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Can we show a sum of symmetrical cosine values is zero by using roots of unity?

Proving complex series $1 + costheta + cos2theta +… + cos ntheta $Zero sum of roots of unity decompositionMinimising a sum of roots of unityAlternating sum of roots of unity $sum_k=0^n-1(-1)^komega^k$A sum of powers of primitive roots of unityCan any triangle be generated using linear transformations of cubic roots of unity?sum of (some) $p$-th roots of unity, $p$ primeSolving $z^5=-16+16sqrt3i$ using the fifth roots of unityProving that $sum cos p theta = 0$ where $theta$ is argument of the primitive roots of unityUsing the fifth root of unity to show the cosine equationSum of nth roots of unity equal to Sqrt[n]

5

$begingroup$

Can we show that

$$cosfracpi7+cosfrac2pi7+cosfrac3pi7+cosfrac4pi7+cosfrac5pi7+cosfrac6pi7=0$$

by considering the seventh roots of unity? If so how could we do it?

Also I have observed that

$$cosfracpi5+cosfrac2pi5+cosfrac3pi5+cosfrac4pi5=0$$

as well, so just out of curiosity, is it true that $$sum_k=1^n-1 cosfrackpin = 0$$

for all $n$ odd?

complex-numbers

share|cite|improve this question

edited May 21 at 21:30

Chase Ryan Taylor

4,53621531

asked May 21 at 21:01

JustWanderingJustWandering

1619

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use complex numbers, something like this
  $endgroup$
  – rtybase
  May 21 at 21:05
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  You could use that $cos (pi-theta)=-cos theta$ instead, and just pair each cosine with its negative.
  $endgroup$
  – Mark Bennet
  May 21 at 21:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, it works for all $n$'s, not just odd ones.
  $endgroup$
  – rtybase
  May 21 at 21:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  because the unpaired one is $cos (pi /2)$
  $endgroup$
  – G Cab
  May 21 at 21:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Acccumulation Even for $n=1$, one has $sum_k=1^n-1cos(kpi/n)=0$.
  $endgroup$
  – Lord Shark the Unknown
  May 22 at 5:39
  

  
  
  
  

 | 
show 2 more comments

5

$begingroup$

Can we show that

$$cosfracpi7+cosfrac2pi7+cosfrac3pi7+cosfrac4pi7+cosfrac5pi7+cosfrac6pi7=0$$

by considering the seventh roots of unity? If so how could we do it?

Also I have observed that

$$cosfracpi5+cosfrac2pi5+cosfrac3pi5+cosfrac4pi5=0$$

as well, so just out of curiosity, is it true that $$sum_k=1^n-1 cosfrackpin = 0$$

for all $n$ odd?

complex-numbers

share|cite|improve this question

edited May 21 at 21:30

Chase Ryan Taylor

4,53621531

asked May 21 at 21:01

JustWanderingJustWandering

1619

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use complex numbers, something like this
  $endgroup$
  – rtybase
  May 21 at 21:05
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  You could use that $cos (pi-theta)=-cos theta$ instead, and just pair each cosine with its negative.
  $endgroup$
  – Mark Bennet
  May 21 at 21:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, it works for all $n$'s, not just odd ones.
  $endgroup$
  – rtybase
  May 21 at 21:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  because the unpaired one is $cos (pi /2)$
  $endgroup$
  – G Cab
  May 21 at 21:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Acccumulation Even for $n=1$, one has $sum_k=1^n-1cos(kpi/n)=0$.
  $endgroup$
  – Lord Shark the Unknown
  May 22 at 5:39
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Can we show that

$$cosfracpi7+cosfrac2pi7+cosfrac3pi7+cosfrac4pi7+cosfrac5pi7+cosfrac6pi7=0$$

by considering the seventh roots of unity? If so how could we do it?

Also I have observed that

$$cosfracpi5+cosfrac2pi5+cosfrac3pi5+cosfrac4pi5=0$$

as well, so just out of curiosity, is it true that $$sum_k=1^n-1 cosfrackpin = 0$$

for all $n$ odd?

complex-numbers

share|cite|improve this question

edited May 21 at 21:30

Chase Ryan Taylor

4,53621531

asked May 21 at 21:01

JustWanderingJustWandering

1619

$endgroup$

share|cite|improve this question

edited May 21 at 21:30

Chase Ryan Taylor

4,53621531

asked May 21 at 21:01

JustWanderingJustWandering

1619

share|cite|improve this question

edited May 21 at 21:30

Chase Ryan Taylor

4,53621531

asked May 21 at 21:01

JustWanderingJustWandering

1619

edited May 21 at 21:30

Chase Ryan Taylor

4,53621531

edited May 21 at 21:30

Chase Ryan Taylor

4,53621531

asked May 21 at 21:01

JustWanderingJustWandering

1619

asked May 21 at 21:01

JustWanderingJustWandering

1619

3 Answers
3

active

oldest

votes

11

$begingroup$

Note that $$cos(pi - alpha)= - cos(alpha)$$ Therefore $$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)+cos(frac4pi7)+cos(frac5pi7)+cos(frac6pi7)=$$

$$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)-cos(frac3pi7)-cos(frac2pi7)-cos(fracpi7)=0$$

