What is $[cos(pi/12)+isin(pi/12)]^16+[cos(pi/12)-isin(pi/12)]^16$?

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What is $[cos(pi/12)+isin(pi/12)]^16+[cos(pi/12)-isin(pi/12)]^16$?



I can use De Moivre's formula for the left part:



$[cos(pi/12)+isin(pi/12)]^16 = cos(4pi/3) + isin(4pi/3) = -dfracsqrt32 + dfraci2$



but I'm stuck at the right part. Thanks in advance.







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  • Use math.stackexchange.com/questions/3510/…
    – lab bhattacharjee
    Aug 26 at 11:45






  • 2




    By the way, you're getting the cosine and sine wrong values in the last part of your computation. It should be $-frac12 - fracsqrt32i$.
    – Henning Makholm
    Aug 26 at 11:53







  • 1




    I can use De Moivre's formula for the left part Call that $z$, then the right part is just $bar z$, so their sum is $z + bar z = 2 operatornameRe(z)$. But you need to fix the left part, first.
    – dxiv
    Aug 27 at 5:25















up vote
2
down vote

favorite












What is $[cos(pi/12)+isin(pi/12)]^16+[cos(pi/12)-isin(pi/12)]^16$?



I can use De Moivre's formula for the left part:



$[cos(pi/12)+isin(pi/12)]^16 = cos(4pi/3) + isin(4pi/3) = -dfracsqrt32 + dfraci2$



but I'm stuck at the right part. Thanks in advance.







share|cite|improve this question




















  • Use math.stackexchange.com/questions/3510/…
    – lab bhattacharjee
    Aug 26 at 11:45






  • 2




    By the way, you're getting the cosine and sine wrong values in the last part of your computation. It should be $-frac12 - fracsqrt32i$.
    – Henning Makholm
    Aug 26 at 11:53







  • 1




    I can use De Moivre's formula for the left part Call that $z$, then the right part is just $bar z$, so their sum is $z + bar z = 2 operatornameRe(z)$. But you need to fix the left part, first.
    – dxiv
    Aug 27 at 5:25













up vote
2
down vote

favorite









up vote
2
down vote

favorite











What is $[cos(pi/12)+isin(pi/12)]^16+[cos(pi/12)-isin(pi/12)]^16$?



I can use De Moivre's formula for the left part:



$[cos(pi/12)+isin(pi/12)]^16 = cos(4pi/3) + isin(4pi/3) = -dfracsqrt32 + dfraci2$



but I'm stuck at the right part. Thanks in advance.







share|cite|improve this question












What is $[cos(pi/12)+isin(pi/12)]^16+[cos(pi/12)-isin(pi/12)]^16$?



I can use De Moivre's formula for the left part:



$[cos(pi/12)+isin(pi/12)]^16 = cos(4pi/3) + isin(4pi/3) = -dfracsqrt32 + dfraci2$



but I'm stuck at the right part. Thanks in advance.









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 26 at 11:44









Surzilla

1087




1087











  • Use math.stackexchange.com/questions/3510/…
    – lab bhattacharjee
    Aug 26 at 11:45






  • 2




    By the way, you're getting the cosine and sine wrong values in the last part of your computation. It should be $-frac12 - fracsqrt32i$.
    – Henning Makholm
    Aug 26 at 11:53







  • 1




    I can use De Moivre's formula for the left part Call that $z$, then the right part is just $bar z$, so their sum is $z + bar z = 2 operatornameRe(z)$. But you need to fix the left part, first.
    – dxiv
    Aug 27 at 5:25

















  • Use math.stackexchange.com/questions/3510/…
    – lab bhattacharjee
    Aug 26 at 11:45






  • 2




    By the way, you're getting the cosine and sine wrong values in the last part of your computation. It should be $-frac12 - fracsqrt32i$.
    – Henning Makholm
    Aug 26 at 11:53







