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Thread: Another Trigonometry Root Proof

  1. #1
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    I have another similar question which i am stucked half way through. I applied the same technique to it as in the other thread:

    Prove that $\displaystyle 2(\cos (\frac{4\pi}{19})+\cos (\frac{6\pi}{19})+\cos (\frac{10\pi}{19})) $ is a root of the equation $\displaystyle \sqrt{4+\sqrt{4+\sqrt{4-x}}}=x$

    Let $\displaystyle B_k=\frac{2k\pi}{19}$ where $\displaystyle 0\leq k \leq 18$

    Since $\displaystyle \sum^{18}_{k=0} \cos B_k=0$, it follows that $\displaystyle \sum^{9}_{k=1} \cos B_k=-\frac{1}{2}$ **

    $\displaystyle (\cos B_2+\cos B_3+\cos B_5)^2=\cos^2 B_2+\cos^2 B_3+ \cos^2 B_5 + 2(\cos B_2\cos B_3+\cos B_2\cos B_5+ \cos B_3\cos B_5)$

    $\displaystyle = \frac{1}{2}(\cos B_4+\cos B_6+\cos B_{10}) - (\cos B_2+\cos B_3+\cos B_5)+1$

    Let $\displaystyle \cos B_4+\cos B_6+\cos B_{10}$ be Q and $\displaystyle \cos B_2+\cos B_3+\cos B_5$ be P

    $\displaystyle P^2=\frac{1}{2}Q-P+1 $ --- 1

    This is where i am stucked, i tried to form the second equation from ** but there are 3 extra terms which give me P+Q+cos B1+ cos B7+ cosB8 and probably i will need more equations here since the roots of the weird square root equation will definitely be more than 2.
    Last edited by Ackbeet; Jun 11th 2011 at 02:22 AM.
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  2. #2
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    Quote Originally Posted by hooke View Post
    I have another similar question which i am stucked half way through. I applied the same technique to it as in the other thread:

    Prove that $\displaystyle 2(\cos (\frac{4\pi}{19})+\cos (\frac{6\pi}{19})+\cos (\frac{10\pi}{19})) $ is a root of the equation $\displaystyle \sqrt{4+\sqrt{4+\sqrt{4-x}}}=x$

    Let $\displaystyle B_k=\frac{2k\pi}{19}$ where $\displaystyle 0\leq k \leq 18$

    Since $\displaystyle \sum^{18}_{k=0} \cos B_k=0$, it follows that $\displaystyle \sum^{9}_{k=1} \cos B_k=-\frac{1}{2}$ **

    $\displaystyle (\cos B_2+\cos B_3+\cos B_5)^2=\cos^2 B_2+\cos^2 B_3+ \cos^2 B_5 + 2(\cos B_2\cos B_3+\cos B_2\cos B_5+ \cos B_3\cos B_5)$

    $\displaystyle = \frac{1}{2}(\cos B_4+\cos B_6+\cos B_{10}) - (\cos B_2+\cos B_3+\cos B_5)+1$

    Let $\displaystyle \cos B_4+\cos B_6+\cos B_{10}$ be Q and $\displaystyle \cos B_2+\cos B_3+\cos B_5$ be P

    $\displaystyle P^2=\frac{1}{2}Q-P+1 $ --- 1

    This is where i am stucked, i tried to form the second equation from ** but there are 3 extra terms which give me P+Q+cos B1+ cos B7+ cosB8 and probably i will need more equations here since the roots of the weird square root equation will definitely be more than 2.
    You are on the right lines there, but you need a third quantity $\displaystyle R = \cos B_1+ \cos B_7+ \cos B_8$ in addition to P and Q. You then have $\displaystyle P+Q+R=-\tfrac12$, and you can also show that

    $\displaystyle P^2 = 1-\tfrac12Q,\quad Q^2 = 1-\tfrac12R,\quad R^2 = 1-\tfrac12P.$

    Next, you need to know that $\displaystyle P>0,\ Q<0$ and $\displaystyle R<0$. I don't see any slick way of showing that analytically, but it's easy to check on a calculator. It follows that

    $\displaystyle P = \sqrt{1-\tfrac12Q},\quad Q = -\sqrt{1-\tfrac12R},\quad R = -\sqrt{1-\tfrac12P}.$

    Therefore $\displaystyle P = \sqrt{1+\tfrac12\sqrt{1+\tfrac12\sqrt{1-\tfrac12P}}}.$ Finally, put $\displaystyle x = 2P$ and the result follows.
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