User:JsfasdF252/sandbox

< User:JsfasdF252 yOuR tExT  

Further angles
Values outside the [0°, 45°] angle range are trivially derived from these values, using circle axis reflection symmetry. (See List of trigonometric identities.)

In the entries below, when a certain number of degrees is related to a regular polygon, the relation is that the number of degrees in each angle of the polygon is (n – 2) times the indicated number of degrees (where n is the number of sides). This is because the sum of the angles of any n-gon is 180° × (n – 2) and so the measure of each angle of any regular n-gon is 180° × (n – 2) ÷ n. Thus for example the entry "45°: square" means that, with n = 4, 180° ÷ n = 45°, and the number of degrees in each angle of a square is (n – 2) × 45° = 90°.

 0°: fundamental

 * $$\sin 0=\cos \frac{\pi}{2}=\cos 90^\circ=0\,$$
 * $$\cos 0=\sin \frac{\pi}{2}=\sin 90^\circ=1\,$$
 * $$\tan 0=\cot \frac{\pi}{2}=\cot 90^\circ=0\,$$
 * $$\cot 0=\tan \frac{\pi}{2}=\tan 90^\circ\text{ is undefined}\,$$

1.5°: regular hecatonicosagon (120-sided polygon)

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

1.875°: regular enneacontahexagon (96-sided polygon)

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

2.25°: regular octacontagon (80-sided polygon)

 * $$\sin\left(\frac{\pi}{80}\right) = \sin\left(2.25^\circ\right) =\frac{1}{2} \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}}}$$
 * $$\cos\left(\frac{\pi}{80}\right) = \cos\left(2.25^\circ\right) =\frac{1}{2} \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}}}$$

2.8125°: regular hexacontatetragon (64-sided polygon)

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

3°: regular hexacontagon (60-sided polygon)

 * $$\sin\left(\frac{\pi}{60}\right) = \sin\left(3^\circ\right) = \frac{2\left(1-\sqrt3\right)\sqrt{5+\sqrt5}+\left(\sqrt{10}-\sqrt2\right)\left(\sqrt3+1\right)}{16}\,$$
 * $$\cos\left(\frac{\pi}{60}\right) = \cos\left(3^\circ\right) = \frac{2\left(1+\sqrt3\right)\sqrt{5+\sqrt5}+\left(\sqrt{10}-\sqrt2\right)\left(\sqrt3-1\right)}{16}\,$$
 * $$\tan\left(\frac{\pi}{60}\right) = \tan\left(3^\circ\right) = \frac{\left[\left(2-\sqrt3\right)\left(3+\sqrt5\right)-2\right]\left[2-\sqrt{10-2\sqrt5}\right]}{4}\,$$
 * $$\cot\left(\frac{\pi}{60}\right) = \cot\left(3^\circ\right) = \frac{\left[\left(2+\sqrt3\right)\left(3+\sqrt5\right)-2\right]\left[2+\sqrt{10-2\sqrt5}\right]}{4}\,$$

3.75°: regular tetracontaoctagon (48-sided polygon)

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

4.5°: regular tetracontagon (40-sided polygon)

 * $$\sin\left(\frac{\pi}{40}\right) = \sin\left(4.5^\circ\right) =\frac{1}{2} \sqrt{2-\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}}$$
 * $$\cos\left(\frac{\pi}{40}\right) = \cos\left(4.5^\circ\right) =\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}}$$

5.625°: regular triacontadigon (32-sided polygon)

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

6°: regular triacontagon (30-sided polygon)

 * $$\sin\frac{\pi}{30}=\sin 6^\circ=\frac{\sqrt{30-\sqrt{180}}-\sqrt5-1}{8}\,$$
 * $$\cos\frac{\pi}{30}=\cos 6^\circ=\frac{\sqrt{10-\sqrt{20}}+\sqrt3+\sqrt{15}}{8}\,$$
 * $$\tan\frac{\pi}{30}=\tan 6^\circ=\frac{\sqrt{10-\sqrt{20}}+\sqrt3-\sqrt{15}}{2}\,$$
 * $$\cot\frac{\pi}{30}=\cot 6^\circ=\frac{\sqrt{27}+\sqrt{15}+\sqrt{50+\sqrt{2420}}}{2}\,$$

7.5°: regular icositetragon (24-sided polygon)

 * $$\sin\left(\frac{\pi}{24}\right)=\sin\left(7.5^\circ\right)=\frac12\sqrt{2-\sqrt{2+\sqrt3}}

