三角函數精確值

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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,並以數學根式分數表示

根式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
\mathbf{0} \tfrac{1}{12} \tfrac{1}{8} \tfrac{1}{6} \tfrac{1}{4} \tfrac{1}{2} \tfrac{3}{4} \mathbf{1}
角度 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[编辑]

基於常識[编辑]

例如:0°、30°、45°

單位圓

經由半角公式的計算[编辑]

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\frac{1}{3}\,[编辑]

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 \,

\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[编辑]

  • \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)[1]

經由合角公式的計算[编辑]

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[编辑]

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[编辑]

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[编辑]

\sin 0=0\,
\cos 0=1\,
\tan 0=0\,

1°:2°的一半[编辑]

\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}[2]

2°:6°的三分之一[编辑]

\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°:正六十邊形[编辑]

\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]\,

4°:12°的三分之一[编辑]

\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)

5°:15°的三分之一、正三十六邊形[编辑]

\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}\,

6°:正三十邊形[编辑]

\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[编辑]

\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[编辑]

\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[编辑]

{\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}}\,

12°:正十五邊形[编辑]

\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[编辑]

\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[编辑]

\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、60°的三分之一[编辑]

\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[编辑]

\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-(2+\sqrt3)(3-\sqrt5)\right]\left[2-\sqrt{2(5+\sqrt5)}\right]\,

360/17°,\mathbf{\left(21\frac{3}{17}\right)^{\circ}}\mathbf{\left(\frac{360}{17}\right)^{\circ}}正十七邊形[编辑]

\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[编辑]

\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°:12°的二倍[编辑]

\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[编辑]

\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[编辑]

\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[编辑]

\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[编辑]

\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}(2\sqrt3-\sqrt5-1)(2\sqrt{5+2\sqrt5}+3+\sqrt5)\,
\cot\frac{11\pi}{60}=\cot33^\circ=\tfrac{1}{4}(2\sqrt3+\sqrt5+1)(2\sqrt{5+2\sqrt5}-3-\sqrt5)\,

36°:正五邊形[编辑]

\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°与21°的和[编辑]

\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[(2-\sqrt3)(3-\sqrt5)-2\right]\left[2-\sqrt{2(5+\sqrt5)}\right]\,

42°:21°的2倍[编辑]

\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=\frac1{2}(\sqrt3+\sqrt{15}-\sqrt{10+2\sqrt5})\,
\cot\frac{7\pi}{30}=\cot 42^\circ=\frac1{2}(3\sqrt3-\sqrt{15}+\sqrt{50-22\sqrt5})\,
\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[编辑]

\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

相關[编辑]

參見[编辑]

參考文獻[编辑]

注释[编辑]

  1. ^ Wolfram Alpha验算:[1]
  2. ^ 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