Evaluating $sinfrac{pi}{2}sinfrac{pi}{2^2}sinfrac{pi}{2^3}cdotssinfrac{pi}{2^{11}}cos frac{pi}{2^{12}}$












2












$begingroup$


Evaluate :
$$sinfrac{pi}{2} times sin frac{pi}{2^2} times sin frac{pi}{2^3} cdots times sinfrac{pi}{2^{11}} times cos frac{pi}{2^{12}}$$

I tried to solve it by using double angle formula, by replacing $sindisplaystylefrac{pi}2$ with $2 sindisplaystylefrac{pi}4 cosdisplaystylefrac{pi}4$ and similarly replacing $sindisplaystylefrac{pi}{2^2}$ with double angle formula. But I'm not getting anywhere from this, can you please suggest me how can I approach to this problem, especially that cosine term at last is making everything so weird.

The thing which is most important to ask is : although I know every theory which applies over here but then also why I'm unable to solve this question? What is the reason for my failure?






trigonometry problem-solving







share|cite|improve this question











$endgroup$












  • $begingroup$
    Let me suggest that to improve the readability of your question, you set your formulas in mathjax. I did part of it for you.
    $endgroup$
    – Lee Mosher
    Jan 31 at 15:01












  • $begingroup$
    @Jan What about the $cos{{piover12}}$ at the end?
    $endgroup$
    – saulspatz
    Jan 31 at 15:04






  • 2




    $begingroup$
    Also, the reason for failing at something is usually that you haven't yet found the path to success. You could give up (sometimes a realistic choice), or you could keep trying.
    $endgroup$
    – Lee Mosher
    Jan 31 at 15:08










  • $begingroup$
    One way is to convert the product into a sum. Hopefully, a summable series (i.e., expressible in closed form) would result.
    $endgroup$
    – Allawonder
    Jan 31 at 15:09










  • $begingroup$
    Viete helpful ?
    $endgroup$
    – Cosmas Zachos
    Jan 31 at 15:24
















2












$begingroup$


Evaluate :
$$sinfrac{pi}{2} times sin frac{pi}{2^2} times sin frac{pi}{2^3} cdots times sinfrac{pi}{2^{11}} times cos frac{pi}{2^{12}}$$

I tried to solve it by using double angle formula, by replacing $sindisplaystylefrac{pi}2$ with $2 sindisplaystylefrac{pi}4 cosdisplaystylefrac{pi}4$ and similarly replacing $sindisplaystylefrac{pi}{2^2}$ with double angle formula. But I'm not getting anywhere from this, can you please suggest me how can I approach to this problem, especially that cosine term at last is making everything so weird.

The thing which is most important to ask is : although I know every theory which applies over here but then also why I'm unable to solve this question? What is the reason for my failure?






trigonometry problem-solving







share|cite|improve this question











$endgroup$












  • $begingroup$
    Let me suggest that to improve the readability of your question, you set your formulas in mathjax. I did part of it for you.
    $endgroup$
    – Lee Mosher
    Jan 31 at 15:01












  • $begingroup$
    @Jan What about the $cos{{piover12}}$ at the end?
    $endgroup$
    – saulspatz
    Jan 31 at 15:04






  • 2




    $begingroup$
    Also, the reason for failing at something is usually that you haven't yet found the path to success. You could give up (sometimes a realistic choice), or you could keep trying.
    $endgroup$
    – Lee Mosher
    Jan 31 at 15:08










  • $begingroup$
    One way is to convert the product into a sum. Hopefully, a summable series (i.e., expressible in closed form) would result.
    $endgroup$
    – Allawonder
    Jan 31 at 15:09










  • $begingroup$
    Viete helpful ?
    $endgroup$
    – Cosmas Zachos
    Jan 31 at 15:24














2












2








2





$begingroup$


Evaluate :
$$sinfrac{pi}{2} times sin frac{pi}{2^2} times sin frac{pi}{2^3} cdots times sinfrac{pi}{2^{11}} times cos frac{pi}{2^{12}}$$

I tried to solve it by using double angle formula, by replacing $sindisplaystylefrac{pi}2$ with $2 sindisplaystylefrac{pi}4 cosdisplaystylefrac{pi}4$ and similarly replacing $sindisplaystylefrac{pi}{2^2}$ with double angle formula. But I'm not getting anywhere from this, can you please suggest me how can I approach to this problem, especially that cosine term at last is making everything so weird.

