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

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$
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?
 A: This answer makes use of the Small Angle Approximation for $\sin$ and radian measures for angles. The latter was not known to the Greeks. I'm unsure if the small angle approximation was known to them in some way. 
Our approximation method has the advantage that it can very easily be done without a calculator, though the error term would not have been known until the classical period of Indian mathematics and astronomy
We seek an $\alpha$ such that $\cos(\alpha) = -\frac{1}{4}$. A quick investigation shows us that $\alpha$ lies in the second quadrant (i.e. $\frac{\pi}{2} < \alpha < \pi$). Let $\alpha = \frac{\pi}{2} + \beta$. Let's use the identity $\cos(\frac{\pi}{2} + \beta) = -\sin(\beta)$.
We then seek a $\beta$ such that $\sin(\beta) = \frac{1}{4}$. Making use of the linear approximation for $\sin$, we find $\beta \approx \frac{1}{4}$. Thus $\alpha \approx \frac{\pi}{2} + \frac{1}{4}$.
Converting to degrees gives
$\alpha \approx 90^{\circ} + \frac{45}{\pi}^{\circ}$. We can use the approximation $\pi \approx \frac{22}{7}$ and long division to get
$$\frac{45}{\pi} \approx \frac{315}{22} = 14.3\overline{18}$$
Giving us the approximation
$$\alpha \approx 104.32^{\circ}$$
This is not far from the true value
$$\alpha = 104.4775\ldots^{\circ}$$
A: 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 $0 \le k < n$.
A: If I were an ancient Greek, I would carefully draw a large circle $K$ in the sand with a fixed stake at the center $O$ connected to a fixed length of string with a marker at the other end. I would make it as large as I needed it to be. Perhaps a few cubits in diameter would be enough for my purposes. I would draw a diameter of the circle $K$ from a fixed point $A$ through $O$ to the opposite side, and, using bisection twice, mark the point $B$ one quarter of the way from $O$ to the opposite side. I now use a simple geometric construction with the fixed length of string to find two points that determine the line perpendicular to the diameter and passing through point $B$. I now draw that line and find where it intersects the circle $K$ at a point $\Gamma$. 
I am now interested in the arc of the circle $K$ from point $A$ to point $\Gamma$ and the central angle $\Theta$ subtended by this arc from the center $O$. The measure of the distance from $A$ to $\Gamma$, which is the chord corresponding to the arc and central angle, is one possible measure of the angle $\Theta$ I am intereseted in. Alternatively, depending on my need for $\Theta$, I could lay a length of string along the arc of the circle $K$ from point $A$ to point $\Gamma$ and measure its length. Now using 360 degrees for the length of the circumference of the circle $K$, I can convert the arc length to degrees and minutes of arc for $\Theta$. This practical procedure would work as well today as it would in those ancient times.
Admittedly, this is a solution by geometric construction and measuring, and not by calculating with numbers (except perhaps for converting to degree measure), but for all practical purposes, and also considering the technology of those times, this method has a few advantages. Measuring was just a relatively simple and well known procedure. Doing all but simple calculations with Greek numerals and operations like taking square roots was harder and much less well known. Alternatively, If I were somebody like Archimedes, I might use other methods like his approximate calculation of $\pi$.
A: If I were a disciple of Archimedes, having learnt his Exhaustion Method and his computation of $\pi$, I would have probably
approached the problem with that technique, thus arriving to use the same
algorithm developed by the chinese mathematician Liu Hui some
5 centuries later.

