Prove that x has a maximum value when the degree is 45 degree?

Question

'The horizontal range,x, of a proiectile launched at an angle. degree, with the horizontal at an initial veiocitv, vi, can be determined by using:
x= (vi/g)*sin 2* degree
Prove that x has a maximum value when the degree is 45 degree. Give some examples.
sin 2* degree = 2 sin* degree* cos* degree.

i don't really know how to prove it, please show some examples

Answer 1

Take the derivative and set equal to 0 to find the maximum
Dx/Ddegree=0
the derivative of sin 2 degree =2 cos 2 degree =0 when 2 degree =90

Answer 2

x = *vi/g * sin2 degree =>

for X max dx/ddegree =0 =>

vi/g * 2cos2degree =0 =>

cos2degree =0 => degree = 90/2 = 45