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Sin^n(theta) as product of sin^1, cos^1 (m*theta)

  1. Nov 27, 2009 #1
    Hello,

    Somewhat urgent question, I would normally try and do this myself but I have a feeling it will take a while, and I sort of need to be working through this pretty quickly, so any help much appreciated. Plus I might end up wasting half a day trying formulas on this.

    I am expanding a function f(x + a sin(t)) as a Taylor series, so I get:

    f(x + a sin (t)) = f(x) + (a sin t)f'(x) + (a sin t)^2/2! *f''(x) + ...

    Now I am trying to decompose this into some sort of Fourier series (not exactly clear on final goal or method yet), so I need this in terms of:

    f(x + a sin (t)) = [ ... ] + [ ... ]*sin(t) + [ ... ]*sin(2t) + ... + [ ... ]*cos(t) + [ ... ]*cos(2t) + ...

    I am wondering if there is any simple way to do this. I have worked out the first one, eg:
    sin^2(t) = 1/2 - 1/2 *cos(t), and was wondering if there are any simple recursion formulas or anything out there to help me find an expression for sin^n(t) in terms of single powers of sines and cosines of multiples of t?

    Thanks for any help,
    Mike
     
  2. jcsd
  3. Nov 27, 2009 #2
    Ok, a little progress has been made:

    I am now working with cos^n (t) as that can be put into only cosine terms. I have also found the formula for this in terms of cos(t), cos(2t), etc. but it is not so helpful for finding what my bracket terms [ ... ] are, since there are many overlaps, as the expansion of f is a series of increasing powers of cos(t) each with a different weighting by a.

    For example, the coefficient of cos(t) would be: [a*f'(x) + (a^3)*f''(x)/2 + ... ]
     
  4. Nov 27, 2009 #3
    First of all note that the powers of trig function can be found from
    [tex]\cos^n(x)=\Re\left(e^{ix}\left(\frac{e^{ix}+e^{-ix}}{2}\right)^{n-1}\right)[/tex]
    and similarly for sine.
    I have to leave now, but I can think about the full problem later.
     
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