De Moivre's Theorem Simplification [closed]
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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
edited Nov 20 at 9:09
Tianlalu
3,0051938
asked Nov 20 at 9:06
ultralight
396
closed as off-topic by heropup, Claude Leibovici, Nosrati, Rebellos, amWhy Nov 20 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.
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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
edited Nov 20 at 9:09
Tianlalu
3,0051938
asked Nov 20 at 9:06
ultralight
396
closed as off-topic by heropup, Claude Leibovici, Nosrati, Rebellos, amWhy Nov 20 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 at 9:30
Yes, $k in mathbb{N}$
– ultralight
Nov 20 at 10:22
Any attempts to the question?
– Tianlalu
Nov 20 at 10:56
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
edited Nov 20 at 9:09
Tianlalu
3,0051938
asked Nov 20 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
edited Nov 20 at 9:09
Tianlalu
3,0051938
asked Nov 20 at 9:06
ultralight
396
edited Nov 20 at 9:09
Tianlalu
3,0051938
asked Nov 20 at 9:06
ultralight
396
edited Nov 20 at 9:09
Tianlalu
3,0051938
edited Nov 20 at 9:09
Tianlalu
3,0051938
edited Nov 20 at 9:09
Tianlalu
3,0051938
3,0051938
asked Nov 20 at 9:06
ultralight
396
asked Nov 20 at 9:06
ultralight
396
asked Nov 20 at 9:06
ultralight
396
396
closed as off-topic by heropup, Claude Leibovici, Nosrati, Rebellos, amWhy Nov 20 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 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 at 9:30
Yes, $k in mathbb{N}$
– ultralight
Nov 20 at 10:22
Any attempts to the question?
– Tianlalu
Nov 20 at 10:56
add a comment |
Is $kinBbb N$?
– Tianlalu
Nov 20 at 9:30
Yes, $k in mathbb{N}$
– ultralight
Nov 20 at 10:22
Any attempts to the question?
– Tianlalu
Nov 20 at 10:56
Is $kinBbb N$?
– Tianlalu
Nov 20 at 9:30
Is $kinBbb N$?
– Tianlalu
Nov 20 at 9:30
Yes, $k in mathbb{N}$
– ultralight
Nov 20 at 10:22
Yes, $k in mathbb{N}$
– ultralight
Nov 20 at 10:22
Any attempts to the question?
– Tianlalu
Nov 20 at 10:56
Any attempts to the question?
– Tianlalu
Nov 20 at 10:56
add a comment |
1 Answer
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2
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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
answered Nov 20 at 11:03
TheD0ubleT
37718
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
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
answered Nov 20 at 11:03
TheD0ubleT
37718
add a comment |
up vote
2
down vote
accepted
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
answered Nov 20 at 11:03
TheD0ubleT
37718
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
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
answered Nov 20 at 11:03
TheD0ubleT
37718
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
answered Nov 20 at 11:03
TheD0ubleT
37718
answered Nov 20 at 11:03
TheD0ubleT
37718
answered Nov 20 at 11:03
TheD0ubleT
37718
answered Nov 20 at 11:03
TheD0ubleT
37718
37718
add a comment |
add a comment |
Is $kinBbb N$?
– Tianlalu
Nov 20 at 9:30
Yes, $k in mathbb{N}$
– ultralight
Nov 20 at 10:22
Any attempts to the question?
– Tianlalu
Nov 20 at 10:56