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    Hello

    I am well and truly stuck on a question:

    Use cos n0 = 0.5((z^n)+(z^-n)) to express cos0+cos30+cos50+...+cos(2n-1)0 (where I have used 0 to represent theta) as a geometric series in terms of z. Hence find this sum in terms of 0.

    I have this series being equal to
    0.5((z^(1-2n))+(z^(3-2n))+...+(z^(2n-3))+(z^(2n-1))
    which is equal to
    0.5(((z^(1-2n))((z^(4n))-1))/((z^(2))-1)).

    If this is correct, then how do I then write this in its simplest form? (in terms of theta)

    (P.S. Please quote me)
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    (Original post by Magu1re)
    Hello

    I am well and truly stuck on a question:

    Use cos n0 = 0.5((z^n)+(z^-n)) to express cos0+cos30+cos50+...+cos(2n-1)0 (where I have used 0 to represent theta) as a geometric series in terms of z. Hence find this sum in terms of 0.

    I have this series being equal to
    0.5((z^(1-2n))+(z^(3-2n))+...+(z^(2n-3))+(z^(2n-1))
    which is equal to
    0.5(((z^(1-2n))((z^(4n))-1))/((z^(2))-1)).

    If this is correct, then how do I then write this in its simplest form? (in terms of theta)

    (P.S. Please quote me)
    Not sure why you got negged for asking a question so I compensated for it. Anyway to the question, Your first line is fine, it's just when you sum it, you simplify it so much that it gets a bit harder to see how to convert it. Go back down your working and see if you can find somewhere where you can convert  \dfrac{1}{2} (z^{2n-1} - \dfrac{1}{z^{2n-1}}) . Also check you've summed both series z + z^3 + z^5 +...+z^(2n-1) and 1/z + 1/z^3 + 1/z^5 +...+z^(2n-1) to the correct term. To convert z^2-1 in terms of theta, it should be quite clear as z =  \cos \theta + i \sin \theta .
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    Totally did not see that this was a double posted problem!
 
 
 
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