查看“︁三角函數精確值”︁的源代码

{{三角学}}
'''三角函數精確值'''是利用[[三角恆等式|三角函數的公式]]將特定的[[三角函數]]值加以化簡，並以數學[[方根|根式]]或[[分數]]表示。

用[[方根|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，<math>\pm\frac{1}{2}</math>，±1。

{|class = "wikitable" width=45% style="text-align: center; vertical-align: middle"
|+ 相同角度的轉換表
! 角度單位 !! colspan=8 | 值
|-
! [[轉 (角)|轉]]
| <math>0</math>
| <math>\frac{1}{12}</math>
| <math>\frac{1}{8}</math>
| <math>\frac{1}{6}</math>
| <math>\frac{1}{4}</math>
| <math>\frac{1}{2}</math>
| <math>\frac{3}{4}</math>
| <math>1</math>
|-
! [[角度]]
| <math>0^{\circ}</math>
| <math>30^{\circ}</math>
| <math>45^{\circ}</math>
| <math>60^{\circ}</math>
| <math>90^{\circ}</math>
| <math>180^{\circ}</math>
| <math>270^{\circ}</math>
| <math>360^{\circ}</math>
|-
! [[弧度]]
| <math>0</math> 
| <math>\frac{\pi}{6}</math>
| <math>\frac{\pi}{4}</math>
| <math>\frac{\pi}{3}</math>
| <math>\frac{\pi}{2}</math>
| <math>\pi</math>
| <math>\frac{3\pi}{2}</math>
| <math>2\pi</math>
|-
! [[梯度 (角)|梯度]]
| <math>0^g</math>
| <math>33\frac{1}{3}^g</math>
| <math>50^g</math>
| <math>66\frac{2}{3}^g</math>
| <math>100^g</math>
| <math>200^g</math>
| <math>300^g</math>
| <math>400^g</math>
|}

==計算方式==
===基於常識===
例如：0°、30°、45°
[[File:Unit_circle_angles.svg|center|500px|單位圓]]

===經由半角公式的計算===
{{see also2|[[三角恒等式#雙倍角、三倍角和半角公式|半角公式]]}}
例如：15°、22.5°
:<math>\sin\left(\frac{x}{2}\right) =  \pm\, \sqrt{\tfrac{1}{2}(1 - \cos x)}</math>

:<math>\cos\left(\frac{x}{2}\right) =  \pm\, \sqrt{\tfrac{1}{2}(1 + \cos x)}</math>

===利用三倍角公式求<math>\frac13\,</math>角===
{{see also2|[[三角恒等式#雙倍角、三倍角和半角公式|三倍角公式]]}}
例如：10°、20°、7°......等等，非三的倍數的角的精確值。
*<math>\sin 3\theta = 3 \sin \theta- 4 \sin^3\theta \,</math>

*<math>\cos 3\theta = 4 \cos^3\theta - 3 \cos \theta \,</math>
把它改為
*<math>\sin \theta = 3 \sin \frac{1}{3}\theta- 4 \sin^3\frac{1}{3}\theta \,</math>
*<math>\cos \theta = 4 \cos^3\frac{1}{3}\theta - 3 \cos \frac{1}{3}\theta \,</math>
把<math>\cos \frac{1}{3}\theta \,</math>當成未知數，<math>\cos \theta \,</math>當成常數項
解[[三次方程|一元三次方程式]]即可求出

例如：<math>\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}}}</math>

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===
{{see|歐拉公式}}
*<math>\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)</math>
*<math>\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)</math>
例如：
:<math>\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)</math>
::<math>= \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]}</math>
:::<math>-\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\}</math><ref>由[[Wolfram Alpha]]验算：[http://www.wolframalpha.com/input/?i=N%5BIm%5B1%2F%282*2%5E%281%2F3%29%29*%28%282%281%2BSqrt%5B3%5D%29*Sqrt%5B5%2BSqrt%5B5%5D%5D%2BSqrt%5B2%5D%28Sqrt%5B5%5D-1%29%28Sqrt%5B3%5D-1%29%29%2Bi%282%281-Sqrt%5B3%5D%29*Sqrt%5B5%2BSqrt%5B5%5D%5D%2BSqrt%5B2%5D%28Sqrt%5B5%5D-1%29%28Sqrt%5B3%5D%2B1%29%29%29%5E%281%2F3%29%5D-Sin%5BPi%2F180%5D%5D%5D] {{Wayback|url=http://www.wolframalpha.com/input/?i=N%5BIm%5B1%2F%282*2%5E%281%2F3%29%29*%28%282%281%2BSqrt%5B3%5D%29*Sqrt%5B5%2BSqrt%5B5%5D%5D%2BSqrt%5B2%5D%28Sqrt%5B5%5D-1%29%28Sqrt%5B3%5D-1%29%29%2Bi%282%281-Sqrt%5B3%5D%29*Sqrt%5B5%2BSqrt%5B5%5D%5D%2BSqrt%5B2%5D%28Sqrt%5B5%5D-1%29%28Sqrt%5B3%5D%2B1%29%29%29%5E%281%2F3%29%5D-Sin%5BPi%2F180%5D%5D%5D |date=20160304134914 }}</ref>

