It is well-known that to maximize the horizontal distance traveled by a projectile fired from the ground at a given speed, one should fire it at a $45$-degree 45^\circ$ angle. What's less-known, though not too difficult to prove with some manipulation of derivatives, is that if one is on an incline of angle $\phi$ below the horizontal, then to maximize horizontal distance traveled, one should fire the projectile at an angle of $\frac{\pi}{4}-\frac{\phi}{2}$ above the horizontal.
To see this, suppose we fire it at an angle of $\theta$ at speed $v_0$ with gravitational constant $g$. Then the equation of the path is $(v_0 t \cos{\theta}, -gt^2+v_0 t \sin{\theta})$, so we are looking for the $t$ at which $v_0 t \cos{\theta} \sin{\phi}-gt^2 \cos{\phi}+v_0 t \sin{\theta} \cos{\phi} = 0$, or $t = \frac{v_0(\cos{\theta}\sin{\phi}+\sin{\theta}\cos{\phi})}{g} = \frac{v_0}{g} \sin(\theta+\phi)$. Then we wish to maximize $\cos{\theta} t$ at this point, or equivalently $\cos{\theta} \sin{(\theta+\phi)}$. Taking the derivative with respect to $\theta$, we find $-\sin{\theta} \sin{\theta+\phi}+\cos{\theta} \cos(\theta+\phi) = \cos{(2\theta+\phi)} = 0$. Thus $2\theta+\phi = \frac{\pi}{2}$, giving us our answer.
Since this formula looks so nice, I'm wondering whether there is a nicer proof of this fact, possibly using symmetry and/or with more physical intuition. In particular, it might involve some kind of rotation (maybe by $\frac{\phi}{2}$?).
