# ចំនួនថេរត្រីកោណមាត្រពិត

## តារាងចំនួនថេរ

### 0°: មូលដ្ឋានគ្រឹះ

$\sin 0=0\,$
$\cos 0=1\,$
$\tan 0=0\,$
$\cot 0=\infty\,$​មិនកំនត់

### 3°: ពហុកោណមានជ្រុង៦០

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

### 6°: ពហុកោណមានជ្រុង៣០

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

### 9°: ពហុកោណមានជ្រុង២០

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

### 12°: ពហុកោណមានជ្រុង១៥

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

### 15°: ពហុកោណមានជ្រុង១២

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

### 18°: ពហុកោណមានជ្រុង១០

$\sin\frac{\pi}{10}=\sin 18^\circ=\frac{\sqrt5-1}{4}=\frac{\varphi-1}{2}=\frac{1}{2\varphi}\,$
$\cos\frac{\pi}{10}=\cos 18^\circ=\frac{\sqrt{2(5+\sqrt5)}}{4}\,$
$\tan\frac{\pi}{10}=\tan 18^\circ=\frac{\sqrt{5(5-2\sqrt5)}}{5}\,$
$\cot\frac{\pi}{10}=\cot 18^\circ=\sqrt{5+2\sqrt 5}\,$

### 21°: ផលបូក 9° + 12°

$\sin\frac{7\pi}{60}=\sin 21^\circ=\frac{2(\sqrt3+1)\sqrt{5-\sqrt5}-\sqrt2(\sqrt3-1)(1+\sqrt5)}{16}\,$
$\cos\frac{7\pi}{60}=\cos 21^\circ=\frac{2(\sqrt3-1)\sqrt{5-\sqrt5}+\sqrt2(\sqrt3+1)(1+\sqrt5)}{16}\,$
$\tan\frac{7\pi}{60}=\tan 21^\circ=\frac{\left(2-(2+\sqrt3)(3-\sqrt5)\right)\left(2-\sqrt{2(5+\sqrt5)}\right)}{4}\,$
$\cot\frac{7\pi}{60}=\cot 21^\circ=\frac{\left(2-(2-\sqrt3)(3-\sqrt5)\right)\left(2+\sqrt{10\sqrt5}\right)}{4}\,$

### 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°

$\sin\frac{2\pi}{15}=\sin 24^\circ=\frac{\sqrt3(\sqrt5+1)-\sqrt2\sqrt{5-\sqrt5}}{8}\,$
$\cos\frac{2\pi}{15}=\cos 24^\circ=\frac{\sqrt6\sqrt{5-\sqrt5}+\sqrt5+1}{8}\,$
$\tan\frac{2\pi}{15}=\tan 24^\circ=\frac{\sqrt{50+22\sqrt5}-\sqrt3(3+\sqrt5)}{2}\,$
$\cot\frac{2\pi}{15}=\cot 24^\circ=\frac{\sqrt2\sqrt{5-\sqrt5}+\sqrt3(\sqrt5-1)}{2}\,$

### 27°: ផលបូក 12° + 15°

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

### 30°:ពហុកោណមានជ្រុង៦

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

### 33°: ផលបូក 15° + 18°

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

### 36°: ពហុកោណមានជ្រុង៥

$\sin\frac{\pi}{5}=\sin 36^\circ=\frac{\sqrt{2(5-\sqrt5)}}{4}\,$
$\cos\frac{\pi}{5}=\cos 36^\circ=\frac{1+\sqrt5}{4}=\frac{\varphi}{2}\,$
$\tan\frac{\pi}{5}=\tan 36^\circ=\sqrt{5-2\sqrt5}\,$
$\cot\frac{\pi}{5}=\cot 36^\circ=\frac{\sqrt{5(5+2\sqrt5)}}{5}\,$

### 39°: ផលបូក 18°+ 21°

$\sin\frac{13\pi}{60}=\sin 39^\circ=\frac{2(1-\sqrt3)\sqrt{5-\sqrt5}+\sqrt2(\sqrt3+1)(\sqrt5+1)}{16}\,$
$\cos\frac{13\pi}{60}=\cos 39^\circ=\frac{2(1+\sqrt3)\sqrt{5-\sqrt5}+\sqrt2(\sqrt3-1)(\sqrt5+1)}{16}\,$
$\tan\frac{13\pi}{60}=\tan 39^\circ=\frac{\left((2-\sqrt3)(3-\sqrt5)-2\right)\left(2-\sqrt{2(5+\sqrt5)}\right)}{4}\,$
$\cot\frac{13\pi}{60}=\cot 39^\circ=\frac{\left((2+\sqrt3)(3-\sqrt5)-2\right)\left(2+\sqrt{2(5+\sqrt5)}\right)}{4}\,$

### 42°: ផលបូក 21° + 21°

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

### 45°: ការេ

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

### 60°: ត្រីកោណ

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

## ចំនាំ

### ការប្រើប្រាស់ចំពោះចំនួនថេរ

មាឌនៃសូលីតដែលបង្កើតដោយពហុកោណមានជ្រុង៥(បញ្ចកោណ) ហើយ $a\,$ ជាប្រវែងនៃជ្រុងរបស់បញ្ចកោណ

សំដែងដោយ

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

ដោយប្រើ

$\cos 36^\circ=\frac{\sqrt5+1}{4}\,$
$\tan 36^\circ=\sqrt{5-2\sqrt5}\,$

វាក្លាយទៅជា

$V=\frac{a^3(15+7\sqrt5)}{4}\,$