An infinite nested radical problem From this link, problem 36, I found that
$$\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4+\sqrt{4-...}}}}}}=2\left(\cos{\dfrac{4\pi}{19}}+\cos{\dfrac{6\pi}{19}}+\cos{\dfrac{10\pi}{19}}\right).$$
The signs : + + - + + - + + - ... .
How to prove it?
Furthermore, how to represent $\sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt{7+2\sqrt{7-2\sqrt{7-...}}}}}}$ by trigonometric function ?
The signs : + - - + - - + - - ... .
Thanks for helping.

I. Solution
You wish to find the exact value of $x,y$ for,
$$x=\sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt{7+2\sqrt{7-2\sqrt{7-...}}}}}}\tag1$$
$$y=\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4+\sqrt{4-...}}}}}}\tag2$$
and specify that both have period $3$. So all you do is substitute at the correct point,
$$x=\sqrt{7+2\sqrt{7-2\sqrt{7-2x}}}\tag3$$
$$y=\sqrt{4+\sqrt{4+\sqrt{4-y}}}\tag4$$
One can easily form a non-radical equation from $(3)$ and $(4)$ by repeated squaring as H.R. has done in his answer. Since $2^3 = 8$, then $x,y$ are at most roots of octics. Fortunately, both octics factor, and the correct $x,y$ are roots of just cubics, given by,
$$x^3 - x^2 - 9x + 1 = 0\tag5$$
$$y^3 + y^2 - 6y - 7 = 0\tag6$$
Using the positive case of the square roots of $(1),(2)$, then the correct values are,
$$x =  4\cos \tfrac{2\pi}{7} +1 =3.4939\dots$$ 
$$y =  2\left(\cos{\tfrac{4\pi}{19}}+\cos{\tfrac{6\pi}{19}}+\cos{\tfrac{10\pi}{19}}\right) = 2.50701\dots$$
II. Some Nice Things
The three roots of $(5)$ are,
$$ 4\cos \tfrac{2\pi}{7} +1,\quad  4\cos \tfrac{4\pi}{7} +1,\quad  4\cos \tfrac{8\pi}{7} +1$$
Ramanujan found the beautiful relation,
$$\sqrt[3]{\cos\bigl(\tfrac{2\pi}7\bigr)}+
\sqrt[3]{\cos\bigl(\tfrac{4\pi}7\bigr)}+
\sqrt[3]{\cos\bigl(\tfrac{8\pi}7\bigr)}=
\sqrt[3]{\tfrac{5-3\sqrt[3]7}2}$$
However, we can generalize that. Using the three roots of $(6)$, then,
$$\sqrt[3]{1+\cos\bigl(\tfrac{4\pi}{19}\bigr)+\cos\bigl(\tfrac{6\pi}{19}\bigr)+\cos\bigl(\tfrac{10\pi}{19}\bigr)}+\\
\sqrt[3]{1+\cos\bigl(\tfrac{2\pi}{19}\bigr)+\cos\bigl(\tfrac{14\pi}{19}\bigr)+\cos\bigl(\tfrac{16\pi}{19}\bigr)}+\\
\sqrt[3]{1+\cos\bigl(\tfrac{8\pi}{19}\bigr)+\cos\bigl(\tfrac{12\pi}{19}\bigr)+\cos\bigl(\tfrac{20\pi}{19}\bigr)}=\\ \sqrt[3]{\frac{-1+3\sqrt[3]{19}}{2}}=1.51867\dots$$
Note the third multiplier is the sum of the first two. One can find relations like these for all cubics. If interested, see this post.
