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Homework Help: Sum to Infinity of a Geometric Series

  1. Aug 17, 2011 #1
    1. The problem statement, all variables and given/known data

    Q.: A geometric series has first term 1 and common ratio [itex]\frac{1}{2}[/itex]sin2[itex]\theta[/itex]. Find the sum of the first 10 terms when [itex]\theta[/itex] = [itex]\frac{\pi}{4}[/itex], giving your answer in the form h - [itex]\frac{1}{2^k}[/itex], where h, k [itex]\in[/itex] N.

    2. Relevant equations

    Sn = [itex]\frac{a(1 - r^n)}{1 - r}[/itex], when IrI < 1

    3. The attempt at a solution

    S10 = [itex]\frac{a(1 - r^n)}{1 - r}[/itex]

    S10 = [itex]\frac{1(1 - (1/2)sin2(\pi/ 4) to the power of 10}{1 - (1/2)sin2(\pi/4)}[/itex] ... The tags did not take, for some reason, hence the 'to the power of 10'. Sorry about that.

    I'm stuck here because if I solve for [itex]\frac{1}{2}[/itex]sin2[itex]\theta[/itex], I get a decimal answer = 0.013706. The question wants the answer defined as whole numbers but I'm unable to work out the next step in the sequence. Can someone help and give me a tip on what to do next? Thanks.

    Ans.: From text book: 2 - (1/ 29)
  2. jcsd
  3. Aug 17, 2011 #2
    How did you get that?


    has a nice answer, no?
  4. Aug 17, 2011 #3
    Ok, so the sum is now...

    [itex]\frac{1 - (1/2)sin(\pi/ 2)tothepowerof10}{1 - (1/2)sin(\pi/ 2)}[/itex]


    1 - (1/2)sin([itex]\pi[/itex]/ 2)10 = 1 - (1/2)sin(\pi/ 2)
    1 - 1 - (1/2)sin([itex]\pi[/itex]/ 2)10 + (1/2)sin(\pi/ 2)
    - (1/2)sin([itex]\pi[/itex]/ 2)9

    Ok, obviously I'm wrong but this is what my attempt looks like. Thanks for the help so far but can you help me out a little bit more...please? Thanks.
  5. Aug 17, 2011 #4
    What is [itex]\sin(\pi /2)[/itex]?? Isn't that a nice value?
  6. Aug 17, 2011 #5
    Err...In a right angled triangle, Sin( [itex]\theta[/itex] ) = Opposite/ Hypotenuse...?
    Last edited: Aug 17, 2011
  7. Aug 17, 2011 #6
    Go check your trigonometric formula's...

    You should really know things like [itex]\sin(0)[/itex] and [itex]\sin(\pi/2)[/itex] by heart.
  8. Aug 17, 2011 #7
    Ok. Sorry about all this. It's literally been years since I last studied maths and I'm having to begin from scratch all over again. You'll have to be patient with this old man. I'll see what I can find out from your advice and show my work tomorrow. Good night.
  9. Aug 17, 2011 #8
    It's alright. This might be helpful: http://en.wikipedia.org/wiki/Trigonometric_identities
  10. Aug 18, 2011 #9
    Ok, so here is, hopefully, my final attempt...

    S10 = [itex]\frac{1(1 - (1/2)sin2(\pi/ 4)tothepowerof10)}{1 - (1/2)sin2(\pi/ 4)}[/itex]

    S10 = [itex]\frac{1 - (1/2)sin(2\pi/ 4)tothepowerof10}{1 - (1/2)sin(2\pi/ 4)}[/itex]

    S10 = [itex]\frac{1 - (1/2)sin(\pi/ 2)tothepowerof10}{1 - (1/2)sin(\pi/ 2)}[/itex]

    S10 = [itex]\frac{1 - (1/2)(\sqrt{4}/2)tothepowerof10}{1 - (1/2)(\sqrt{4}/2)}[/itex]

    S10 = [itex]\frac{1 - (2/ 4)tothepowerof10}{1 - (2/ 4)}[/itex]

    S10 = [itex]\frac{1 - (1/2)tothepowerof10}{(1/2)}[/itex]

    S10 = 2 - [itex]\frac{1}{2^9}[/itex]
  11. Aug 18, 2011 #10


    User Avatar
    Science Advisor

    Are you talking about [itex]sin^{10}(\pi/2)[/itex] or [itex]sin((\pi/2)^{10})[/itex]?

    [itex]sin(\pi/2)= 1[/itex], but why in the world would you write that as "[itex]\sqrt{4}/2[/itex]"?
  12. Aug 18, 2011 #11


    Staff: Mentor

    I've seen some texts present the trig functions in this way as a mnemnonic device.

    [itex]sin(0)= \sqrt{0}/2[/itex]
    [itex]sin(\pi/6)= \sqrt{1}/2[/itex]
    [itex]sin(\pi/4)= \sqrt{2}/2[/itex]
    [itex]sin(\pi/3)= \sqrt{3}/2[/itex]
    [itex]sin(\pi/2)= \sqrt{4}/2[/itex]

    The OP might have been referring to something like this, and didn't realize that [itex]\sqrt{4}/2[/itex] could and should be replaced with 1.
  13. Aug 18, 2011 #12
    Based on the answer, I solved the question as sin(([itex]\pi[/itex]/ 2)10).

    As regards, expressing sin([itex]\pi[/itex]/ 4) as [itex]\frac{\sqrt{4}}{2}[/itex]; yes, I was using the mnemonic but, as I indicated above, this is the first time I've dealt with these kinds of equations so I'll express it as 1, henceforth.

    Thanks for helping, guys. I really do appreciate it.
    Last edited: Aug 18, 2011
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