===經由和角公式的計算===
例如：21° = 9° + 12°
:<math>\sin(x \pm y) = \sin(x) \cos(y) \pm \cos(x) \sin(y)\,</math>

:<math>\cos(x \pm y) = \cos(x) \cos(y) \mp \sin(x) \sin(y)\,</math>

===經由托勒密定理的計算===
{{see|托勒密定理|弦函數}}
[[File:Ptolemy Pentagon.svg|thumb|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，
:<math>a^2+ab=b^2</math>
:<math>\left(\frac{a}{b}\right)^2+\frac{a}{b}=1</math>
:<math>\mathrm{crd}\ {36^\circ}=\mathrm{crd}\left(\angle\mathrm{ADB}\right)=\frac{a}{b}=\frac{\sqrt{5}-1}{2}</math>
:<math>\mathrm{crd}\ {\theta}=2\sin{\frac{\theta}{2}}\,</math>
:<math>\sin{18^\circ}=\frac{\sqrt5-1}{4}</math>

== 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===
: <math>\sin 0=0\,</math>
: <math>\cos 0=1\,</math>
: <math>\tan 0=0\,</math>

=== 1°：2°的一半===
:<math>\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}+</math>
::::<math>\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}</math><ref>使用Mathematica驗算，代碼為N[ArcSin[(1 + Sqrt[3] I)/16 Power[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, (3)^-1] + (1 - Sqrt[3] I)/16 Power[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, (3)^-1]], 100]/Degree結果為1與原角度無誤差</ref>

=== 1.5°：正一百二十边形===
:<math>\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}</math>
:<math>\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}</math>

=== 1.875°：正九十六边形===
:<math>\sin\left(\frac{\pi}{96}\right) = \sin\left(1.875^\circ\right) = \frac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}</math>
:<math>\cos\left(\frac{\pi}{96}\right) = \cos\left(1.875^\circ\right) = \frac12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}</math>
:<math>\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}}</math>

=== 2°：6°的三分之一===
:<math>\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)</math>
::<math>= \frac{1}{4i}\Bigg\{\sqrt[3]{\left[\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)\right]+i\left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]}</math>
:::<math>-\sqrt[3]{\left[\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)\right]-i\left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]}\Bigg\}</math>
:<math>\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)</math>
::<math>= \frac{1}{4}\Bigg\{\sqrt[3]{\left[\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)\right]+i\left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]}</math>
:::<math>+\sqrt[3]{\left[\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)\right]-i\left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]}\Bigg\}</math>

=== 2.25°：正八十边形 ===
:<math>\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}</math>
:<math>\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}</math>
:<math>\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}}</math>
:<math>\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}}</math>
:<math>\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}}</math>
:<math>\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}}</math>

=== 2.8125°：正六十四边形 ===
:<math>\sin\left(\frac{\pi}{64}\right) = \sin\left(2.8125^\circ\right) = \frac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}</math>
:<math>\cos\left(\frac{\pi}{64}\right) = \cos\left(2.8125^\circ\right) = \frac12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}</math>

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

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

=== 4°：12°的三分之一 ===
:<math>\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)</math>
::<math>= \frac{1}{4i}\Bigg\{\sqrt[3]{\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]+i\left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]}</math>
:::<math>-\sqrt[3]{\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]-i\left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]}\Bigg\}</math>
:<math>\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)</math>
::<math>= \frac{1}{4}\Bigg\{\sqrt[3]{\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]+i\left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]}</math>
:::<math>+\sqrt[3]{\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]-i\left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]}\Bigg\}</math>

