Thread: Trig identities!!!!!!!!!
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Old June 18th, 2008, 09:48 PM
Soroban Soroban is offline
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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}}
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