Let's start with a general scheme:
given the point $P_0$ on the unit circle, we know one of its coordinates (e.g. $P_x$) and want to determine
the angle $\beta$ defined with the $x$ axis.
Of course, at the times of Archimedes we would have spoken about the triangle $O-P_0-P_x$, etc., but let's use now a more 
agile Cartesian representation.
At that time I was supposed to know also the Pythagorean theorem and so I would have no difficulty in 
computing 
$$ \bbox[lightyellow] {  
\eqalign{
  & y_{\,0}  = \overline {OP_{\,y} }  = \sqrt {1 - \overline {OP_{\,x} } ^{\,2} }  = \sqrt {1 - x_{\,0} ^{\,2} }   \cr 
  & Q_{\,0}  = \left( {1,\;y_{\,0} /x_{\,0} } \right)  \cr 
  & h_{\,0}  = y_{\,0} /x_{\,0}   \cr 
  & s_{\,0}  = \overline {UP_{\,0} }  = \sqrt {\left( {1 - x_{\,0} } \right)^{\,2}  + y_{\,0} ^{\,2} }   \cr 
  & M_{\,1}  = \left( {\left( {1 + x_{\,0} } \right)/2,\;y_{\,0} /2} \right)  \cr 
  & m_{\,1}  = \overline {OM_{\,1} }  = {1 \over 2}\sqrt {\left( {1 + x_{\,0} } \right)^{\,2}  + y_{\,0} ^{\,2} }   \cr 
  &  \cr} 
}$$
then apply proportions to get
$$ \bbox[lightyellow] {  
{1 \over {m_{\,1} }} = {{x_{\,1} } \over {\left( {1 + x_{\,0} } \right)/2}} = {{y_{\,1} } \over {y_{\,0} /2}}
}$$
i.e.
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  x_{\,1}  = \left( {1 + x_{\,0} } \right)\;/\;\sqrt {\left( {1 + x_{\,0} } \right)^{\,2}  + y_{\,0} ^{\,2} }  \hfill \cr 
  y_{\,1}  = y_{\,0} \;/\;\sqrt {\left( {1 + x_{\,0} } \right)^{\,2}  + y_{\,0} ^{\,2} }  \hfill \cr 
  h_{\,1}  = y_{\,1} /x_{\,1}  = y_{\,0} /\left( {1 + x_{\,0} } \right) \hfill \cr 
  s_{\,1}  = \overline {UP_{\,1} }  = \sqrt {\left( {1 - x_{\,1} } \right)^{\,2}  + y_{\,1} ^{\,2} }  \hfill \cr}  \right.
}$$
Then I can iterate the recurrence until obtaining
that the difference between $y_n$ and $h_n$ is
acceptably small (also considering the number of significant digits that my avatar could manage).
From my master I know infact that the length of the arc
is comprised among the sides of the inscribed and circumscribed polygon, so
I know that
$$ \bbox[lightyellow] {  
y_{\,n}  < s_{\,n}  < {\beta  \over {2^{\,n} }} < h_{\,n} 
}$$
In the example you proposed we get, just after $4$ iterations
$$ \bbox[lightyellow] {  
y_4=0.082289  \quad   s_4=0.082359 \quad  h_4= 0.082569
}$$
that is
$$ \bbox[lightyellow] {  
1.31774\approx 75.5^\circ \; <\;  \beta \; < \; 1.3211\approx 75.7^\circ
}$$
and actually it is $\beta= \arccos(1/4) \approx 1.318 \approx 75.52^\circ$
Concerning the comment from Fractional Inquirer, note that
the recursion above can be reverted to give
$$ \bbox[lightyellow] {  
\eqalign{
  & x_{\,n}  = \left( {1 + x_{\,n - 1} } \right)\;/\;\sqrt {\left( {1 + x_{\,n - 1} } \right)^{\,2}  + 1 - x_{\,n - 1} ^{\,2} }   \cr 
  & \quad \quad  \Downarrow   \cr 
  & 2\left( {1 + x_{\,n - 1} } \right)x_{\,n} ^{\,2}  = \left( {1 + x_{\,n - 1} } \right)^{\,2}   \cr 
  & \quad \quad  \Downarrow   \cr 
  & x_{\,n - 1}  = 2x_{\,n} ^{\,2}  - 1\quad  \Leftrightarrow \quad \cos \left( {2\alpha } \right) = 2\cos ^{\,2} \alpha  - 1 \cr} 
}$$
and similarly for $y_n$, $h_n$, that is the Duplication formulas 
were essentially known at that time (and in fact those given for the direct recurrence ahead are just the half-angle wrt full angle formulas ).   
So, after halving the angle a sufficient number of times to assume it to be approximately equal
to the sine (tangent), I could have go back through the duplication formula to compute the
cosine (or sine or tangent) of the original angle.
That means that, as to avoid having to repeatedly take the square root, I could have proceeded as follows
$$ \bbox[lightyellow] {  
\eqalign{
  & \beta \; \to \;\beta /2^{\,n}  \approx \sin \left( {\beta /2^{\,n} } \right)  \cr 
  & \cos ^2 \left( {\beta /2^{\,n} } \right) \approx 1 - \left( {\beta /2^{\,n} } \right)^2   \cr 
  & \cos \left( {\beta /2^{\,n - 1} } \right) = 2\cos ^2 \left( {\beta /2^{\,n} } \right) - 1 = 1 - 2\left( {\beta /2^{\,n} } \right)^2   \cr 
  & \cos \left( {\beta /2^{\,n - 2} } \right) = 2\left( {1 - 2\left( {\beta /2^{\,n} } \right)^2 } \right)^2  - 1 = 2\left( {2\left( {\beta /2^{\,n} } \right)^2  - 1} \right)^2  - 1  \cr 
  & f(x) = 2x^2  - 1\quad  \to \quad \cos \beta  = f^{\left( n \right)} (\beta /2^{\,n} ) \cr} 
}$$
But
$$ \bbox[lightyellow] {  
T_{2n} (x) = 2T_n ^2 (x) - 1
}$$
is one of the relations obeyed by the Chebishev Polynomials of  1st kind,
so that the iteration in the recurrence above, interestingly, translates into
$$ \bbox[lightyellow] {  
\cos \beta  = T_{2^{\,n} } \left( {\beta /2^{\,n} } \right)
}$$
A: 
Now using a 'linear-crunch' interpolation method with the 'Hipparchus Trig
  Table Simulation'. Also - how about just interpolating on your favorite $30°$ right triangle!

