Range of a parabolic shot Prove that the range of a parabolic shoted from a height $h$, speed $v$ and angle $\alpha$ with the floor is maximum when $\alpha$ satisfies
$\cos(2\alpha) = \frac{g h}{v^2  + gh}$ 
This is my attemp. I need to have this done soon, and I have been trying to figure out where did I make a mistake. It may be a silly mistake I am missing.
I start by writing down the trayectory:
$\alpha(t) = (tv \cos(\alpha), h +t v\sin(\alpha)- \frac{1}{2}gt^2)$
Now let's get the time it takes the proyectile to land:
$\alpha_y(t) = 0 \iff  h +t v\sin(\alpha)- \frac{1}{2}gt^2 =0 \iff t^*=\frac{v\sin(\alpha) + \sqrt{v^2\sin^2\alpha + 2gh}}{g}$
And we put that time in $\alpha_x$ to see how far the proyectile lands:
$\alpha_x(t^*) = \frac{v^2\cos\alpha}{g}(\sin{\alpha} + \sqrt{\sin^2\alpha + \frac{2gh}{v^2}})$
Now I take the derivative respect to $\alpha$
$\frac{\partial}{\partial\alpha}\alpha_x(t^*) = \frac{v^2}{g}(\cos^2\alpha - \sin^2\alpha + \frac{\sin\alpha\cos^2\alpha}{\sqrt{\sin^2\alpha + \frac{2gh}{v^2}}} - \sin\alpha\sqrt{\sin\alpha + \frac{2gh}{v^2}})$
And then, It follows:
$\frac{\partial}{\partial\alpha}\alpha_x(t^*) = 0 \iff \cos{2\alpha}\sqrt{\sin^2\alpha + \frac{2gh}{v^2}} + \sin\alpha\cos^2\alpha - \sin\alpha(\sin^2\alpha + \frac{2gh}{v^2}) = 0$
Thus
$\cos{2\alpha} = \frac{\sin\alpha(\sin^2\alpha + \frac{2gh}{v^2}) - \sin\alpha\cos^2\alpha}{\sqrt{\sin^2\alpha + \frac{2gh}{v^2}}}$
Which is not the result I am supposed to get
 A: Using a CAS, one realizes that your solution doesn't work: in fact it is well known that, in this case, one must proceed implicitly.
You start from the cartesian equation of the trajectory$$y=h+x \tan \alpha-x^2 \frac {g\,(1+\tan^2 \alpha)}{2v^2}$$The non-zero $x$-solution of the equation $y=0$ gives the range of the projection.
Because $y=0$ defines (implicitly) the range $x$ as a function $x(\alpha)$, differentiate the equation with respect to $\alpha$ and solve with respect to $x'(\alpha)$ putting this equal to zero to obtain the stationary value of $x(\alpha)$.
You have $$g\,x\tan\alpha-v^2=0$$Then couple the latter with $y=0$.
You find $$x=\frac vg \sqrt {v^2+2gh}$$$$\tan\alpha=\frac v{\sqrt {v^2+2gh}}$$
Using the relation $$\cos {2\alpha}=\frac {1-\tan^2 \alpha}{1+\tan^2 \alpha}$$you obtain the condition.
A: Everything looks good down to the part where you are taking the derivative. One thing that could really simplify your work (and which will also put your answer in the form you want) is to use the double angle identity for sin: cos($\alpha$) sin($\alpha$) = sin(2*$\alpha$)/2.
The other thing also is that you have to solve for cos(2$\alpha$), that is you should cos(2$\alpha$) on the left hand side, but no alphas on the right hand side.
