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Maths Olympiad Problem

  1. Oct 19, 2004 #1
    The following is a problem I got in a Maths Olympiad, I had to solve it without a calculator, although I couldn't solve it:

    sin 1 + sin 2 + sin 3 + ... + sin 90

    If anyone could show me how to solve this I would really appreciate it.
     
  2. jcsd
  3. Oct 19, 2004 #2

    matt grime

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    write sin(n) = sin(1)cos(n-1)+cos(1)sin(n-1), and use the fact that sin(x) = cos(90-x) and you're done.
     
  4. Oct 19, 2004 #3
    I don't know will this method work ,

    but i will try to find

    (sin 1 + sin 2 + sin 3 + ... + sin 90)(cos 1)

    = (sin 1)(cos 1) + (sin 2)(cos 1) + (sin 3)(cos 1) + ... + (sin 90)(cos 1)


    or (sin 1 + sin 2 + sin 3 + ... + sin 90)(sin 1)

    this is just a suggection , may not work. :smile:
     
  5. Oct 19, 2004 #4
    Or simply use euler's rule
    and u r done

    -- AI
     
  6. Oct 19, 2004 #5
    There is another way to do this, but you have to know how to use complex numbers, series, etc. It gives you a general formula.

    Basically, you can say that this is a sum from 1 to N of sin(n). This is hard in itself to evaluate, so you use Euler's formula where e^(i*x)=cos(x)+i*sin(x). You then use the sum from 1 to N of e^(i*n*x) which is a geometric series. The nth sum is then equal to (1-e^iN*x)/(1-e^i*x).

    If you can get this far, figure out a way to rewrite the (1-e^iN*x)/(1-e^i*x) so that it looks like a different function. If you have no idea what I'm talking about (I don't know what level math olympiad is at), I don't know method without using complex numbers/series/functions.
     
  7. Oct 19, 2004 #6

    uart

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    Are you sure they didn't ask you merely to approximate it rather than actually solve it exactly?

    I've got a feeling that they are expecting you to realize that it is a close approximation to,
    [tex]180/\pi \, \int_{0}^{\pi/2}\sin(x) \,dx = 180/\pi[/tex]

    or numerically about 57.
     
    Last edited: Oct 19, 2004
  8. Oct 19, 2004 #7

    uart

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    Actually that integral (above) is an ok approximation but not all that good because it's just an upper sum. It would be a lot closer if you averaged the upper and lower sums as in

    Integral ~= 1/2 { (sin 0 + sin 1 + ... sin 89) + (sin 1 + sin 2 + ..sin 90) }

    = 1/2 { S -1 + S} = S - 1/2

    So a better approximation to the given Sum (S) would be,

    [tex]S \simeq 180/\pi \, \int_{0}^{\pi/2}\sin(x) \,dx + 1/2 = 180/\pi +1/2[/tex]
     
  9. Oct 19, 2004 #8
    I don't really understand how you arrived at that, and how it'd help solve the problem. Can you please elaborate?
     
  10. Oct 20, 2004 #9
    All of this stuff is great, thanks guys!
    This still doesn't solve the problem without the use of a calculator ...
    except if you can calculate the sines of numbers in your head!
     
  11. Oct 20, 2004 #10

    matt grime

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    For other people who might not be aware, maths olympiad questions are multiple choice.

    Perhaps if you posted the options we might be able to help you decide which decimal answer was the most likely.

    I'd presumed that the possible answers were in exact forms.

    As for devious:

    sin(n)=sin(n-1+1) = sin(n-1)cos(1) + sin(1)cos(n-1)


    Let S be the original sum, then

    S= sin(1)(1+cos(1)+cos(2)+cos(3)+...+cos(89)) + cos(1)(sin(1)+sin(2)+..+sin(89))

    S=sin(1)S + cos(1)S - cos(1)

    since the sums of the cos's is the same as the sum of the sin's.
    from which we can find S.
     
  12. Oct 20, 2004 #11
    This was a 3rd round maths olympiad, and it wasn't multiple choice.
     
  13. Oct 20, 2004 #12

    matt grime

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    Ah, sorry. thought it was one of the first two.

    DId it explicitly ask you to work out the decimal expansion of the sum, or evaluate it? surely an exact solution involving sin(1) and cos(1) is far better than any decimal expansion? This is the Olympiad, so I'd presume it uses proper mathematical standards ie an answer of 2pi is written and not "6.28 to 2 dp", and hence that the solutions offered here which are 'exact' constitute the one that they wanted.
     
  14. Oct 20, 2004 #13
    I'm looking at this and I think that the way to handle this is the way I above mentioned.

    If you follow the steps above, you get the value (1-e^iN*x)/(1-e^i*x).

    This can then be factored into (you take e^(iN*x/2) out of the top and e^(i*x/2) out of the bottom)

    e^( iN*x/2 ) * ( e^(-iN*x/2) - e^(iN*x/2) ) / ( e^(ix/2) * ( e^(-ix/2) - e^(ix/2) )

    Recognize that e^(-iN*x/2) - e^(iN*x/2) is = -2i*sin(Nx/2) and that e^(-ix/2) - e^(ix/2) = -2i*sin(x/2)

    Also, this was a combined expansion of cos(nx)+isin(nx) so the value for sin(nx) will be the imaginary part of the above.

    You get:

    sum sin(nx)= sin(Nx/2)/sin(x/2) * sin((N-1)x/2)
    Plugging in N=90 and x=1, you get the value 56.7943 (using degrees not radians), the same as mentioned above, and you now have a general formula for all problems of this type.
     
  15. Oct 20, 2004 #14
    I don't think the value needs to evaluated ....
    if one gets sin(pi/35) as an answer , thats an answer enough .......
    (if u catch my drift...)

    -- AI
     
  16. Oct 20, 2004 #15
    I don't really get this one. :shy:
    Shouldn't it be: S= sin(1)S + cos(1)S + sin(1)?
     
  17. Oct 20, 2004 #16
    cos(1)[sin(1)+sin(2)+..+sin(89)]
    =cos(1)[sin(1)+sin(2)+..+sin(89)+ sin(90) - sin(90)]
    =cos(1)[S - sin(90)]
    =cos(1)[S - 1]
    =cos(1)S - cos(1)

    if I'm right
    :tongue2:
     
  18. Oct 20, 2004 #17
    You are. I messed up some stuff when I typed it out. :approve:
     
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