Proof of the product formula for \(\dfrac{\pi}{2\sqrt{3}}\)

  Hello everyone.  Today we will prove the product formula for  \(\dfrac{\pi}{2\sqrt{3}}\).  We will use the method of direct proof as a proof method.  Theorem 1:  \[\frac{\pi}{2\sqrt{3}}=\displaystyle\sum_{n=1}^{\infty}\frac{\chi(n)}{n}\]\[\text{where} \quad \chi(n)=\begin{cases} 1, & \text{if } n \equiv 1 \pmod{6}\\-1, & \text{if } n \equiv -1 \pmod{6}\\0, & \text{otherwise}\end{cases}\] Theorem 2:  We have\[\frac{\pi}{2\sqrt{3}}=\frac{5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdots}{6 \cdot 6 \cdot 12 \cdot 12 \cdot 18 \cdot 18 \cdot 24 \cdot 30 \cdots}\]expression whose numerators are the sequence of the odd prime numbers greater than \(3\) and whose denominators are even–even numbers one unit more or less than the corresponding numerators. Proof: By Theorem 1 we know that \[\frac{\pi}{2\sqrt{3}}=1-\frac{1}{5}+\frac{1}{7}-\frac{1}{11}+\frac{1}{13}-\frac{1}{17}+\frac{1}{19}-\cdots\] we will have \[\frac{1}{5} \cdot \frac{\pi}{2\sqrt{3}}=\frac{1}{5
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Proof that the central angle of the circle is equal to twice the corresponding inscribed angle

Hello everyone. Today we will prove that the central angle of a circle is equal to twice the corresponding inscribed angle of the circle. We will use the method of direct proof as a proof method. 

Theorem: The central angle of a circle is equal to twice the corresponding inscribed angle of the circle. 

Proof: 

We can reformulate the statement as follows: 

Let \(P, Q, R \) be three arbitrary points on the circumference of a circle \(k (O, r) \). Then, \(\angle QOP = 2 \angle QRP \). 

We will construct the proof by proving three separate possible cases: 

First case: The center of the circle is on a side of the inscribed angle. 

We can write the following equations : \[\angle POR + \angle ORP + \angle RPO = 180 ^ {\circ} \] \[\angle QOP = 180 ^{\circ} - \angle POR \] \[\angle ORP = \angle QRP \] Combining the first and second equation we get: \[\angle QOP = \angle ORP + \angle RPO \] Since triangle \(\triangle POR \) is isosceles we have that \(\angle RPO = \angle ORP \), so: \[\angle QOP = \angle ORP + \angle ORP \] \[\angle QOP = 2 \angle ORP \] That is, since \(\angle ORP = \angle QRP \) we get: \[ \angle QOP = 2 \angle QRP \] 

Second case: The center of the circle is in the inner area of ​​the inscribed angle. 

Consider the following diagram:


Since the points \(S, O, R \) lie on the same line, based on the results from the first case, we conclude that \(\angle SOP = 2 \angle SRP \) and \(\angle QOS = 2 \angle QRS \) . Since \(\angle QOP = \angle SOP + \angle QOS \) we have that: \[\angle QOP = 2 \angle SRP + 2 \angle QRS \] \[\angle QOP = 2 (\angle SRP + \angle QRS) \] \[\angle QOP = 2 \angle QRP \] 

Third case: The center of the circle is outside the inner area of ​​the inscribed angle. 

Consider the following diagram:


Since the points \(S, O, R \) lie on the same line, based on the results from the first case, we conclude that \(\angle POS = 2 \angle PRS \) and \(\angle QOS = 2 \angle QRS \) . Since \(\angle QOP = \angle QOS- \angle POS \) we have that: \[\angle QOP = 2 \angle QRS-2 \angle PRS \] \[\angle QOP = 2 ( \angle QRS- \angle PRS) \] \[\angle QOP = 2 \angle QRP \] 

\(\blacksquare\)

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