How can I calculate $\alpha=\arccos\left(-\frac{1}{4}\right)$ without using a calculator? How can I calculate $\alpha$, without using a calculator?

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I know $x = -\frac{1}{4} \implies y= \frac{\sqrt{15}}{4}, $ now how can I calculate $$\arccos\left(-\frac{1}{4}\right) = \alpha,\quad \arcsin\left(\frac{\sqrt{15}}{4}\right) = (180° - \alpha),$$  without using a calculator? How did the Greeks to calculate the angle?

You can read the binary digits of $\arccos(x)/\pi$ off the signs of $2\cos(2^kx)$, which is an easy to compute sequence defined recursively with $x_{n+1} = x_n^2-2$.
More precisely, you put a $1$ digit when the product of the signs so far is negative, and  a $0$ otherwise :
$\begin{matrix}x_0 & -1/2 & - & - \\ 
x_1 &-7/4 & - & + \\ 
x_2 & 17/16 & + & + \\
x_3 & -223/256 & - & - \end{matrix}$
Now this starts getting hard because squareing $3$ digits number is a lot of hard work, so let me roughly approximate the fractions with $2$ digit numerators and denominators.
$\begin{matrix} -23/25 & & & \le x_3 \le & & & -11/13 & - & - \\ 
-11/8 & \le & -217/169 & \le x_4 \le & -721/625 & \le& -8/7 & - & + \\ 
-34/49 & \le & -34/49 & \le x_5 \le & -7/64 & \le & -7/64 & - & - \\
-2 & \le & -8143/4096 & \le x_6 \le & -3646/2401 & \le & -36/25 & - & + \\
4/63 & \le & 46/625 & \le x_7 \le & 2 & \le & 2 & + & + \\
 \end{matrix}$
And now this is too imprecise to continue.
So far I got the cumulative sign sequence $(-,+,+,-,+,-,+,+)$ and so the angle is between $(2^{-1}+2^{-4}+2^{-6})\pi$ and $(2^{-1}+2^{-4}+2^{-6}+2^{-8})\pi$
In degrees you replace $\pi$ with $180$, so those are $104.06\ldots$ and $104.77\ldots$

The recurrence follows from the addition formula : 
$2\cos(2x) = 2\cos^2(x)-2\sin^2(x) = 4\cos^2(x)-2 = (2\cos(x))^2-2$
Suppose you call $a_n \in [0 ; \pi]$ the angle whose cosine is $2x_n$.
If $x_n\ge 0$ then $a_n \in [0 ; \pi/2] $ and then $a_{n+1} = 2a_n$, so the binary digits of $a_n/\pi$ are $.0$ followed with the binary digits of $a_{n+1}/\pi$
If $x_n \le 0$ then $a_n \in [\pi/2 ; \pi]$ and then $a_{n+1} = 2\pi-2a_n$, so the binary digits of $a_n/\pi$ are $.1$ followed with the inverted binary digits of $a_{n+1}/\pi$ 
Thus $a_{n+1} = \pm 2 a_n \mod {2\pi}$, and by induction, $a_n = \pm 2^n a_0 \pmod {2\pi}$ where the sign
depends on the parity of the number of negative $x_k$ encountered for $0 \le k < n$. The $n$th digit is $0$ if and only if $2^n a_0 \in [0 ; \pi] \pmod {2\pi}$, which means $\pm a_n \in [0;\pi] \pmod {2\pi}$ with the same sign. But since $a_n \in [0;\pi]$, the digit is $0$ if the sign was $+$ and it is $1$ is the sign was $-$.
And so the $n$th binary digit correspond to the parity of the number of negative cosines encountered for $0 \le k < n$.
