Given that $cosfrackpi7$ for $k = 1,3,5$ are the roots of the equation $8x^3-4 x^2 - 4 x + 1=0$, find values of $sinfrackpi14$ [closed]

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Given that $cosfracpi7, cosfrac3pi7 cosfrac5pi7$ are the roots of the equation $8x^3-4 x^2 - 4x + 1=0$ . The value of $sinfracpi14 ;sinfrac3pi14 ;sinfrac5pi14$



1. I was trying to solve this problem using theory of equations taking the product of the roots to be $-1$.




trigonometry roots




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closed as off-topic by Xander Henderson, John Ma, Sahiba Arora, Jean-Claude Arbaut, Adrian Keister Aug 18 at 0:21


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  • "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." – Xander Henderson, John Ma, Sahiba Arora, Jean-Claude Arbaut, Adrian Keister
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  • I asked a similar question here:math.stackexchange.com/questions/2878704/… Hope it is helpful.
    – James Warthington
    Aug 17 at 2:03














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Given that $cosfracpi7, cosfrac3pi7 cosfrac5pi7$ are the roots of the equation $8x^3-4 x^2 - 4x + 1=0$ . The value of $sinfracpi14 ;sinfrac3pi14 ;sinfrac5pi14$



1. I was trying to solve this problem using theory of equations taking the product of the roots to be $-1$.




trigonometry roots




share|cite|improve this question














closed as off-topic by Xander Henderson, John Ma, Sahiba Arora, Jean-Claude Arbaut, Adrian Keister Aug 18 at 0:21


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." – Xander Henderson, John Ma, Sahiba Arora, Jean-Claude Arbaut, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.












  • I asked a similar question here:math.stackexchange.com/questions/2878704/… Hope it is helpful.
    – James Warthington
    Aug 17 at 2:03












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Given that $cosfracpi7, cosfrac3pi7 cosfrac5pi7$ are the roots of the equation $8x^3-4 x^2 - 4x + 1=0$ . The value of $sinfracpi14 ;sinfrac3pi14 ;sinfrac5pi14$



1. I was trying to solve this problem using theory of equations taking the product of the roots to be $-1$.




trigonometry roots




share|cite|improve this question














Given that $cosfracpi7, cosfrac3pi7 cosfrac5pi7$ are the roots of the equation $8x^3-4 x^2 - 4x + 1=0$ . The value of $sinfracpi14 ;sinfrac3pi14 ;sinfrac5pi14$



1. I was trying to solve this problem using theory of equations taking the product of the roots to be $-1$.





trigonometry roots





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edited Aug 17 at 3:37









Ahmad Bazzi

3,2131420




3,2131420










asked Aug 17 at 1:02









user584880

6




6




closed as off-topic by Xander Henderson, John Ma, Sahiba Arora, Jean-Claude Arbaut, Adrian Keister Aug 18 at 0:21


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." – Xander Henderson, John Ma, Sahiba Arora, Jean-Claude Arbaut, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Xander Henderson, John Ma, Sahiba Arora, Jean-Claude Arbaut, Adrian Keister Aug 18 at 0:21


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." – Xander Henderson, John Ma, Sahiba Arora, Jean-Claude Arbaut, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.











  • I asked a similar question here:math.stackexchange.com/questions/2878704/… Hope it is helpful.
    – James Warthington
    Aug 17 at 2:03
















  • I asked a similar question here:math.stackexchange.com/questions/2878704/… Hope it is helpful.
    – James Warthington
    Aug 17 at 2:03















I asked a similar question here:math.stackexchange.com/questions/2878704/… Hope it is helpful.
– James Warthington
Aug 17 at 2:03




I asked a similar question here:math.stackexchange.com/questions/2878704/… Hope it is helpful.
– James Warthington
Aug 17 at 2:03










2 Answers
2






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Hint:



$$dfracpi2-dfrac(2k+1)pi7=?$$ $k=0,1,2$






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  • Thanks sir.can you please tell me from where I can get more questions of this type.
    – user584880
    Aug 17 at 4:39

















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take $w = e^k pi i / 14$ for any of $k = 1,3,5.$
Any satisfies $w^14+ 1 = 0.$ More precisely, $w neq i$ and
$$ w^12 - w^10 + w^8 - w^6 + w^4 - w^2 + 1 = 0. $$



Then $2 sin frack pi14 = fracw - frac1wi = fracw^2 - 1i w;$ let us name
$$ x = fracw^2 - 1i w. $$
A little fiddling shows
$$ -x^6 + 5 x^4 - 6 x^2 + 1 = fracw^12 - w^10 + w^8 - w^6 + w^4 - w^2 + 1w^6 $$ which is actually $0.$ Our double sines are roots of $x^6 - 5 x^4 + 6 x^2 - 1.$ Taking $x=2s$ tells us that the original sines (as well as their negatives) are roots of
$$ 64s^6 - 80 s^4 + 24 s^2 - 1 $$
enter image description here






