqrt{2+\sqrt2}\right)\left(\sqrt{15}+\sqrt3-\sqrt{10-2\sqrt5}\right) - \left(\sqrt{2-\sqrt2}\right)\left(\sqrt{30-6\sqrt5}+\sqrt5+1\right)}{16}
:\cos\left(\frac{\pi}{120}\right) = \cos\left(1.5^\circ\right) = \frac{\left(\sqrt{2+\sqrt2}\right)\left(\sqrt{30-6\sqrt5}+\sqrt5+1\right) + \left(\sqrt{2-\sqrt2}\right)\left(\sqrt{15}+\sqrt3-\sqrt{10-2\sqrt5}\right)}{16}

1.875°：正九十六边形
:\sin\left(\frac{\pi}{96}\right) = \sin\left(1.875^\circ\right) = \frac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}
:\cos\left(\frac{\pi}{96}\right) = \cos\left(1.875^\circ\right) = \frac12\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-\sqrt5)}+\sqrt3(\sqrt5+1)\right]+i\left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]}
:::-\sqrt[3]{\left[\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)\right]-i\left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\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-\sqrt5)}+\sqrt3(\sqrt5+1)\right]+i\left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]}
:::+\sqrt[3]{\left[\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)\right]-i\left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]}\Bigg\}

2.25°：正八十边形
:\sin\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°：正六十四边形
:\sin\left(\frac{\pi}{64}\right) = \sin\left(2.8125^\circ\right) = \frac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}
:\cos\left(\frac{\pi}{64}\right) = \cos\left(2.8125^\circ\right) = \frac12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}

3°：正六十边形
: \sin\frac{\pi}{60}=\sin 3^\circ=\tfrac{1}{4} \sqrt{8-\sqrt3-\sqrt{15}-\sqrt{10-2\sqrt5}}\,
: \cos\frac{\pi}{60}=\cos 3^\circ=\tfrac{1}{4} \sqrt{8+\sqrt3+\sqrt{15}+\sqrt{10-2\sqrt5}}\,
: \tan\frac{\pi}{60}=\tan 3^\circ=\tfrac{1}{4} \left[(2-\sqrt3)(3+\sqrt5)-2\right]\left[2-\sqrt{2(5-\sqrt5)}\right]\,

3.75°：正四十八边形
:\sin\left(\frac{\pi}{48}\right) = \sin\left(3.75^\circ\right) = \frac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}
:\cos\left(\frac{\pi}{48}\right) = \cos\left(3.75^\circ\right) = \frac12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}

4°：12°的三分之一
:\sin{4^\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+\sqrt5)}+\sqrt5-1\right]+i\left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]}
:::-\sqrt[3]{\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]-i\left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\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+\sqrt5)}+\sqrt5-1\right]+i\left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]}
:::+\sqrt[3]{\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]-i\left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]}\Bigg\}

4.5°：正四十边形
:\sin\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
:\sin\left(\frac{\pi}{32}\right) = \sin\left(5.625^\circ\right) = \frac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}
:\cos\left(\frac{\pi}{32}\right) = \cos\left(5.625^\circ\right) = \frac12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}

6°：正三十边形
: \sin\frac{\pi}{30}=\sin 6^\circ=\tfrac{1}{8} \left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]\,
: \cos\frac{\pi}{30}=\cos 6^\circ=\tfrac{1}{8} \left[\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)\right]\,
: \tan\frac{\pi}{30}=\tan 6^\circ=\tfrac{1}{2} \left[\sqrt{2(5-\sqrt5)}-\sqrt3(\sqrt5-1)\right]\,
: \cot\frac{\pi}{30}=\cot 6^\circ=\tfrac{1}{2} \left(\sqrt{50+22\sqrt5}+3\sqrt3+\sqrt{15}\right)\,
: \sec\frac{\pi}{30}=\sec 6^\circ=\sqrt3-\sqrt{5-2\sqrt5}\,
: \csc\frac{\pi}{30}=\csc 6^\circ=2+\sqrt5+\sqrt{15+6\sqrt5}\,

7.5°：正二十四边形
: \sin\frac{\pi}{24}=\sin 7.5^\circ=\tfrac{1}{4} \sqrt{8-2\sqrt6-2\sqrt2}\,
: \cos\frac{\pi}{24}=\cos 7.5^\circ=\tfrac{1}{4} \sqrt{8+2\sqrt6+2\sqrt2}\,
: \tan\frac{\pi}{24}=\tan 7.5^\circ=\sqrt6+\sqrt2-2-\sqrt3\,
: \cot\frac{\pi}{24}=\cot 7.5^\circ=\sqrt6+\sqrt2+2+\sqrt3\,
: \sec\frac{\pi}{24}=\sec 7.5^\circ=\sqrt{16-6\sqrt6-10\sqrt2+8\sqrt3}\,
: \csc\frac{\pi}{24}=\csc 7.5^\circ=\sqrt{16+6\sqrt6+10\sqrt2+8\sqrt3}\,