= \frac14\sqrt{8-2\sqrt6-2\sqrt2}$$
 * $$\cos\left(\frac{\pi}{24}\right)=\cos\left(7.5^\circ\right)=\frac12\sqrt{2+\sqrt{2+\sqrt3}}

= \frac14\sqrt{8+2\sqrt6+2\sqrt2} $$
 * $$\tan\left(\frac{\pi}{24}\right)=\tan\left(7.5^\circ\right)=\sqrt6-\sqrt3+\sqrt2-2\ = \left(\sqrt2-1\right)\left(\sqrt3-\sqrt2\right)$$
 * $$\cot\left(\frac{\pi}{24}\right)=\cot\left(7.5^\circ\right)=\sqrt6+\sqrt3+\sqrt2+2\ = \left(\sqrt2+1\right)\left(\sqrt3+\sqrt2\right)$$

9°: regular icosagon (20-sided polygon)

 * $$\sin\frac{\pi}{20}=\sin 9^\circ=\frac12 \sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}$$
 * $$\cos\frac{\pi}{20}=\cos 9^\circ=\frac12 \sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}$$
 * $$\tan\frac{\pi}{20}=\tan 9^\circ=\sqrt5+1-\sqrt{5+2\sqrt5}\,$$
 * $$\cot\frac{\pi}{20}=\cot 9^\circ=\sqrt5+1+\sqrt{5+2\sqrt5}\,$$

11.25°: regular hexadecagon (16-sided polygon)

 * $$\sin\frac{\pi}{16}=\sin 11.25^\circ=\frac{1}{2}\sqrt{2-\sqrt{2+\sqrt{2}}}$$
 * $$\cos\frac{\pi}{16}=\cos 11.25^\circ=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2}}}$$
 * $$\tan\frac{\pi}{16}=\tan 11.25^\circ=\sqrt{4+2\sqrt{2}}-\sqrt{2}-1$$
 * $$\cot\frac{\pi}{16}=\cot 11.25^\circ=\sqrt{4+2\sqrt{2}}+\sqrt{2}+1$$

12°: regular pentadecagon (15-sided polygon)

 * $$\sin\frac{\pi}{15}=\sin 12^\circ=\tfrac{1}{8} \left[\sqrt{2\left(5+\sqrt5\right)}+\sqrt3-\sqrt{15}\right]\,$$
 * $$\cos\frac{\pi}{15}=\cos 12^\circ=\tfrac{1}{8} \left[\sqrt{6\left(5+\sqrt5\right)}+\sqrt5-1\right]\,$$
 * $$\tan\frac{\pi}{15}=\tan 12^\circ=\tfrac{1}{2} \left[3\sqrt3-\sqrt{15}-\sqrt{2\left(25-11\sqrt5\right)}\,\right]\,$$
 * $$\cot\frac{\pi}{15}=\cot 12^\circ=\tfrac{1}{2} \left[\sqrt{15}+\sqrt3+\sqrt{2\left(5+\sqrt5\right)}\,\right]\,$$

15°: regular dodecagon (12-sided polygon)

 * $$\sin\frac{\pi}{12}=\sin 15^\circ=\frac{1}{4}\left(\sqrt6-\sqrt2\right) = \frac12\sqrt{2-\sqrt3}$$
 * $$\cos\frac{\pi}{12}=\cos 15^\circ=\frac{1}{4}\left(\sqrt6+\sqrt2\right)= \frac12\sqrt{2+\sqrt3}$$
 * $$\tan\frac{\pi}{12}=\tan 15^\circ=2-\sqrt3\,$$
 * $$\cot\frac{\pi}{12}=\cot 15^\circ=2+\sqrt3\,$$

75°: sum 30° + 45°

 * $$\sin\frac{5\pi}{12}=\sin 75^\circ=\tfrac{1}{4}\left(\sqrt6+\sqrt2\right)\,$$
 * $$\cos\frac{5\pi}{12}=\cos 75^\circ=\tfrac{1}{4}\left(\sqrt6-\sqrt2\right)\,$$
 * $$\tan\frac{5\pi}{12}=\tan 75^\circ=2+\sqrt3\,$$
 * $$\cot\frac{5\pi}{12}=\cot 75^\circ=2-\sqrt3\,$$

===18°: regular decagon (10-sided polygon) ===
 * $$\sin\frac{\pi}{10}=\sin 18^\circ=\tfrac{1}{4}\left(\sqrt5-1\right)=\frac{1}{1+\sqrt 5}\,$$
 * $$\cos\frac{\pi}{10}=\cos 18^\circ=\tfrac{1}{4}\sqrt{2\left(5+\sqrt5\right)}\,$$
 * $$\tan\frac{\pi}{10}=\tan 18^\circ=\tfrac{1}{5}\sqrt{5\left(5-2\sqrt5\right)}\,$$
 * $$\cot\frac{\pi}{10}=\cot 18^\circ=\sqrt{5+2\sqrt 5}\,$$

