Galileo,in his book $Two \text{ } new \text{ } sciences$,stated that "for elevations which exceed or fall short of $45^{\circ}$ by equal amounts,the ranges are equal". Prove this statement.

(N/A) The horizontal range $R$ of a projectile launched with initial velocity $v_{o}$ at an angle $\theta_{o}$ is given by the formula:
$R = \frac{v_{o}^{2} \sin(2\theta_{o})}{g}$
Consider two angles of projection,$\theta_{1} = 45^{\circ} + \alpha$ and $\theta_{2} = 45^{\circ} - \alpha$,where $\alpha$ is the equal amount by which they exceed or fall short of $45^{\circ}$.
For $\theta_{1} = 45^{\circ} + \alpha$,the range $R_{1}$ is:
$R_{1} = \frac{v_{o}^{2} \sin(2(45^{\circ} + \alpha))}{g} = \frac{v_{o}^{2} \sin(90^{\circ} + 2\alpha)}{g} = \frac{v_{o}^{2} \cos(2\alpha)}{g}$
For $\theta_{2} = 45^{\circ} - \alpha$,the range $R_{2}$ is:
$R_{2} = \frac{v_{o}^{2} \sin(2(45^{\circ} - \alpha))}{g} = \frac{v_{o}^{2} \sin(90^{\circ} - 2\alpha)}{g} = \frac{v_{o}^{2} \cos(2\alpha)}{g}$
Since $R_{1} = R_{2}$,it is proven that for elevations which exceed or fall short of $45^{\circ}$ by equal amounts,the ranges are equal.