Question

Find $\sin \frac{x}{2},\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ for $\sin x=\frac{1}{4},x$ in quadrant $II.$

Answer 1

Step $1:$ Check sign for $\sin \frac{x}{2},\cos \frac{x}{2}$ and $\tan \frac{x}{2}$
Given that $x$ is in quadrant $II.$
$\Rightarrow {90}^{\circ }<x<{180}^{\circ }$
$\Rightarrow \frac{{90}^{\circ }}{2}<\frac{x}{2}<\frac{{180}^{\circ }}{2}$
$\Rightarrow {45}^{\circ }<\frac{x}{2}<{90}^{\circ }$.
$\Rightarrow \frac{x}{2}$ lies in $1^{st}$ quadrant.
$\therefore \sin \frac{x}{2},\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are positive.

Step $2:$
Solve for the value of $\cos \frac{x}{2}$
${\sin }^{2}x+{\cos }^{2}x=1$
$\Rightarrow {\left(\frac{1}{4}\right)}^2+{\cos }^{2}x=1$
$\Rightarrow {\cos }^{2}x=1-\frac{1}{16}$
$\Rightarrow {\cos }^{2}x=\frac{15}{16}$
$\Rightarrow \cos x=\pm \frac{\sqrt{15}}{4}$
Given $x$ is in quadrant $II.$
So $\cos x$ is negative.
$\therefore \cos x=-\frac{\sqrt{15}}{4}$
$\cos x=2{\cos }^2\frac{x}{2}-1$
$\Rightarrow -\frac{\sqrt{15}}{4}=2{\cos }^2\left(\frac{x}{2}\right)-1$
$\Rightarrow 2{\cos }^2\left(\frac{x}{2}\right)=1-\frac{\sqrt{15}}{4}$
$\Rightarrow {\cos }^2\left(\frac{x}{2}\right)=\frac{4-\sqrt{15}}{8}\Rightarrow \cos \frac{x}{2}=\pm \sqrt{\frac{4-\sqrt{15}}{8}}$
$\Rightarrow \cos \frac{x}{2}=\pm \sqrt{\frac{2(4-\sqrt{15})}{16}}=\pm \frac{\sqrt{8-2\sqrt{15}}}{4}=\frac{\sqrt{8-2\sqrt{15}}}{4}$

Step $3:$
Solve for the value of $\sin \frac{x}{2}$
${\sin }^{2}x+{\cos }^{2}x=1$ Replacing $x$ by $\frac{x}{2}$
${\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}=1$
$\Rightarrow {\sin }^2\frac{x}{2}+\frac{8-2\sqrt{15}}{16}=1$
$\Rightarrow {\sin }^2\frac{x}{2}=1-\frac{8-2\sqrt{15}}{16}$
$\Rightarrow \sin \frac{x}{2}=\frac{\sqrt{8+2\sqrt{15}}}{4}$

Step $4:$
$\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$
$\Rightarrow \tan \frac{x}{2}=\frac{\frac{\sqrt{8+2\sqrt{15}}}{4}}{\frac{\sqrt{8-2\sqrt{15}}}{4}}=\frac{\sqrt{8+2\sqrt{15}}}{4}\times \frac{4}{\sqrt{8-2\sqrt{15}}}$
$\Rightarrow \tan \frac{x}{2}=\sqrt{\frac{8+2\sqrt{15}}{8-2\sqrt{15}}}$
$\Rightarrow \tan \frac{x}{2}=\sqrt{\frac{8+2\sqrt{15}}{8-2\sqrt{15}}\times \frac{8+2\sqrt{15}}{8+2\sqrt{15}}}$
$\Rightarrow \tan \frac{x}{2}=\sqrt{\frac{(8+2\sqrt{15}{)}^2}{64-4\times 15}}=\frac{8+2\sqrt{15}}{2}$
$\Rightarrow \tan \frac{x}{2}=4+\sqrt{15}$