Projectile motion, 45 degree angle gives the biggest displacement, prove it?

Question

so somehow i am suppose to prove that the 45 degree angle gives the biggest displacement. my teacher walked us through half of the steps and left us from there, help me figure out why the 45 degree angle gives the biggest displacement

it starts like this
x= the angle
Vyi [vetical initial velocity] V sin x
Vx- V cos x
Vyf - 0
T= ?
a[ acceleration due to gravity] =g
dx- horizontal displacement

to find the time in terms of v
Vf= Vi+at and from there we got t= [-Vsin x ]/ g

then to find the displacement
dx= V*t
dx= V cos x * [-Vsin x ]/ g
dx=[ -V^2] g * cos x * sin x

and we are suppose to take it from there , i understand that -[V^2] g part is a constant and the whole displacement basically depends on cos x * sin x and this is what is suppose to be the biggest. please show how u got that 45 degree angle gives biggest displacement. thanx

Answer 1

Nah...... I'm not gonna 'give' this one away. It's too good an exercise. But look at the -definitions- of sine and cosine (in terms of the sides of a triangle) and then look real hard at Pythagorean Theorem and you'll see how to prove it.

Doug