9°：正二十边形
: \sin\frac{\pi}{20}=\sin 9^\circ=\tfrac{1}{4} \sqrt{8-2\sqrt{10+2\sqrt5}}\,
: \cos\frac{\pi}{20}=\cos 9^\circ=\tfrac{1}{4} \sqrt{8+2\sqrt{10+2\sqrt5}}\,
: \tan\frac{\pi}{20}=\tan 9^\circ=\sqrt5+1-\sqrt{5+2\sqrt5}\,
: \cot\frac{\pi}{20}=\cot9^\circ=\sqrt5+1+\sqrt{5+2\sqrt5}\,

10°：正十八边形
:{\tan10^\circ=-\frac{-1-\sqrt{3}{\rm{i}}}{6}\sqrt[3]{-12\sqrt3 + 36{\rm{i}}}-\frac{-1+\sqrt{3}{\rm{i}}}{6}\sqrt[3]{-12\sqrt3 - 36{\rm{i}}} + \frac{1}{\sqrt3}}\,

11.25°：正十六边形
:\sin\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°：正十五边形
: \sin\frac{\pi}{15}=\sin 12^\circ=\tfrac{1}{8} \left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]\,
: \cos\frac{\pi}{15}=\cos 12^\circ=\tfrac{1}{8} \left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]\,
: \tan\frac{\pi}{15}=\tan 12^\circ=\tfrac{1}{2} \left[\sqrt3(3-\sqrt5)-\sqrt{2(25-11\sqrt5)}\right]\,

15°：正十二边形
: \sin\frac{\pi}{12}=\sin 15^\circ=\frac{1}{4}\sqrt2\left(\sqrt3-1\right)\,
: \cos\frac{\pi}{12}=\cos 15^\circ=\frac{1}{4}\sqrt2\left(\sqrt3+1\right)\,
: \tan\frac{\pi}{12}=\tan 15^\circ=2-\sqrt3\,
: \cot\frac{\pi}{12}=\cot 15^\circ=2+\sqrt3\,

18°：正十边形
: \sin\frac{\pi}{10}=\sin 18^\circ=\tfrac{1}{4}\left(\sqrt5-1\right)=\tfrac{1}{2}\varphi^{-1}\,
: \cos\frac{\pi}{10}=\cos 18^\circ=\tfrac{1}{4}\sqrt{2\left(5+\sqrt5\right)}\,
: \tan\frac{\pi}{10}=\tan 18^\circ=\tfrac{1}{5}\sqrt{5\left(5-2\sqrt5\right)}\,

20°：正九边形、60°的三分之一
: \sin\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°与12°的和
: \sin\frac{7\pi}{60}=\sin 21^\circ=\tfrac{1}{4}\sqrt{8+\sqrt3-\sqrt{15}-\sqrt{10+2\sqrt5}}\,
: \cos\frac{7\pi}{60}=\cos 21^\circ=\tfrac{1}{4}\sqrt{8-\sqrt3+\sqrt{15}+\sqrt{10+2\sqrt5}}\,
: \tan\frac{7\pi}{60}=\tan 21^\circ=\tfrac{1}{4}\left[2-\left(2+\sqrt3\right)\left(3-\sqrt5\right)\right]\left[2-\sqrt{2\left(5+\sqrt5\right)}\right]\,

360/17°，\mathbf{\left(21\frac{3}{17}\right)^{\circ}}，\mathbf{\left(\frac{360}{17}\right)^{\circ}}：正十七边形
:\operatorname{cos}{2\pi\over17}=\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}}}{16}

22.5°：正八边形
: \sin\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[\sqrt3(\sqrt5+1)-\sqrt2\sqrt{5-\sqrt5}\right]\,
: \cos\frac{2\pi}{15}=\cos 24^\circ=\tfrac{1}{8}\left(\sqrt6\sqrt{5-\sqrt5}+\sqrt5+1\right)\,
: \tan\frac{2\pi}{15}=\tan 24^\circ=\tfrac{1}{2}\left[\sqrt{2(25+11\sqrt5)}-\sqrt3(3+\sqrt5)\right]\,

180/7°，\mathbf{\left(25\frac{5}{7}\right)^{\circ}}，\mathbf{\left(\frac{180}{7}\right)^{\circ}}：正七边形
: \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+\sqrt5}-\sqrt2\;(\sqrt5-1)\right]\,
: \cos\frac{3\pi}{20}=\cos 27^\circ=\tfrac{1}{8}\left[2\sqrt{5+\sqrt5}+\sqrt2\;\left(\sqrt5-1\right)\right]\,
: \tan\frac{3\pi}{20}=\tan 27^\circ=\sqrt5-1-\sqrt{5-2\sqrt5}\,

30°：正六边形
: \sin\frac{\pi}{6}=\sin 30^\circ=\tfrac{1}{2}\,
: \cos\frac{\pi}{6}=\cos 30^\circ=\tfrac{1}{2}\sqrt3\,
: \tan\frac{\pi}{6}=\tan 30^\circ=\tfrac{1}{3}\sqrt3\,