The thing which is most important to ask is : although I know every theory which applies over here but then also why I'm unable to solve this question? What is the reason for my failure?






trigonometry problem-solving







share|cite|improve this question











$endgroup$




Evaluate :
$$sinfrac{pi}{2} times sin frac{pi}{2^2} times sin frac{pi}{2^3} cdots times sinfrac{pi}{2^{11}} times cos frac{pi}{2^{12}}$$

I tried to solve it by using double angle formula, by replacing $sindisplaystylefrac{pi}2$ with $2 sindisplaystylefrac{pi}4 cosdisplaystylefrac{pi}4$ and similarly replacing $sindisplaystylefrac{pi}{2^2}$ with double angle formula. But I'm not getting anywhere from this, can you please suggest me how can I approach to this problem, especially that cosine term at last is making everything so weird.

The thing which is most important to ask is : although I know every theory which applies over here but then also why I'm unable to solve this question? What is the reason for my failure?






trigonometry problem-solving




trigonometry problem-solving






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Feb 1 at 5:40









Blue

49.5k870157




49.5k870157










asked Jan 31 at 14:56









adesh mishraadesh mishra

274




274












  • $begingroup$
    Let me suggest that to improve the readability of your question, you set your formulas in mathjax. I did part of it for you.
    $endgroup$
    – Lee Mosher
    Jan 31 at 15:01












  • $begingroup$
    @Jan What about the $cos{{piover12}}$ at the end?
    $endgroup$
    – saulspatz
    Jan 31 at 15:04






  • 2




    $begingroup$
    Also, the reason for failing at something is usually that you haven't yet found the path to success. You could give up (sometimes a realistic choice), or you could keep trying.
    $endgroup$
    – Lee Mosher
    Jan 31 at 15:08










  • $begingroup$
    One way is to convert the product into a sum. Hopefully, a summable series (i.e., expressible in closed form) would result.
    $endgroup$
    – Allawonder
    Jan 31 at 15:09










  • $begingroup$
    Viete helpful ?
    $endgroup$
    – Cosmas Zachos
    Jan 31 at 15:24


















  • $begingroup$
    Let me suggest that to improve the readability of your question, you set your formulas in mathjax. I did part of it for you.
    $endgroup$
    – Lee Mosher
    Jan 31 at 15:01












  • $begingroup$
    @Jan What about the $cos{{piover12}}$ at the end?
    $endgroup$
    – saulspatz
    Jan 31 at 15:04






  • 2




    $begingroup$
    Also, the reason for failing at something is usually that you haven't yet found the path to success. You could give up (sometimes a realistic choice), or you could keep trying.
    $endgroup$
    – Lee Mosher
    Jan 31 at 15:08










  • $begingroup$
    One way is to convert the product into a sum. Hopefully, a summable series (i.e., expressible in closed form) would result.
    $endgroup$
    – Allawonder
    Jan 31 at 15:09










  • $begingroup$
    Viete helpful ?
    $endgroup$
    – Cosmas Zachos
    Jan 31 at 15:24
















$begingroup$
Let me suggest that to improve the readability of your question, you set your formulas in mathjax. I did part of it for you.
$endgroup$
– Lee Mosher
Jan 31 at 15:01






$begingroup$
Let me suggest that to improve the readability of your question, you set your formulas in mathjax. I did part of it for you.
$endgroup$
– Lee Mosher
Jan 31 at 15:01














$begingroup$
@Jan What about the $cos{{piover12}}$ at the end?
$endgroup$
– saulspatz
Jan 31 at 15:04




$begingroup$
@Jan What about the $cos{{piover12}}$ at the end?
$endgroup$
– saulspatz
Jan 31 at 15:04




2




2




$begingroup$
Also, the reason for failing at something is usually that you haven't yet found the path to success. You could give up (sometimes a realistic choice), or you could keep trying.
$endgroup$
– Lee Mosher
Jan 31 at 15:08




$begingroup$
Also, the reason for failing at something is usually that you haven't yet found the path to success. You could give up (sometimes a realistic choice), or you could keep trying.
$endgroup$
– Lee Mosher
Jan 31 at 15:08












$begingroup$
One way is to convert the product into a sum. Hopefully, a summable series (i.e., expressible in closed form) would result.
$endgroup$
– Allawonder
Jan 31 at 15:09




