Question

Find sin $\frac{x}{2}$, cos $\frac{x}{2}$ and tan $\frac{x}{2}$ in each of the following:

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

Answer 1

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

$\therefore {\text{cos}}^2$ x = 1 - ${\text{sin}}^2$ x

$\Rightarrow {\text{cos}}^{2}x=1-{\left(\frac{1}{4}\right)}^2=1-\frac{1}{16}=\frac{15}{16}$

$\therefore$ cos x = $\pm \frac{\sqrt{15}}{4}$

But x lies in second quadrant.

$\therefore$ cos x =- $\frac{\sqrt{15}}{4}$

Also $\frac{\pi }{2}<x<\pi \Rightarrow \frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2}$

So $\frac{x}{2}$ lies in first quadrant.

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

Now cos $\frac{\pi }{2}=\sqrt{\frac{\text{1+cos x}}{2}}=\sqrt{\frac{\frac{4-\sqrt{15}}{4}}{2}}$

= $\frac{4-\sqrt{15}}{8}=\frac{\sqrt{8-2\sqrt{15}}}{4}$

sin $\frac{x}{2}=\sqrt{\frac{\text{1-cos x}}{2}}=\sqrt{\frac{1+\frac{\sqrt{15}}{4}}{2}}$

= $\frac{\sqrt{4+\sqrt{15}}}{8}$ = $\frac{\sqrt{8+2\sqrt{15}}}{4}\text{tan}\frac{x}{2}=\frac{\text{sin}\frac{x}{2}}{\text{cos}\frac{x}{2}}=\frac{\frac{\sqrt{8+2\sqrt{15}}}{4}}{\frac{\sqrt{8-2\sqrt{15}}}{4}}$

= $\frac{\sqrt{8+2\sqrt{15}}}{\sqrt{8-2\sqrt{15}}}=\frac{\sqrt{4+\sqrt{15}}}{\sqrt{4-\sqrt{15}}}$

= $\frac{4+\sqrt{15}}{\sqrt{16-15}}=4+\sqrt{15}.$