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Numeral value of this Sin(1)*Sin(2)*Sin(3)* *Sin(89)

  1. Jun 6, 2008 #1
    find the numeral value of this
    Note:the numbers in brackets are degree
  2. jcsd
  3. Jun 6, 2008 #2

    matt grime

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    Is this posted as a challenge for others, or do you want to try to solve it for yourself with some help?
  4. Jun 6, 2008 #3
    challenge of course!!!!!!!!!!!!!!!!!
  5. Jun 6, 2008 #4
    haha, this is a pretty cool little problem.
  6. Jun 6, 2008 #5
    Pair products of sines whose degrees add up to 90 i.e.
    Sin(1)*Sin(2)*Sin(3)*...*Sin(89)=sin(1)*sin(89) * sin(2)*sin(88) * .... * sin(44)*sin(46) *sin(45)

    Use the trigonometric product to sum formula, and cos(90)=0 to obtain

    2^(-44) * cos(88)*cos(86)*cos(84)* ... *cos(2) * sin(45) = 2^(-44.5) * cos(88)*cos(2) * cos(86)* cos(4) * ... cos(46)*cos(44)

    I guess that successive application of the product to sum forumula can now be used, but am too lazy to investigate further.

    Numerical calculation gives an answer of about ~ 2^(-85.75) if it's any help.
  7. Jun 6, 2008 #6
    It's not quite so simple as successive applications of the exact same trick. If you went one further, then you would get pairings that don't quite add up to 90.

    86 and 2; 82 and 6; 78 and 10, etc all add to 88
  8. Jun 6, 2008 #7
    Yes, the next step could be

    2^(-66.5) * sin(4) * sin(8) * sin(12) ... * sin(84) * sin(88)

    And then....?

  9. Jun 7, 2008 #8
    this problem does not need a calculator
    just some formulas and a creative brain
  10. Jun 7, 2008 #9
    But the answer is 0?
  11. Jun 7, 2008 #10

    matt grime

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    The answer is 0 if and only if one of the terms of the product is zero. sin is not zero for any of the angles 1,2,3,..,89 degrees.
  12. Jun 7, 2008 #11
  13. Jun 8, 2008 #12

    Gib Z

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    For some reason I swear there is a nifty way to simplify [tex] \left( x - \frac{1}{x} \right)\left( x^2 - \frac{1}{x^2} \right)\left( x^3 - \frac{1}{x^3} \right).....\left( x^{89} - \frac{1}{x^{89}} \right)[/tex] but I can't see it right now. If so, then this problem is easily solved.
  14. Jun 8, 2008 #13
    That would be completely ignoring the possibly useful fact that cos(90) = 0. It is possible there is a nifty simplification for that product, but it would provide a much more general solution than needed.
  15. Jun 12, 2008 #14


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    I got the same 22 term simplification as Kittel Knight. Has anyone found a better solution? Does someone want to post a clue as to the form of the answer we can expect?
  16. Jun 12, 2008 #15


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    Look at the link yenchin posted. Another way to obtain the result in there is to consider the roots of the equation x^n-1=0, i.e. the nth roots of unity given by 1, z, z^2, ..., z^(n-1), where z=e^(2pi*i/n). This leads us to the equation 1+x+...+x^(n-1)=(x-z)(x-z^2)...(x-z^(n-1)). Plug in x=1 and take absolute values. Finally, try to find a nice expression for |1-z^k|. What do you get?
  17. Jun 13, 2008 #16


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    Ok thanks morphism, somehow I missed that link before - very interesting result.

    Anyway it follows immediately from that result that the sine product posed in this thread can be simplified to [itex]\sqrt{179/2} \, \times \, 2^{-89} [/itex]

    Last edited: Jun 13, 2008
  18. Jun 14, 2008 #17


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    Ok let me get it right this time.

    [itex]\sqrt{90} \, \times \, 2^{-89} [/itex]
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