Two nested radicals with similar result

Question

Sorry for my last post it's my bad .So I ask to this (the true ^^)nested radical :

$$\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{\cdots }}}}}}}}=\frac{1+\sqrt{5}+\sqrt{30-6\sqrt{5}}}{4}$$

The period is $4$ and the related equation is :

$$\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+x}}}}=x$$

There is a big similarity with this How to prove $\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$

Following the answer of Tito Piezas III can solve this nested radical (with $2$).

My question: Can someone explain these similarities between these two nested radicals ?

Any helps is highly appreciated .

Thanks a lot .

Answer 1

We can solve the equation $\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+x}}}}=x$ as follows.

Since $2+x\ge 0$, we have $x\ge -2$. Since $\sqrt{2+x}\le 2$, $x\le 2$. So we can put $x=2\cos 32\alpha$ for some $0\le\alpha\le \tfrac{\pi}{32}$. Then

$$\sqrt{2+x}=\sqrt{2+2\cos 32\alpha}=2\cos 16\alpha$$

$$\sqrt{2-2\cos 16\alpha}=2\sin 8\alpha$$

$$\sqrt{2-2\sin 8\alpha}=\sqrt{2-2\cos\left(\frac{\pi}2- 8\alpha\right)}=2\sin\left(\frac{\pi}4- 4\alpha\right)$$

$$\sqrt{2+2\sin\left(\frac{\pi}4-4\alpha\right)}=\sqrt{2+2\cos\left(\frac{\pi}4+4\alpha\right)}=2\cos\left(\frac{\pi}8+2\alpha\right)$$

So $$2\cos\left(\frac{\pi}8+2\alpha\right)=2\cos 32\alpha $$

$$\sin\left(17\alpha+\frac{\pi}{16}\right)\sin\left(15\alpha-\frac{\pi}{16}\right)=0$$

$$17\alpha+\frac{\pi}{16}=\pi n\mbox{ or }15\alpha-\frac{\pi}{16}=\pi n,\,\, n\in\Bbb Z$$

For the range $0\le\alpha\le \tfrac{\pi}{32}$ fits only $15\alpha-\frac{\pi}{16}=0$, so $\alpha=\frac{\pi}{15\cdot 16}$ and

$$x=2\cos 32\alpha=2\cos\frac{2\pi}{15}.$$

Answer 2

As a sequence to answer by above https://math.stackexchange.com/users/71850/alex-ravsky

$$2\cos(\frac{2\pi}{15}) = 2\cos24°= 2\cos(60-36)°$$ $2(\cos60°\cdot\cos36° + \sin60°\cdot\sin36°)$ = $2(\frac{1}{2}\cdot\frac{\sqrt5 +1}{4} +\frac{\sqrt3}{2}\cdot\frac{\sqrt10-2\sqrt5}{4}) $ Further simplification leads to $$\frac{1+\sqrt5+\sqrt{30-6\sqrt5}}{4}$$

Therefore value of cyclic infinite nested square roots of 2 and value of finite nested radical are same. That is $2\cos24°$