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 cosine theorem

Hello everyone. Today we will prove the cosine theorem. We will use the method of direct proof as a proof method. This theorem was first formulated by the Persian mathematician Kashani. 

Theorem: Let \(a, b, c \) be the sides of any triangle and let the angles \(\alpha, \beta, \gamma \) be the angles opposite the sides \(a, b, c \), respectively. Then the following equations hold: \[c ^ 2 = a ^ 2 + b ^ 2-2ab \cos \gamma \] \[b ^ 2 = a ^ 2 + c ^ 2-2ac \cos \beta \] \[a ^ 2 = b ^ 2 + c ^ 2-2bc \cos \alpha \] Proof: 

We will now prove that the first equation holds i.e. : \[c ^ 2 = a ^ 2 + b ^ 2-2ab \cos \gamma \] Other two equations can be proved in an analogous way. 

There are three possible cases, and these are: \(\gamma \) is a right angle, \(\gamma \) is an acute angle, and \(\gamma \) is an obtuse angle. 

First case: \(\gamma = 90 ^ {\circ} \) 

According to the Pythagoras' theorem, we know that for a right triangle with hypotenuse \(c \) the following equation holds: \[c ^ 2 = a ^ 2 + b ^ 2 \] On the other hand, since \(\cos \gamma = \cos 90 ^ {\circ} = 0 \) it is true that: \[c ^ 2 = a ^ 2 + b ^ 2-0 \] \[c ^ 2 = a ^ 2 + b ^ 2-2ab \cos 90 ^ {\circ} \] \[c ^ 2 = a ^ 2 + b ^ 2-2ab \cos \gamma \] Second case: \(\gamma <90 ^ {\circ} \) 

Consider the following diagram of an acute triangle \(\triangle ABC \) with height \(h \).

According to the Pythagoras' theorem, we can write the following two equations: \[a ^ 2 = h ^ 2 + p ^ 2 \] \[c ^ 2 = h ^ 2 + (bp) ^ 2 \] Combining these two equations we get: \[ c ^ 2 = a ^ 2-p ^ 2 + (bp) ^ 2 \] \[c ^ 2 = a ^ 2-p ^ 2 + b ^ 2-2bp + p ^ 2 \] \[c ^ 2 = a ^ 2 + b ^ 2-2bp \] Since \(p = a \cos \gamma \) it follows: \[c ^ 2 = a ^ 2 + b ^ 2-2ab \cos \gamma \]Third case: \(\gamma> 90 ^ {\circ} \) 

Consider the following diagram of an obtuse triangle \(\triangle ABC \) with height \(h \).

According to the Pythagoras' theorem, we can write the following two equations: \[a ^ 2 = h ^ 2 + p ^ 2 \] \[c ^ 2 = h ^ 2 + (b + p) ^ 2 \] Combining these two equations we get: \[c ^ 2 = a ^ 2-p ^ 2 + (b + p) ^ 2 \] \[c ^ 2 = a ^ 2-p ^ 2 + b ^ 2 + 2bp + p ^ 2 \] \[ c ^ 2 = a ^ 2 + b ^ 2 + 2bp \] Since \(p = a \cos \left (180 ^{\circ}-\gamma \right) \) it follows: \[c ^ 2 = a ^ 2 + b ^ 2 + 2ab \cos \left (180 ^ {\circ}-\gamma \right) \] Applying identity \(\cos \left (180 ^ {\circ}-\gamma \right) = - \cos \gamma \) we get: \[c ^ 2 = a ^ 2 + b ^ 2-2ab \cos \gamma \]

\(\blacksquare\)

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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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