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.
 A: 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.
A: Hint
First let us give a name to the nested radicals
$$x=\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4+\sqrt{4-...}}}}}}$$
Then you can observe that
$$((x^2-4)^2-4)^2=4-x$$
A: HINT:
Let's try to show that the RHS satisfies the same equation as the LHS. 
Consider 
$$z = \cos \frac{2\pi}{19} + i \sin \frac{2\pi}{19}$$
Then $z^{19}-1 = 0$ and $z \ne 1$ so 
$$\sum_{k=0}^{18} z^k = 0$$
On the right had side you have 
$$ z^2 + z^{-2} +  z^3 + z^{-3} + z^5 + z^{-5} = z^2 + z^{17} + z^3 + z^{16} + z^5 + z^{14}$$ 
Take the polynomial equation $P(x) = 0$ obtained for the LHS and show that 
$$P(z^2 + z^{17} + z^3 + z^{16} + z^5 + z^{14})  \vdots \sum_{k=0}^{18} z^k $$
A: I can´t solve in detail this question, however I want to write a remark about because I think it might be of interest to some people (we disregard the convergence whose existence is implicit by the statement assuming to be true).
(1) Let $E$ be equal to the RHS; we have the challenging equality $$E^3+E^2-6E-7=0$$ (you can calculate the numerical value of $E$ and verify the corresponding approximation; I give below the source of this relation).
(2) The answer given by H.R. is quite “natural” but without solving the equation this appears almost as the question itself (where $E$ must be a root of the equation).
(3) The equation in (2), of 8-degree, factorizes as $$(x^2+x-4)(x^3-2x^2-3x+5)(x^3+x^2-6x-7)=0$$ so $E$ must be root of one of the two cubic factors (why?).
(4) One finds that the third factor is what suits. We add here, JUST FOR THE BEGINNERS, the trigonometrical solution of this equation with some pertinent discussion. $$x^3+x^2-6x-7=0\iff X^3+ax^2+bx+c=0 \space\text{where} \space a=1;\space b=-6;\space c=-7$$
$$p={3b-a^2\over 9}={-19\over 9}$$  $$q={9ab-27c-2a^3\over 54}=\frac{133}{54}$$  $$\Delta=p^3+q^2<0$$ There are three distinct real roots because the discriminant $\Delta$ is negative. Making $$\cos(\theta)=\frac{q}{\sqrt{-p^3}}=\frac{7\sqrt{171}}{114}$$
the solutions are given by $$\begin{cases}x_1=2\sqrt{-p}\cos (\frac{\theta}{3})-\frac a3\\ x_2=2\sqrt{-p}\cos (\frac{\theta+2\pi}{3})-\frac a3\\x_3=2\sqrt{-p}\cos (\frac{\theta+4\pi}{3})-\frac a3\end{cases}$$
These three roots are $$x_1\approx 2,5077;\space x_2\approx -1,2219;\space x_3\approx -2,2851$$ We get an approximation for the angle $\theta\approx 36,5867$ degrees so $\frac{\theta}{3}\approx 12,1955$ degrees. This calculation allows us to say that
$$\color{red}{E=x_1}$$
