Find sinx/2 , cosx/2 and tanx/2 in sinx is equal to 1/4 , x in quadrant two

Find $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$ in

Q (10)

$\small \sin x = \frac{1}{4}$ ,x in quadrant II

Answers (1)

$\frac{\pi}{2} < x < \pi\\ \\ \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$ All functions are positive in this range

We know that
   $\cos^{2}x = 1 - \sin^{2}x$
= 1 - $\left ( \frac{1}{4} \right )^{2}$ = $1 - \frac{1}{16}$ = $\frac{15}{16}$

cos x = $\sqrt\frac{15}{16} = \pm \frac{\sqrt15}{4} = - \frac{\sqrt15}{4}$ (cos x is -ve in II quadrant)

We know that
cosx = $2\cos^{2}\frac{x}{2} - 1$
   $2\cos^{2}\frac{x}{2} = \cos x + 1 = -\frac{\sqrt15}{4} + 1 = \frac{-\sqrt15+4}{4}$

$\cos^{2}\frac{x}{2} = \frac{-\sqrt15+4}{8}$
   $\cos\frac{x}{2} = \pm \sqrt\frac{-\sqrt15+4}{8} = \frac{\sqrt{-\sqrt15+4}}{2\sqrt2} = \frac{\sqrt{8-2\sqrt15}}{4}$ (because all functions are positive in given range)

similarly,
cos x = $1-2\sin^{2}\frac{x}{2}$
   $2\sin^{2}\frac{x}{2} = 1 - \cos x\\ \\ 2\sin^{2}\frac{x}{2} = 1 -\left (\frac{-\sqrt15}{4} \right ) = \frac{4+\sqrt15}{4}$
$\sin\frac{x}{2} = \pm \sqrt\frac{\sqrt15+4}{8} = \frac{\sqrt{\sqrt15+4}}{2\sqrt2} = \frac{\sqrt{8+2\sqrt15}}{4}$ (because all functions are positive in given range)
$\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{\sqrt{8+2\sqrt15}}{4}}{\frac{\sqrt{8-2\sqrt15}}{4}} = \frac{{8+2\sqrt15}}{\sqrt{64 - 15\times4}} = \frac{{8+2\sqrt15}}{\sqrt{4}} = 4 + \sqrt15$