The ancient Greeks did not use the concept of a function, nor did they use Cartesian Coordinates for the Euclidean Plane, but that did not stop them from making great advances in science.
Hipparchus was a Greek astronomer, and is now called the "the father of trigonometry"; he was the first to tabulate the corresponding values of arc and chord for a series of angles. Searching the web, we can describe his contribution with this high-level (modern) summary:

Hipparchus created trigonometry tables by inscribing a 48 sided
  regular polygon into the unit circle, allowing scientists to use
  linear interpolation to approximate angles.


The 48-gon has central angles of $7.5°$ ($\frac{30°}{4}$) with $1,080$ diagonals (chords); see Wolfram. 
So, given the OP's question Hipparchus would look as the data in his table to find the corresponding diagonals and then approximate the angle.
His works/tables have been lost so we can't use them here.

I thought it would be fun to create a 48-gon '$\mathbb R \times \mathbb R$ coordinate table' and to solve the OP's problem using linear interpolation. I used google sheets to create a (really accurate) table and looked at the data to find that the angle $\alpha$ was between 97.5° and 105°. I approximated the angle using "lazy $x$ only" interpolation and got $\alpha \approx 104.4844°$.
Here is a screenshot:


Here we interpolated using just the x-coordinate. This approach gave a pretty good approximation since the 'meat of the movement' was in the $x$ coordinate. 
The coordinate $(-.25, \frac{\sqrt{15}}{4})$ lies between two points of the 48-gon and you can calculate the lengths of the two corresponding 'division' segments. In the 'linear $x$' interpolation, the 'interpolation factor' was $0.9312585$, but when you combine/crunch both the $x$ and $y$ coordinates (see below) you get a factor of $0.9302951$, and that gives you a very precise approximation of
$\alpha \approx 104.4772°$
compared to the true value of $\alpha = 104.4775\ldots °$.
If you want to see the (new and improved) spreadsheet interpolation formulas in action, click on this; here is the linear-crunch interpolation math:
Let $C_0$, $C_1$ be adjacent vertices of the 48-gon inscribed in the unit circle of $\mathbb R^2$, with corresponding angles of $\alpha_0$ and $\alpha_1$ with $\alpha_1 - \alpha_0 = 7.5°$ and let $(x,y)$ be on the unit circle and between $C_0$ and $C_1$.
Let $S_0$ be the length of the chord line segment joining $C_0$ to $(x,y)$.
Let $S_1$ be the length of the chord line segment joining $C_1$ to $(x,y)$.
Let $S = S_0 + S_1$
Let $R_0 = \frac{S_0}{S}$
Let $R_1 = \frac{S_1}{S}$
Then approximate the angle $\alpha$ for $(x,y)$ with this interpolation:
$\alpha \approx R_1 \, \alpha_0 + R_0 \, \alpha_1$
It should not be surprising that to calculate $arccos$ ($x$ coordinate) using a trig table, better results are obtained by looking for $arcsin$ ($y$ coordinate) at the same time, using circle geometry.
Notice that this 'linear crunch' interpolation gives the exact answer when $S_0 = S_1$. But the angle we are looking for is around $15°$ away from the axis - or around $1/2$ of $30°$. So we apply this general math to the first quadrant:
$C_0 = (1,0) \text{ marking } 0°$
$C_1 = (\frac{\sqrt 3}{2},\frac{1}{2}) \text{ marking } 30°$
$(x,y) =  (\frac{\sqrt15}{4},\frac{1}{4})$
Here are the calcs:
$S_0 = 0.2520086$
$S_1 = 0.2700908$
$S = 0.5220993$
$R_0 = 0.4826832$
$R_1 = 0.5173168$
$\alpha \approx R_1 \, 0° + R_0 \, 30° = 14.4804962°$
I wonder if ancient Greek astronomers, like Ptolemy, used similar linear-crunch  interpolation techniques, calculating square roots in their angle approximation algorithm. Most likely they found 'good enough' approximations to match with their observations and make predictions.

