# Thread: Another Trigonometry Root Proof

1. 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.

2. Originally Posted by hooke
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.