The same goes for other natural numbers $n$ instead of $7$.

share|cite|improve this answer

edited May 22 at 3:22

answered May 21 at 21:16

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.7k42163

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  and even number also
  $endgroup$
  – G Cab
  May 21 at 21:41
  

  
  
  
  

add a comment | 

5

$begingroup$

I think you can use Euler's Formula.

The Nth roots of unity = $e^2pi k i/N$ for values of k between $0$ and $N-1$ inclusive.

There sum from k to $N-1$ is a geometric series.

$S= sum_k=0^N-1 e^2pi i k/N=frac1cdot e^(2pi i /N)N-1e^2pi i /N-1$

The numerator is zero for any N.

But the real part of $S$ is the real part of the individual terms of the sum, i.e. the cosines. The real part of the sum is zero so the sum of the real parts of the roots of unity is 0.

share|cite|improve this answer

answered May 21 at 21:24

TurlocTheRedTurlocTheRed

1,216412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  <pedant>This works only for integer N greater than 1</pedant>
  $endgroup$
  – Acccumulation
  May 22 at 5:16
  

  
  
  
  

add a comment | 

2

$begingroup$

Pointing at the link I left in the comments

$$1 + costheta + cos2theta +... + cos ntheta = frac12 + fracsin[(n+frac12)theta]2sin(fractheta2)$$

Then for $forall ninmathbbN, n>0$
$$cosfracpin+1+ cosfrac2pin+1 +... + cos fracnpin+1 = fracsinleft[(n+frac12)fracpin+1right]2sinleft(fracpi2(n+1)right)-frac12=\
fracsinleft[frac2n+12(n+1)piright]2sinleft(fracpi2(n+1)right)-frac12=
fracsinleft[pi-fracpi2(n+1)right]2sinleft(fracpi2(n+1)right)-frac12=frac12-frac12=0$$

share|cite|improve this answer

answered May 21 at 21:29

rtybasertybase

12.2k31634

$endgroup$

add a comment | 

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11

$begingroup$

Note that $$cos(pi - alpha)= - cos(alpha)$$ Therefore $$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)+cos(frac4pi7)+cos(frac5pi7)+cos(frac6pi7)=$$

$$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)-cos(frac3pi7)-cos(frac2pi7)-cos(fracpi7)=0$$

The same goes for other natural numbers $n$ instead of $7$.

share|cite|improve this answer

edited May 22 at 3:22

answered May 21 at 21:16

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.7k42163

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  and even number also
  $endgroup$
  – G Cab
  May 21 at 21:41
  

  
  
  
  

add a comment | 

11

$begingroup$

Note that $$cos(pi - alpha)= - cos(alpha)$$ Therefore $$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)+cos(frac4pi7)+cos(frac5pi7)+cos(frac6pi7)=$$

$$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)-cos(frac3pi7)-cos(frac2pi7)-cos(fracpi7)=0$$

The same goes for other natural numbers $n$ instead of $7$.

share|cite|improve this answer

edited May 22 at 3:22

answered May 21 at 21:16

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.7k42163

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  and even number also
  $endgroup$
  – G Cab
  May 21 at 21:41
  

  
  
  
  

add a comment | 

$begingroup$

Note that $$cos(pi - alpha)= - cos(alpha)$$ Therefore $$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)+cos(frac4pi7)+cos(frac5pi7)+cos(frac6pi7)=$$

$$cos(fracpi7)+cos(frac2pi7)+cos(frac3pi7)-cos(frac3pi7)-cos(frac2pi7)-cos(fracpi7)=0$$

The same goes for other natural numbers $n$ instead of $7$.

share|cite|improve this answer

edited May 22 at 3:22

answered May 21 at 21:16

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.7k42163

$endgroup$

share|cite|improve this answer

edited May 22 at 3:22

answered May 21 at 21:16

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.7k42163

answered May 21 at 21:16

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.7k42163

answered May 21 at 21:16

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.7k42163

5

$begingroup$

I think you can use Euler's Formula.

The Nth roots of unity = $e^2pi k i/N$ for values of k between $0$ and $N-1$ inclusive.

There sum from k to $N-1$ is a geometric series.

$S= sum_k=0^N-1 e^2pi i k/N=frac1cdot e^(2pi i /N)N-1e^2pi i /N-1$

The numerator is zero for any N.

