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

Concept: Projectile range symmetry — the range $R = \dfrac{u^2\sin2\theta}{g}$ depends on $\sin2\theta$, which is symmetric about $\theta=45^\circ$.

1. Let the two angles be $45^\circ+\alpha$ and $45^\circ-\alpha$.
2. $\sin[2(45^\circ+\alpha)] = \sin(90^\circ+2\alpha) = \cos 2\alpha$.
3. $\sin[2(45^\circ-\alpha)] = \sin(90^\circ-2\alpha) = \cos 2\alpha$. …

The idea

Throw a stone steeply and it goes high but lands close; throw it shallowly and it stays low but also lands close. Somewhere in between, at $45^\circ$, the range is maximum. Galileo's claim is that this trade-off is perfectly symmetric: any two angles equally spaced above and below $45^\circ$ give exactly the same range.

Step 1 — The range formula

For a projectile launched with speed $u$ at angle $\theta$ above the horizontal (landing at the same height it was launched from):

Step 2 — Two angles symmetric about $45^\circ$

Let the two angles be $45^\circ + \alpha$ and $45^\circ - \alpha$, where $0^\circ \le \alpha \le 45^\circ$.

Step 3 — Range at $45^\circ+\alpha$

Using $\sin(90^\circ+\beta) = \cos\beta$:

Step 4 — Range at $45^\circ-\alpha$

Using $\sin(90^\circ-\beta) = \cos\beta$:

Step 5 — Compare

$R_1 = R_2 = \frac{u^2\cos 2\alpha}{g}$ …

Concept: A General Complementary-Angle Symmetry, Applied as a Corollary

Method: Prove the General "Complementary Angles Give Equal Range" Theorem First

Rather than substituting $45^\circ+\alpha$ and $45^\circ-\alpha$ directly into the range formula and simplifying each separately, this method first proves a completely general fact -- any two complementary launch angles give the same range -- and then observes that $45^\circ+\alpha$ and $45^\circ-\alpha$ are automatically complementary (they sum to $90^\circ$ for every $\alpha$), so Galileo's statement follows as an immediate, one-line corollary.

Step 1 -- The range formula

For a projectile launched at speed $u$, angle $\theta$, landing at the same height:

Step 2 -- Prove the general lemma: $R(\theta) = R(90^\circ-\theta)$ for every $\theta$

Consider the range at the complementary angle $90^\circ-\theta$:

Using the identity $\sin(180^\circ-x) = \sin x$ (valid for any $x$, not just this problem's specific angles):

Step 3 -- Recognize that $45^\circ+\alpha$ and $45^\circ-\alpha$ are complementary

Check their sum, for any $\alpha$:

So $45^\circ-\alpha = 90^\circ - (45^\circ+\alpha)$ -- these two angles are exactly a complementary pair, with $\theta = 45^\circ+\alpha$.

Step 4 -- Apply the lemma directly

By Step 2, applied with $\theta = 45^\circ+\alpha$:

$R(45^\circ+\alpha) = R\big(90^\circ-(45^\circ+\alpha)\big) = R(45^\circ-\alpha)$ …