We have to find sin x/2, given that sinx = 1/4 and x lies between pi/2 and pi.

Now cos x = 1- 2 (sin x/2)^2

(cos x)^2 = 1 - (sin x)^2

=> (cos x)^2 = 1 - (1/4)^2

=> cos x = sqrt [ 15/ 16]

=> 2...

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We have to find sin x/2, given that sinx = 1/4 and x lies between pi/2 and pi.

Now cos x = 1- 2 (sin x/2)^2

(cos x)^2 = 1 - (sin x)^2

=> (cos x)^2 = 1 - (1/4)^2

=> cos x = sqrt [ 15/ 16]

=> 2 (sin x/2)^2 = 1 - cos x

=> 2 (sin x/2)^2 = 1 - sqrt [ 15/ 16]

=> (sin x/2)^2 = [(1 - sqrt (15/16))/2]

=> sin x/2 = sqrt [ [(1 - sqrt (15/16))/2]]

**Therefore sin x/2 = sqrt [ [(1 - sqrt (15/16))/2]].**