But the real part of $S$ is the real part of the individual terms of the sum, i.e. the cosines. The real part of the sum is zero so the sum of the real parts of the roots of unity is 0.

share|cite|improve this answer

answered May 21 at 21:24

TurlocTheRedTurlocTheRed

1,216412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  <pedant>This works only for integer N greater than 1</pedant>
  $endgroup$
  – Acccumulation
  May 22 at 5:16
  

  
  
  
  

add a comment | 

5

$begingroup$

I think you can use Euler's Formula.

The Nth roots of unity = $e^2pi k i/N$ for values of k between $0$ and $N-1$ inclusive.

There sum from k to $N-1$ is a geometric series.

$S= sum_k=0^N-1 e^2pi i k/N=frac1cdot e^(2pi i /N)N-1e^2pi i /N-1$

The numerator is zero for any N.

But the real part of $S$ is the real part of the individual terms of the sum, i.e. the cosines. The real part of the sum is zero so the sum of the real parts of the roots of unity is 0.

share|cite|improve this answer

answered May 21 at 21:24

TurlocTheRedTurlocTheRed

1,216412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  <pedant>This works only for integer N greater than 1</pedant>
  $endgroup$
  – Acccumulation
  May 22 at 5:16
  

  
  
  
  

add a comment | 

$begingroup$

I think you can use Euler's Formula.

The Nth roots of unity = $e^2pi k i/N$ for values of k between $0$ and $N-1$ inclusive.

There sum from k to $N-1$ is a geometric series.

$S= sum_k=0^N-1 e^2pi i k/N=frac1cdot e^(2pi i /N)N-1e^2pi i /N-1$

The numerator is zero for any N.

But the real part of $S$ is the real part of the individual terms of the sum, i.e. the cosines. The real part of the sum is zero so the sum of the real parts of the roots of unity is 0.

share|cite|improve this answer

answered May 21 at 21:24

TurlocTheRedTurlocTheRed

1,216412

$endgroup$

share|cite|improve this answer

answered May 21 at 21:24

TurlocTheRedTurlocTheRed

1,216412

answered May 21 at 21:24

TurlocTheRedTurlocTheRed

1,216412

answered May 21 at 21:24

TurlocTheRedTurlocTheRed

1,216412

2

$begingroup$

Pointing at the link I left in the comments

$$1 + costheta + cos2theta +... + cos ntheta = frac12 + fracsin[(n+frac12)theta]2sin(fractheta2)$$

Then for $forall ninmathbbN, n>0$
$$cosfracpin+1+ cosfrac2pin+1 +... + cos fracnpin+1 = fracsinleft[(n+frac12)fracpin+1right]2sinleft(fracpi2(n+1)right)-frac12=\
fracsinleft[frac2n+12(n+1)piright]2sinleft(fracpi2(n+1)right)-frac12=
fracsinleft[pi-fracpi2(n+1)right]2sinleft(fracpi2(n+1)right)-frac12=frac12-frac12=0$$

share|cite|improve this answer

answered May 21 at 21:29

rtybasertybase

12.2k31634

$endgroup$

add a comment | 

2

$begingroup$

Pointing at the link I left in the comments

$$1 + costheta + cos2theta +... + cos ntheta = frac12 + fracsin[(n+frac12)theta]2sin(fractheta2)$$

Then for $forall ninmathbbN, n>0$
$$cosfracpin+1+ cosfrac2pin+1 +... + cos fracnpin+1 = fracsinleft[(n+frac12)fracpin+1right]2sinleft(fracpi2(n+1)right)-frac12=\
fracsinleft[frac2n+12(n+1)piright]2sinleft(fracpi2(n+1)right)-frac12=
fracsinleft[pi-fracpi2(n+1)right]2sinleft(fracpi2(n+1)right)-frac12=frac12-frac12=0$$

share|cite|improve this answer

answered May 21 at 21:29

rtybasertybase

12.2k31634

$endgroup$

add a comment | 

$begingroup$

Pointing at the link I left in the comments

$$1 + costheta + cos2theta +... + cos ntheta = frac12 + fracsin[(n+frac12)theta]2sin(fractheta2)$$

Then for $forall ninmathbbN, n>0$
$$cosfracpin+1+ cosfrac2pin+1 +... + cos fracnpin+1 = fracsinleft[(n+frac12)fracpin+1right]2sinleft(fracpi2(n+1)right)-frac12=\
fracsinleft[frac2n+12(n+1)piright]2sinleft(fracpi2(n+1)right)-frac12=
fracsinleft[pi-fracpi2(n+1)right]2sinleft(fracpi2(n+1)right)-frac12=frac12-frac12=0$$

share|cite|improve this answer

answered May 21 at 21:29

rtybasertybase

12.2k31634

$endgroup$

share|cite|improve this answer

answered May 21 at 21:29

rtybasertybase

12.2k31634

answered May 21 at 21:29

rtybasertybase

12.2k31634

answered May 21 at 21:29

rtybasertybase

12.2k31634