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Is there an easy way to prove the identity?

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

While solving one question, I am stuck, which looks obvious but without any feasible way to approach.

Few observations, not sure if it would help $$ \begin{align} \dfrac{\dfrac{\pi}{7}+\dfrac{3\pi}{7}}{2} &= \dfrac{2\pi}{7}\\\\ \dfrac{\pi}{7} + \dfrac{3\pi}{7} + \dfrac{2\pi}{7} &= \pi - \dfrac{\pi}{7} \end{align} $$

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Related : math.stackexchange.com/questions/140388/… as $\cos(\pi-x)=-\cos x$ –  lab bhattacharjee Aug 24 '13 at 17:03

3 Answers 3

up vote 3 down vote accepted

Yes, This problem in 1963 IMO.http://www.artofproblemsolving.com/Forum/viewtopic.php?p=346908&sid=8ad587e18dd5fa9dd5456496a8daadfd#p346908

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Nice finding :-). BTW, what do you mean by This problem in 1963 –  Abhijit Mar 31 '13 at 8:52
    
Yes. my other problem is from 1963, my frend. –  math110 Mar 31 '13 at 8:59
    
IMO: International Mathematical Olympiad (see link for archives of problems, including 1963). –  Jean-Claude Arbaut Mar 31 '13 at 9:14

Let $w = \cos \left ( \frac{2\pi}{7} \right ) + i\sin \left ( \frac{2\pi}{7} \right )$ so that $w^7 = 1$. Thus

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

Since $w + w^{-1} = \cos \left ( \frac{2\pi}{7} \right ) + i\sin \left ( \frac{2\pi}{7} \right ) + \cos \left ( - \frac{2\pi}{7} \right ) + i\sin \left ( - \frac{2\pi}{7} \right ) = 2\cos \left ( \frac{2\pi}{7} \right )$, using de Moivre's theroem:

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

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

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

And hence

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

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Hint

  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. $$ - \cos \left( \dfrac{2 \pi}{7} \right) = \cos { \left( \pi -\dfrac{2\pi}{7} \right) } = \cos { \left( \dfrac{5\pi}{7} \right) } $$
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