Question 1209455
.
find value of: cos(2π/7)+cos(4π/7)+cos(6π/7)
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            This is a nice problem of the level slightly higher than ordinary school assignment,

            and its beauty is four times greater than anything you've ever seen before.



<pre>
Consider the doubled sum

    2S = 2cos(2π/7) + 2cos(4π/7) + 2cos(6π/7).


It is nothing else as the sum of six complex numbers


    2S =   (cos(2π/7)  + isin(2π/7))  + (cos(4π/7)  + isin(6π/7))  + (cos(6π/7) + isin(6π/7)) + 

         + (cos(12π/7) + isin(12π/7)) + (cos(10π/7) + isin(10π/7)) + (cos(8π/7) + isin(8π/7)).    (2)


( think one minute - why it is so.  I leave it to you. Recall the signs of sine and cosine ! ).


Consider 1 + 2S.  This expression, 1 + 2S,  represents the sum of all complex roots of equation


    z^7 - 1 = 0,   (or  z^7 = 1, which is the same).     (3)


According to Vieta's theorem, the sum of the roots of this equation is equal to the coefficient at z^6
in equation (3), which is zero.  So, we conclude that 

    1 + 2S = 0,   or   S = -1/2.


At this point, the problem is solved completely.


<U>ANSWER</U>.  cos(2π/7)+cos(4π/7)+cos(6π/7) = -1/2.


You can check it using your calculator .
</pre>

Solved.


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In regular school, they do not teach such tricks - only at Math circles or Math schools.


Or in special Math textbooks - collections of problems for Math circles and Math Olympiads.


Or in web-sites like this one.



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Tutor @math_tutor in his post re-told my solution in his own words.


Thank you for popularizing my ideas.



//  Actually, I am not very happy, when somebody comes and re-tells my solution/solutions
in his own words.


It is the same as if somebody, let say, @math_tutor2020, comes to re-tell Hemingway prose in his own words.



As far as I know, no one in healthy mind has ever tried to retell Hemingway or another writer in their own words - 
it's somehow not accepted in the world. But for some reason it is thriving at this forum, UNFORTUNATELY.



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