$\sqrt{\frac{h^{2}}{4}+d^{2}-h d \cos x}-\sqrt{\frac{h^{2}}{4}+d^{2}+h d \cos x}$ equals $h\cos x$? Trying to simplify the expression, I observed: $y=\sqrt{\frac{h^{2}}{4}+d^{2}-h d \cos x}-\sqrt{\frac{h^{2}}{4}+d^{2}+h d \cos x}$ graphically equals $y=h\cos x$  when pluging in arbitrary values of $h$ and $d$. The result can be seen here.
I have tried but I haven't been able to prove it mathematically. Please help me prove it!
 A: I'm going to take $y = \sqrt{\frac{h^{2}}{4}+d^{2}+h d \cos x}-\sqrt{\frac{h^{2}}{4}+d^{2}-h d \cos x}\,\,\,\,$ and consider only positive values for $d$ and $h$ (it's identical to what you put into wolfram alpha, the $y$ in your question looks more like $- \cos$).
Now, $y$ and $\cos$ are not equal. Try adding $- \cos(x)$ in wolfram alpha and you'll see it's not identically 0.
Also your equation implies that  $y$ is independent of $d$. But if we set $x = \pi/4$, $h = 2$, we see that $y$ is not constant with respect to $d$. However we also see that it approaches a constant (even the correct one) as $d \to \infty$.
Therefore we set out to investigate the limit of $y$ as $d$ goes to infinity. I'm going to do a geometrical sketch in hopes that someone else can fill in the gaps in rigour.
Set $a = \sqrt{\frac{h^{2}}{4}+d^{2}+h d \cos x}$ and $ b = \sqrt{\frac{h^{2}}{4}+d^{2}-h d \cos x} $. Then (for $x \in (0, \pi)$) we have the below geometrical picture: 
This follows from the law of cosines and the fact that $\cos(\pi - x) = - \cos(x)$. Observe that as $d$ grows, $a$ and $b$ will become more and more parallel to $d$. In the limit we will end up with something that looks like this: 
From this we gather:
$$ \lim_{d\to \infty} \sqrt{\frac{h^{2}}{4}+d^{2}+h d \cos x}-\sqrt{\frac{h^{2}}{4}+d^{2}-h d \cos x} = \lim_{d\to \infty} a - b = 2\frac{h}{2}\cos x = h\cos x, $$
which is what we wanted, hurray!
(Note that for $x \notin (0, \pi)$ we can use some properties of cosine to show that the above argument still works.)

Extra:
To evaluate $\lim_{d\to \infty}y=\lim_{d\to \infty} \sqrt{\frac{h^{2}}{4}+d^{2}+h d t}-\sqrt{\frac{h^{2}}{4}+d^{2}-h d t}$ for any $t$, we can use the following argument.
Rewrite the limit as
$$\lim_{d\to \infty}y =\lim_{d\to \infty} \sqrt{\left(d + \frac{ht}{2}\right)^2 + \frac{h^2}{4}(1-t^2)} - \sqrt{\left(d - \frac{ht}{2}\right)^2 + \frac{h^2}{4}(1-t^2)}.$$
Note that these square roots will be well defined for sufficiently large values of $d$. Also note that they have asymptotes $d + \frac{ht}{2}$ and $d - \frac{ht}{2}$, respectively. This is true for the first one because
$$\lim_{d\to \infty}\sqrt{\left(d + \frac{ht}{2}\right)^2 + \frac{h^2}{4}(1-t^2)} - (d + \frac{ht}{2}) =  \lim_{d\to \infty} \frac{\frac{h^2}{4}(1-t^2)}{\sqrt{\left(d + \frac{ht}{2}\right)^2 + \frac{h^2}{4}(1-t^2)} + (d + \frac{ht}{2})} = 0,$$
and similarly for the other one.
This means that
$$\lim_{d\to \infty}y = \lim_{d\to \infty} (d + \frac{ht}{2}) + f(d) - (d - \frac{ht}{2}) - g(d),$$
where $f$ and $g$ go to zero as $d$ goes to infinity. Therefore we have the desired result!
A: try $h=2,d=1$ and you will get $$1-\cos(x)^2=\sqrt{1-\cos(x)^2}$$ what is i.g. not true.