72°: sum 36° + 36°

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


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

21°: sum 9° + 12°

 * $$\sin\frac{7\pi}{60}=\sin 21^\circ=\frac{1}{16}\left(2\left(\sqrt3+1\right)\sqrt{5-\sqrt5}-\left(\sqrt6-\sqrt2\right)\left(1+\sqrt5\right)\right)\,$$
 * $$\cos\frac{7\pi}{60}=\cos 21^\circ=\frac{1}{16}\left(2\left(\sqrt3-1\right)\sqrt{5-\sqrt5}+\left(\sqrt6+\sqrt2\right)\left(1+\sqrt5\right)\right)\,$$
 * $$\tan\frac{7\pi}{60}=\tan 21^\circ=\frac{1}{4}\left(2-\left(2+\sqrt3\right)\left(3-\sqrt5\right)\right)\left(2-\sqrt{2\left(5+\sqrt5\right)}\right)\,$$
 * $$\cot\frac{7\pi}{60}=\cot 21^\circ=\frac{1}{4}\left(2-\left(2-\sqrt3\right)\left(3-\sqrt5\right)\right)\left(2+\sqrt{2\left(5+\sqrt5\right)}\right)\,$$

22.5°: regular octagon

 * $$\sin\frac{\pi}{8}=\sin 22.5^\circ=\frac{1}{2}\sqrt{2-\sqrt{2}},$$
 * $$\cos\frac{\pi}{8}=\cos 22.5^\circ=\frac{1}{2}\sqrt{2+\sqrt{2}}\,$$
 * $$\tan\frac{\pi}{8}=\tan 22.5^\circ=\sqrt{2}-1\,$$
 * $$\cot\frac{\pi}{8}=\cot 22.5^\circ=\sqrt{2}+1=\delta_S\,$$, the silver ratio

67.5°: sum 7.5° + 60°

 * $$\sin\frac{3\pi}{8}=\sin 67.5^\circ=\tfrac{1}{2}\sqrt{2+\sqrt{2}}\,$$
 * $$\cos\frac{3\pi}{8}=\cos 67.5^\circ=\tfrac{1}{2}\sqrt{2-\sqrt{2}}\,$$
 * $$\tan\frac{3\pi}{8}=\tan 67.5^\circ=\sqrt{2}+1\,$$
 * $$\cot\frac{3\pi}{8}=\cot 67.5^\circ=\sqrt{2}-1\,$$

24°: sum 12° + 12°

 * $$\sin\frac{2\pi}{15}=\sin 24^\circ=\tfrac{1}{8}\left[\sqrt{15}+\sqrt3-\sqrt{2\left(5-\sqrt5\right)}\right]\,$$
 * $$\cos\frac{2\pi}{15}=\cos 24^\circ=\tfrac{1}{8}\left(\sqrt{6\left(5-\sqrt5\right)}+\sqrt5+1\right)\,$$
 * $$\tan\frac{2\pi}{15}=\tan 24^\circ=\tfrac{1}{2}\left[\sqrt{50+22\sqrt5}-3\sqrt3-\sqrt{15}\right]\,$$
 * $$\cot\frac{2\pi}{15}=\cot 24^\circ=\tfrac{1}{2}\left[\sqrt{15}-\sqrt3+\sqrt{2\left(5-\sqrt5\right)}\right]\,$$

27°: sum 12° + 15°

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

30°: regular hexagon

 * $$\sin\frac{\pi}{6}=\sin 30^\circ=\frac{1}{2}\,$$
 * $$\cos\frac{\pi}{6}=\cos 30^\circ=\frac{\sqrt3}{2}\,$$
 * $$\tan\frac{\pi}{6}=\tan 30^\circ=\frac{\sqrt3}{3}=\frac{1}{\sqrt3}\,$$
 * $$\cot\frac{\pi}{6}=\cot 30^\circ=\sqrt3\,$$

60°: equilateral triangle

 * $$\sin\frac{\pi}{3}=\sin 60^\circ=\frac{\sqrt3}{2}\,$$
 * $$\cos\frac{\pi}{3}=\cos 60^\circ=\frac{1}{2}\,$$
 * $$\tan\frac{\pi}{3}=\tan 60^\circ=\sqrt3\,$$
 * $$\cot\frac{\pi}{3}=\cot 60^\circ=\frac{\sqrt3}{3}=\frac{1}{\sqrt3}\,$$

