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

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

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

    (\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)

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

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

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

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

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

    (\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)

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

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

    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 R = \cos B_1+ \cos B_7+ \cos B_8 in addition to P and Q. You then have P+Q+R=-\tfrac12, and you can also show that

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

    Next, you need to know that P>0,\ Q<0 and 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

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

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