三角函數精確值

三角函數精確值

Table of constants

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°: fundamental

$\sin 0=0\,$

$\cos 0=1\,$

$\tan 0=0\,$

3°: 60-sided polygon

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

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

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

6°: 30-sided polygon

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

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

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

9°: 20-sided polygon

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

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

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

12°: 15-sided polygon

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

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

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

15°: dodecagon

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

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

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

18°: decagon

$\sin {\frac {\pi }{10}}=\sin 18^{\circ }={\tfrac {1}{4}}\left({\sqrt {5}}-1\right)={\tfrac {1}{2}}\varphi ^{-1}\,$

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

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

21°: sum 9° + 12°

$\sin {\frac {7\pi }{60}}=\sin 21^{\circ }={\tfrac {1}{16}}\left[2({\sqrt {3}}+1){\sqrt {5-{\sqrt {5}}}}-{\sqrt {2}}({\sqrt {3}}-1)(1+{\sqrt {5}})\right]\,$

$\cos {\frac {7\pi }{60}}=\cos 21^{\circ }={\tfrac {1}{16}}\left[2({\sqrt {3}}-1){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {3}}+1)(1+{\sqrt {5}})\right]\,$

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

22.5°: octagon

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

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

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

24°: sum 12° + 12°

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

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

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

27°: sum 12° + 15°

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

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

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

30°: hexagon

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

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

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

33°: sum 15° + 18°

$\sin {\frac {11\pi }{60}}=\sin 33^{\circ }={\tfrac {1}{16}}\left[2({\sqrt {3}}-1){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}(1+{\sqrt {3}})({\sqrt {5}}-1)\right]\,$

$\cos {\frac {11\pi }{60}}=\cos 33^{\circ }={\tfrac {1}{16}}\left[2({\sqrt {3}}+1){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}(1-{\sqrt {3}})({\sqrt {5}}-1)\right]\,$

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

36°: pentagon

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

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

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

39°: sum 18° + 21°

$\sin {\frac {13\pi }{60}}=\sin 39^{\circ }={\tfrac {1}{16}}[2(1-{\sqrt {3}}){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {3}}+1)({\sqrt {5}}+1)]\,$

$\cos {\frac {13\pi }{60}}=\cos 39^{\circ }={\tfrac {1}{16}}[2(1+{\sqrt {3}}){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {3}}-1)({\sqrt {5}}+1)]\,$

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

42°: sum 21° + 21°

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

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

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

45°: square

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

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

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

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