=== 4.5°：正四十边形 ===
:<math>\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}}}}</math>
:<math>\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}}}}</math>

=== 5°：15°的三分之一、正三十六边形 ===
: <math>\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}\,</math>

=== 5.625°：正三十二边形===
:<math>\sin\left(\frac{\pi}{32}\right) = \sin\left(5.625^\circ\right) = \frac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}</math>
:<math>\cos\left(\frac{\pi}{32}\right) = \cos\left(5.625^\circ\right) = \frac12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}</math>

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

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

=== 9°：正二十边形===

{{see|二十边形}}
: <math>\sin\frac{\pi}{20}=\sin 9^\circ=\tfrac{1}{4} \sqrt{8-2\sqrt{10+2\sqrt5}}\,</math>
: <math>\cos\frac{\pi}{20}=\cos 9^\circ=\tfrac{1}{4} \sqrt{8+2\sqrt{10+2\sqrt5}}\,</math>
: <math>\tan\frac{\pi}{20}=\tan 9^\circ=\sqrt5+1-\sqrt{5+2\sqrt5}\,</math>
: <math>\cot\frac{\pi}{20}=\cot9^\circ=\sqrt5+1+\sqrt{5+2\sqrt5}\,</math>

=== 10°：正十八边形 ===

{{see|十八边形}}
:<math>{\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}}\,</math>

=== 11.25°：正十六边形 ===
:<math>\sin\frac{\pi}{16}=\sin 11.25^\circ=\frac{1}{2}\sqrt{2-\sqrt{2+\sqrt{2}}}</math>
:<math>\cos\frac{\pi}{16}=\cos 11.25^\circ=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2}}}</math>
:<math>\tan\frac{\pi}{16}=\tan 11.25^\circ=\sqrt{4+2\sqrt{2}}-\sqrt{2}-1</math>
:<math>\cot\frac{\pi}{16}=\cot 11.25^\circ=\sqrt{4+2\sqrt{2}}+\sqrt{2}+1</math>

=== 12°：正十五边形 ===

{{see|十五边形}}
: <math>\sin\frac{\pi}{15}=\sin 12^\circ=\tfrac{1}{8} \left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]\,</math>
: <math>\cos\frac{\pi}{15}=\cos 12^\circ=\tfrac{1}{8} \left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]\,</math>
: <math>\tan\frac{\pi}{15}=\tan 12^\circ=\tfrac{1}{2} \left[\sqrt3(3-\sqrt5)-\sqrt{2(25-11\sqrt5)}\right]\,</math>

=== 15°：正十二边形 ===

{{see|十二边形}}
: <math>\sin\frac{\pi}{12}=\sin 15^\circ=\frac{1}{4}\sqrt2\left(\sqrt3-1\right)\,</math>
: <math>\cos\frac{\pi}{12}=\cos 15^\circ=\frac{1}{4}\sqrt2\left(\sqrt3+1\right)\,</math>
: <math>\tan\frac{\pi}{12}=\tan 15^\circ=2-\sqrt3\,</math>
: <math>\cot\frac{\pi}{12}=\cot 15^\circ=2+\sqrt3\,</math>

=== 18°：正十边形 ===

{{see|十边形}}
: <math>\sin\frac{\pi}{10}=\sin 18^\circ=\tfrac{1}{4}\left(\sqrt5-1\right)=\tfrac{1}{2}\varphi^{-1}\,</math>
: <math>\cos\frac{\pi}{10}=\cos 18^\circ=\tfrac{1}{4}\sqrt{2\left(5+\sqrt5\right)}\,</math>
: <math>\tan\frac{\pi}{10}=\tan 18^\circ=\tfrac{1}{5}\sqrt{5\left(5-2\sqrt5\right)}\,</math>

=== 20°：正九边形、60°的三分之一 ===

{{see|九边形}}
: <math>\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}}}=</math>
:: <math>2^{-\frac{4}{3}}\left(\sqrt[3]{i-\sqrt{3}}-\sqrt[3]{i+\sqrt{3}}\right)</math>
: <math>\cos\frac{\pi}{9}=\cos 20^\circ=</math>
:: <math>2^{-\frac{4}{3}}\left(\sqrt[3]{1+i\sqrt{3}}+\sqrt[3]{1-i\sqrt{3}}\right)</math>