$begingroup$
One way is to convert the product into a sum. Hopefully, a summable series (i.e., expressible in closed form) would result.
$endgroup$
– Allawonder
Jan 31 at 15:09












$begingroup$
Viete helpful ?
$endgroup$
– Cosmas Zachos
Jan 31 at 15:24




$begingroup$
Viete helpful ?
$endgroup$
– Cosmas Zachos
Jan 31 at 15:24










1 Answer
1






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oldest

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0












$begingroup$

Whatever you would do, the result would be a very small number very close to $0$
$$sinfrac{pi}{2} times sin frac{pi}{2^2} times sin frac{pi}{2^3}times cdots times sinfrac{pi}{2^{11}} times cos frac{pi}{2^{12}}$$



We know the value of the sine for some of the angles. The first are
$${1 , frac {sqrt 2} 2,frac{sqrt{2-sqrt{2}}}{2},frac{sqrt{2-sqrt{2+sqrt{2}}}}{2},frac{sqrt{2-sqrt{2+sqrt{2+sqrt{2}}}}}{2},frac{sqrt{2-sqrt{2+sqrt{2+sqrt{2+sqrt{2}}}}} }{2} }$$ and the pattern continue for ever.



For the next one, corresponding to $sin frac{pi}{2^8}$ the exact value is
$$frac{sqrt{2-sqrt{2+sqrt{2+sqrt{2+sqrt{2+sqrt{2+sqrt{2}}}}}}}}{2}approx0.012271538$$ while
$$frac{pi}{2^8} approx 0.012271846$$ that is to say that you can approximate all the sines by the argument. Concerning the cosine, the argument is so small that you can assume it is almost equal to $1$.



If you take into account the fact that $sin(x)<x$, an upper bound for the product of sines would be
$$ frac{ pi^9 ,sqrt 2}{2^{64}}approx 2.285 times 10^{-15}$$ while it should be $approx 2.208 times 10^{-15}$.



Concerning
$$cos frac{pi}{2^{12}}approx 0.99999970586$$



All of that makes the value of the expression quite close to the machine epsilon.






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    1 Answer
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    1 Answer
    1






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    oldest

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    active

    oldest

    votes









    0












    $begingroup$

    Whatever you would do, the result would be a very small number very close to $0$
    $$sinfrac{pi}{2} times sin frac{pi}{2^2} times sin frac{pi}{2^3}times cdots times sinfrac{pi}{2^{11}} times cos frac{pi}{2^{12}}$$



    We know the value of the sine for some of the angles. The first are
    $${1 , frac {sqrt 2} 2,frac{sqrt{2-sqrt{2}}}{2},frac{sqrt{2-sqrt{2+sqrt{2}}}}{2},frac{sqrt{2-sqrt{2+sqrt{2+sqrt{2}}}}}{2},frac{sqrt{2-sqrt{2+sqrt{2+sqrt{2+sqrt{2}}}}} }{2} }$$ and the pattern continue for ever.



    For the next one, corresponding to $sin frac{pi}{2^8}$ the exact value is
    $$frac{sqrt{2-sqrt{2+sqrt{2+sqrt{2+sqrt{2+sqrt{2+sqrt{2}}}}}}}}{2}approx0.012271538$$ while
    $$frac{pi}{2^8} approx 0.012271846$$ that is to say that you can approximate all the sines by the argument. Concerning the cosine, the argument is so small that you can assume it is almost equal to $1$.



    If you take into account the fact that $sin(x)<x$, an upper bound for the product of sines would be
    $$ frac{ pi^9 ,sqrt 2}{2^{64}}approx 2.285 times 10^{-15}$$ while it should be $approx 2.208 times 10^{-15}$.



    Concerning
    $$cos frac{pi}{2^{12}}approx 0.99999970586$$



    All of that makes the value of the expression quite close to the machine epsilon.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Whatever you would do, the result would be a very small number very close to $0$
      $$sinfrac{pi}{2} times sin frac{pi}{2^2} times sin frac{pi}{2^3}times cdots times sinfrac{pi}{2^{11}} times cos frac{pi}{2^{12}}$$



      We know the value of the sine for some of the angles. The first are
      $${1 , frac {sqrt 2} 2,frac{sqrt{2-sqrt{2}}}{2},frac{sqrt{2-sqrt{2+sqrt{2}}}}{2},frac{sqrt{2-sqrt{2+sqrt{2+sqrt{2}}}}}{2},frac{sqrt{2-sqrt{2+sqrt{2+sqrt{2+sqrt{2}}}}} }{2} }$$ and the pattern continue for ever.



