We have to prove that sin (x/2) = sqrt [(1 - cos x)/2]

sqrt [(1 - cos x)/2]

use the relation cos x = 1 - 2(sin (x/2))^2

=> sqrt [(1 - (1 - 2*(sin x/2)^2))/2]

=> sqrt [(1 - 1 + 2*(sin x/2)^2)/2]

=> sqrt [(2*(sin x/2)^2)/2]

=> sqrt [(sin x/2)^2]

=> sin (x/2)

**This proves that sin (x/2) = sqrt [(1 - cos x)/2]**

The first step in the process of proving the half angle identity is to raise to square both side, to eliminate the square root.

[sin (x/2)]^2 = (1 - cos x)/2

We'll multiply by 2 both sides:

2[sin (x/2)]^2 = (1 - cos x)

We'll write cos x = cos 2*(x/2) = 1 - 2[sin (x/2)]^2

We'll replace cos x by the equivalent expression:

2[sin (x/2)]^2 = 1 - {1 - 2[sin (x/2)]^2}

2[sin (x/2)]^2 = 1 - 1 + 2[sin (x/2)]^2

We'll eliminate like terms:

2[sin (x/2)]^2 = 2[sin (x/2)]^2

**Since the LHS = RHS, the identity sin (x/2) = sqrt [(1 - cos x)/2] is verified.**