Question #b6b30

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Answer 1 · 2017-02-21T01:43:25

Given

$sin x = \frac{1}{4}$ $sinx=1/4$

$and 0 < x < \frac{π}{2} \to x \in 1st quadrant$ $and 0< x < pi/2-> x in "1st quadrant"$

So $cos x = \sqrt{1 - {sin}^{2} x} = \sqrt{1 - \frac{1}{16}} = \frac{\sqrt{15}}{4}$ $cosx=sqrt(1-sin^2x)=sqrt(1-1/16)=sqrt15/4$

Now

$sin (\frac{x}{2}) = \sqrt{\frac{1}{2} (1 - cos x)}$ $sin(x/2)=sqrt(1/2(1-cosx))$

$= \sqrt{\frac{1}{2} (1 - \frac{\sqrt{15}}{4})}$ $=sqrt(1/2(1-sqrt15/4))$

$= \sqrt{\frac{1}{8} (4 - \sqrt{15})}$ $=sqrt(1/8(4-sqrt15)$