=== 21°：9°与12°的和 ===
: <math>\sin\frac{7\pi}{60}=\sin 21^\circ=\tfrac{1}{4}\sqrt{8+\sqrt3-\sqrt{15}-\sqrt{10+2\sqrt5}}\,</math>
: <math>\cos\frac{7\pi}{60}=\cos 21^\circ=\tfrac{1}{4}\sqrt{8-\sqrt3+\sqrt{15}+\sqrt{10+2\sqrt5}}\,</math>
: <math>\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]\,</math>

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

{{see|十七边形}}
:<math>\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}</math>

=== 22.5°：正八边形 ===

{{see|八边形}}
: <math>\sin\frac{\pi}{8}=\sin 22.5^\circ=\tfrac{1}{2} \left( \sqrt{2-\sqrt{2}} \right)</math> 
: <math>\cos\frac{\pi}{8}=\cos 22.5^\circ=\tfrac{1}{2} \left( \sqrt{2+\sqrt{2}} \right)\,</math>
: <math>\tan\frac{\pi}{8}=\tan 22.5^\circ=\sqrt{2}-1\,</math>

=== 24°：12°的二倍 ===
: <math>\sin\frac{2\pi}{15}=\sin 24^\circ=\tfrac{1}{8}\left[\sqrt3(\sqrt5+1)-\sqrt2\sqrt{5-\sqrt5}\right]\,</math>
: <math>\cos\frac{2\pi}{15}=\cos 24^\circ=\tfrac{1}{8}\left(\sqrt6\sqrt{5-\sqrt5}+\sqrt5+1\right)\,</math>
: <math>\tan\frac{2\pi}{15}=\tan 24^\circ=\tfrac{1}{2}\left[\sqrt{2(25+11\sqrt5)}-\sqrt3(3+\sqrt5)\right]\,</math>

=== 180/7°，<math>\mathbf{\left(25\frac{5}{7}\right)^{\circ}}</math>，<math>\mathbf{\left(\frac{180}{7}\right)^{\circ}}</math>：正七边形===

{{see|七边形}}
: <math>\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}</math>

=== 27°：12°与15°的和===
: <math>\sin\frac{3\pi}{20}=\sin 27^\circ=\tfrac{1}{8}\left[2\sqrt{5+\sqrt5}-\sqrt2\;(\sqrt5-1)\right]\,</math>
: <math>\cos\frac{3\pi}{20}=\cos 27^\circ=\tfrac{1}{8}\left[2\sqrt{5+\sqrt5}+\sqrt2\;\left(\sqrt5-1\right)\right]\,</math>
: <math>\tan\frac{3\pi}{20}=\tan 27^\circ=\sqrt5-1-\sqrt{5-2\sqrt5}\,</math>

=== 30°：正六边形 ===

{{see|六边形}}
: <math>\sin\frac{\pi}{6}=\sin 30^\circ=\tfrac{1}{2}\,</math>
: <math>\cos\frac{\pi}{6}=\cos 30^\circ=\tfrac{1}{2}\sqrt3\,</math>
: <math>\tan\frac{\pi}{6}=\tan 30^\circ=\tfrac{1}{3}\sqrt3\,</math>

=== 33°：15°与18°的和 ===
: <math>\sin\frac{11\pi}{60}=\sin 33^\circ=\tfrac{1}{4}\sqrt{8-\sqrt3-\sqrt{15}+\sqrt{10-2\sqrt5}}\,</math>
: <math>\cos\frac{11\pi}{60}=\cos 33^\circ=\tfrac{1}{4}\sqrt{8+\sqrt3+\sqrt{15}-\sqrt{10-2\sqrt5}}\,</math>
: <math>\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)\,</math>
: <math>\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)\,</math>

=== 36°：正五边形 ===

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

=== 39°：18°与21°的和 ===
: <math>\sin\frac{13\pi}{60}=\sin 39^\circ=\tfrac1{4}\sqrt{8-\sqrt3+\sqrt{15}+\sqrt{10+2\sqrt5}}\,</math>
: <math>\cos\frac{13\pi}{60}=\cos 39^\circ=\tfrac1{4}\sqrt{8+\sqrt3-\sqrt{15}+\sqrt{10+2\sqrt5}}\,</math>
: <math>\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]\,</math>

