De Moivre's Theorem Simplification [closed]

0

Prove that
$$(1 + operatorname{cis}(x))^k + (1 + operatorname{cis}(-x))^k = 2^{k+1}cosleft(frac{kx}{2}right)cos^kleft(frac{x}{2}right).$$

Any guidance would be greatly appreciated.

Thanks.

complex-numbers complex-geometry

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edited Nov 20 '18 at 9:09

Tianlalu

3,09621038

asked Nov 20 '18 at 9:06

ultralight

396

closed as off-topic by heropup, Claude Leibovici, Nosrati, Rebellos, amWhy Nov 20 '18 at 19:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Claude Leibovici, Nosrati, Rebellos, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is $kinBbb N$?
  – Tianlalu
  Nov 20 '18 at 9:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, $k in mathbb{N}$
  – ultralight
  Nov 20 '18 at 10:22
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Any attempts to the question?
  – Tianlalu
  Nov 20 '18 at 10:56
  
  
  

  
  
  
  

add a comment | 

0

Prove that
$$(1 + operatorname{cis}(x))^k + (1 + operatorname{cis}(-x))^k = 2^{k+1}cosleft(frac{kx}{2}right)cos^kleft(frac{x}{2}right).$$

Any guidance would be greatly appreciated.

Thanks.

complex-numbers complex-geometry

share|cite|improve this question

edited Nov 20 '18 at 9:09

Tianlalu

3,09621038

asked Nov 20 '18 at 9:06

ultralight

396

closed as off-topic by heropup, Claude Leibovici, Nosrati, Rebellos, amWhy Nov 20 '18 at 19:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Claude Leibovici, Nosrati, Rebellos, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is $kinBbb N$?
  – Tianlalu
  Nov 20 '18 at 9:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, $k in mathbb{N}$
  – ultralight
  Nov 20 '18 at 10:22
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Any attempts to the question?
  – Tianlalu
  Nov 20 '18 at 10:56
  
  
  

  
  
  
  

add a comment | 

0

0

0

Prove that
$$(1 + operatorname{cis}(x))^k + (1 + operatorname{cis}(-x))^k = 2^{k+1}cosleft(frac{kx}{2}right)cos^kleft(frac{x}{2}right).$$

Any guidance would be greatly appreciated.

Thanks.

complex-numbers complex-geometry

share|cite|improve this question

edited Nov 20 '18 at 9:09

Tianlalu

3,09621038

asked Nov 20 '18 at 9:06

ultralight

396

Prove that
$$(1 + operatorname{cis}(x))^k + (1 + operatorname{cis}(-x))^k = 2^{k+1}cosleft(frac{kx}{2}right)cos^kleft(frac{x}{2}right).$$

Any guidance would be greatly appreciated.

Thanks.

complex-numbers complex-geometry

complex-numbers complex-geometry

share|cite|improve this question

edited Nov 20 '18 at 9:09

Tianlalu

3,09621038

asked Nov 20 '18 at 9:06

ultralight

396

share|cite|improve this question

edited Nov 20 '18 at 9:09

Tianlalu

3,09621038

asked Nov 20 '18 at 9:06

ultralight

396

share|cite|improve this question

share|cite|improve this question

edited Nov 20 '18 at 9:09

Tianlalu

3,09621038

edited Nov 20 '18 at 9:09

Tianlalu

3,09621038

edited Nov 20 '18 at 9:09

Tianlalu

3,09621038

3,09621038

asked Nov 20 '18 at 9:06

ultralight

396

asked Nov 20 '18 at 9:06

ultralight

396

asked Nov 20 '18 at 9:06

ultralight

396

396

closed as off-topic by heropup, Claude Leibovici, Nosrati, Rebellos, amWhy Nov 20 '18 at 19:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Claude Leibovici, Nosrati, Rebellos, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by heropup, Claude Leibovici, Nosrati, Rebellos, amWhy Nov 20 '18 at 19:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Claude Leibovici, Nosrati, Rebellos, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is $kinBbb N$?
  – Tianlalu
  Nov 20 '18 at 9:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, $k in mathbb{N}$
  – ultralight
  Nov 20 '18 at 10:22
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Any attempts to the question?
  – Tianlalu
  Nov 20 '18 at 10:56
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is $kinBbb N$?
  – Tianlalu
  Nov 20 '18 at 9:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, $k in mathbb{N}$
  – ultralight
  Nov 20 '18 at 10:22
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Any attempts to the question?
  – Tianlalu
  Nov 20 '18 at 10:56
  
  
  

  
  
  
  

Is $kinBbb N$?
– Tianlalu
Nov 20 '18 at 9:30

Is $kinBbb N$?
– Tianlalu
Nov 20 '18 at 9:30

Yes, $k in mathbb{N}$
– ultralight
Nov 20 '18 at 10:22

Yes, $k in mathbb{N}$
– ultralight
Nov 20 '18 at 10:22

Any attempts to the question?
– Tianlalu
Nov 20 '18 at 10:56

Any attempts to the question?
– Tianlalu
Nov 20 '18 at 10:56

add a comment | 

1 Answer
1

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2

Start by dividing both sides by $2^k$

Then calculate $frac{(1+cis(x))^k}{2^k} = cis(frac{x}{2})^kfrac{(cis(frac{-x}{2})+cis(frac{x}{2}))^k}{2^k}=cis(frac{x}{2})^kcos^k(frac{x}{2})$

The same for $frac{(1+cis(-x))^k}{2^k} = cis(frac{-x}{2})^kcos^k(frac{x}{2})$


Sum both terms and factor by $cos^k(frac{x}{2})$

That gives you $cos^k(frac{x}{2})(cis(frac{x}{2})^k + cis(frac{-x}{2})^k) = cos^k(frac{x}{2})(cis(frac{kx}{2}) + cis(frac{-kx}{2}))$ using Moivre formula beacuse $kinBbb N$

