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Math Help - De Moivre's theorem.

  1. #1
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    De Moivre's theorem.

    Find tan5x in terms of tanx.

    Not sure if I'm meant to manipulate z=cosx+isinx to get z/cosx=1+itanx?
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  2. #2
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    Hello, Erghhh!

    It can be done, but I don't know if my method is what they want.


    Use DeMoivre's Theorem to find \tan5x in terms of \tan x
    DeMoivre's Theorem says: . \left(\cos\theta + i\sin\theta\right)^5 \:=\:\cos5\theta + i\sin5\theta


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

    . . \cos5x + i\sin5x \:=\:\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

    . . \cos5x + i\sin 5x \:=\:\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

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



    We have: . \tan5x \:=\:\frac{\sin5x}{\cos5x} \;=\;\frac{5\cos^4\!x\sin x - 10\cos^2\!x\sin^3\!x + \sin^5\!x} {\cos^5\!x - 10\cos^3\!x\sin^2\!x + 5\cos x\sin^4\!x}


    Divide top and bottom by \cos^5\!x\!:\;\;\tan 5x \;=\; \frac{5\,\dfrac{\sin x}{\cos x} - 10\,\dfrac{\sin^3\!x}{\cos^3\!x} + \dfrac{\sin^5\!x}{\cos^5\!x}} {1 - 10\,\dfrac{\sin^2\!x}{\cos^2\!x} + 5\,\dfrac{\sin^4\!x}{\cos^4\!x}}

    . . Therefore: . \boxed{\tan5x \;=\;\frac{5\tan x - 10\tan^3\!x + \tan^5\!x}{1 - 10\tan^2\!x + 5\cos^4\!x}}

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  3. #3
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    We never have to go through all that ever again!
    These multiple-angle formulas can be written with a rhythmic pattern.


    Suppose we want: . \sin6x,\;\cos6x,\;\tan6x


    Expand: . (\cos x + \sin x)^6
    . . \cos^6\!x + 6\cos^5\!x\sin x + 15\cos^4\!x\sin^2\!x + 20\cos^3\!x\sin^3\!x + 15\cos^2\!x\sin^4\!x +  6\cos x\sin^5\!x + \sin^6\!x


    Write these terms alternately in the denominator and numerator of a fraction.

    Start with the denominator: . \frac{\qquad6\cos^5\!x\sin x \qquad 20\cos^3\!x\sin^3\!x \qquad 6\cos x\sin^5\!x}{\cos^6\!x \qquad 15\cos^4\!x\sin^2\!x \qquad 15\cos^2\!x\sin^4\!x \qquad \sin^6\!x}


    Insert alternating signs in the numerator and in the denominator:

    . . . . . . . \frac{6\cos^5\!x\sin x \:{\color{red}-}\: 20\cos^3\!x\sin^3\!x \:{\color{red}+} \:6\cos x\sin^5\!x}{\cos^6\!x \:{\color{red}-}\: 15\cos^4\!x\sin^2\!x \:{\color{red}+}\: 15\cos^2\!x\sin^4\!x \:{\color{red}-}\:\sin^6\!x}


    And we have two of the multiple-angle identities:

    Numerator: . . \boxed{\sin 6x \;=\;6\cos^5\!\sin x - 20\cos^3\!x\sin^3\!x + 6\cos x\sin^5\!x}

    Denominator: . \boxed{\cos6x \;=\;\cos^6\!x - 15\cos^4\!x\sin^2\!x + 15\cos^2\!x\sin^4\!x - \sin^6\!x}


    For \tan6x divide top and bottom of the fraction by \cos^6\!x

    . . \boxed{\tan6x \;=\;\frac{6\tan x - 20\tan^3\!x + 6\tan^5\!x}{1 - 16\tan^2\!x + 15\tan^4\!x - \tan^6\!x}}

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