មើលកូដរបស់ រូបមន្តត្រីកោណមាត្រត្រង់ kπ/7

== តំលៃប្រហែល ==
យើងមានតំលៃប្រហែល
:<math> cos(\frac{\pi}{7}) \simeq 0,9009688... \simeq \frac{9}{10} ~</math>
តំលៃនេះអាចអោយយើងសង់[[បន្ទាត់]] និង កំប៉ានៃមុំ ដែលមានរង្វាស់ជិតស្មើនឹង  <math> \frac{\pi}{7} </math> ។

[[រូបភាព:Pi sur 7.JPG|ស្តាំ]]
យើងគូស[[អង្កត់]] <math>\ [AB] </math>  និង ចំនុច P មួយដែល
:<math> AP = \frac{9}{10}AB ~</math>

គេមានចំនុច C មួយជាចំនុចប្រសព្វរវាង[[រង្វង់]]ផ្ចិត A [[កាំ]] AB ជាមួយនឹងបន្ទាត់កែង AB ត្រង់ P ។

ហេតុនេះមុំ  <math> \widehat{BAC} ~</math> មានរង្វាស់ប្រហែលនឹង <math> \frac{\pi}{7} </math> ។

== ដំណោះស្រាយសមីការមួយចំនួន ==
* [[សមីការ]]៖

<math> X^3 - X^2 -2X +1 = 0 ~</math>

មានឫស៖

<math> \{2\cos(\frac{\pi}{7}), -2\cos(\frac{2\pi}{7}), 2\cos(\frac{3\pi}{7})\} ~</math>


* សមីការ៖

<math> X^3 + X^2 -2X - 1 = 0 ~</math>

មានឫស៖

<math> \{-2\cos(\frac{\pi}{7}),2\cos(\frac{2\pi}{7}),-2\cos(\frac{3\pi}{7})\} ~</math>


* សមីការ៖

<math> X^3 + \sqrt{7}X^2 - \sqrt{7} = 0 ~</math>

មានឫស៖
<math> \{2\sin(\frac{\pi}{7}),-2\sin(\frac{2\pi}{7}),-2\sin(\frac{3\pi}{7})\} ~</math>


* សមីការ៖

<math> X^3 - \sqrt{7}X^2 + \sqrt{7} = 0 ~</math>

a pour racines :

<math> \{-2\sin(\frac{\pi}{7}),2\sin(\frac{2\pi}{7}),2\sin(\frac{3\pi}{7})\} ~</math>


* សមីការ៖

<math> X^3 + \sqrt{7}X^2 - 7X + \sqrt{7} = 0 ~</math>

មានឫស៖

<math> \{\tan(\frac{\pi}{7}),\tan(\frac{2\pi}{7}),-\tan(\frac{3\pi}{7})\} ~</math>


* សមីការ៖

<math> X^3 - \sqrt{7}X^2 - 7X - \sqrt{7} = 0 ~</math>

មានឫស៖

<math> \{-\tan(\frac{\pi}{7}),-\tan(\frac{2\pi}{7}),\tan(\frac{3\pi}{7})\} ~</math>

== រូបមន្តអូម៉ូសែន ==
<math> \cos(\frac{\pi}{7}) - \cos(\frac{2\pi}{7}) + \cos(\frac{3\pi}{7}) = \frac{1}{2}~</math>

<math> \cos(\frac{\pi}{7}) . \cos(\frac{2\pi}{7}) . \cos(\frac{3\pi}{7}) = -\frac{1}{8}~</math>

<math> \cos(\frac{\pi}{7}).\cos(\frac{2\pi}{7})  - \cos(\frac{\pi}{7}).\cos(\frac{3\pi}{7}) + \cos(\frac{2\pi}{7}).\cos(\frac{3\pi}{7}) = -\frac{1}{2}~</math>

<math> \sin(\frac{\pi}{7}) - \sin(\frac{2\pi}{7}) - \sin(\frac{3\pi}{7}) = -\frac{\sqrt{7}}{2}~</math>

<math> \sin(\frac{\pi}{7}) . \sin(\frac{2\pi}{7}) . \sin(\frac{3\pi}{7}) = \frac{\sqrt{7}}{8}~</math>

<math> \sin(\frac{\pi}{7}).\sin(\frac{2\pi}{7})  + \sin(\frac{\pi}{7}).\sin(\frac{3\pi}{7}) - \sin(\frac{2\pi}{7}).\sin(\frac{3\pi}{7}) = 0 ~</math>

<math> \tan(\frac{\pi}{7}) + \tan(\frac{2\pi}{7}) - \tan(\frac{3\pi}{7}) = -\sqrt{7}~</math>

<math> \tan(\frac{\pi}{7}) . \tan(\frac{2\pi}{7}) . \tan(\frac{3\pi}{7}) = \sqrt{7}~</math>

<math> \tan(\frac{\pi}{7}).\tan(\frac{2\pi}{7})  - \tan(\frac{\pi}{7}).\tan(\frac{3\pi}{7}) - \tan(\frac{2\pi}{7}).\tan(\frac{3\pi}{7}) = -7~</math>

== រូបមន្តលីនែអ៊ែរ ==

<math> \cos(\frac{\pi}{7})\cos(\frac{2\pi}{7}) = \frac{1}{2}\cos(\frac{2\pi}{7}) + \frac{1}{4} ~</math>

