三角函數精確值

三角学
历史 三角函数 广义三角函数 反三角函数
参考
恒等式 精确值 三角表
定理
正弦定理 余弦定理 正切定理 余切定理 勾股定理
微积分
三角换元法 三角函数的积分 三角函数的微分 反三角函数的积分

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。

注意：以下為相同角度的轉換表：

相同角度的轉換表

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

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$\frac{1}{3}\,$角

例如：10°、20°、7°......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：$\sin\frac{\pi}{9}=\sin 20^\circ=\sqrt[3]{-\frac{\sqrt{3}}{16}+\sqrt{-\frac{1}{256}}}+\sqrt[3]{-\frac{\sqrt{3}}{16}-\sqrt{-\frac{1}{256}}}$

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：

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

$= \frac{1}{4\sqrt[3]{2}i}\Bigg(\sqrt[3]{\left[2(1+\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3-1)\right]+i\left[2(1-\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3+1)\right]}$

$-\sqrt[3]{\left[2(1+\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3-1)\right]-i\left[2(1-\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3+1)\right]}\Bigg)$

經由合角公式的計算

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/f, from 托勒密定理

例如：18°

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

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

$\sin{18^\circ}=\frac{1}{1+\sqrt{5}}=\tfrac{1}{4}\left(\sqrt5-1\right)$

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

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

2°:6°的三分之一

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

$= \frac{1}{4i}\Bigg(\sqrt[3]{\left[\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)\right]+i\left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]}$

$-\sqrt[3]{\left[\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)\right]-i\left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]}\Bigg)$

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

$= \frac{1}{4}\Bigg(\sqrt[3]{\left[\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)\right]+i\left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]}$

$+\sqrt[3]{\left[\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)\right]-i\left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]}\Bigg)$

3°: 正60邊形

$\sin\frac{\pi}{60}=\sin 3^\circ=\tfrac{1}{16} \left[2(1-\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3+1)\right]\,$

$\cos\frac{\pi}{60}=\cos 3^\circ=\tfrac{1}{16} \left[2(1+\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3-1)\right]\,$

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

4°:12°的三分之一

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

$= \frac{1}{4i}\Bigg(\sqrt[3]{\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]+i\left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]}$

$-\sqrt[3]{\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]-i\left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]}\Bigg)$

$\cos{4^\circ} = \frac{1}{2}\left(\sqrt[12]{\cos{6^\circ}+i\sin{12^\circ}}+\sqrt[3]{\cos{12^\circ}-i\sin{12^\circ}}\right)$

$= \frac{1}{4}\Bigg(\sqrt[3]{\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]+i\left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]}$

$+\sqrt[3]{\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]-i\left[\sqrt{2(5+\sqrt5)}-\sqrt3(\sqrt5-1)\right]}\Bigg)$

6°: 正30邊形

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

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

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

9°: 正20邊形

$\sin\frac{\pi}{20}=\sin 9^\circ=\tfrac{1}{8} \left[\sqrt2(\sqrt5+1)-2\sqrt{5-\sqrt5}\right]\,$

$\cos\frac{\pi}{20}=\cos 9^\circ=\tfrac{1}{8} \left[\sqrt2(\sqrt5+1)+2\sqrt{5-\sqrt5}\right]\,$

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

10°: 正18邊形

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

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

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

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

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

18°: 正十邊形

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

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

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

20°: 正九邊形 和 60°的三分之一($\frac{1}{3}\,$ 60°)

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

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

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

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

21°: 9° 與 12°的和

$\sin\frac{7\pi}{60}=\sin 21^\circ=\tfrac{1}{16}\left[2(\sqrt3+1)\sqrt{5-\sqrt5}-\sqrt2(\sqrt3-1)(1+\sqrt5)\right]\,$

$\cos\frac{7\pi}{60}=\cos 21^\circ=\tfrac{1}{16}\left[2(\sqrt3-1)\sqrt{5-\sqrt5}+\sqrt2(\sqrt3+1)(1+\sqrt5)\right]\,$

$\tan\frac{7\pi}{60}=\tan 21^\circ=\tfrac{1}{4}\left[2-(2+\sqrt3)(3-\sqrt5)\right]\left[2-\sqrt{2(5+\sqrt5)}\right]\,$

$\mathbf{\left(21\frac{3}{17}\right)^{\circ}}$，$\mathbf{\left(\frac{360}{17}\right)^{\circ}}$：正17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

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

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

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

24°: 兩倍的 12° 角

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

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

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

25(5/7)°，(180/7)°：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

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

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

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

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° 之和

$\sin\frac{11\pi}{60}=\sin 33^\circ=\tfrac{1}{16}\left[2(\sqrt3-1)\sqrt{5+\sqrt5}+\sqrt2(1+\sqrt3)(\sqrt5-1)\right]\,$

$\cos\frac{11\pi}{60}=\cos 33^\circ=\tfrac{1}{16}\left[2(\sqrt3+1)\sqrt{5+\sqrt5}+\sqrt2(1-\sqrt3)(\sqrt5-1)\right]\,$

$\tan\frac{11\pi}{60}=\tan 33^\circ=\tfrac{1}{4}\left[2-(2-\sqrt3)(3+\sqrt5)\right]\left[2+\sqrt{2(5-\sqrt5)}\right]\,$

36°: 正五邊形

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

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

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

39°: 18°角加21°角

$\sin\frac{13\pi}{60}=\sin 39^\circ=\tfrac1{16}[2(1-\sqrt3)\sqrt{5-\sqrt5}+\sqrt2(\sqrt3+1)(\sqrt5+1)]\,$

$\cos\frac{13\pi}{60}=\cos 39^\circ=\tfrac1{16}[2(1+\sqrt3)\sqrt{5-\sqrt5}+\sqrt2(\sqrt3-1)(\sqrt5+1)]\,$

$\tan\frac{13\pi}{60}=\tan 39^\circ=\tfrac14\left[(2-\sqrt3)(3-\sqrt5)-2\right]\left[2-\sqrt{2(5+\sqrt5)}\right]\,$

42°: 21°的兩倍

$\sin\frac{7\pi}{30}=\sin 42^\circ=\frac{\sqrt6\sqrt{5+\sqrt5}-\sqrt5+1}{8}\,$

$\cos\frac{7\pi}{30}=\cos 42^\circ=\frac{\sqrt2\sqrt{5+\sqrt5}+\sqrt3(\sqrt5-1)}{8}\,$

$\tan\frac{7\pi}{30}=\tan 42^\circ=\frac{\sqrt3(\sqrt5+1)-\sqrt2\sqrt{5+\sqrt5}}{2}\,$

45°: 正方形

$\sin\frac{\pi}{4}=\sin 45^\circ=\frac{\sqrt2}{2}=\frac{1}{\sqrt2}\,$

$\cos\frac{\pi}{4}=\cos 45^\circ=\frac{\sqrt2}{2}=\frac{1}{\sqrt2}\,$

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

參見

• 可作图多边形
• 三角數
• 十七邊形

參考文獻

• 埃里克·韦斯坦因, Constructible polygon at MathWorld
• 埃里克·韦斯坦因, Trigonometry angles at MathWorld
  • 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
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