Question 1209455: find value of: cos(2π/7)+cos(4π/7)+cos(6π/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(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.
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.
ANSWER. cos(2π/7)+cos(4π/7)+cos(6π/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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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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