A: I have an idea. it is known that the ancient Greeks were able to calculate the bisect an angle with compass and straightedge or ruler.
Through this method you can find an optical approximation to Punt $-\frac{1}{4}$ by interpolating:


In this way you can easily calculate by hand a good approximation to the angle $\beta$
\begin{align}\beta&\approx 90\left[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} - \frac{1}{16} + \frac{1}{32} - \frac{1}{64}\right]\\
&= 90\left[\left(\frac12+\frac14+\frac18 +\frac1{32}\right)-\left(\frac1{16} + \frac1{64}\right)\right]\\
&=90\left[\frac{29}{32} - \frac{5}{64} \right]\\
&=90\left[\frac{53}{64}\right] \\
&=\quad\frac{4770}{64}\\
&=\quad 74,53^°
\end{align}

$$\alpha \approx (180^° - 74,53^°) = 105.46^°$$


I do not think that this simple method has been unknown to the Greeks.
A: Assuming we know the addition formulas
$$\sin(x\pm y) = \sin(x)\cos(y)\pm \sin(y)\cos(x) $$
$$\cos(x\pm y) = \cos(x)\cos(y) \mp \sin(x)\sin(y)$$
we are able to also calculate $\sin(\frac{x}{2})$ and $\sin(nx)$ for $\in \mathbb N$ assuming that $\sin(x)$ is known. (Same with $\cos$) 
Consequently we can find an approximate solution to $\cos(\phi) = a$ for any $a\in[-1,+1]$ by choosing some big $N$ and then finding $n\in \mathbb N$ such that:
$$ \cos\bigg(\frac{n \pi}{2^N}\bigg) \approx a$$
A: HINT.- Getting into the skin of an ancient Greek geometry (no Archimedes of course), I would proceed as follows:
1) Construct the angle $\beta$ whose cosinus is equal to $\dfrac14$ (clearly with a right triangle having $1$ and $\sqrt{15}$ as legs and $4$ as hypotenuse).The searched angle $\alpha$ obviously satisfies $$\alpha=\dfrac{\pi}{2}+\beta$$
2) I notice that $\cos15^{\circ}=\sqrt{\dfrac{1-\cos 30^{\circ}}{2}}=\dfrac{\sqrt{2-\sqrt3}}{2}\approx0.2588190\approx\dfrac14$
Since I am not Archimede by hypothesis, I take for $\beta$ what seems to me a good approximation so$$\beta\approx\frac{\pi}{12}\Rightarrow \alpha\approx \frac{\pi}{2}+\frac{\pi}{12}=\frac{7\pi}{12}$$
$$\color{red}{\alpha\approx\frac{7\pi}{12}}$$
A: We (as did the Greeks too) can use trig to express the required angles as $ \alpha_1=(\pi-\tan^{-1} 4) ,\,\alpha_2 = (\pi+\tan^{-1} 4),$ when terminator radius vectors lie in quadrants $ 2,3$.
When a chord perpendicular to a diameter is cut inside an eccentric circle radius $4$, the  geometric mean of unequal segments so formed is the length of bisected segment = $\sqrt{15} $ for unequal segments $3,5$ in the present case. Such circle construction is well known to Greeks, and is one of Euclid's theorems. That circle not drawn as the sides $(\sqrt 15 ,1,4) $ also make a more obvious Pythagorean triplet.

