Help me prove this trig identity

2sin^2(x/2) == sin(x)tan(x/2)
I can't get this and I have test coming up. Can anyone help me?

2 sin²(x/2)

= 2 sin(x/2) sin(x/2)

= 2 sin(x/2) cos(x/2) sin(x/2) / cos(x/2)

= (2 sin(x/2) cos(x/2)) * (sin(x/2) / cos(x/2))

= sin(x) tan(x/2)

Given:

2sin²(x/2) = sin(x)tan(x/2)

Use tan(u) = sin(u)/cos(u):

2sin²(x/2) = sin(x)sin(x/2)/cos(x/2)

Remove common factor:

2sin(x/2) = sin(x)/cos(x/2)

2sin(x/2)cos(x/2) = sin(x)

let 2u = x, then x/2 = u:

sin(2u) = 2sin(u)cos(u)

sin(2u) = sin(2u)
QED