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Math Help - De Moivre's theorem

  1. #1
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    De Moivre's theorem

    By comparing expressions for (\cos\theta+i\sin\theta)^5 given by De Moivre's theorem and by the binomial theorem, prove that
    \cos 5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta.
    By considering the equation \cos 5\theta=0, prove that
    \cos (\frac{\pi}{10}).\cos (\frac{3\pi}{10})=\frac{1}{4}\sqrt{5}

    My problem is in the second part.
    \cos 5\theta=0 means that 16\cos^5\theta-20\cos^3\theta+5\cos\theta=0, but after this I don't see how \cos (\frac{\pi}{10}).\cos (\frac{3\pi}{10})=\frac{1}{2}\sqrt{5} relates to it.
    Thanks!
    Last edited by arze; September 30th 2010 at 05:31 AM. Reason: typo
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  2. #2
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    5(\pi/10)= \pi/2 and 5(3\pi/10)= 3\pi/2, both of which have cosine 0 so both cos(\pi/10) and cos(3\pi/10) satisfy that equation.
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  3. #3
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    I think you sovle equation and get your 5 roots:
    which are:
    cos (2k(pi)), where k = 0,1,2,3,4,

    Then you manipulate it trigonoemtrically. But i cant see how currently
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  4. #4
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    \displaystyle \cos\frac{\pi}{10}\cdot\cos\frac{3\pi}{10} = \frac{1}{2}\cos{\frac{\pi}{5}+\frac{1}{2}\cos{\fra  c{2\pi}{5} = \frac{1}{4}\sqrt{5} \ne \frac{1}{2}\sqrt{5}.

    The book needs a reality check.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    5(\pi/10)= \pi/2 and 5(3\pi/10)= 3\pi/2, both of which have cosine 0 so both cos(\pi/10) and cos(3\pi/10) satisfy that equation.
    Umm, I understand that part. My problem is with the proof. Do you mean find the roots of the equation 16\cos 5\theta-20\cos 3\theta+5\cos\theta=0?
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  6. #6
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    Quote Originally Posted by TheCoffeeMachine View Post
    \displaystyle \cos\frac{\pi}{10}\cdot\cos\frac{3\pi}{10} = \frac{1}{2}\cos{\frac{\pi}{5}+\frac{1}{2}\cos{\fra  c{2\pi}{5} = \frac{1}{4}\sqrt{5} \ne \frac{1}{2}\sqrt{5}.

    The book needs a reality check.
    sorry, my bad, its a typo
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  7. #7
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    Ok. Here we go - we have:

    \cos{5\theta} = 16\cos^5\theta-20\cos^3\theta+5\cos\theta

    \displaystyle \Rightarrow \frac{\cos{5\theta}}{\cos{\theta}} = 16\cos^4\theta-20\cos^2\theta+5.

    So \cos{5\theta} = 0 \Rightarrow 16\cos^4\theta-20\cos^2\theta+5 = 0.

    Now, \displaystyle \cos{5\theta} = 0 \Rightarrow \theta = \frac{\pi}{10}, \frac{3\pi}{10}, \frac{5\pi}{10}, \frac{7\pi}{10}, \frac{9\pi}{10}, ...

    Obviously we cannot have  \displaystyle{\theta} = \frac{5\pi}{10} because it yields \cos{\theta} = 0.

    Also \displaystyle \cos\frac{9\pi}{10} = -\cos{\frac{\pi}{10}} and \displaystyle \cos\frac{7\pi}{10} = -\cos{\frac{3\pi}{10}}.

    So [tex]\displaystyle \cos\frac{\pi}{10}[/Math], [tex]\displaystyle \cos\frac{3\pi}{10}[/Math], [tex]\displaystyle-\cos\frac{3\pi}{10}[/Math], and \displaystyle-\cos\frac{\pi}{10}

    are the roots of \displaystyle 16\cos^4\theta-20\cos^2\theta+5 = 0.

    Consider \displaystyle f(\theta) = 16\cos^4\theta-20\cos^2\theta+5 .

    By Vieta's relations if \alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}, are the roots of f(\theta), then:

    \displaystyle \alpha_{1}\alpha_{2}\alpha_{3}\alpha_{4} = \frac{5}{16}.

    But we already know that these roots are  \displaystyle \cos\frac{\pi}{10}, \cos\frac{3\pi}{10}, -\cos\frac{3\pi}{10}, -\cos\frac{\pi}{10},

    so we have [tex]\displaystyle \cos\frac{\pi}{10}\cos\frac{3\pi}{10}\cos\frac{3\p i}{10}\cos\frac{\pi}{10} = \frac{5}{16} \Rightarrow \cos^2\frac{\pi}{10}\cos^2\frac{3\pi}{10} = \frac{5}{16}[/Math].

    \displaystyle \therefore \boxed{\cos\frac{\pi}{10}\cos\frac{3\pi}{10} = \frac{1}{4}\sqrt{5}}.
    Last edited by TheCoffeeMachine; September 30th 2010 at 09:19 AM.
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