33°: sum 15° + 18°

 * $$\sin\frac{11\pi}{60}=\sin 33^\circ=\tfrac{1}{16}\left[2\left(\sqrt3-1\right)\sqrt{5+\sqrt5}+\sqrt2\left(1+\sqrt3\right)\left(\sqrt5-1\right)\right]\,$$
 * $$\cos\frac{11\pi}{60}=\cos 33^\circ=\tfrac{1}{16}\left[2\left(\sqrt3+1\right)\sqrt{5+\sqrt5}+\sqrt2\left(1-\sqrt3\right)\left(\sqrt5-1\right)\right]\,$$
 * $$\tan\frac{11\pi}{60}=\tan 33^\circ=\tfrac{1}{4}\left[2-\left(2-\sqrt3\right)\left(3+\sqrt5\right)\right]\left[2+\sqrt{2\left(5-\sqrt5\right)}\,\right]\,$$
 * $$\cot\frac{11\pi}{60}=\cot 33^\circ=\tfrac{1}{4}\left[2-\left(2+\sqrt3\right)\left(3+\sqrt5\right)\right]\left[2-\sqrt{2\left(5-\sqrt5\right)}\,\right]\,$$

36°: regular pentagon

 * $$\sin\frac{\pi}{5}=\sin 36^\circ=\frac14\sqrt{10-2\sqrt{5}}$$
 * $$\cos\frac{\pi}{5}=\cos 36^\circ=\frac{\sqrt5+1}{4}=\frac{\varphi}{2},$$ where $φ$ is the golden ratio;
 * $$\tan\frac{\pi}{5}=\tan 36^\circ=\sqrt{5-2\sqrt{5}}\,$$
 * $$\cot\frac{\pi}{5}=\cot 36^\circ=\frac15\sqrt{25+10\sqrt{5}}$$
 * $$\cot\frac{\pi}{5}=\cot 36^\circ=\frac15\sqrt{25+10\sqrt{5}}$$

54°: sum 27° + 27°

 * $$\sin\frac{3\pi}{10}=\sin 54^\circ=\frac{\sqrt5+1}{4}\,\!$$
 * $$\cos\frac{3\pi}{10}=\cos 54^\circ=\frac{\sqrt{10-2\sqrt{5}}}{4}$$
 * $$\tan\frac{3\pi}{10}=\tan 54^\circ=\frac{\sqrt{25+10\sqrt{5}}}{5}\,$$
 * $$\cot\frac{3\pi}{10}=\cot 54^\circ=\sqrt{5-2\sqrt{5}}\,$$

39°: sum 18° + 21°

 * $$\sin\frac{13\pi}{60}=\sin 39^\circ=\tfrac1{16}\left[2\left(1-\sqrt3\right)\sqrt{5-\sqrt5}+\sqrt2\left(\sqrt3+1\right)\left(\sqrt5+1\right)\right]\,$$
 * $$\cos\frac{13\pi}{60}=\cos 39^\circ=\tfrac1{16}\left[2\left(1+\sqrt3\right)\sqrt{5-\sqrt5}+\sqrt2\left(\sqrt3-1\right)\left(\sqrt5+1\right)\right]\,$$
 * $$\tan\frac{13\pi}{60}=\tan 39^\circ=\tfrac14\left[\left(2-\sqrt3\right)\left(3-\sqrt5\right)-2\right]\left[2-\sqrt{2\left(5+\sqrt5\right)}\,\right]\,$$
 * $$\cot\frac{13\pi}{60}=\cot 39^\circ=\tfrac14\left[\left(2+\sqrt3\right)\left(3-\sqrt5\right)-2\right]\left[2+\sqrt{2\left(5+\sqrt5\right)}\,\right]\,$$

42°: sum 21° + 21°

 * $$\sin\frac{7\pi}{30}=\sin 42^\circ=\frac{\sqrt{30+6\sqrt{5}}-\sqrt5+1}{8}\,$$
 * $$\cos\frac{7\pi}{30}=\cos 42^\circ=\frac{\sqrt{15}-\sqrt3+\sqrt{10+2\sqrt{5}}}{8}\,$$
 * $$\tan\frac{7\pi}{30}=\tan 42^\circ=\frac{\sqrt{15}+\sqrt3-\sqrt{10+2\sqrt{5}}}{2}\,$$
 * $$\cot\frac{7\pi}{30}=\cot 42^\circ=\frac{\sqrt{50-22\sqrt{5}}+3\sqrt{3}-\sqrt{15}}{2}\,$$

45°: square

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

Foo
Baz

