Question

If $\sin x=\frac{1}{4}$ and $x$ is in quadrant II, then
I. $\sin \frac{x}{2}=\frac{\sqrt{15}}{2}$
II. $\cos \frac{x}{2}=\frac{\sqrt{8-2\sqrt{15}}}{4}$
III. $\tan \frac{x}{2}=4\pm \sqrt{15}$

• A. I is correct
• B. II is correct
• C. III is correct
• D. All are correct

Answer 1

Answer: (B)

$\sin x=\frac{1}{4}$
Given that, $x$ is in second quadrant.
i.e., $\frac{\pi }{2}<x<\pi$
$\because \sin x=\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}$
$\therefore \frac{1}{4}=\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}$
$\Rightarrow 1+{\tan }^2\frac{x}{2}=8\tan \frac{x}{2}$
$\Rightarrow {\tan }^2\frac{x}{2}-8\tan \frac{x}{2}+1=0$
$\Rightarrow \tan \frac{x}{2}=\frac{8\pm \sqrt{64-4}}{2}$
$\Rightarrow \tan \frac{x}{2}=\frac{8\pm 2\sqrt{15}}{2}\Rightarrow \tan \frac{x}{2}=4\pm \sqrt{15}$
$\because \frac{\pi }{2}<x<\pi \Rightarrow \frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2}$
i.e., $\frac{x}{2}$ lies in Ist quadrant and as
$\frac{x}{2}\in \left(\frac{\pi }{4},\frac{\pi }{2}\right),\tan \frac{x}{2}>1$
$\Rightarrow \tan \frac{x}{2}=4+\sqrt{15}$
Now, ${\sec }^2\frac{x}{2}=1+{\tan }^2\frac{x}{2}=1+(4+\sqrt{15}{)}^2$
$=1+16+15+8\sqrt{15}$
$\Rightarrow {\sec }^2\frac{x}{2}=32+8\sqrt{15}$
$\Rightarrow {\sec }^2\frac{x}{2}=8(4+\sqrt{15})$
$\Rightarrow {\cos }^2\frac{x}{2}=\frac{1}{{\sec }^2\frac{x}{2}}$
$=\frac{1}{8(4+\sqrt{15})}\times \frac{(4-\sqrt{15})}{(4-\sqrt{15})}$
$\Rightarrow {\cos }^2\frac{x}{2}=\frac{(4-\sqrt{15})}{8}$
$\sin x=\frac{1}{4},\cos x=\frac{-\sqrt{15}}{4}$
$2{\sin }^2\frac{x}{2}=1+\frac{\sqrt{15}}{4}=\frac{4+\sqrt{15}}{4}$
${\sin }^2\frac{x}{2}=\frac{4+\sqrt{15}}{8}$
$\Rightarrow \sin \frac{x}{2}=\sqrt{\frac{4+\sqrt{15}}{8}}$