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Math Help - De Moivre's formula and Binomial theorem

  1. #1
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    Question De Moivre's formula and Binomial theorem

    Use de Moivre's formula and the Binomial theorem to derive a formula that expresses cos5x as a polynomial in cosx, or more precisely determine the constants ABC.

    This is what I have done.. is it correct or im I totally wrong?
    De Moivre's formula and Binomial theorem-picture-1.png
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  2. #2
    Senior Member I-Think's Avatar
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    The question says to use De' Moivre's Theorem and I didn't see you use it (I could be wrong), so I'll start it.

    (cosx+isinx)^5=cos5x+isin5x
    Expanding
    cos^5x+i5cos^4x{sinx}-10cos^3x{sin^2x}-i10cos^2x{sin^3x}+5cosx{sin^4x}+isin^5x
    =cos5x+isin5x

    Now compare terms with real coefficients, these terms are those that will make cos5x
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  3. #3
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    Question

    Quote Originally Posted by I-Think View Post
    The question says to use De' Moivre's Theorem and I didn't see you use it (I could be wrong), so I'll start it.

    (cosx+isinx)^5=cos5x+isin5x
    Expanding
    cos^5x+i5cos^4x{sinx}-10cos^3x{sin^2x}-i10cos^2x{sin^3x}+5cosx{sin^4x}+isin^5x
    =cos5x+isin5x

    Now compare terms with real coefficients, these terms are those that will make cos5x

    Do you mean that A = 1, B = -10, C= 5 ???
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  4. #4
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    Hello, sf1903!

    Use de Moivre's formula and the Binomial theorem to derive a formula
    that expresses \cos5x as a polynomial in \cos x

    DeMoivre's formula says: . \cos5x + i\sin5x \:=\:(\cos x + i\sin x)^5 .[1]


    Expand the right side:

    . . \cos5x + i\sin5x \:=\:(\cos x + i\sin x)^5

    . . . . =\; \cos^5\!x + 5i\cos^4\!x\sin x - 10\cos^3\!x\sin^2\!x - 10i\cos^2\!x\sin^3\!x + 5\cos x\sin^4\!x + i\sin^5\!x

    . . . . =\; \left(\cos^5\!x -10\cos^3\!x\sin^2\!x + 5\cos x\sin^4\!x\right) + i\left(5\cos^4\!x\sin x - 10\cos^2\!x\sin^3\!x + \sin^5\!x\right)


    Equate real and imaginary components:


    \cos5x \:=\: \cos^5\!x - 10\cos^3\!x\sin^2\!x + 5\cos x\sin^4\!x

    . . . . =\;\cos^5\!x - 10\cos^3\!x(1-\cos^2\!x) + 5\cos x(1-\cos^2\!x)^2

    . . . . = \;\cos^5\!x - 10\cos^3\!x + 10\cos^5\!x + 5\cos x - 10\cos^3\!x + 5\cos^5\!x

    \boxed{\cos5x \:=\:16\cos^5\!x - 20\cos^3\!x + 5\cos x}



    \sin5x \:=\:5\cos^4\!x\sin x - 10\cos^2\!x\sin^3\!x + \sin^5\!x

    . . . . =\; 5(1-\sin^2\!x)^2\sin x - 10(1-\sin^2\!x)\sin^3\!x + \sin^5\!x

    . . . . =\; 5\sin x - 10\sin^3\!x + 5\sin^5\!x - 10\sin^3\!x + 10\sin^5\!x + \sin^5\!x

    \boxed{\sin 5x \:=\:5\sin x - 20\sin^3\!x + 16\sin^5\!x}

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