<math> \cos(\frac{\pi}{7})\cos(\frac{3\pi}{7}) = \frac{1}{2}\cos(\frac{\pi}{7}) - \frac{1}{4} ~</math>

<math> \cos(\frac{2\pi}{7})\cos(\frac{3\pi}{7}) = -\frac{1}{2}\cos(\frac{3\pi}{7}) + \frac{1}{4} ~</math>

<math> \cos^2(\frac{\pi}{7}) =\frac{1}{2} + \frac{1}{2}\cos(\frac{2\pi}{7}) ~</math>

<math> \cos^2(\frac{2\pi}{7}) =\frac{1}{2} - \frac{1}{2}\cos(\frac{3\pi}{7}) ~</math>

<math> \cos^2(\frac{3\pi}{7}) =\frac{1}{2} - \frac{1}{2}\cos(\frac{\pi}{7}) ~</math>


== រូបមន្តបន្ថយ ==
ចំពោះតំលៃផ្សេងៗនៃ k ក្នុង kπ/7 គេអាចត្រលប់ទៅរូបមន្តមុន

<math> \cos(\frac{4\pi}{7}) = -\cos(\frac{3\pi}{7}) ~</math>

<math> \cos(\frac{5\pi}{7}) = -\cos(\frac{2\pi}{7}) ~</math>
 
<math> \cos(\frac{6\pi}{7}) = -\cos(\frac{\pi}{7}) ~</math>

<math> \sin(\frac{4\pi}{7}) = \sin(\frac{3\pi}{7}) ~</math>

<math> \sin(\frac{5\pi}{7}) = \sin(\frac{2\pi}{7}) ~</math>
 
<math> \sin(\frac{6\pi}{7}) = \sin(\frac{\pi}{7}) ~</math>

<math> \tan(\frac{4\pi}{7}) = -\tan(\frac{3\pi}{7}) ~</math>

<math> \tan(\frac{5\pi}{7}) = -\tan(\frac{2\pi}{7}) ~</math>
 
<math> \tan(\frac{6\pi}{7}) = -\tan(\frac{\pi}{7}) ~</math>

== លក្ខណៈគួរកត់សំគាល់ ==
យើងមាន

<math> \forall k \in \mathbb{Z}, \quad2^k\left(\cos^k(\frac{\pi}{7}) + \cos^k(\frac{3\pi}{7}) + \cos^k(\frac{5\pi}{7}) \right) \in \mathbb{N}^* ~</math>


* ចំពោះតំលៃដំបូងនៃ k វិជ្ជមាន គេទទួលបាន

<math> 2\left(\cos(\frac{\pi}{7}) + \cos(\frac{3\pi}{7}) + \cos(\frac{5\pi}{7})\right) = 1 ~</math>

<math> 2^2\left(\cos^2(\frac{\pi}{7}) + \cos^2(\frac{3\pi}{7}) + \cos^2(\frac{5\pi}{7})\right) = 5 ~</math>

<math> 2^3\left(\cos^3(\frac{\pi}{7}) + \cos^3(\frac{3\pi}{7}) + \cos^3(\frac{5\pi}{7})\right) = 4 ~</math>

<math> 2^4\left(\cos^4(\frac{\pi}{7}) + \cos^4(\frac{3\pi}{7}) + \cos^4(\frac{5\pi}{7})\right) = 13 ~</math>

<math> 2^5\left(\cos^5(\frac{\pi}{7}) + \cos^5(\frac{3\pi}{7}) + \cos^5(\frac{5\pi}{7})\right) = 16 ~</math>

<math> 2^6\left(\cos^6(\frac{\pi}{7}) + \cos^6(\frac{3\pi}{7}) + \cos^6(\frac{5\pi}{7})\right) = 38 ~</math>

<math> 2^7\left(\cos^7(\frac{\pi}{7}) + \cos^7(\frac{3\pi}{7}) + \cos^7(\frac{5\pi}{7})\right) = 57 ~</math>

។ល។

* ចំពោះតំលៃដំបូងនៃ k អវិជ្ជមាន គេទទួលបាន

<math> \frac{1}{2}\left(\frac{1}{\cos(\frac{\pi}{7})} + \frac{1}{\cos(\frac{3\pi}{7})} + \frac{1}{\cos(\frac{5\pi}{7})}\right) = 2 ~</math>

<math> \frac{1}{2^2}\left(\frac{1}{\cos^2(\frac{\pi}{7})} + \frac{1}{\cos^2(\frac{3\pi}{7})} + \frac{1}{\cos^2(\frac{5\pi}{7})}\right) = 6 ~</math>

<math> \frac{1}{2^3}\left(\frac{1}{\cos^3(\frac{\pi}{7})} + \frac{1}{\cos^3(\frac{3\pi}{7})} + \frac{1}{\cos^3(\frac{5\pi}{7})}\right) = 11 ~</math>

<math> \frac{1}{2^4}\left(\frac{1}{\cos^4(\frac{\pi}{7})} + \frac{1}{\cos^4(\frac{3\pi}{7})} + \frac{1}{\cos^4(\frac{5\pi}{7})}\right) = 26 ~</math>

<math> \frac{1}{2^5}\left(\frac{1}{\cos^5(\frac{\pi}{7})} + \frac{1}{\cos^5(\frac{3\pi}{7})} + \frac{1}{\cos^5(\frac{5\pi}{7})}\right) = 57 ~</math>

។ល។

[[Category:ត្រីកោណមាត្រ]]
[[Category:គណិតវិទ្យា]]