Challenging Recreational Mathematics: ▪ Γωνίες κανονικών πολυγώνων

Παρασκευή 5 Αυγούστου 2011

▪ Γωνίες κανονικών πολυγώνων

ΤΡΙΓΩΝΟΜΕΤΡΙΚΕΣ ΣΤΑΘΕΡΕΣ

Γωνία 3°: Κανονικό 60-γωνο

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

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

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

Γωνία 6°: Κανονικό 30-γωνο

$\sin\frac{\pi}{30}=\sin 6^\circ=\tfrac{1}{8} \left[\sqrt{6(5-\sqrt5)}-\sqrt5-1\right]\,$ $\cos\frac{\pi}{30}=\cos 6^\circ=\tfrac{1}{8} \left[\sqrt{2(5-\sqrt5)}+\sqrt3(\sqrt5+1)\right]\,$ $\tan\frac{\pi}{30}=\tan 6^\circ=\tfrac{1}{2} \left[\sqrt{2(5-\sqrt5)}-\sqrt3(\sqrt5-1)\right]\,$ $\cot\frac{\pi}{30}=\cot 6^\circ=\tfrac{1}{2} \left[\sqrt3(3+\sqrt5)+\sqrt{2(25+11\sqrt5)}\right]\,$

Γωνία 9°: Κανονικό 20-γωνο

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

Γωνία 12°: Κανονικό 15-γωνο

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

Γωνία 15°: Κανονικό 12-γωνο

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

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

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

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

Γωνία 18°: Κανονικό 10-γωνο

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

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

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

$\cot\frac{\pi}{10}=\cot 18^\circ=\sqrt{5+2\sqrt 5}\,$

Γωνία 21°: Άθροισμα: 9° + 12°

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

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

Γωνία 22.5°: Κανονικό οκτάγωνο

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

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

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

$\cot\frac{\pi}{8}=\cot 22.5^\circ=\sqrt{2}+1\,$

Γωνία 24°: Άθροισμα: 12° + 12°

$\sin\frac{2\pi}{15}=\sin 24^\circ=\tfrac{1}{8}\left[\sqrt3(\sqrt5+1)-\sqrt2\sqrt{5-\sqrt5}\right]\,$ $\cos\frac{2\pi}{15}=\cos 24^\circ=\tfrac{1}{8}\left(\sqrt6\sqrt{5-\sqrt5}+\sqrt5+1\right)\,$

$\tan\frac{2\pi}{15}=\tan 24^\circ=\tfrac{1}{2}\left[\sqrt{2(25+11\sqrt5)}-\sqrt3(3+\sqrt5)\right]\,$ $\cot\frac{2\pi}{15}=\cot 24^\circ=\tfrac{1}{2}\left[\sqrt2\sqrt{5-\sqrt5}+\sqrt3(\sqrt5-1)\right]\,$

Γωνία 27°: Άθροισμα: 12° + 15°

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

Γωνία 30°: Κανονικό εξάγωνο

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

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

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

$\cot\frac{\pi}{6}=\cot 30^\circ=\sqrt3\,$

Γωνία 33°: Άθροισμα: 15° + 18°

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

Γωνία 36°: Κανονικό πεντάγωνο

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

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

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

$\cot\frac{\pi}{5}=\cot 36^\circ=\tfrac15[\sqrt{5(5+2\sqrt5)}]\,$

Γωνία 39°: Άθροισμα: 18° + 21°

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

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

Γωνία 42°: Άθροισμα: 21° + 21°

$\sin\frac{7\pi}{30}=\sin 42^\circ=\frac{\sqrt6\sqrt{5+\sqrt5}-\sqrt5+1}{8}\,$ $\cos\frac{7\pi}{30}=\cos 42^\circ=\frac{\sqrt2\sqrt{5+\sqrt5}+\sqrt3(\sqrt5-1)}{8}\,$ $\tan\frac{7\pi}{30}=\tan 42^\circ=\frac{\sqrt3(\sqrt5+1)-\sqrt2\sqrt{5+\sqrt5}}{2}\,$ $\cot\frac{7\pi}{30}=\cot 42^\circ=\frac{\sqrt{2(25-11\sqrt5)}+\sqrt3(3-\sqrt5)}{2}\,$