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。

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°

### 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。

### 經由歐拉公式的計算

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

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

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

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

### 1.875°：正九十六邊形

$$\sin\left(\frac{\pi}{96}\right) = \sin\left(1.875^\circ\right) = \frac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}$$

$$\cos\left(\frac{\pi}{96}\right) = \cos\left(1.875^\circ\right) = \frac12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}$$

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

### 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

$$\sin\left(\frac{\pi}{64}\right) = \sin\left(2.8125^\circ\right) = \frac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}$$

$$\cos\left(\frac{\pi}{64}\right) = \cos\left(2.8125^\circ\right) = \frac12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}$$

### 3°：正六十邊形

$$\sin\frac{\pi}{60}=\sin 3^\circ=\tfrac{1}{4} \sqrt{8-\sqrt3-\sqrt{15}-\sqrt{10-2\sqrt5}}\,$$

$$\cos\frac{\pi}{60}=\cos 3^\circ=\tfrac{1}{4} \sqrt{8+\sqrt3+\sqrt{15}+\sqrt{10-2\sqrt5}}\,$$

$$\tan\frac{\pi}{60}=\tan 3^\circ=\tfrac{1}{4} \left[(2-\sqrt3)(3+\sqrt5)-2\right]\left[2-\sqrt{2(5-\sqrt5)}\right]\,$$

### 3.75°：正四十八邊形

$$\sin\left(\frac{\pi}{48}\right) = \sin\left(3.75^\circ\right) = \frac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}$$

$$\cos\left(\frac{\pi}{48}\right) = \cos\left(3.75^\circ\right) = \frac12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}$$

### 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

$$\sin\left(\frac{\pi}{32}\right) = \sin\left(5.625^\circ\right) = \frac12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}$$

$$\cos\left(\frac{\pi}{32}\right) = \cos\left(5.625^\circ\right) = \frac12\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}$$

### 6°：正三十邊形

$$\sin\frac{\pi}{30}=\sin 6^\circ=\tfrac{1}{8} \left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]\,$$

$$\cos\frac{\pi}{30}=\cos 6^\circ=\tfrac{1}{8} \left[\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)\right]\,$$

$$\tan\frac{\pi}{30}=\tan 6^\circ=\tfrac{1}{2} \left[\sqrt{2(5-\sqrt5)}-\sqrt3(\sqrt5-1)\right]\,$$

$$\cot\frac{\pi}{30}=\cot 6^\circ=\tfrac{1}{2} \left(\sqrt{50+22\sqrt5}+3\sqrt3+\sqrt{15}\right)\,$$

$$\sec\frac{\pi}{30}=\sec 6^\circ=\sqrt3-\sqrt{5-2\sqrt5}\,$$

$$\csc\frac{\pi}{30}=\csc 6^\circ=2+\sqrt5+\sqrt{15+6\sqrt5}\,$$

### 7.5°：正二十四邊形

$$\sin\frac{\pi}{24}=\sin 7.5^\circ=\tfrac{1}{4} \sqrt{8-2\sqrt6-2\sqrt2}\,$$

$$\cos\frac{\pi}{24}=\cos 7.5^\circ=\tfrac{1}{4} \sqrt{8+2\sqrt6+2\sqrt2}\,$$

$$\tan\frac{\pi}{24}=\tan 7.5^\circ=\sqrt6+\sqrt2-2-\sqrt3\,$$

$$\cot\frac{\pi}{24}=\cot 7.5^\circ=\sqrt6+\sqrt2+2+\sqrt3\,$$

$$\sec\frac{\pi}{24}=\sec 7.5^\circ=\sqrt{16-6\sqrt6-10\sqrt2+8\sqrt3}\,$$

$$\csc\frac{\pi}{24}=\csc 7.5^\circ=\sqrt{16+6\sqrt6+10\sqrt2+8\sqrt3}\,$$

### 9°：正二十邊形

$$\sin\frac{\pi}{20}=\sin 9^\circ=\tfrac{1}{4} \sqrt{8-2\sqrt{10+2\sqrt5}}\,$$

$$\cos\frac{\pi}{20}=\cos 9^\circ=\tfrac{1}{4} \sqrt{8+2\sqrt{10+2\sqrt5}}\,$$

$$\tan\frac{\pi}{20}=\tan 9^\circ=\sqrt5+1-\sqrt{5+2\sqrt5}\,$$

$$\cot\frac{\pi}{20}=\cot9^\circ=\sqrt5+1+\sqrt{5+2\sqrt5}\,$$

### 10°：正十八邊形

$${\tan10^\circ=-\frac{-1-\sqrt{3}{\rm{i}}}{6}\sqrt[3]{-12\sqrt3 + 36{\rm{i}}}-\frac{-1+\sqrt{3}{\rm{i}}}{6}\sqrt[3]{-12\sqrt3 - 36{\rm{i}}} + \frac{1}{\sqrt3}}\,$$

### 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

$$\sin\frac{\pi}{15}=\sin 12^\circ=\tfrac{1}{8} \left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]\,$$

$$\cos\frac{\pi}{15}=\cos 12^\circ=\tfrac{1}{8} \left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]\,$$

$$\tan\frac{\pi}{15}=\tan 12^\circ=\tfrac{1}{2} \left[\sqrt3(3-\sqrt5)-\sqrt{2(25-11\sqrt5)}\right]\,$$

