Sinus issues

1: sin(2a)=f(x)
2: sin^2(a)=g(x)

Any way to rewrite those two so the left hand sides becomes the same, without introducing any trigonometric term (sin, cos, tan) nor any dependence on a to the right hand sides?

(If anyone wants the physical background, I'm stuck on calculating the initial velocity for a thrown item where the height and length of the throw is known, but the angle isn't.)

f(x) = sin(2a) = 2sin(a)cos(a)
= 2sin(a)sqrt[1 - sin^2(a)]
= 2sqrt[g(x)(1 - g(x))]

Thanks. Wasn't fun beating everything into shape with that, but it was doable.

Without using trig I managed to get that the initial velocity squared is equal to:
 = 2*g*H + g*D^2/(8*H)

Where g is the graviational constant, H is the max height, and D is the length of the throw.  I was surprised how nice it turned out.