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Math Help - Equation

  1. #1
    Junior Member
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    Equation

    Solve \cos^2x+\cos^22x+\cos^23x=1
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  2. #2
    Senior Member
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    Cos(n*x)

    First, you need to express \cos(2x) and \cos(3x) in terms of \cos x. ( \cos (nx) is for positive integer n always an nth degree polynomial of \cos x; they're called the Chebyshev_polynomials)
    For \cos(2x) you should have the double angle identity \cos(2x) = 2\cos^2 x - 1.
    For \cos(3x), use the cosine sum formula with 3x=2x+x to get:
    \cos(3x)=\cos(2x+x)
    =\cos(2x)\cos x - \sin(2x)\sin x
    =(2\cos^2 x - 1)\cos x - 2\sin^2 x \cos x (from the double angle identities)
    =2\cos^3 x - \cos x - 2(1-\cos^2 x) \cos x
    =4\cos^3 x - 3\cos x

    Thus:
    \cos^2 x + \cos^2 2x + \cos^2 3x = 1
    \cos^2 x + (2\cos^2 x - 1)^2 + (4\cos^3 x - 3\cos x)^2 = 1
    \cos^2 x + 4\cos^4 x - 4\cos^2 x + 1 + 16\cos^6 x - 24\cos^4 x + 9\cos^2 x= 1
    16\cos^6 x - 20\cos^4 x + 6\cos^2 x= 0 (simplifying, and subtracting one from both sides)
    8\cos^6 x - 10\cos^4 x + 3\cos^2 x= 0 (dividing by two)

    This is an easily factorable polynomial, and should give you three equations of the form cos^2 x = a for some a, which you can then solve.

    --Kevin C.
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Complex numbers can also solve the equation.

    \cos x = \frac{{e^{ix} + e^{ - ix} }}<br />
{2}.

    This one becomes nicer.
    Last edited by Krizalid; December 13th 2007 at 09:02 AM.
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