33°：15°与18°的和
: \sin\frac{11\pi}{60}=\sin 33^\circ=\tfrac{1}{4}\sqrt{8-\sqrt3-\sqrt{15}+\sqrt{10-2\sqrt5}}\,
: \cos\frac{11\pi}{60}=\cos 33^\circ=\tfrac{1}{4}\sqrt{8+\sqrt3+\sqrt{15}-\sqrt{10-2\sqrt5}}\,
: \tan\frac{11\pi}{60}=\tan 33^\circ=\tfrac{1}{4}\left(2\sqrt3-\sqrt5-1\right)\left(2\sqrt{5+2\sqrt5}+3+\sqrt5\right)\,
: \cot\frac{11\pi}{60}=\cot33^\circ=\tfrac{1}{4}\left(2\sqrt3+\sqrt5+1\right)\left(2\sqrt{5+2\sqrt5}-3-\sqrt5\right)\,

36°：正五边形
: \sin\frac{\pi}{5}=\sin 36^\circ=\tfrac14\left[\sqrt{2\left(5-\sqrt5\right)}\right]\,
: \cos\frac{\pi}{5}=\cos 36^\circ=\frac{1+\sqrt5}{4}=\tfrac{1}{2}\varphi\,
: \tan\frac{\pi}{5}=\tan 36^\circ=\sqrt{5-2\sqrt5}\,

39°：18°与21°的和
: \sin\frac{13\pi}{60}=\sin 39^\circ=\tfrac1{4}\sqrt{8-\sqrt3+\sqrt{15}+\sqrt{10+2\sqrt5}}\,
: \cos\frac{13\pi}{60}=\cos 39^\circ=\tfrac1{4}\sqrt{8+\sqrt3-\sqrt{15}+\sqrt{10+2\sqrt5}}\,
: \tan\frac{13\pi}{60}=\tan 39^\circ=\tfrac14\left[\left(2-\sqrt3\right)\left(3-\sqrt5\right)-2\right]\left[2-\sqrt{2\left(5+\sqrt5\right)}\right]\,

42°：21°的2倍
: \sin\frac{7\pi}{30}=\sin 42^\circ=\frac{\sqrt6\sqrt{5+\sqrt5}-\sqrt5+1}{8}\,
: \cos\frac{7\pi}{30}=\cos 42^\circ=\frac{\sqrt2\sqrt{5+\sqrt5}+\sqrt3\left(\sqrt5-1\right)}{8}\,
: \tan\frac{7\pi}{30}=\tan 42^\circ=\frac1{2}\left(\sqrt3+\sqrt{15}-\sqrt{10+2\sqrt5}\right)\,
: \cot\frac{7\pi}{30}=\cot 42^\circ=\frac1{2}\left(3\sqrt3-\sqrt{15}+\sqrt{50-22\sqrt5}\right)\,
: \sec\frac{7\pi}{30}=\sec 42^\circ=\sqrt{5+2\sqrt5}-\sqrt3\,
: \sec\frac{7\pi}{30}=\sec 42^\circ=\sqrt{15-6\sqrt5}+\sqrt5-2\,

45°：正方形
: \sin\frac{\pi}{4}=\sin 45^\circ=\frac{\sqrt2}{2}=\frac{1}{\sqrt2}\,
: \cos\frac{\pi}{4}=\cos 45^\circ=\frac{\sqrt2}{2}=\frac{1}{\sqrt2}\,
: \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°与27°的和
:\sin\frac{3\pi}{10}=\sin 54^\circ=\frac{\sqrt5+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°：等边三角形
:\sin\frac{\pi}{3}=\sin 60^\circ=\frac{\sqrt3}{2}\,
:\cos\frac{\pi}{3}=\cos 60^\circ=\frac{1}{2}\,
:\tan\frac{\pi}{3}=\tan 60^\circ=\sqrt3\,
:\cot\frac{\pi}{3}=\cot 60^\circ=\frac{\sqrt3}{3}=\frac{1}{\sqrt3}\,

67.5°：7.5°与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+\sqrt5\right)}\,
:\cos\frac{2\pi}{5}=\cos 72^\circ=\tfrac{1}{4}\left(\sqrt5-1\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\sqrt5\right)}\,

75°: 30°与45°的和
:\sin\frac{5\pi}{12}=\sin 75^\circ=\tfrac{1}{4}\left(\sqrt6+\sqrt2\right)\,
:\cos\frac{5\pi}{12}=\cos 75^\circ=\tfrac{1}{4}\left(\sqrt6-\sqrt2\right)\,
:\tan\frac{5\pi}{12}=\tan 75^\circ=2+\sqrt3\,
:\cot\frac{5\pi}{12}=\cot 75^\circ=2-\sqrt3\,
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}。

相關
參見
*可作图多边形
*
*十七邊形

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