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De Moivre identity problem

  1. Jan 16, 2014 #1
    Use de Moivre's identity to find real values of a and b in the equation below such that the equation is valid.

    [itex]cos^6(x)+sin^6(x)+a(cos^4(x)+sin^4(x))+b=0[/itex]

    Hint: Write [itex]cos(x)[/itex] & [itex]sin(x)[/itex] in terms of [itex]e^{ix}[/itex] & [itex]e^{-ix}[/itex].

    Check your values of [itex]a[/itex] and [itex]b[/itex] are valid by substituting in a value of [itex]x[/itex]. State, with explanation, two values of [itex]x[/itex]which would not have been sufficient checks on your values of [itex]a[/itex] and [itex]b[/itex].


    I've managed to obtain the following expression:

    [itex]\frac{3}{8}cos(4x)+\frac{a}{4}cos(4x)+\frac{3a}{4}+\frac{5}{8}+b=0[/itex], and I checked the model solution, and this is the expression they've got too. But then they simply state [itex]a=\frac{-3}{2}[/itex], and work out [itex]b[/itex] from there. But I don't understand how they got that value for [itex]a[/itex]. Can someone explain to me what I'm missing?
     
    Last edited: Jan 16, 2014
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  3. Jan 16, 2014 #2

    AlephZero

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    The equation [itex]\frac{3}{8}cos(4x)+\frac{a}{4}cos(4x)+\frac{3a}{4}+\frac{5}{8}+b=0[/itex] has to be true for all values of ##x##.

    So you can get two equations from it, similar to "equating coefficients" of polynomials, as in http://en.wikipedia.org/wiki/Equating_coefficients
     
  4. Jan 16, 2014 #3

    Ray Vickson

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    You have
    [tex] \left( \frac{3}{8} + \frac{a}{4} \right) \cos(4x) + \frac{5}{8} + b \equiv 0[/tex]
    where ##\equiv## means that it is an identity that holds for all ##x## (much more than just an equation!). So, the coefficient of ##\cos(4x)## must vanish.
     
  5. Jan 17, 2014 #4
    Okay, so I understand what AlephZero said. I picked 2 random values of x, obtained simultaneousness equations and solved them for a and b. I got a=-3/2 and b=1/2. But I don't understand the emboldened part. Why must the coefficient of ##cos(4x)## vanish?

    Also, for the last part of the question where it asks for 2 values of x which would not have been sufficient checks, do I just write down the 2 random values I used before to get my simultaneous equations?
     
    Last edited: Jan 17, 2014
  6. Jan 17, 2014 #5

    Ray Vickson

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    The coefficient of cos(4x) must vanish because the equation must hold for ALL values of x; If the coefficient of cos(4x) did not vanish you could find infinitely many different values for x that would make the equation false.

    Look at it this way: you could re-write the equation as
    [tex] \left( \frac{3}{8} + \frac{a}{4} \right) \cos(4x) = - \frac{5}{8} - b [/tex]
    The right-hand-side is a constant (not dependent on x), so the left-hand-side must also not depend on x. The only way you can make that happen is to set the coefficient of cos(4x) to zero.
     
  7. Jan 17, 2014 #6
    Got it. Thanks
     
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