(Dis)Proof of function to find the distance to the horizon. I had an assignment not too long ago that went like this:
Say you're standing on a tower of 605 meters tall (neglect your own height for the purpose of this question), how far can you see to the horizon?
In this question you can assume the earth to be a perfect sphere with a radius of $6.4 * 10^6$ meters.
I tried to solve this problem using the following method:
Let us graph the surface of earth as a cosine function with radius $6.4*10^6$, and draw a tangent line to this function at $x=0$:

We'll then have 2 functions: $g(x)=6.4*10^6$ and $f(x)=6.4*10^6 cos(x)$. We'll then be looking for the point where $g(x)-f(x)=605$, or: $6.4*10^6(1-cos(x))=605$
Some algebra:
$\frac{605}{6.4*10^6}=1-cos(x)$
Therefore $x=arccos(1-\frac{605}{6.4*10^6})$
Since $x$ is in radians, we'll be able to convert to curve length by multiplying by the radius. Or in other words:
$d=6.4*10^6 *arccos(1-\frac{605}{6.4*10^6})$ Where d is the distance one can see.
Now I've had multiple people telling me this is not true, so I've been trying to set up a formal proof or disproof for my equation but have been unable to do so. Is anyone here able to do that?
Edit: To clarify, we're talking about the arclength here. The alternate solution that gets mentioned is: $r(1-\frac{1}{cos(x)})=605$ as opposed to $r(1-cos(x))=605$
Thanks in advance!
 A: The surface of the earth is not a cosine.
It is a sphere, which would be modeled as a circle.
Start at a point 605 meters above ground.
From there, find the tangent to the sphere from that point.
The length of the tangent is the distance.
Use Pythagoras.
If you can't figure it out, I'll help you.

Here's my way.
Let $r$ be the radius of the earth
and $d$ the distance above ground.
Let $C$ be the center of the earth,
$P$ the point above ground,
and $T$ the point of tangency.
You want to find $PT$.
Since $PT$ is tangent to the earth,
$TC$ is perpendicular to $PT$.
Therefore
$CPT$ is a right triangle
with $CP$ the hypotenuse.
Therefore
$PT^2+TC^2 = PC^2$.
We know that
$TC = r$
and
$PC = r+d$.
Therefore
$PT^2
=PC^2-TC^2
=(r+d)^2-r^2
=2rd+d^2
$
so that
$PT
=\sqrt{2rd+d^2}
$.
I get
$88006.23884134578$ for this.
Since
$r$ is much larger than $d$,
we can use the approximation
valid for small $x$
of
$\sqrt{1+x}
\approx 1+x/2
$.
Using this,
$\begin{array}\\
PT
&=\sqrt{2rd+d^2}\\
&=\sqrt{2rd}\sqrt{1+d/(2r)}\\
&\approx\sqrt{2rd}(1+d/(4r))\\
&=\sqrt{2rd}+\sqrt{d^3/(8r)}\\
&\approx\sqrt{2rd}\\
\end{array}
$
I get
$88000$ for this.

Added after a comment:
To get the arc length.
The angle $DCP$ satisfies
$\tan(DCP)
=\dfrac{PT}{TC}
=\dfrac{\sqrt{2rd+d^2}}{r}
=\sqrt{2d/r+(d/r)^2}
$.
The arc length is, therefore,
since
$d/r$ is very small,
$\begin{array}\\
r\,\arctan(\sqrt{2d/r+(d/r)^2})
&\approx r\,\arctan(\sqrt{2d/r})\\
&\approx r\,\sqrt{2d/r}\\
&= \sqrt{2rd}\\
\end{array}
$
This is the same approximation
as for the straight line
since the arc is very flat.
You can also use the approximation
$\arctan(x)
\approx x-\dfrac{x^3}{3}
$
to get a more accurate answer.
