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find value of: cos(2Ï€/7)+cos(4Ï€/7)+cos(6Ï€/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.
Consider the doubled sum
2S = 2cos(2Ï€/7) + 2cos(4Ï€/7) + 2cos(6Ï€/7).
It is nothing else as the sum of six complex numbers
2S = (cos(2Ï€/7) + isin(2Ï€/7)) + (cos(4Ï€/7) + isin(6Ï€/7)) + (cos(6Ï€/7) + isin(6Ï€/7)) +
+ (cos(12Ï€/7) + isin(12Ï€/7)) + (cos(10Ï€/7) + isin(10Ï€/7)) + (cos(8Ï€/7) + isin(8Ï€/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.
ANSWER. cos(2Ï€/7)+cos(4Ï€/7)+cos(6Ï€/7) = -1/2.
You can check it using your calculator .
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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Hello, @math_tuitor2020, please remove all your insinuations
about me from all your posts - WITHOUT DELAY.
Jan. 20, 2025.
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