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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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


















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















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
















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















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















share|cite|improve this question













share|cite|improve this question




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edited Nov 20 '18 at 9:09









Tianlalu

3,09621038




3,09621038










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




















  • 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












1 Answer
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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






share|cite|improve this answer




























    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


























      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
























        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












        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












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 20 '18 at 11:03









        TheD0ubleT

        39218




        39218















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