Cómo probar $\cos\left(\pi\over7\right)-\cos\left({2\pi}\over7\right)+\cos\left({3\pi}\over7\right)=\cos\left({\pi}\over3 \right)$

Question

¿Hay alguna forma fácil de demostrar la identidad?

$$\cos \left ( \frac{\pi}{7} \right ) - \cos \left ( \frac{2\pi}{7} \right ) + \cos \left ( \frac{3\pi}{7} \right ) = \cos \left (\frac{\pi}{3} \right )$$

Mientras se resuelve una pregunta Estoy atascado, lo que parece obvio pero sin ninguna forma factible de abordarlo.

Algunas observaciones, no estoy seguro de que sirvan de algo $$ \begin{align} \dfrac{\dfrac{\pi}{7}+\dfrac{3\pi}{7}}{2} &= \dfrac{2\pi}{7}\\\\ \dfrac{\pi}{7} + \dfrac{3\pi}{7} + \dfrac{2\pi}{7} &= \pi - \dfrac{\pi}{7} \end{align} $$

Answer 1

Dejemos que $w = \cos \left ( \frac{2\pi}{7} \right ) + i\sin \left ( \frac{2\pi}{7} \right )$ para que $w^7 = 1$ . Así,

$$\begin{align*} w^7 - 1 &= 0\\ (w-1)(w^6 + w^5 + w^4 + w^3 + w^2 + w + 1) &= 0\\ w^6 + w^5 + w^4 + w^3 + w^2 + w + 1 &= 0 &&\text{since } w \ne 1\\ \left ( w^3 + w^{-3} \right ) + \left ( w^2 + w^{-2} \right ) + \left ( w + w^{-1} \right ) &= -1 &&\text{since } w \ne 0\\ \end{align*}$$

Desde $w + w^{-1} = \cos \left ( \frac{2\pi}{7} \right ) + i\sin \left ( \frac{2\pi}{7} \right ) + \cos \left ( - \frac{2\pi}{7} \right ) + i\sin \left ( - \frac{2\pi}{7} \right ) = 2\cos \left ( \frac{2\pi}{7} \right )$ utilizando el teroema de De Moivre:

$$\begin{align*} 2\cos \left ( 3\times \frac{2\pi}{7} \right ) + 2\cos \left ( 2\times \frac{2\pi}{7} \right ) + 2\cos \left ( \frac{2\pi}{7} \right ) &= -1\\ \cos \left ( \frac{6\pi}{7} \right ) + \cos \left ( \frac{4\pi}{7} \right ) + \cos \left ( \frac{2\pi}{7} \right ) &= -\frac{1}{2}= -\cos \left (\frac{\pi}{3} \right ) \end{align*}$$

Utilizando $\cos(\theta) = -\cos \left (\pi - \theta \right )$ :

$$-\cos \left ( \frac{\pi}{7} \right ) - \cos \left ( \frac{3\pi}{7} \right ) + \cos \left ( \frac{2\pi}{7} \right ) = -\cos \left (\frac{\pi}{3} \right )$$

Y por lo tanto

$$\cos \left ( \frac{\pi}{7} \right ) - \cos \left ( \frac{2\pi}{7} \right ) + \cos \left ( \frac{3\pi}{7} \right ) = \cos \left (\frac{\pi}{3} \right )$$

Answer 2

Sí, este problema en 1963 IMO. http://www.artofproblemsolving.com/Forum/viewtopic.php?p=346908&sid=8ad587e18dd5fa9dd5456496a8daadfd#p346908

Answer 3

Porque $$\cos \frac{\pi}{7} - \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7} = \frac{2\sin\frac{\pi}{7}\cos \frac{\pi}{7} - 2\sin\frac{\pi}{7}\cos\frac{2\pi}{7} + 2\sin\frac{\pi}{7}\cos\frac{3\pi}{7}}{2\sin\frac{\pi}{7}}=$$ $$=\frac{\sin\frac{2\pi}{7}- \sin\frac{3\pi}{7}+\sin\frac{\pi}{7} + \sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}}{2\sin\frac{\pi}{7}}=\frac{\sin\frac{\pi}{7}}{2\sin\frac{\pi}{7}}=\frac{1}{2}=\cos\frac{\pi}{3}.$$

Answer 4

Sugerencia

1. $$ \cos A + \cos B = 2 \cos \left( \dfrac{A + B}{2} \right) \cos \left( \dfrac{A - B}{2} \right) $$
2. $$ \cos A - \cos B = - 2 \sin \left( \dfrac{A + B}{2} \right) \cos \left( \dfrac{A - B}{2} \right) $$
3. $$ \cos \left( \dfrac{2 \pi}{7} \right) = \cos { \left( 2 \theta \right) } \tag{ $ \theta = \dfrac{\pi}{7} $ } $$
4. $$ - \cos \left( \dfrac{2 \pi}{7} \right) = \cos { \left( \pi -\dfrac{2\pi}{7} \right) } = \cos { \left( \dfrac{5\pi}{7} \right) } $$