A: The equation of the unity circle is $$x^2+y^2=1$$
then you have a function $f(x)=\sqrt{1-x^2}$ defined on $[-1/4,1]$ the length of a curve by defintion is $$ \int_{-1/4}^1\sqrt{1+(f'(x))^2}dx$$ 
We have $f'(x)=\frac{-x}{\sqrt{1-x^2}}$, then $$\alpha=\int_{-1/4}^1\sqrt{1+\frac{x^2}{1-x^2}}dx$$
After simplification: $$\alpha=\int_{-1/4}^1\frac{1}{\sqrt{1-x^2}}dx=\arcsin(1)-\arcsin(-1/4)$$
Hence $$\alpha=\frac{\pi}{2}-\arcsin(-1/4)$$
The problematic consists in how to approximate $\arcsin(-1/4)$;
we can use a Taylor series: decomposition of $\arcsin(x)$, one has $$\arcsin(x)= x+ 1/2 (x^3/3) + (1/2)(3/4)(x^5/5) + (1/2)(3/4)(5/6)(x^7/7) ...$$
$$\arcsin(-1/4)= -1/4+ 1/2 ((-1/4)^3/3) + (1/2)(3/4)((-1/4)^5/5) + (1/2)(3/4)(5/6)((-1/4)^7/7)...=   -    0.252$$
Then $$\alpha=  1.822 rad$$
A: In this answer we want to use Ptolemy's table of chords. The next section deals with the theory and the last section answers the OP's question.

First some references,
History of trigonometry / Classical antiquity
Chord (geometry) / In trigonometry
Ptolemy's table of chords
Aristarchus's inequality
Chord Tables of Hipparchus and Ptolemy
On a circle of radius $r$ we have
$\tag 1 \text{sin}(\frac{\theta}{2}) = \frac{\text{chord}(\theta)}{2r}$
and Ptolemy calculated his circle chord lengths with high precision ranging from $0°$ to $180°$ in increments of ${\frac{1}{2}}°$. In any case, up to a multiplicative factor, he had all the values for $\text{sin}(\theta)$ with $\theta$ ranging from $0°$ to $90°$ in increments of ${\frac{1}{4}}°$.
We will frame Ptolemy's calculations in terms of the sine function.
By accounting for errors and using formulas for adding/subtracting angles and halving angles, in principle Ptolemy could calculate the table entries to any desired degree of precision. He had some 'rock solid' values like $\text{sin}(30°)$ and other values not requiring taking multiple formula steps.
For example, he had a 'one step' formula to calculate $\text{sin}(18°)$. He could then use another step to get the sine of $30° - 18° = 12°$. He can then proceed using half-angle formulas to get values for $6°$, $3°$, $1.5°$ and $.75°$. To get the lowest unit of $.25°$ Ptolemy used Aristarchus's inequality to get estimates. He was now ready to fill in all the table entries.
Since there are many 'systems of equations' (ways of getting values using identities) that the table must satisfy, if necessary Ptolemy could adjust the value of $\text{sin}(0.25°)$ or any other table entry to get a better fit and compensate for accuracy 'drift' when engaging in multi-step calculations.
His table was Chapter 11 in Book 1 of his treatise Almagest.

For the OP's question, 
Since $\text{arccos}(-\frac{1}{4}) = 90° + \text{arcsin}(\frac{1}{4})$ the answer can be found in Ptolemy's 'sine table'.
Assume that all entries in the table are in decimal rounded to $5$ places of precision.
To get $\text{arcsin}(\frac{1}{4})$ you look at the column values of the table and work backwards, finding this
$\quad \text{sin}(14.50°) = 0.25038$
$\quad \text{sin}(14.25°) = 0.24615$
Using interpolation,
$\quad \displaystyle \text{arcsin}(\frac{1}{4}) \approx 14.25+(\frac{0.25-0.24615}{0.25038-0.24615})*(0.25) = 14.47754137°$
Since all calculation have been rounded to $5$ decimal places, Ptolemy gives an answer of
$\quad 104.47754°$
to the OP's question, but not completely trusting the last digit.
Note that the exact answer rounded to five places is
$\quad 14.47751°$.
