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 of the sine theorem

Hello everyone. Today we will prove the sine theorem. We will use the method of direct proof as a proof method. This proof was first constructed by the Persian mathematician Tusi. 

Theorem: Let \(a, b, c \) be the sides of any triangle, and \(R \) the radius of the circumscribed circle around that triangle and let the angles \(\alpha, \beta, \gamma \) be the angles opposite the sides \(a, b, c \), respectively. Then the following equations hold: \[\frac {a} {\sin \alpha} = \frac {b} {\sin \beta} = \frac {c} {\sin \gamma} = 2R \] Proof: 

Now we will prove that the following equation holds: \(\frac {a} {\sin \alpha} = 2R \). The equations \(\frac {b} {\sin \beta} = 2R \) and \(\frac {c} {\sin \gamma} = 2R \) can be proved in an analogous way. 

Consider the following diagram showing the triangle \(\triangle ABC \) with the circumscribed circle of radius \(R \).

Since angles inscribed in a circle and subtended by the same chord are equal we have that \(\angle CA'B = \alpha \). We also know that the angle inscribed in a circle and subtended by the diameter of the circle is equal to \(90 ^ {\circ} \), so it follows that \(\angle A'BC = 90 ^ {\circ} \). Further, according to the definition of the sine function, we have that \(a = 2R \sin \angle CA'B \), or \(a = 2R \sin \alpha \). Hence,  \(\frac {a} {\sin \alpha} = 2R \). 

\(\blacksquare \)

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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 set of prime numbers is infinite

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