三角函數精確值

Template:三角学 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。

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

Template:See also2 例如：15°、22.5°

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

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

利用三倍角公式求$\frac{1}{3}$角

Template:See also2 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

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

例如：

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

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

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

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

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，

$a^2+ab=b^2$

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

$\mathrm{c}\mathrm{r}\mathrm{d}36^{\circ }=\mathrm{c}\mathrm{r}\mathrm{d}\left(\mathrm{\angle }\mathrm{A}\mathrm{D}\mathrm{B}\right)=\frac{a}{b}=\frac{\sqrt{5}-1}{2}$

$\mathrm{c}\mathrm{r}\mathrm{d}\theta =2\sin \frac{\theta }{2}$

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

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

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

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

1.5°：正一百二十边形

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

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

1.875°：正九十六边形

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

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

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

2°：6°的三分之一

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

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

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

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

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

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

2.25°：正八十边形

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

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

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

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

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

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

2.8125°：正六十四边形

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

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

3°：正六十边形

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

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

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

3.75°：正四十八边形

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

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

4°：12°的三分之一

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

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

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

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

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

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

4.5°：正四十边形

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

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

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

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

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

6°：正三十边形

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

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

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

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

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

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

7.5°：正二十四边形

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

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

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

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

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

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

9°：正二十边形

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

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

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

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

10°：正十八边形

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

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

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

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

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

15°：正十二边形

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

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

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

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

18°：正十边形

$\sin \frac{\pi }{10}=\sin 18^{\circ }=\frac{1}{4}(\sqrt{5}-1)=\frac{1}{2}{\phi }^{-1}$

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

$\tan \frac{\pi }{10}=\tan 18^{\circ }=\frac{1}{5}\sqrt{5(5-2\sqrt{5})}$

20°：正九边形、60°的三分之一

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

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

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

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

21°：9°与12°的和

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

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

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

360/17°，${\left(21\frac{3}{17}\right)}^{\mathbf{\circ }}$，${\left(\frac{360}{17}\right)}^{\mathbf{\circ }}$：正十七边形

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

22.5°：正八边形

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

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

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

24°：12°的二倍

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

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

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

180/7°，${\left(25\frac{5}{7}\right)}^{\mathbf{\circ }}$，${\left(\frac{180}{7}\right)}^{\mathbf{\circ }}$：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

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

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

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

30°：正六边形

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

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

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

33°：15°与18°的和

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

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

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

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

36°：正五边形

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

$\cos \frac{\pi }{5}=\cos 36^{\circ }=\frac{1+\sqrt{5}}{4}=\frac{1}{2}\phi$

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

39°：18°与21°的和

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

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

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

42°：21°的2倍

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

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

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

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

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

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

45°：正方形

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

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

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

48°

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

54°：27°与27°的和

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

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

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

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

60°：等边三角形

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

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

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

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

67.5°：7.5°与60°的和

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

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

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

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

72°：36°的二倍

$\sin \frac{2\pi }{5}=\sin 72^{\circ }=\frac{1}{4}\sqrt{2(5+\sqrt{5})}$

$\cos \frac{2\pi }{5}=\cos 72^{\circ }=\frac{1}{4}(\sqrt{5}-1)$

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

$\cot \frac{2\pi }{5}=\cot 72^{\circ }=\frac{1}{5}\sqrt{5(5-2\sqrt{5})}$

75°: 30°与45°的和

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

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

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

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

81°

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

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

參見

• 可作图多边形
• Template:Tsl
• 十七邊形

參考文獻

• Template:MathWorld
• Template:MathWorld
  • π/3 (60°)Template:Wayback—π/6 (30°)Template:Wayback—π/12 (15°)Template:Wayback—π/24 (7.5°)Template:Wayback
  • π/4 (45°)Template:Wayback—π/8 (22.5°)Template:Wayback—π/16 (11.25°)Template:Wayback—π/32 (5.625°)Template:Wayback
  • π/5 (36°)Template:Wayback—π/10 (18°)Template:Wayback—π/20 (9°)Template:Wayback
  • π/7Template:Wayback—π/14
  • π/9 (20°)Template:Wayback—π/18 (10°)Template:Wayback
  • π/11Template:Wayback
  • π/13Template:Wayback
  • π/15 (12°)Template:Wayback—π/30 (6°)Template:Wayback
  • π/17Template:Wayback
  • π/19
  • π/23Template:Wayback
• Template:Cite journal
• Template:Cite arxiv
• Template:Cite journal Template:MR
• Template:Cite journal Template:MR
• Template:Cite journal Template:MR
• Template:Cite journal Template:MR Template:JSTOR