  • 1




    I can use De Moivre's formula for the left part Call that $z$, then the right part is just $bar z$, so their sum is $z + bar z = 2 operatornameRe(z)$. But you need to fix the left part, first.
    – dxiv
    Aug 27 at 5:25
















Use math.stackexchange.com/questions/3510/…
– lab bhattacharjee
Aug 26 at 11:45




Use math.stackexchange.com/questions/3510/…
– lab bhattacharjee
Aug 26 at 11:45




2




2




By the way, you're getting the cosine and sine wrong values in the last part of your computation. It should be $-frac12 - fracsqrt32i$.
– Henning Makholm
Aug 26 at 11:53





By the way, you're getting the cosine and sine wrong values in the last part of your computation. It should be $-frac12 - fracsqrt32i$.
– Henning Makholm
Aug 26 at 11:53





1




1




I can use De Moivre's formula for the left part Call that $z$, then the right part is just $bar z$, so their sum is $z + bar z = 2 operatornameRe(z)$. But you need to fix the left part, first.
– dxiv
Aug 27 at 5:25





I can use De Moivre's formula for the left part Call that $z$, then the right part is just $bar z$, so their sum is $z + bar z = 2 operatornameRe(z)$. But you need to fix the left part, first.
– dxiv
Aug 27 at 5:25











4 Answers
4






active

oldest

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up vote
4
down vote



accepted










Note that $$cos(pi/12)-isin(pi/12)=cos(-pi/12)+isin(-pi/12)$$






share|cite|improve this answer



























    up vote
    2
    down vote













    $$(cos y-isin y)^n=dfrac1(cos y+isin y)^n=dfrac1cos(ny)+isin(ny)=cos(ny)-isin(ny)$$



    as $(cos x+isin x)(cos x-isin x)=1$






    share|cite|improve this answer





























      up vote
      2
      down vote













      $(cos y-isin y)^n=(cos(-y)+isin(-y))^n=cos(n(-y))+isin(n(-y))=cos(ny)-isin(ny)$






      share|cite|improve this answer



























        up vote
        0
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        It's much simpler with the exponential notation:
        beginalign
        (cospi/12+isinpi/12) ^16&+(cospi/12-isinpi/12 )^16\&=mathrm e^tfrac4ipi3+mathrm e^tfrac-4ipi3=2cosfrac4pi3=2cdotfrac12
        endalign






        share|cite|improve this answer






















        • @dxiv: Fixed. Thanks for pointing it!
          – Bernard
          Aug 27 at 9:20










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        4 Answers
        4






        active

        oldest

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        4 Answers
        4






        active

        oldest

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        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        4
        down vote



        accepted










        Note that $$cos(pi/12)-isin(pi/12)=cos(-pi/12)+isin(-pi/12)$$






        share|cite|improve this answer
























          up vote
          4
          down vote



          accepted










          Note that $$cos(pi/12)-isin(pi/12)=cos(-pi/12)+isin(-pi/12)$$






          share|cite|improve this answer






















            up vote
            4
            down vote



            accepted







            up vote
            4
            down vote



            accepted






            Note that $$cos(pi/12)-isin(pi/12)=cos(-pi/12)+isin(-pi/12)$$






            share|cite|improve this answer












            Note that $$cos(pi/12)-isin(pi/12)=cos(-pi/12)+isin(-pi/12)$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 26 at 11:45









            Suzet

            2,421527




            2,421527




















                up vote
                2
                down vote













                $$(cos y-isin y)^n=dfrac1(cos y+isin y)^n=dfrac1cos(ny)+isin(ny)=cos(ny)-isin(ny)$$



                as $(cos x+isin x)(cos x-isin x)=1$






                share|cite|improve this answer


























                  up vote
                  2
                  down vote













                  $$(cos y-isin y)^n=dfrac1(cos y+isin y)^n=dfrac1cos(ny)+isin(ny)=cos(ny)-isin(ny)$$



                  as $(cos x+isin x)(cos x-isin x)=1$






                  share|cite|improve this answer
























                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    $$(cos y-isin y)^n=dfrac1(cos y+isin y)^n=dfrac1cos(ny)+isin(ny)=cos(ny)-isin(ny)$$



                    as $(cos x+isin x)(cos x-isin x)=1$






                    share|cite|improve this answer














                    $$(cos y-isin y)^n=dfrac1(cos y+isin y)^n=dfrac1cos(ny)+isin(ny)=cos(ny)-isin(ny)$$



                    as $(cos x+isin x)(cos x-isin x)=1$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Aug 26 at 11:50









