ចំនួនថេរត្រីកោណមាត្រពិត

តារាងចំនួនថេរ

0°: មូលដ្ឋានគ្រឹះ

$\sin 0=0$

$\cos 0=1$

$\tan 0=0$

$\cot 0=\mathrm{\infty }$​មិនកំនត់

3°: ពហុកោណមានជ្រុង៦០

$\sin \frac{\pi }{60}=\sin 3^{\circ }=\frac{2(1-\sqrt{3})\sqrt{5+\sqrt{5}}+\sqrt{2}(\sqrt{5}-1)(\sqrt{3}+1)}{16}$

$\cos \frac{\pi }{60}=\cos 3^{\circ }=\frac{2(1+\sqrt{3})\sqrt{5+\sqrt{5}}+\sqrt{2}(\sqrt{5}-1)(\sqrt{3}-1)}{16}$

$\tan \frac{\pi }{60}=\tan 3^{\circ }=\frac{((2-\sqrt{3})(3+\sqrt{5})-2)(2-\sqrt{2(5-\sqrt{5})})}{4}$

$\cot \frac{\pi }{60}=\cot 3^{\circ }=\frac{((2+\sqrt{3})(3+\sqrt{5})-2)(2+\sqrt{2(5-\sqrt{5})})}{4}$

6°: ពហុកោណមានជ្រុង៣០

$\sin \frac{\pi }{30}=\sin 6^{\circ }=\frac{\sqrt{6(5-\sqrt{5})}-\sqrt{5}-1}{8}$

$\cos \frac{\pi }{30}=\cos 6^{\circ }=\frac{\sqrt{2(5-\sqrt{5})}+\sqrt{3}(\sqrt{5}+1)}{8}$

$\tan \frac{\pi }{30}=\tan 6^{\circ }=\frac{\sqrt{2(5-\sqrt{5})}-\sqrt{3}(\sqrt{5}-1)}{2}$

$\cot \frac{\pi }{30}=\cot 6^{\circ }=\frac{\sqrt{3}(3+\sqrt{5})+\sqrt{2(25+11\sqrt{5})}}{2}$

9°: ពហុកោណមានជ្រុង២០

$\sin \frac{\pi }{20}=\sin 9^{\circ }=\frac{\sqrt{2}(\sqrt{5}+1)-2\sqrt{5-\sqrt{5}}}{8}$

$\cos \frac{\pi }{20}=\cos 9^{\circ }=\frac{\sqrt{2}(\sqrt{5}+1)+2\sqrt{5-\sqrt{5}}}{8}$

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

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

12°: ពហុកោណមានជ្រុង១៥

$\sin \frac{\pi }{15}=\sin 12^{\circ }=\frac{\sqrt{2(5+\sqrt{5})}-\sqrt{3}(\sqrt{5}-1)}{8}$

$\cos \frac{\pi }{15}=\cos 12^{\circ }=\frac{\sqrt{6(5+\sqrt{5})}+\sqrt{5}-1}{8}$

$\tan \frac{\pi }{15}=\tan 12^{\circ }=\frac{\sqrt{3}(3-\sqrt{5})-\sqrt{2(25-11\sqrt{5})}}{2}$

$\cot \frac{\pi }{15}=\cot 12^{\circ }=\frac{\sqrt{3}(\sqrt{5}+1)+\sqrt{2(5+\sqrt{5})}}{2}$

15°: ពហុកោណមានជ្រុង១២

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

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

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

$\cot \frac{\pi }{12}=\cot 15^{\circ }=2+\sqrt{3}$

18°: ពហុកោណមានជ្រុង១០

$\sin \frac{\pi }{10}=\sin 18^{\circ }=\frac{\sqrt{5}-1}{4}=\frac{\phi -1}{2}=\frac{1}{2\phi }$

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

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

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

21°: ផលបូក 9° + 12°

$\sin \frac{7\pi }{60}=\sin 21^{\circ }=\frac{2(\sqrt{3}+1)\sqrt{5-\sqrt{5}}-\sqrt{2}(\sqrt{3}-1)(1+\sqrt{5})}{16}$

$\cos \frac{7\pi }{60}=\cos 21^{\circ }=\frac{2(\sqrt{3}-1)\sqrt{5-\sqrt{5}}+\sqrt{2}(\sqrt{3}+1)(1+\sqrt{5})}{16}$

$\tan \frac{7\pi }{60}=\tan 21^{\circ }=\frac{(2-(2+\sqrt{3})(3-\sqrt{5}))(2-\sqrt{2(5+\sqrt{5})})}{4}$

$\cot \frac{7\pi }{60}=\cot 21^{\circ }=\frac{(2-(2-\sqrt{3})(3-\sqrt{5}))(2+\sqrt{10\sqrt{5}})}{4}$

