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

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3

$begingroup$

Can we show that

$$cosfrac{pi}{7}+cosfrac{2pi}{7}+cosfrac{3pi}{7}+cosfrac{4pi}{7}+cosfrac{5pi}{7}+cosfrac{6pi}{7}=0$$

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

Also I have observed that

$$cosfrac{pi}{5}+cosfrac{2pi}{5}+cosfrac{3pi}{5}+cosfrac{4pi}{5}=0$$

as well, so just out of curiosity, is it true that $$sum_{k=1}^{n-1} cosfrac{kpi}{n} = 0$$

for all $n$ odd?

complex-numbers

share|cite|improve this question

edited 8 hours ago

Chase Ryan Taylor

4,51121531

asked 8 hours ago

JustWanderingJustWandering

1337

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use complex numbers, something like this
  $endgroup$
  – rtybase
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You could use that $cos (pi-theta)=-cos theta$ instead, and just pair each cosine with its negative.
  $endgroup$
  – Mark Bennet
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, it works for all $n$'s, not just odd ones.
  $endgroup$
  – rtybase
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  because the unpaired one is $cos (pi /2)$
  $endgroup$
  – G Cab
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You should research how complex numbers can also be represented as vectors
  $endgroup$
  – Chase Ryan Taylor
  6 hours ago
  

  
  
  
  

 | 
show 2 more comments

3

$begingroup$

Can we show that

$$cosfrac{pi}{7}+cosfrac{2pi}{7}+cosfrac{3pi}{7}+cosfrac{4pi}{7}+cosfrac{5pi}{7}+cosfrac{6pi}{7}=0$$

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

Also I have observed that

$$cosfrac{pi}{5}+cosfrac{2pi}{5}+cosfrac{3pi}{5}+cosfrac{4pi}{5}=0$$

as well, so just out of curiosity, is it true that $$sum_{k=1}^{n-1} cosfrac{kpi}{n} = 0$$

for all $n$ odd?

complex-numbers

share|cite|improve this question

edited 8 hours ago

Chase Ryan Taylor

4,51121531

asked 8 hours ago

JustWanderingJustWandering

1337

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use complex numbers, something like this
  $endgroup$
  – rtybase
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You could use that $cos (pi-theta)=-cos theta$ instead, and just pair each cosine with its negative.
  $endgroup$
  – Mark Bennet
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, it works for all $n$'s, not just odd ones.
  $endgroup$
  – rtybase
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  because the unpaired one is $cos (pi /2)$
  $endgroup$
  – G Cab
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You should research how complex numbers can also be represented as vectors
  $endgroup$
  – Chase Ryan Taylor
  6 hours ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Can we show that

$$cosfrac{pi}{7}+cosfrac{2pi}{7}+cosfrac{3pi}{7}+cosfrac{4pi}{7}+cosfrac{5pi}{7}+cosfrac{6pi}{7}=0$$

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

Also I have observed that

$$cosfrac{pi}{5}+cosfrac{2pi}{5}+cosfrac{3pi}{5}+cosfrac{4pi}{5}=0$$

as well, so just out of curiosity, is it true that $$sum_{k=1}^{n-1} cosfrac{kpi}{n} = 0$$

for all $n$ odd?

complex-numbers

share|cite|improve this question

edited 8 hours ago

Chase Ryan Taylor

4,51121531

asked 8 hours ago

JustWanderingJustWandering

1337

$endgroup$

share|cite|improve this question

edited 8 hours ago

Chase Ryan Taylor

4,51121531

asked 8 hours ago

JustWanderingJustWandering

1337

share|cite|improve this question

edited 8 hours ago

Chase Ryan Taylor

4,51121531

asked 8 hours ago

JustWanderingJustWandering

1337

edited 8 hours ago

Chase Ryan Taylor

4,51121531

edited 8 hours ago

Chase Ryan Taylor

4,51121531

asked 8 hours ago

JustWanderingJustWandering

1337

asked 8 hours ago

JustWanderingJustWandering

1337

3 Answers
3

active

oldest

votes

6

$begingroup$

Note that $$cos(pi - alpha)= - cos(alpha)$$ Therefore $$cos(frac{pi}{7})+cos(frac{2pi}{7})+cos(frac{3pi}{7})+cos(frac{4pi}{7})+cos(frac{5pi}{7})+cos(frac{6pi}{7})=$$

$$cos(frac{pi}{7})+cos(frac{2pi}{7})+cos(frac{3pi}{7})-cos(frac{3pi}{7})-cos(frac{2pi}{7})-cos(frac{pi}{7})=0$$

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

share|cite|improve this answer

edited 2 hours ago

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

43.8k42061

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  and even number also
  $endgroup$
  – G Cab
  7 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

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.