                    Lord Shark the Unknown

                    88.2k955115




                    88.2k955115










                    answered Aug 26 at 11:47









                    lab bhattacharjee

                    216k14153265




                    216k14153265




















                        up vote
                        2
                        down vote













                        $(cos y-isin y)^n=(cos(-y)+isin(-y))^n=cos(n(-y))+isin(n(-y))=cos(ny)-isin(ny)$






                        share|cite|improve this answer
























                          up vote
                          2
                          down vote













                          $(cos y-isin y)^n=(cos(-y)+isin(-y))^n=cos(n(-y))+isin(n(-y))=cos(ny)-isin(ny)$






                          share|cite|improve this answer






















                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            $(cos y-isin y)^n=(cos(-y)+isin(-y))^n=cos(n(-y))+isin(n(-y))=cos(ny)-isin(ny)$






                            share|cite|improve this answer












                            $(cos y-isin y)^n=(cos(-y)+isin(-y))^n=cos(n(-y))+isin(n(-y))=cos(ny)-isin(ny)$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Aug 26 at 11:59









                            drhab

                            88.4k541120




                            88.4k541120




















                                up vote
                                0
                                down vote













                                It's much simpler with the exponential notation:
                                beginalign
                                (cospi/12+isinpi/12) ^16&+(cospi/12-isinpi/12 )^16\&=mathrm e^tfrac4ipi3+mathrm e^tfrac-4ipi3=2cosfrac4pi3=2cdotfrac12
                                endalign






                                share|cite|improve this answer






















                                • @dxiv: Fixed. Thanks for pointing it!
                                  – Bernard
                                  Aug 27 at 9:20














                                up vote
                                0
                                down vote













                                It's much simpler with the exponential notation:
                                beginalign
                                (cospi/12+isinpi/12) ^16&+(cospi/12-isinpi/12 )^16\&=mathrm e^tfrac4ipi3+mathrm e^tfrac-4ipi3=2cosfrac4pi3=2cdotfrac12
                                endalign






                                share|cite|improve this answer






















                                • @dxiv: Fixed. Thanks for pointing it!
                                  – Bernard
                                  Aug 27 at 9:20












                                up vote
                                0
                                down vote










                                up vote
                                0
                                down vote









                                It's much simpler with the exponential notation:
                                beginalign
                                (cospi/12+isinpi/12) ^16&+(cospi/12-isinpi/12 )^16\&=mathrm e^tfrac4ipi3+mathrm e^tfrac-4ipi3=2cosfrac4pi3=2cdotfrac12
                                endalign






                                share|cite|improve this answer














                                It's much simpler with the exponential notation:
                                beginalign
                                (cospi/12+isinpi/12) ^16&+(cospi/12-isinpi/12 )^16\&=mathrm e^tfrac4ipi3+mathrm e^tfrac-4ipi3=2cosfrac4pi3=2cdotfrac12
                                endalign







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Aug 27 at 9:19

























                                answered Aug 26 at 12:30









                                Bernard

                                111k635102




                                111k635102











                                • @dxiv: Fixed. Thanks for pointing it!
                                  – Bernard
                                  Aug 27 at 9:20
















                                • @dxiv: Fixed. Thanks for pointing it!
                                  – Bernard
                                  Aug 27 at 9:20















                                @dxiv: Fixed. Thanks for pointing it!
                                – Bernard
                                Aug 27 at 9:20




                                @dxiv: Fixed. Thanks for pointing it!
                                – Bernard
                                Aug 27 at 9:20

















                                 

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