Question

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

Answer 1

Here, x is in quadrant II.

i.e., $\frac{\pi}{2}<x<{\pi}$

$⇒\frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}$

Therefore, $sin\frac{x}{2}\,and\,,cos\frac{x}{2}\,and\,\,tan\frac{x}{2}\,are\,all\,positive.$

it is given that $sin x=-\frac{1}{4}.$

$cos^2x=1-sin^2x=1-(\frac{1}{4})^2=1-\frac{1}{16}=\frac{15}{16}$

$⇒cosx=-\frac{\sqrt15}{4}$ [cos x is negative in quadrant II]

$sin^2\frac{x}{2}=\frac{1-cosx}{2}=\frac{1-(-\frac{\sqrt15}{4})}{2}=\frac{4+{\sqrt15}}{8}$

$⇒sin\frac{x}{2}=\sqrt\frac{4+\sqrt15}{8}\,[∴\,sin\frac{x}{2}\,is\,positive]$

$=\sqrt\frac{4+\sqrt15}{8}×\frac{2}{2}$

$=\sqrt\frac{8+2\sqrt15}{16}$

$=\sqrt\frac{8+2\sqrt15}{4}$

$cos^2\frac{x}{2}=\frac{1+cosx}{2}=\frac{1+(-\frac{\sqrt15}{4})}{2}=\frac{4-\sqrt15}{8}$

$⇒cos\frac{x}{2}=\sqrt\frac{4-\sqrt15}{8}\,\,\,[∴\,sin\frac{x}{2}\,is\,positive]$

$=\sqrt\frac{4-\sqrt15}{8}×\frac{2}{2}$

$=\sqrt\frac{8-2\sqrt15}{16}$

$=\sqrt\frac{8-2\sqrt15}{4}$

$tan\frac{x}{2}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=\frac{(\frac{\sqrt8+2\sqrt15}{4})}{(\frac{\sqrt8-2\sqrt15}{4})}=\frac{\sqrt8+2\sqrt15}{\sqrt8-2\sqrt15}$

$\sqrt\frac{8+2\sqrt15}{8-2\sqrt15}×\frac{8+2\sqrt15}{8+2\sqrt15}$

$=\sqrt\frac{(8+2\sqrt15)^2}{64-60}=\frac{8+2\sqrt15}{2}=4+\sqrt15$

Thus, the respective values of $sin\frac{x}{2},cos\frac{x}{2}\,and\,\,tan\frac{x}{2}\,are\,\frac{\sqrt8+2\sqrt15}{4},\frac{\sqrt8-2\sqrt15}{4}\,and\,\,4+\sqrt15.$