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

=== 45°：正方形 ===

{{see|正方形}}
: <math>\sin\frac{\pi}{4}=\sin 45^\circ=\frac{\sqrt2}{2}=\frac{1}{\sqrt2}\,</math>
: <math>\cos\frac{\pi}{4}=\cos 45^\circ=\frac{\sqrt2}{2}=\frac{1}{\sqrt2}\,</math>
: <math>\tan\frac{\pi}{4}=\tan 45^\circ=1</math>

=== 48° ===
: <math>\sin 48^\circ=\frac{1}{4}\sqrt{7-\sqrt{5}+\sqrt{6(5-\sqrt{5})}}</math>

===54°：27°与27°的和===
:<math>\sin\frac{3\pi}{10}=\sin 54^\circ=\frac{\sqrt5+1}{4}\,\!</math>
:<math>\cos\frac{3\pi}{10}=\cos 54^\circ=\frac{\sqrt{10-2\sqrt{5}}}{4}</math>
:<math>\tan\frac{3\pi}{10}=\tan 54^\circ=\frac{\sqrt{25+10\sqrt{5}}}{5}\,</math>
:<math>\cot\frac{3\pi}{10}=\cot 54^\circ=\sqrt{5-2\sqrt{5}}\,</math>

===60°：等边三角形===
:<math>\sin\frac{\pi}{3}=\sin 60^\circ=\frac{\sqrt3}{2}\,</math>
:<math>\cos\frac{\pi}{3}=\cos 60^\circ=\frac{1}{2}\,</math>
:<math>\tan\frac{\pi}{3}=\tan 60^\circ=\sqrt3\,</math>
:<math>\cot\frac{\pi}{3}=\cot 60^\circ=\frac{\sqrt3}{3}=\frac{1}{\sqrt3}\,</math>

===67.5°：7.5°与60°的和===
:<math>\sin\frac{3\pi}{8}=\sin 67.5^\circ=\tfrac{1}{2}\sqrt{2+\sqrt{2}}\,</math>
:<math>\cos\frac{3\pi}{8}=\cos 67.5^\circ=\tfrac{1}{2}\sqrt{2-\sqrt{2}}\,</math>
:<math>\tan\frac{3\pi}{8}=\tan 67.5^\circ=\sqrt{2}+1\,</math>
:<math>\cot\frac{3\pi}{8}=\cot 67.5^\circ=\sqrt{2}-1\,</math>

===72°：36°的二倍===
:<math>\sin\frac{2\pi}{5}=\sin 72^\circ=\tfrac{1}{4}\sqrt{2\left(5+\sqrt5\right)}\,</math>
:<math>\cos\frac{2\pi}{5}=\cos 72^\circ=\tfrac{1}{4}\left(\sqrt5-1\right)\,</math>

:<math>\tan\frac{2\pi}{5}=\tan 72^\circ=\sqrt{5+2\sqrt 5}\,</math>
:<math>\cot\frac{2\pi}{5}=\cot 72^\circ=\tfrac{1}{5}\sqrt{5\left(5-2\sqrt5\right)}\,</math>

=== 75°: 30°与45°的和 ===
:<math>\sin\frac{5\pi}{12}=\sin 75^\circ=\tfrac{1}{4}\left(\sqrt6+\sqrt2\right)\,</math>
:<math>\cos\frac{5\pi}{12}=\cos 75^\circ=\tfrac{1}{4}\left(\sqrt6-\sqrt2\right)\,</math>
:<math>\tan\frac{5\pi}{12}=\tan 75^\circ=2+\sqrt3\,</math>
:<math>\cot\frac{5\pi}{12}=\cot 75^\circ=2-\sqrt3\,</math>
=== 81° ===
:<math>\sin 81^\circ=\frac{1}{2}\sqrt{\frac{1}{2}\Big(4+\sqrt{2(5+\sqrt{5})}\Big)}</math>

=== 90°：根本 ===
:<math>\sin \frac{\pi}{2}=\sin 90^\circ=1\,</math>
:<math>\cos \frac{\pi}{2}=\cos 90^\circ=0\,</math>
:<math>\cot \frac{\pi}{2}=\cot 90^\circ=0\,</math>

==列表==

在下表中，<math>i</math>為[[虛數單位]]，<math>\omega=\exp(\frac{\pi i}{3})=-\frac{1}{2}+\frac{1}{2}i\sqrt{3}</math>。

