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Thread: de Moivre question

  1. #1
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    Question de Moivre question

    I need help with this question:
    By applying de Moivre's theorem and by also expanding (cos a + i sin a)^5, express cos 5a as a polynomial in cos a

    Nothing jumps out, and for there must be an easier way that expanding the whole thing. Plus I don't fully understand the question

    Thanks
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  2. #2
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    You can say,
    $\displaystyle (cos x+i\sin x)^5=\cos 5x+i\sin 5x$
    Also, you can expand this,
    $\displaystyle a=\cos x$ and $\displaystyle b=i\sin x$
    Then,
    $\displaystyle (a+b)^5=\sum_{k=0}^5{5\choose k}a^kb^{n-k}$
    By binomal formula.

    Just so you remember,
    $\displaystyle i^2=-1$
    $\displaystyle i^3=-i$
    $\displaystyle i^4=1$
    ....
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  3. #3
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    ~ ~ ~

    Hello, chancey!

    DeMoivre's Theorem: .$\displaystyle (\cos\theta + i\sin\theta)^n\;=\;\cos(n\theta) + i\sin(n\theta) $


    By applying DeMoivre's theorem and by also expanding $\displaystyle (\cos\theta + i\sin\theta)^5$,
    express $\displaystyle \cos5\theta$ as a polynomial in $\displaystyle \cos\theta$.

    From the theorem, we have: .$\displaystyle \cos5\theta + i\sin5\theta \;= \;(\cos\theta + i\sin\theta)^5$

    Expand the right side:
    .$\displaystyle \cos5\theta + i\sin5\theta \;= \;\cos^5\!\theta + 5i\cos^4\!\sin\theta - 10\cos^3\!\theta\sin^2\!\theta$ $\displaystyle - 10i\cos^2\!\theta\sin^3\!\theta + 5\cos\theta\sin^4\!\theta + i\sin^5\!\theta$

    And we have: .$\displaystyle \cos5\theta + i\sin5\theta$
    . . $\displaystyle = \;\left(\cos^5\!\theta - 10\cos^3\!\theta\sin^2\!\theta + 5\cos\theta\sin^4\!\theta\right)$ $\displaystyle + i\left(5\cos^4\!\theta\sin\theta - 10\cos^2\!\theta\sin^3\!\theta + \sin^5\!\theta\right)$

    Equate real components: .$\displaystyle \cos5\theta\;= \;\cos^5\!\theta - 10\cos^3\!\theta\sin^2\!\theta + 5\cos\theta\sin^4\!\theta$ **

    Replace $\displaystyle \sin^2\!\theta$ with $\displaystyle 1 - \cos^2\!\theta:$
    . . $\displaystyle \cos5\theta \;= \;\cos^5\!\theta - 10\cos^3\!\theta\left(1 - \cos^2\!\theta\right) + 5\cos\theta\left(1 - \cos^2\!\theta\right)^2 $

    Therefore: .$\displaystyle \boxed{\cos5\theta \;= \;16\cos^5\!\theta - 20\cos^3\!\theta + 5\cos\!\theta}$

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    ** .If we equate the 'imaginary' components, we have:
    . . . . $\displaystyle \sin5\theta \;= \;5\cos^4\!\theta\sin\theta - 10\cos^2\!\theta\sin^3\!\theta + \sin^5\!\theta$

    . . which simplifies to: .$\displaystyle \boxed{\sin5\theta \;= \;5\sin\theta - 20\sin^3\!\theta + 16\sin^5\!\theta}$

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Another bonus . . .

    $\displaystyle \tan5\theta \;= \;\frac{\sin5\theta}{\cos5\theta} \;= \;\frac{5\cos^4\!\theta\sin\theta - 10\cos^2\!\theta\sin^3\!\theta + \sin^5\!\theta}{\cos^5\!\theta - 10\cos^3\!\theta\sin^2\!\theta + 5\cos\theta\sin^4\!\theta} $


    Divide top and bottom by $\displaystyle \cos^5\!\theta:$

    $\displaystyle \tan5\theta \;= \;\frac{\frac{5\cos^4\!\theta\sin\theta}{\cos^5\!\ theta} - \frac{10\cos^2\!\theta\sin^3\!\theta}{\cos^5\!\the ta} + \frac{\sin^5\!\theta}{\cos^5\!\theta}}{\frac{\cos^ 5\!\theta}{\cos^5\!\theta} - \frac{10\cos^3\!\theta\sin^2\!\theta}{\cos^5\!\the ta} + \frac{5\cos\theta\sin^4\!\theta}{\cos^5\!\theta} } $ .$\displaystyle = \;\frac{5\frac{\sin\theta}{\cos\theta} - 10\frac{\sin^3\!\theta}{\cos^3\!\theta} + \frac{\sin^5\!\theta}{\cos^5\!\theta}}{1 - 10\frac{\sin^2\!\theta}{\cos^2\!\theta} + 5\frac{\sin^4\!\theta}{\cos^4\!\theta} }$


    Therefore: .$\displaystyle \boxed{\tan5\theta \;= \;\frac{5\tan\theta - 10\tan^3\!\theta + \tan^5\!\theta}{1 - 10\tan^2\!\theta + 5\tan^4\!\theta}} $

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