Prove the identity sinx/2=squareroot(1-cosx)/2.
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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]
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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.
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