{|class="wikitable"
|<math>n</math>
|<math>\sin\left(\frac{2\pi}{n}\right)</math>
|<math>\cos\left(\frac{2\pi}{n}\right)</math>
|<math>\tan\left(\frac{2\pi}{n}\right)</math>
|-
|1
|<math>0</math>
|<math>1</math>
|<math>0</math>
|-
|2
|<math>0</math>
|<math>-1</math>
|<math>0</math>
|-
|3
|<math>\frac{1}{2}\sqrt{3}</math>
|<math>-\frac{1}{2}</math>
|<math>-\sqrt{3}</math>
|-
|4
|<math>1</math>
|<math>0</math>
|<math>\pm\infty</math>
|-
|5
|<math>\frac{1}{4}\left(\sqrt{10+2\sqrt{5}}\right)</math>
|<math>\frac{1}{4}\left(\sqrt{5}-1\right)</math>
|<math>\sqrt{5+2\sqrt{5}}</math>
|-
|6
|<math>\frac{1}{2}\sqrt{3}</math>
|<math>\frac{1}{2}</math>
|<math>\sqrt{3}</math>
|-
|7
|<math>\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)}</math>
|<math>\frac{1}{6}\left(-1+\sqrt[3]{\frac{7+21\sqrt{-3}}{2}}+\sqrt[3]{\frac{7-21\sqrt{-3}}{2}}\right)</math>
|
|-
|8
|<math>\frac{1}{2}\sqrt{2}</math>
|<math>\frac{1}{2}\sqrt{2}</math>
|<math>1</math>
|-
|9
|<math>\frac{1}{4}\left(\sqrt[3]{4(-1+\sqrt{-3})}+\sqrt[3]{4(-1-\sqrt{-3})}\right)</math>
|
|
|-
|10
|<math>\frac{1}{4}\left(\sqrt{10-2\sqrt{5}}\right)</math>
|<math>\frac{1}{4}\left(\sqrt{5}+1\right)</math>
|<math>\sqrt{5-2\sqrt{5}}</math>
|-
|11
|
|
|
|-
|12
|<math>\frac{1}{2}</math>
|<math>\frac{1}{2}\sqrt{3}</math>
|<math>\frac{1}{3}\sqrt{3}</math>
|-
|13
|
|
|
|-
|14
|<math>\frac{1}{24}\sqrt{3\left(112-\sqrt[3]{14336+\sqrt{-5549064193}}-\sqrt[3]{14336-\sqrt{-5549064193}}\right)} </math>
|<math>\frac{1}{24}\sqrt{3\left(80+\sqrt[3]{14336+\sqrt{-5549064193}}+\sqrt[3]{14336-\sqrt{-5549064193}}\right)} </math>
|<math>\sqrt{\frac{112-\sqrt[3]{14336+\sqrt{-5549064193}}-\sqrt[3]{14336-\sqrt{-5549064193}}}{80+\sqrt[3]{14336+\sqrt{-5549064193}}+\sqrt[3]{14336-\sqrt{-5549064193}}}}</math>
|-
|15
|<math>\frac{1}{8}\left(\sqrt{15}+\sqrt{3}-\sqrt{10-2\sqrt{5}}\right)</math>
|<math>\frac{1}{8}\left(1+\sqrt{5}+\sqrt{30-6\sqrt{5}}\right)</math>
|<math>\frac{1}{2}\left(-3\sqrt{3}-\sqrt{15}+\sqrt{50+22\sqrt{5}}\right)</math>
|-
|16
|<math>\frac{1}{2}\left(\sqrt{2-\sqrt{2}}\right)</math> 
|<math>\frac{1}{2}\left(\sqrt{2+\sqrt{2}}\right)</math>
|<math>\sqrt{2}-1</math>
|-
|17
|<math>\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)}}</math>
|<math>\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)</math>
|
|-
|18
|<math>\frac{1}{4}\left(\sqrt[3]{4\sqrt{-1}-4\sqrt{3}}-\sqrt[3]{4\sqrt{-1}+4\sqrt{3}}\right)</math>
|<math>\frac{1}{4}\left(\sqrt[3]{4+4\sqrt{-3}}+\sqrt[3]{4-4\sqrt{-3}}\right)</math>
|
|-
|19
|
|
|
|-
|20
|<math>\frac{1}{4}\left(\sqrt{5}-1\right)</math>
|<math>\frac{1}{4}\left(\sqrt{10+2\sqrt{5}}\right)</math>
|<math>\frac{1}{5}\left(\sqrt{25-10\sqrt{5}}\right)</math>
|-
|21
|
|
|
|-
|22
|
|
|
|-
|23
|
|
|
|-
|24
|<math>\frac{1}{4}\left(\sqrt{6}-\sqrt{2}\right)</math>
|<math>\frac{1}{4}\left(\sqrt{6}+\sqrt{2}\right)</math>
|<math>2-\sqrt{3}</math>
|}

