# Trig identities!!!!!!!!!

• Jun 18th 2008, 05:36 PM
andymac
Trig identities!!!!!!!!!
if cotx= -2/3 and pi/2<x<pi

solve for tan(x/2)

Thanks
• Jun 18th 2008, 06:08 PM
TKHunny
$\frac{\pi}{2}
• Jun 18th 2008, 08:48 PM
Soroban
Hello, andymac!

Quote:

If $\cot x\:=\: -\frac{2}{3}\:\text{ and }\;\frac{\pi}{2} < x < \pi,\:\text{ find: }\tan \frac{x}{2}$

$\cot x \:=\:-\frac{2}{3}\quad\Rightarrow\quad \tan x \:=\:-\frac{3}{2} \:=\:\frac{opp}{adj}$

Since $x$ is in Quadrant 2, $opp = 3,\:adj = -2 \quad\Rightarrow\quad hyp = \sqrt{13}$
. . Hence: . $\cos x \:=\:-\frac{2}{\sqrt{13}}$

Half-angle identity: . $\tan\frac{x}{2} \;=\;\pm\sqrt{\frac{1-\cos x}{1 + \cos x}}$

Since $x$ is in Quadrant 2, $\frac{x}{2}$ is in Quadrant 1.

We have: . $\tan\frac{x}{2} \;=\;\sqrt{\frac{1-(-\frac{2}{\sqrt{13}})}{1 + (-\frac{2}{\sqrt{13}})}} \;=\;\sqrt{\frac{\sqrt{13} + 2}{\sqrt{13} - 2}}$

Rationalize: . $\tan\frac{x}{2} \;= \;\sqrt{\frac{\sqrt{13}+2}{\sqrt{13}-2}\cdot{\color{blue}\frac{\sqrt{13}+2}{\sqrt{13}+2 }}} \;=\;\sqrt{\frac{(\sqrt{13} + 2)^2}{9}}$

Therefore: . $\boxed{\tan\frac{x}{2} \;=\;\frac{\sqrt{13}+2}{3}}$