22.5°: ពហុកោណមានជ្រុង៨

$\sin \frac{\pi }{8}=\sin 22.5^{\circ }=\frac{\sqrt{2-\sqrt{2}}}{2}$

$\cos \frac{\pi }{8}=\cos 22.5^{\circ }=\frac{\sqrt{2+\sqrt{2}}}{2}$

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

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

24°: ផលបូល 12° + 12°

$\sin \frac{2\pi }{15}=\sin 24^{\circ }=\frac{\sqrt{3}(\sqrt{5}+1)-\sqrt{2}\sqrt{5-\sqrt{5}}}{8}$

$\cos \frac{2\pi }{15}=\cos 24^{\circ }=\frac{\sqrt{6}\sqrt{5-\sqrt{5}}+\sqrt{5}+1}{8}$

$\tan \frac{2\pi }{15}=\tan 24^{\circ }=\frac{\sqrt{50+22\sqrt{5}}-\sqrt{3}(3+\sqrt{5})}{2}$

$\cot \frac{2\pi }{15}=\cot 24^{\circ }=\frac{\sqrt{2}\sqrt{5-\sqrt{5}}+\sqrt{3}(\sqrt{5}-1)}{2}$

27°: ផលបូក 12° + 15°

$\sin \frac{3\pi }{20}=\sin 27^{\circ }=\frac{(\sqrt{5}+1)\sqrt{5+\sqrt{5}}-\sqrt{2}(\sqrt{5}-1)}{8}$

$\cos \frac{3\pi }{20}=\cos 27^{\circ }=\frac{(\sqrt{5}+1)\sqrt{5+\sqrt{5}}+\sqrt{2}(\sqrt{5}-1)}{8}$

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

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

30°:ពហុកោណមានជ្រុង៦

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

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

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

$\cot \frac{\pi }{6}=\cot 30^{\circ }=\sqrt{3}$

33°: ផលបូក 15° + 18°

$\sin \frac{11\pi }{60}=\sin 33^{\circ }=\frac{2(\sqrt{3}-1)\sqrt{5+\sqrt{5}}+\sqrt{2}(1+\sqrt{3})(\sqrt{5}-1)}{16}$

$\cos \frac{11\pi }{60}=\cos 33^{\circ }=\frac{2(\sqrt{3}+1)\sqrt{5+\sqrt{5}}+\sqrt{2}(1-\sqrt{3})(\sqrt{5}-1)}{16}$

$\tan \frac{11\pi }{60}=\tan 33^{\circ }=\frac{(2-(2-\sqrt{3})(3+\sqrt{5}))(2+\sqrt{2(5-\sqrt{5})})}{4}$

$\cot \frac{11\pi }{60}=\cot 33^{\circ }=\frac{(2-(2+\sqrt{3})(3+\sqrt{5}))(2-\sqrt{2(5-\sqrt{5})})}{4}$

36°: ពហុកោណមានជ្រុង៥

$\sin \frac{\pi }{5}=\sin 36^{\circ }=\frac{\sqrt{2(5-\sqrt{5})}}{4}$

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

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

$\cot \frac{\pi }{5}=\cot 36^{\circ }=\frac{\sqrt{5(5+2\sqrt{5})}}{5}$

39°: ផលបូក 18°+ 21°

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

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

$\tan \frac{13\pi }{60}=\tan 39^{\circ }=\frac{((2-\sqrt{3})(3-\sqrt{5})-2)(2-\sqrt{2(5+\sqrt{5})})}{4}$

$\cot \frac{13\pi }{60}=\cot 39^{\circ }=\frac{((2+\sqrt{3})(3-\sqrt{5})-2)(2+\sqrt{2(5+\sqrt{5})})}{4}$

42°: ផលបូក 21° + 21°

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

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

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

$\cot \frac{7\pi }{30}=\cot 42^{\circ }=\frac{\sqrt{2(25-11\sqrt{5})}+\sqrt{3}(3-\sqrt{5})}{2}$

45°: ការេ

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

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

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

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

60°: ត្រីកោណ

$\sin \frac{\pi }{3}=\sin 60^{\circ }=\frac{\sqrt{3}}{2}$

$\cos \frac{\pi }{3}=\cos 60^{\circ }=\frac{1}{2}$

$\tan \frac{\pi }{3}=\tan 60^{\circ }=\sqrt{3}$

$\cot \frac{\pi }{3}=\cot 60^{\circ }=\frac{\sqrt{3}}{3}$

ចំនាំ

ការប្រើប្រាស់ចំពោះចំនួនថេរ

មាឌនៃសូលីតដែលបង្កើតដោយពហុកោណមានជ្រុង៥(បញ្ចកោណ) ហើយ $a$ ជាប្រវែងនៃជ្រុងរបស់បញ្ចកោណ

សំដែងដោយ

$V=\frac{5a^3\cos 36^{\circ }}{{\tan }^{2}36^{\circ }}$

ដោយប្រើ

$\cos 36^{\circ }=\frac{\sqrt{5}+1}{4}$

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

វាក្លាយទៅជា

$V=\frac{a^3(15+7\sqrt{5})}{4}$