==相關==
{{see|三角函數|三角恆等式}}
{{see|en:Generating trigonometric tables}}

==參見==
*[[可作图多边形]]
*{{tsl|en|Trigonometric_number|三角數}}
*[[十七邊形]]

==參考文獻==
* {{MathWorld|title=Constructible polygon|urlname=ConstructiblePolygon}}
* {{MathWorld|title=Trigonometry angles|urlname=TrigonometryAngles}}
** [http://mathworld.wolfram.com/TrigonometryAnglesPi3.html π/3 (60°)]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi3.html |date=20110212164640 }}—[http://mathworld.wolfram.com/TrigonometryAnglesPi6.html π/6 (30°)]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi6.html |date=20110522183409 }}—[http://mathworld.wolfram.com/TrigonometryAnglesPi12.html π/12 (15°)]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi12.html |date=20100717010251 }}—[http://mathworld.wolfram.com/TrigonometryAnglesPi24.html π/24 (7.5°)]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi24.html |date=20101024141647 }}
** [http://mathworld.wolfram.com/TrigonometryAnglesPi4.html π/4 (45°)]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi4.html |date=20110522183132 }}—[http://mathworld.wolfram.com/TrigonometryAnglesPi8.html π/8 (22.5°)]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi8.html |date=20110522183536 }}—[http://mathworld.wolfram.com/TrigonometryAnglesPi16.html π/16 (11.25°)]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi16.html |date=20110521041823 }}—[http://mathworld.wolfram.com/TrigonometryAnglesPi32.html π/32 (5.625°)]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi32.html |date=20110522185037 }}
** [http://mathworld.wolfram.com/TrigonometryAnglesPi5.html π/5 (36°)]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi5.html |date=20110522183141 }}—[http://mathworld.wolfram.com/TrigonometryAnglesPi10.html π/10 (18°)]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi10.html |date=20101031084302 }}—[http://mathworld.wolfram.com/TrigonometryAnglesPi20.html π/20 (9°)]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi20.html |date=20110522184929 }}
** [http://mathworld.wolfram.com/TrigonometryAnglesPi7.html π/7]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi7.html |date=20101119052456 }}—''π/14''
** [http://mathworld.wolfram.com/TrigonometryAnglesPi9.html π/9 (20°)]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi9.html |date=20110211115235 }}—[http://mathworld.wolfram.com/TrigonometryAnglesPi18.html π/18 (10°)]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi18.html |date=20110522184700 }}
** [http://mathworld.wolfram.com/TrigonometryAnglesPi11.html π/11]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi11.html |date=20110522184232 }}
** [http://mathworld.wolfram.com/TrigonometryAnglesPi13.html π/13]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi13.html |date=20110522184549 }}
** [http://mathworld.wolfram.com/TrigonometryAnglesPi15.html π/15 (12°)]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi15.html |date=20101024141641 }}—[http://mathworld.wolfram.com/TrigonometryAnglesPi30.html π/30 (6°)]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi30.html |date=20101024141651 }}
** [http://mathworld.wolfram.com/TrigonometryAnglesPi17.html π/17]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi17.html |date=20101027111119 }}
** ''π/19''
** [http://mathworld.wolfram.com/TrigonometryAnglesPi23.html π/23]{{Wayback|url=http://mathworld.wolfram.com/TrigonometryAnglesPi23.html |date=20110522185320 }}
* {{Cite journal
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|title=Evaluation of quantum mechanical perturbation sums in terms of quadratic surds and their use in approximation of zeta(3)/pi^3
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* {{cite journal
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* {{cite journal
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==注释==
<references/>

{{三角函數}}
{{無理數導航}}
[[Category:三角學]]
[[Category:三角函数]]