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote













    Hint:



    $$dfracpi2-dfrac(2k+1)pi7=?$$ $k=0,1,2$






    share|cite|improve this answer




















    • Thanks sir.can you please tell me from where I can get more questions of this type.
      – user584880
      Aug 17 at 4:39














    up vote
    1
    down vote













    Hint:



    $$dfracpi2-dfrac(2k+1)pi7=?$$ $k=0,1,2$






    share|cite|improve this answer




















    • Thanks sir.can you please tell me from where I can get more questions of this type.
      – user584880
      Aug 17 at 4:39












    up vote
    1
    down vote










    up vote
    1
    down vote









    Hint:



    $$dfracpi2-dfrac(2k+1)pi7=?$$ $k=0,1,2$






    share|cite|improve this answer












    Hint:



    $$dfracpi2-dfrac(2k+1)pi7=?$$ $k=0,1,2$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Aug 17 at 1:35









    lab bhattacharjee

    215k14152264




    215k14152264











    • Thanks sir.can you please tell me from where I can get more questions of this type.
      – user584880
      Aug 17 at 4:39
















    • Thanks sir.can you please tell me from where I can get more questions of this type.
      – user584880
      Aug 17 at 4:39















    Thanks sir.can you please tell me from where I can get more questions of this type.
    – user584880
    Aug 17 at 4:39




    Thanks sir.can you please tell me from where I can get more questions of this type.
    – user584880
    Aug 17 at 4:39










    up vote
    0
    down vote













    take $w = e^k pi i / 14$ for any of $k = 1,3,5.$
    Any satisfies $w^14+ 1 = 0.$ More precisely, $w neq i$ and
    $$ w^12 - w^10 + w^8 - w^6 + w^4 - w^2 + 1 = 0. $$



    Then $2 sin frack pi14 = fracw - frac1wi = fracw^2 - 1i w;$ let us name
    $$ x = fracw^2 - 1i w. $$
    A little fiddling shows
    $$ -x^6 + 5 x^4 - 6 x^2 + 1 = fracw^12 - w^10 + w^8 - w^6 + w^4 - w^2 + 1w^6 $$ which is actually $0.$ Our double sines are roots of $x^6 - 5 x^4 + 6 x^2 - 1.$ Taking $x=2s$ tells us that the original sines (as well as their negatives) are roots of
    $$ 64s^6 - 80 s^4 + 24 s^2 - 1 $$
    enter image description here






    share|cite|improve this answer
























      up vote
      0
      down vote













      take $w = e^k pi i / 14$ for any of $k = 1,3,5.$
      Any satisfies $w^14+ 1 = 0.$ More precisely, $w neq i$ and
      $$ w^12 - w^10 + w^8 - w^6 + w^4 - w^2 + 1 = 0. $$



      Then $2 sin frack pi14 = fracw - frac1wi = fracw^2 - 1i w;$ let us name
      $$ x = fracw^2 - 1i w. $$
      A little fiddling shows
      $$ -x^6 + 5 x^4 - 6 x^2 + 1 = fracw^12 - w^10 + w^8 - w^6 + w^4 - w^2 + 1w^6 $$ which is actually $0.$ Our double sines are roots of $x^6 - 5 x^4 + 6 x^2 - 1.$ Taking $x=2s$ tells us that the original sines (as well as their negatives) are roots of
      $$ 64s^6 - 80 s^4 + 24 s^2 - 1 $$
      enter image description here






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        take $w = e^k pi i / 14$ for any of $k = 1,3,5.$
        Any satisfies $w^14+ 1 = 0.$ More precisely, $w neq i$ and
        $$ w^12 - w^10 + w^8 - w^6 + w^4 - w^2 + 1 = 0. $$



        Then $2 sin frack pi14 = fracw - frac1wi = fracw^2 - 1i w;$ let us name
        $$ x = fracw^2 - 1i w. $$
        A little fiddling shows
        $$ -x^6 + 5 x^4 - 6 x^2 + 1 = fracw^12 - w^10 + w^8 - w^6 + w^4 - w^2 + 1w^6 $$ which is actually $0.$ Our double sines are roots of $x^6 - 5 x^4 + 6 x^2 - 1.$ Taking $x=2s$ tells us that the original sines (as well as their negatives) are roots of
        $$ 64s^6 - 80 s^4 + 24 s^2 - 1 $$
        enter image description here






        share|cite|improve this answer












        take $w = e^k pi i / 14$ for any of $k = 1,3,5.$
        Any satisfies $w^14+ 1 = 0.$ More precisely, $w neq i$ and
        $$ w^12 - w^10 + w^8 - w^6 + w^4 - w^2 + 1 = 0. $$



        Then $2 sin frack pi14 = fracw - frac1wi = fracw^2 - 1i w;$ let us name
        $$ x = fracw^2 - 1i w. $$
        A little fiddling shows
        $$ -x^6 + 5 x^4 - 6 x^2 + 1 = fracw^12 - w^10 + w^8 - w^6 + w^4 - w^2 + 1w^6 $$ which is actually $0.$ Our double sines are roots of $x^6 - 5 x^4 + 6 x^2 - 1.$ Taking $x=2s$ tells us that the original sines (as well as their negatives) are roots of
        $$ 64s^6 - 80 s^4 + 24 s^2 - 1 $$
        enter image description here







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 17 at 1:56









        Will Jagy

        97.5k595196




        97.5k595196












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