$S= sum_{k=0}^{N-1} e^{2pi i k/N}=frac{1cdot e^{(2pi i /N)N}-1}{e^{2pi i /N}-1}$

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 8 hours ago

TurlocTheRedTurlocTheRed

1,151412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  <pedant>This works only for integer N greater than 1</pedant>
  $endgroup$
  – Acccumulation
  23 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Pointing at the link I left in the comments

$$1 + costheta + cos2theta +... + cos ntheta = frac{1}{2} + frac{sin[(n+frac{1}{2})theta]}{2sin(frac{theta}{2})}$$

Then for $forall ninmathbb{N}, n>0$
$$cosfrac{pi}{n+1}+ cosfrac{2pi}{n+1} +... + cos frac{npi}{n+1} = frac{sinleft[(n+frac{1}{2})frac{pi}{n+1}right]}{2sinleft(frac{pi}{2(n+1)}right)}-frac{1}{2}=\
frac{sinleft[frac{2n+1}{2(n+1)}piright]}{2sinleft(frac{pi}{2(n+1)}right)}-frac{1}{2}=
frac{sinleft[pi-frac{pi}{2(n+1)}right]}{2sinleft(frac{pi}{2(n+1)}right)}-frac{1}{2}=frac{1}{2}-frac{1}{2}=0$$

share|cite|improve this answer

answered 8 hours ago

rtybasertybase

12k31534

$endgroup$

add a comment | 

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6

$begingroup$

Note that $$cos(pi - alpha)= - cos(alpha)$$ Therefore $$cos(frac{pi}{7})+cos(frac{2pi}{7})+cos(frac{3pi}{7})+cos(frac{4pi}{7})+cos(frac{5pi}{7})+cos(frac{6pi}{7})=$$

$$cos(frac{pi}{7})+cos(frac{2pi}{7})+cos(frac{3pi}{7})-cos(frac{3pi}{7})-cos(frac{2pi}{7})-cos(frac{pi}{7})=0$$

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

share|cite|improve this answer

edited 2 hours ago

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

43.8k42061

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  and even number also
  $endgroup$
  – G Cab
  7 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

Note that $$cos(pi - alpha)= - cos(alpha)$$ Therefore $$cos(frac{pi}{7})+cos(frac{2pi}{7})+cos(frac{3pi}{7})+cos(frac{4pi}{7})+cos(frac{5pi}{7})+cos(frac{6pi}{7})=$$

$$cos(frac{pi}{7})+cos(frac{2pi}{7})+cos(frac{3pi}{7})-cos(frac{3pi}{7})-cos(frac{2pi}{7})-cos(frac{pi}{7})=0$$

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

share|cite|improve this answer

edited 2 hours ago

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

43.8k42061

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  and even number also
  $endgroup$
  – G Cab
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Note that $$cos(pi - alpha)= - cos(alpha)$$ Therefore $$cos(frac{pi}{7})+cos(frac{2pi}{7})+cos(frac{3pi}{7})+cos(frac{4pi}{7})+cos(frac{5pi}{7})+cos(frac{6pi}{7})=$$

$$cos(frac{pi}{7})+cos(frac{2pi}{7})+cos(frac{3pi}{7})-cos(frac{3pi}{7})-cos(frac{2pi}{7})-cos(frac{pi}{7})=0$$

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

share|cite|improve this answer

edited 2 hours ago

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

43.8k42061

$endgroup$

share|cite|improve this answer

edited 2 hours ago

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

43.8k42061

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

43.8k42061

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

43.8k42061

4

$begingroup$

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.

$S= sum_{k=0}^{N-1} e^{2pi i k/N}=frac{1cdot e^{(2pi i /N)N}-1}{e^{2pi i /N}-1}$

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 8 hours ago

TurlocTheRedTurlocTheRed

1,151412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  <pedant>This works only for integer N greater than 1</pedant>
  $endgroup$
  – Acccumulation
  23 mins ago
  

  
  
  
  

add a comment | 

4

$begingroup$

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.