Finally $cis(frac{kx}{2}) + cis(frac{-kx}{2}) = 2cos(frac{kx}{2})$

Thus $$frac{(1+cis(x))^k+(1+cis(-x))^k}{2^k} = 2cos(frac{kx}{2})cos^k(frac{x}{2})$$


Hope it helps

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answered Nov 20 '18 at 11:03

TheD0ubleT

39218

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

Start by dividing both sides by $2^k$

Then calculate $frac{(1+cis(x))^k}{2^k} = cis(frac{x}{2})^kfrac{(cis(frac{-x}{2})+cis(frac{x}{2}))^k}{2^k}=cis(frac{x}{2})^kcos^k(frac{x}{2})$

The same for $frac{(1+cis(-x))^k}{2^k} = cis(frac{-x}{2})^kcos^k(frac{x}{2})$


Sum both terms and factor by $cos^k(frac{x}{2})$

That gives you $cos^k(frac{x}{2})(cis(frac{x}{2})^k + cis(frac{-x}{2})^k) = cos^k(frac{x}{2})(cis(frac{kx}{2}) + cis(frac{-kx}{2}))$ using Moivre formula beacuse $kinBbb N$

Finally $cis(frac{kx}{2}) + cis(frac{-kx}{2}) = 2cos(frac{kx}{2})$

Thus $$frac{(1+cis(x))^k+(1+cis(-x))^k}{2^k} = 2cos(frac{kx}{2})cos^k(frac{x}{2})$$


Hope it helps

share|cite|improve this answer

answered Nov 20 '18 at 11:03

TheD0ubleT

39218

add a comment | 

2

Start by dividing both sides by $2^k$

Then calculate $frac{(1+cis(x))^k}{2^k} = cis(frac{x}{2})^kfrac{(cis(frac{-x}{2})+cis(frac{x}{2}))^k}{2^k}=cis(frac{x}{2})^kcos^k(frac{x}{2})$

The same for $frac{(1+cis(-x))^k}{2^k} = cis(frac{-x}{2})^kcos^k(frac{x}{2})$


Sum both terms and factor by $cos^k(frac{x}{2})$

That gives you $cos^k(frac{x}{2})(cis(frac{x}{2})^k + cis(frac{-x}{2})^k) = cos^k(frac{x}{2})(cis(frac{kx}{2}) + cis(frac{-kx}{2}))$ using Moivre formula beacuse $kinBbb N$

Finally $cis(frac{kx}{2}) + cis(frac{-kx}{2}) = 2cos(frac{kx}{2})$

Thus $$frac{(1+cis(x))^k+(1+cis(-x))^k}{2^k} = 2cos(frac{kx}{2})cos^k(frac{x}{2})$$


Hope it helps

share|cite|improve this answer

answered Nov 20 '18 at 11:03

TheD0ubleT

39218

add a comment | 

2

2

2

Start by dividing both sides by $2^k$

Then calculate $frac{(1+cis(x))^k}{2^k} = cis(frac{x}{2})^kfrac{(cis(frac{-x}{2})+cis(frac{x}{2}))^k}{2^k}=cis(frac{x}{2})^kcos^k(frac{x}{2})$

The same for $frac{(1+cis(-x))^k}{2^k} = cis(frac{-x}{2})^kcos^k(frac{x}{2})$


Sum both terms and factor by $cos^k(frac{x}{2})$

That gives you $cos^k(frac{x}{2})(cis(frac{x}{2})^k + cis(frac{-x}{2})^k) = cos^k(frac{x}{2})(cis(frac{kx}{2}) + cis(frac{-kx}{2}))$ using Moivre formula beacuse $kinBbb N$

Finally $cis(frac{kx}{2}) + cis(frac{-kx}{2}) = 2cos(frac{kx}{2})$

Thus $$frac{(1+cis(x))^k+(1+cis(-x))^k}{2^k} = 2cos(frac{kx}{2})cos^k(frac{x}{2})$$


Hope it helps

share|cite|improve this answer

answered Nov 20 '18 at 11:03

TheD0ubleT

39218

Start by dividing both sides by $2^k$

Then calculate $frac{(1+cis(x))^k}{2^k} = cis(frac{x}{2})^kfrac{(cis(frac{-x}{2})+cis(frac{x}{2}))^k}{2^k}=cis(frac{x}{2})^kcos^k(frac{x}{2})$

The same for $frac{(1+cis(-x))^k}{2^k} = cis(frac{-x}{2})^kcos^k(frac{x}{2})$


Sum both terms and factor by $cos^k(frac{x}{2})$

That gives you $cos^k(frac{x}{2})(cis(frac{x}{2})^k + cis(frac{-x}{2})^k) = cos^k(frac{x}{2})(cis(frac{kx}{2}) + cis(frac{-kx}{2}))$ using Moivre formula beacuse $kinBbb N$

Finally $cis(frac{kx}{2}) + cis(frac{-kx}{2}) = 2cos(frac{kx}{2})$

Thus $$frac{(1+cis(x))^k+(1+cis(-x))^k}{2^k} = 2cos(frac{kx}{2})cos^k(frac{x}{2})$$


Hope it helps

share|cite|improve this answer

answered Nov 20 '18 at 11:03

TheD0ubleT

39218

share|cite|improve this answer

share|cite|improve this answer

answered Nov 20 '18 at 11:03

TheD0ubleT

39218

answered Nov 20 '18 at 11:03

TheD0ubleT

39218

answered Nov 20 '18 at 11:03

TheD0ubleT

39218

39218

add a comment | 

add a comment | 

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