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Thread: Trig Identity Question

  1. #1 Trig Identity Question 
    Forum Junior Vroomfondel's Avatar
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    I was working through my physics book when I came upon the following:

    cos<sup>m</sup>(x) = 1/2<sup>m-1</sup>(cos(mx) + m*cos((m-2)x) + ... + B(m,n)*cos((m-2*n)x) + ... + L)

    Where B(m,n) is the binomial coefficient [m!/(n!(m-n)!)] and

    L = m!/(2((m/2)!)<sup>2</sup>) for m even

    L = m!/(2((m+1)/2)!((m-1)/2)!) for m odd

    It seems to me like L should be 0 for m odd. Also, can anyone help me out with a proof of it? I have cos(mx) expanded with the Chebyshev Polynomials, but I am having trouble explicitly expressing an inverse to find cos<sup>m</sup>(x) in terms of cos(mx),cos((m-1)x,...

    Thanks,
    Vroom


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  3. #2  
    Forum Ph.D.
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    Can you check your formula for L even as well? For m = 4, you should have

    ********** cos<sup>4</sup>(x) = 1⁄2<sup>3</sup>[cos(4x)+4cos(2x)+3]

    4!∕(2(4⁄2)!) is not equal to 3. :?


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  4. #3  
    Forum Junior Vroomfondel's Avatar
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    4!/(2(4/2)!) = 4*3*2/(2*2!*2!) = 4*3*2/(2*2*2) = 2*2*3*2/(2*2*2) = 3
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  5. #4  
    Forum Professor serpicojr's Avatar
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    Quote Originally Posted by Vroomfondel
    4!/(2(4/2)!) = 4*3*2/(2*2!*2!)
    Where did the second 2! in the denominator come from?
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  6. #5  
    Forum Junior Vroomfondel's Avatar
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    So many 2's, knew I would miss one. L = m!/(2((m/2)!)<sup>2</sup>). I was just trying to write it the way it was in the book, would have been easier if I wrote L = B(m,m/2)/2. Fixed it in my original post as well.
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  7. #6  
    Forum Professor serpicojr's Avatar
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    So I've started the details of a proof based on the following fact. Recall the addition formula for cosine:

    cos(a+b) = cos(a)cos(b)-sin(a)sin(b)

    So applying this to a = mx, b = x, we get:

    cos((m+1)x) = cos(mx)cos(x)-sin(mx)sin(x)

    Now applying this to a = mx, b = -x, we get:

    cos((m-1)x) = cos(mx)cos(-x)-sin(mx)sin(-x)

    But recall that cos(-x) = cos(x) (cosine is even) and sin(-x) = -sin(x) (sine is odd), and so:

    cos((m-1)x) = cos(mx)cos(x)+sin(mx)sin(x)

    Now add this equation to the one for cos((m+1)x), and you get:

    cos((m+1)x)+cos((m-1)x) = 2cos(mx)cos(x)

    Dividing by 2 and swapping sides, we now have a formula for the product cos(mx)cos(x) in terms of functions of the form cos(kx):

    cos(mx)cos(x) = (cos((m+1)x)+cos((m-1)x))/2

    You can use this to deduce some recursive formulas for your coefficients, and then using these properties, you should be able to show that the coefficients have to be equal to what you claim they are.

    (Note: I think in your formula for m odd, you want Lcos(x) and not just L.)
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