      For the next one, corresponding to $sin frac{pi}{2^8}$ the exact value is
      $$frac{sqrt{2-sqrt{2+sqrt{2+sqrt{2+sqrt{2+sqrt{2+sqrt{2}}}}}}}}{2}approx0.012271538$$ while
      $$frac{pi}{2^8} approx 0.012271846$$ that is to say that you can approximate all the sines by the argument. Concerning the cosine, the argument is so small that you can assume it is almost equal to $1$.



      If you take into account the fact that $sin(x)<x$, an upper bound for the product of sines would be
      $$ frac{ pi^9 ,sqrt 2}{2^{64}}approx 2.285 times 10^{-15}$$ while it should be $approx 2.208 times 10^{-15}$.



      Concerning
      $$cos frac{pi}{2^{12}}approx 0.99999970586$$



      All of that makes the value of the expression quite close to the machine epsilon.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Whatever you would do, the result would be a very small number very close to $0$
        $$sinfrac{pi}{2} times sin frac{pi}{2^2} times sin frac{pi}{2^3}times cdots times sinfrac{pi}{2^{11}} times cos frac{pi}{2^{12}}$$



        We know the value of the sine for some of the angles. The first are
        $${1 , frac {sqrt 2} 2,frac{sqrt{2-sqrt{2}}}{2},frac{sqrt{2-sqrt{2+sqrt{2}}}}{2},frac{sqrt{2-sqrt{2+sqrt{2+sqrt{2}}}}}{2},frac{sqrt{2-sqrt{2+sqrt{2+sqrt{2+sqrt{2}}}}} }{2} }$$ and the pattern continue for ever.



        For the next one, corresponding to $sin frac{pi}{2^8}$ the exact value is
        $$frac{sqrt{2-sqrt{2+sqrt{2+sqrt{2+sqrt{2+sqrt{2+sqrt{2}}}}}}}}{2}approx0.012271538$$ while
        $$frac{pi}{2^8} approx 0.012271846$$ that is to say that you can approximate all the sines by the argument. Concerning the cosine, the argument is so small that you can assume it is almost equal to $1$.



        If you take into account the fact that $sin(x)<x$, an upper bound for the product of sines would be
        $$ frac{ pi^9 ,sqrt 2}{2^{64}}approx 2.285 times 10^{-15}$$ while it should be $approx 2.208 times 10^{-15}$.



        Concerning
        $$cos frac{pi}{2^{12}}approx 0.99999970586$$



        All of that makes the value of the expression quite close to the machine epsilon.






        share|cite|improve this answer









        $endgroup$



        Whatever you would do, the result would be a very small number very close to $0$
        $$sinfrac{pi}{2} times sin frac{pi}{2^2} times sin frac{pi}{2^3}times cdots times sinfrac{pi}{2^{11}} times cos frac{pi}{2^{12}}$$



        We know the value of the sine for some of the angles. The first are
        $${1 , frac {sqrt 2} 2,frac{sqrt{2-sqrt{2}}}{2},frac{sqrt{2-sqrt{2+sqrt{2}}}}{2},frac{sqrt{2-sqrt{2+sqrt{2+sqrt{2}}}}}{2},frac{sqrt{2-sqrt{2+sqrt{2+sqrt{2+sqrt{2}}}}} }{2} }$$ and the pattern continue for ever.



        For the next one, corresponding to $sin frac{pi}{2^8}$ the exact value is
        $$frac{sqrt{2-sqrt{2+sqrt{2+sqrt{2+sqrt{2+sqrt{2+sqrt{2}}}}}}}}{2}approx0.012271538$$ while
        $$frac{pi}{2^8} approx 0.012271846$$ that is to say that you can approximate all the sines by the argument. Concerning the cosine, the argument is so small that you can assume it is almost equal to $1$.



        If you take into account the fact that $sin(x)<x$, an upper bound for the product of sines would be
        $$ frac{ pi^9 ,sqrt 2}{2^{64}}approx 2.285 times 10^{-15}$$ while it should be $approx 2.208 times 10^{-15}$.



        Concerning
        $$cos frac{pi}{2^{12}}approx 0.99999970586$$



        All of that makes the value of the expression quite close to the machine epsilon.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 1 at 7:07









        Claude LeiboviciClaude Leibovici

        125k1158135




        125k1158135






























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