SOLUTION: find value of: cos(2Ï€/7)+cos(4Ï€/7)+cos(6Ï€/7)

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Question 1209455: find value of: cos(2Ï€/7)+cos(4Ï€/7)+cos(6Ï€/7)
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(53521)   (Show Source): You can put this solution on YOUR website!
.
find value of: cos(2Ï€/7)+cos(4Ï€/7)+cos(6Ï€/7)
~~~~~~~~~~~~~~~~~~~


            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.

-------------------------

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.


///////////////////////////


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.


/\/\/\/\/\/\/\/\/\/\/\/


***************************************************************************

        Hello,  @math_tuitor2020,  please remove all your insinuations
            about me from all your posts   -   WITHOUT  DELAY.

                                                    Jan. 20, 2025.

***************************************************************************



Answer by math_tutor2020(3828)   (Show Source): You can put this solution on YOUR website!

Answer: -1/2

Explanation

Each term of
cos(2pi/7)+cos(4pi/7)+cos(6pi/7)
is of the form
cos(2k*pi/7)
where k is from the set {1,2,3}

It might not be entirely obvious, but the expression cos(2k*pi/7) suggests we'll be using the nth roots of unity.
In this case n = 7.

The roots of z^n = 1 are of the form cis(2k*pi/n)
cis is shorthand for "cosine i sine"
for example, cis(2pi) = cos(2pi) + i*sin(2pi)
k is from the set {0,1,2,...,n-1}
Since the expression we're calculating involves cosines only, we'll focus on the real parts of cis(2k*pi/7) and ignore the imaginary parts.

I'll be using the trig identity cos(x) = cos(2pi - x) to be able to say something like cos(8pi/7) = cos(6pi/7)
Here's a bit of scratch work to show what I mean
cos(8pi/7) = cos(2pi - 8pi/7)
= cos(14pi/7 - 8pi/7)
= cos( (14pi-8pi)/7 )
= cos(6pi/7)
This will be useful to show that we have symmetry going on.
Similar steps are applied to demonstrate cos(10pi/7) = cos(4pi/7) and cos(12pi/7) = cos(2pi/7)

Here are the real parts of the 7 roots for z^7 = 1.
cos(0) = 1
cos(2pi/7) = a
cos(4pi/7) = b
cos(6pi/7) = c
cos(8pi/7) = cos(6pi/7) = c
cos(10pi/7) = cos(4pi/7) = b
cos(12pi/7) = cos(2pi/7) = a
Once again note the symmetry.

Those 7 values add to
1+(a+b+c)+(c+b+a) = 1+2*(a+b+c)
The goal is to find a+b+c.

The roots of z^n = 1 add to 0 which is a result of Vieta's formulas
This means the real parts of the roots must add to 0.
1+2*(a+b+c) = 0
a+b+c = -1/2 is the final answer.

Verification using WolframAlpha
You can also use Desmos and GeoGebra among many other online tools.
Keep in mind that -1/2 = -0.5 of course.

ikleyn your "solution" was complete garbage which I cleaned up. You're welcome. You need to stop being so arrogant with such a head of hot air. And you do NOT own Vieta's formulas nor anything you presented in your solution. You are a complete joke. Stop wasting everyone's time. And I just found out that you use AI in nearly all of your solutions. That explains a lot.

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