For reference:

x = vt cos θ
y = vt sin θ - ½ g t²

You can find when it returns to it's original height by setting y=0:

y = vt sin θ - ½ g t² = 0
rearranging: vt sin θ = ½ g t²
divide by t: v sin θ = ½ g t
solving for t: t = v sin θ / ½ g = 2(v sin θ)/g

that is the time when it hits the ground, so it tells you when the max "x" value is:

max x = v cos θ × 2(v sin θ)/g = (2v²/g) cos θ sin θ

More neatly:

Now, are you asking for the maximum

a) x (horizontal distance), which I have just shown, or
b) y (height), or
c) actual distance from origin, which would be √(x²+y²)

OK, so you want a formula like:

d_max = f(θ)

But I still don't know if displacement is in terms of x, y or distance from origin (the hardest!), so I will just demonstrate with x. As I showed before:

x_max = (2v²/g) cos θ sin θ

Now, we are lucky because the double-angle formula for sin is: sin 2θ = 2 sin θ cos θ, so we can substitute:

x_max = (v²/g) sin 2θ

And because you have fixed "v" and "g" (gravity), they could be combined into, say: C = v²/g

x_max = C sin 2θ

And that is maximum range in terms of angle.

But you are now going to say that you want to know maximum actual distance, ie the maximum might be somewhere in the sky during it's path, is that right?

If I'm right, you want to find the value of θ that gives the highest value of y, in terms of v and x.

Start with the equation that you gave in the other trajectory topic:

y = x tanθ - gx²(1+ tan² θ)/2v²                           

We need to find the maximum of y by changing θ. This can be found by differentiating with respect to θ and solving for when that expression is equal to 0. dy/dx = x sec² θ - 2gx² (tan θsec² θ)/2v².  [Note: sec θ = 1/cos θ]

The value of θ for which this expression is equal to 0 is the value which will give the highest value of y.
So, x sec² θ - 2gx² (tan θsec² θ)/2v²
Rearrange: x sec² θ = 2gx² (tan θsec² θ)/2v²
Cancel various things: 1 = (gx tan θ)/v²
Solve: θ = tan-¹ (v²/gx)

Happily, the only time θ appears in your original equation is as a function of tan, which means that the tan and tan-¹ will always cancel out.

Your final equation, after a bit of rearrangement, is y = 3v²/2g - gx²/2v², where v is the initial velocity, x is the distance from start point and g is acceleration due to gravity. The equation only works when it returns a value of y that is positive.