### 15°：正十二邊形

$$\sin\frac{\pi}{12}=\sin 15^\circ=\frac{1}{4}\sqrt2\left(\sqrt3-1\right)\,$$

$$\cos\frac{\pi}{12}=\cos 15^\circ=\frac{1}{4}\sqrt2\left(\sqrt3+1\right)\,$$

$$\tan\frac{\pi}{12}=\tan 15^\circ=2-\sqrt3\,$$

$$\cot\frac{\pi}{12}=\cot 15^\circ=2+\sqrt3\,$$

### 18°：正十邊形

$$\sin\frac{\pi}{10}=\sin 18^\circ=\tfrac{1}{4}\left(\sqrt5-1\right)=\tfrac{1}{2}\varphi^{-1}\,$$

$$\cos\frac{\pi}{10}=\cos 18^\circ=\tfrac{1}{4}\sqrt{2\left(5+\sqrt5\right)}\,$$

$$\tan\frac{\pi}{10}=\tan 18^\circ=\tfrac{1}{5}\sqrt{5\left(5-2\sqrt5\right)}\,$$

### 20°：正九邊形、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°與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°，$\mathbf{\left(21\frac{3}{17}\right)^{\circ}}$，$\mathbf{\left(\frac{360}{17}\right)^{\circ}}$：正十七邊形

$$\operatorname{cos}{2\pi\over17}=\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}}}{16}$$

### 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

$$\sin\frac{2\pi}{15}=\sin 24^\circ=\tfrac{1}{8}\left[\sqrt3(\sqrt5+1)-\sqrt2\sqrt{5-\sqrt5}\right]\,$$

$$\cos\frac{2\pi}{15}=\cos 24^\circ=\tfrac{1}{8}\left(\sqrt6\sqrt{5-\sqrt5}+\sqrt5+1\right)\,$$

$$\tan\frac{2\pi}{15}=\tan 24^\circ=\tfrac{1}{2}\left[\sqrt{2(25+11\sqrt5)}-\sqrt3(3+\sqrt5)\right]\,$$

### 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

$$\sin\frac{3\pi}{20}=\sin 27^\circ=\tfrac{1}{8}\left[2\sqrt{5+\sqrt5}-\sqrt2\;(\sqrt5-1)\right]\,$$

$$\cos\frac{3\pi}{20}=\cos 27^\circ=\tfrac{1}{8}\left[2\sqrt{5+\sqrt5}+\sqrt2\;\left(\sqrt5-1\right)\right]\,$$

$$\tan\frac{3\pi}{20}=\tan 27^\circ=\sqrt5-1-\sqrt{5-2\sqrt5}\,$$

### 30°：正六邊形

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

$$\cos\frac{\pi}{6}=\cos 30^\circ=\tfrac{1}{2}\sqrt3\,$$

$$\tan\frac{\pi}{6}=\tan 30^\circ=\tfrac{1}{3}\sqrt3\,$$

### 33°：15°與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°：正五邊形

$$\sin\frac{\pi}{5}=\sin 36^\circ=\tfrac14\left[\sqrt{2\left(5-\sqrt5\right)}\right]\,$$

$$\cos\frac{\pi}{5}=\cos 36^\circ=\frac{1+\sqrt5}{4}=\tfrac{1}{2}\varphi\,$$

$$\tan\frac{\pi}{5}=\tan 36^\circ=\sqrt{5-2\sqrt5}\,$$

### 39°：18°與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°與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°：等邊三角形

$$\sin\frac{\pi}{3}=\sin 60^\circ=\frac{\sqrt3}{2}\,$$

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

$$\tan\frac{\pi}{3}=\tan 60^\circ=\sqrt3\,$$

$$\cot\frac{\pi}{3}=\cot 60^\circ=\frac{\sqrt3}{3}=\frac{1}{\sqrt3}\,$$

### 67.5°：7.5°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

$$\sin\frac{2\pi}{5}=\sin 72^\circ=\tfrac{1}{4}\sqrt{2\left(5+\sqrt5\right)}\,$$

$$\cos\frac{2\pi}{5}=\cos 72^\circ=\tfrac{1}{4}\left(\sqrt5-1\right)\,$$

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

$$\cot\frac{2\pi}{5}=\cot 72^\circ=\tfrac{1}{5}\sqrt{5\left(5-2\sqrt5\right)}\,$$

### 75°: 30°與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，$\omega=\exp(\frac{\pi i}{3})=-\frac{1}{2}+\frac{1}{2}i\sqrt{3}$。

$n$｜$\sin\left(\frac{2\pi}{n}\right)$｜$\cos\left(\frac{2\pi}{n}\right)$｜$\tan\left(\frac{2\pi}{n}\right)$

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｜$\frac{1}{4}\left(\sqrt{6}-\sqrt{2}\right)$｜$\frac{1}{4}\left(\sqrt{6}+\sqrt{2}\right)$｜$2-\sqrt{3}$

## 參考文獻

- π/3 (60°)— π/6 (30°)— π/12 (15°)— π/24 (7.5°)
- π/4 (45°)— π/8 (22.5°)— π/16 (11.25°)— π/32 (5.625°)
- π/5 (36°)— π/10 (18°)— π/20 (9°)
- π/7—π/14
- π/9 (20°)— π/18 (10°)
- π/11
- π/13
- π/15 (12°)— π/30 (6°)
- π/17
- π/19
- π/23