Answer
The proof is below.
Work Step by Step
We know the following equation:
$2gh = v_0^2 sin^2\theta$
We do not need the $v_0^2$ term, so we want to find a way to cancel it out. Thus, we use the following equation:
$x = \frac{v_0^2 sin(2\theta)}{g}$
Solving for $v_0^2$ gives:
$v_0^2 = \frac{gx}{sin(2\theta)}$
We substitute this into the first equation to obtain:
$2gh = \frac{gx}{sin(2\theta)} sin^2\theta$
$2h = \frac{x}{sin(2\theta)} sin^2\theta$
$\frac{2hsin(2\theta)}{sin^2\theta} = x$
$\frac{4hsin(\theta)cos(\theta)}{sin^2\theta} = x$
$x=\frac{4h}{tan\theta}$