$S= sum_{k=0}^{N-1} e^{2pi i k/N}=frac{1cdot e^{(2pi i /N)N}-1}{e^{2pi i /N}-1}$

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 8 hours ago

TurlocTheRedTurlocTheRed

1,151412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  <pedant>This works only for integer N greater than 1</pedant>
  $endgroup$
  – Acccumulation
  23 mins ago
  

  
  
  
  

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.

$S= sum_{k=0}^{N-1} e^{2pi i k/N}=frac{1cdot e^{(2pi i /N)N}-1}{e^{2pi i /N}-1}$

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 8 hours ago

TurlocTheRedTurlocTheRed

1,151412

$endgroup$

share|cite|improve this answer

answered 8 hours ago

TurlocTheRedTurlocTheRed

1,151412

answered 8 hours ago

TurlocTheRedTurlocTheRed

1,151412

answered 8 hours ago

TurlocTheRedTurlocTheRed

1,151412

1

$begingroup$

Pointing at the link I left in the comments

$$1 + costheta + cos2theta +... + cos ntheta = frac{1}{2} + frac{sin[(n+frac{1}{2})theta]}{2sin(frac{theta}{2})}$$

Then for $forall ninmathbb{N}, n>0$
$$cosfrac{pi}{n+1}+ cosfrac{2pi}{n+1} +... + cos frac{npi}{n+1} = frac{sinleft[(n+frac{1}{2})frac{pi}{n+1}right]}{2sinleft(frac{pi}{2(n+1)}right)}-frac{1}{2}=\
frac{sinleft[frac{2n+1}{2(n+1)}piright]}{2sinleft(frac{pi}{2(n+1)}right)}-frac{1}{2}=
frac{sinleft[pi-frac{pi}{2(n+1)}right]}{2sinleft(frac{pi}{2(n+1)}right)}-frac{1}{2}=frac{1}{2}-frac{1}{2}=0$$

share|cite|improve this answer

answered 8 hours ago

rtybasertybase

12k31534

$endgroup$

add a comment | 

1

$begingroup$

Pointing at the link I left in the comments

$$1 + costheta + cos2theta +... + cos ntheta = frac{1}{2} + frac{sin[(n+frac{1}{2})theta]}{2sin(frac{theta}{2})}$$

Then for $forall ninmathbb{N}, n>0$
$$cosfrac{pi}{n+1}+ cosfrac{2pi}{n+1} +... + cos frac{npi}{n+1} = frac{sinleft[(n+frac{1}{2})frac{pi}{n+1}right]}{2sinleft(frac{pi}{2(n+1)}right)}-frac{1}{2}=\
frac{sinleft[frac{2n+1}{2(n+1)}piright]}{2sinleft(frac{pi}{2(n+1)}right)}-frac{1}{2}=
frac{sinleft[pi-frac{pi}{2(n+1)}right]}{2sinleft(frac{pi}{2(n+1)}right)}-frac{1}{2}=frac{1}{2}-frac{1}{2}=0$$

share|cite|improve this answer

answered 8 hours ago

rtybasertybase

12k31534

$endgroup$

add a comment | 

$begingroup$

Pointing at the link I left in the comments

$$1 + costheta + cos2theta +... + cos ntheta = frac{1}{2} + frac{sin[(n+frac{1}{2})theta]}{2sin(frac{theta}{2})}$$

Then for $forall ninmathbb{N}, n>0$
$$cosfrac{pi}{n+1}+ cosfrac{2pi}{n+1} +... + cos frac{npi}{n+1} = frac{sinleft[(n+frac{1}{2})frac{pi}{n+1}right]}{2sinleft(frac{pi}{2(n+1)}right)}-frac{1}{2}=\
frac{sinleft[frac{2n+1}{2(n+1)}piright]}{2sinleft(frac{pi}{2(n+1)}right)}-frac{1}{2}=
frac{sinleft[pi-frac{pi}{2(n+1)}right]}{2sinleft(frac{pi}{2(n+1)}right)}-frac{1}{2}=frac{1}{2}-frac{1}{2}=0$$

share|cite|improve this answer

answered 8 hours ago

rtybasertybase

12k31534

$endgroup$

share|cite|improve this answer

answered 8 hours ago

rtybasertybase

12k31534

answered 8 hours ago

rtybasertybase

12k31534

answered 8 hours ago

rtybasertybase

12k31534