# Find the exact value of sinx/2 if sinx=1/4 and x is such that pi/2<x<pi.

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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]].**

We'll determine sin (x/2), using the half angle formula

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

We know, from enunciation, that:

Pi < x < Pi / 2

We'll divide by 2 the inequality:

Pi / 2 < x / 2 < Pi / 4

From the above inequality, the angle x/2 is in the 1st quadrant and the value of sin (x/2) is positive.

Since sin x = 1/4, we'll apply the trigonometric identity

(sin x)^2 + (cos x)^2 = 1 to determine cos x,

We'll recall that x is in 2nd quadrant where cos x is negative.

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

cos x = - sqrt(1 - 1/16)

cos x = - sqrt(15) / 4

We'll substitute cos x by its value in the formula for sin x/2. **sin x/2 = sqrt [ (1 - sqrt(15)/4) / 2 ] **