Question

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

tan x = - $\frac{4}{3}$, x in quadrant II.

Answer 1

Here tan x = $\frac{-4}{3}$, x in quardrant II

$\therefore {\text{sec}}^2$ x = 1 + ${\text{tan}}^2$ x

$\Rightarrow {\text{sec}}^2$ x =1 + ${\left(\frac{-4}{3}\right)}^2=1+\frac{16}{9}=\frac{25}{9}$

$\therefore$ sec x = $\pm \frac{5}{3}\Rightarrow$ cos x = $\pm \frac{3}{5}$

But x lies in the second quadrant.

$\therefore$ cos x = $\frac{-3}{5}$

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

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

Now, cos $\frac{x}{2}=\sqrt{\frac{1+\text{cos x}}{2}}=\sqrt{\frac{1-\frac{3}{5}}{2}}$

= $\sqrt{\frac{1}{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{5}}{5}$

sin $\frac{x}{2}=\sqrt{\frac{1-\text{cos x}}{2}}=\frac{\sqrt{1+\frac{3}{5}}}{2}$

= $\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$

tan $\frac{x}{2}=\frac{\text{sin}\frac{x}{2}}{\text{cos}\frac{x}{2}}=\frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}$ = 2.